The ball drops quicker as a result of the topspin, which might make it more challenging for the opponent to strike the ball at the appropriate moment. Topspin may be used to produce shots that are more difficult to return because of this.
What is Topspin and Underspin?Generally, The Magnus effect is caused by the difference in air pressure on the top and bottom of a spinning object.
On a ball with a topspin, the air pressure is greater on the top of the ball than on the bottom, which creates a lifting force that causes the ball to move upward.
This can cause the ball to have more height or to curve in a specific direction, making it more difficult for the opponent to return.
The topspin also causes the ball to drop faster than it would otherwise, which can make it more difficult for the opponent to hit the ball at the right time. This makes topspin a useful tool for making shots that are harder to return.
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g assume that there are no other targets between moon and earth, and therefore range aliasing can be tolerated. what is the maximum pulse repetition frequency that can be used for the purpose of mapping the (half-sphere) surface of the moon?
Assuming that range aliasing can be tolerated and there are no other targets between the moon and Earth, a pulse repetition frequency of around 10-20 MHz would likely be sufficient for mapping the moon surface.
The maximum pulse repetition frequency (PRF) that can be used for mapping the surface of the moon is determined by the need to avoid range ambiguities and to achieve the desired resolution. The PRF is the number of pulses emitted per second. A high PRF allows for high resolution mapping but also increases the chance of range ambiguities.
If range ambiguities can be tolerated, a high PRF can be used to achieve a high resolution map. The altitude of the spacecraft, the size of the antenna, and the desired resolution of the map are also factors that determine the maximum PRF. A general estimate is that a PRF of around 10-20 MHz would be sufficient for mapping the surface of the moon with a high resolution.
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michael's scale measures the mass of objects as consistently 2kg less than their actual mass. how would you describe the scale? group of answer choices
The scale used by Michael is precise but not accurate when it measures the mass.
While measuring the quantities, we use different devices. These devices has to be precise as well as accurate.
Being precise means that it has to give the same reading, which is the case for us because we are having the scale that is measuring the value of the scale consistently that is actually 2 kg less than the actual mass of the object.
Being accurate means that the reading should be either very very close to the actual reading or equal to the actual reading, which is not the case to us because the scale is not showing the correct value of the mass.
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If an astronaut weighs 981 N on Earth and only 160 N on the
Moon, then what is his mass on Earth?
Answer: The mass on Moon will be 98. 1 kg.
Explanation:
As per the known fact, the mass is independent of acceleration due to gravity of planet.
suppose that the acceleration of a model rocket is proportional to the difference between 260 ft/sec and the rocket's velocity. if it is initially at rest and its initial acceleration is 260 ft/sec, how long will it take to accelerate to 208 ft/s?
It will take the model rocket 0.2 seconds to accelerate from rest to a velocity of 208 ft/sec.
In order to answer the question of how long it will take a model rocket to accelerate from rest to a velocity of 208 ft/sec, we need to understand the acceleration of the rocket.
According to the given parameters, the acceleration of the model rocket is proportional to the difference between 260 ft/sec and the rocket's velocity.
Therefore, when the rocket is initially at rest and its initial acceleration is 260 ft/sec, the time it takes to accelerate to 208 ft/s can be calculated as follows:
Acceleration = (260 ft/sec - 208 ft/sec) / time
Rearranging the equation, we get:
Time = (260 ft/sec - 208 ft/sec) / acceleration
Plugging in the values given, we get:
Time = (260 ft/sec - 208 ft/sec) / 260 ft/sec
Time = 0.2 seconds
Therefore, it will take the model rocket 0.2 seconds to accelerate from rest to a velocity of 208 ft/sec.
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8. (a) a spherical mirror is cut in half horizontally and the top half removed. will an image be formed by the bottom half of the mirror? how does the image formed by the half mirror compare to the image formed before cutting the mirror in half?
As the radius of curvature and focal length of the mirror is bottom half of the mirror is same as earlier, so the image formation by this mirror will not change. As a result, the bottom half of the spherical mirror will reflect an image that is an exact replica of the image it projected prior to being chopped.
Only the aperture area will be cut in half when a mirror is cut in half horizontally; the radius of curvature remains unchanged.
The spherical mirror has a radius of curvature that is twice its focal length. Because the radius of curvature remains unchanged when the mirror is cut horizontally, the focal length of the bottom portion will be the same as in the initial situation.
The location of the image created by the lower portion of the mirror is therefore the same as in the preceding situation according to the mirror equation.
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we are experiencing a gravitational pull from the sun and from the laptop in front of us. true or false
Answer:
the answer it's flas
Explanation:
I wish I help you
how does the magnitude of the electric field at the origin for the quarter-circle arc you have just calculated comnpare to the electric field at the origin produced by a point charge q
although the electric field at the origin owing to the quarter-circle arc is zero, the electric field at the origin due to a point charge q is non-zero.
An electric field is a physical field that surrounds electrically charged particles and acts as an attractor or repellent to all other charged particles in the vicinity. It can also refer to a system of charged particles' physical field. The force per unit charge exerted on a positive test charge that is at rest at a given position is the force per unit charge that is used to define the electric field analytically. Electric charge or magnetic fields with variable amplitudes can produce an electric field. The area of space surrounding an electrically charged particle or object in which the charge body perceives force is known as the electric field.
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you stand on a tall cliff and throw a ball so that it is initially moving purely horizontally. if we neglect the effect of air friction, what is the motion of the ball a short time later.?
If we neglect the effect of air friction, then the motion of the ball a short time later will be a combination of horizontal and vertical motion.
What is the motion of the ball after a short time?If you stand on tall cliff and throw ball so that it is initially moving purely horizontally and neglecting the effect of air friction, motion of the ball a short time later will be combination of horizontal and vertical motion. This is because the only force acting on ball is gravity, which will pull the ball downwards. Therefore, ball will experience a vertical acceleration due to gravity, while continuing to move horizontally at constant velocity.
Motion of the ball is a parabolic path, which is the trajectory of projectile under the influence of gravity. Horizontal component of the motion will remain constant, while vertical component will be affected by gravity and will decrease until ball reaches maximum height and then it will increase again until the ball hit the ground.
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calculate the amgnitued of the cetripetal force acting on earth as it orbits the sun, assuming a circular orbit and an orital speed of 3.00 times 10^4 meters per second
Answer:
Approximately [tex]3.59\times 10^{22}\; {\rm N}[/tex] (assuming that the orbit is circular.)
Explanation:
When an object is in a circular orbit of radius [tex]r[/tex] with a (linear) speed of [tex]v[/tex], the centripetal acceleration of this object would be [tex]a = (v^{2} / r)[/tex]. If the mass of the object is [tex]m[/tex], magnitude of the net force on this object would be [tex]F = m\, a = (m\, v^{2} / r)[/tex].
Look up the mass of the Earth: [tex]m \approx 5.9722\times 10^{24}\; {\rm kg}[/tex].
Look up the radius of the orbit (mean distance between the Earth and the Sun): [tex]r \approx 1.4960\times 10^{11}\; {\rm m}[/tex] (one Astronomical Unit.)
It is given that the linear orbital speed is [tex]v = 3.00 \times 10^{4}\; {\rm m\cdot s^{-1}}[/tex].
Therefore, magnitude of the centripetal force (net force) on the Earth would be:
[tex]\begin{aligned}F &= m\, a = \frac{m\, v^{2}}{r} \\ &\approx \frac{(5.9722 \times 10^{24}\; {\rm kg})\, (3.00 \times 10^{4}\; {\rm m\cdot s^{-1}})^{2}}{1.4960 \times 10^{11}\; {\rm m}} \\ &\approx 3.59 \times 10^{22}\; {\rm N} \end{aligned}[/tex].
a doubly charged ion is accelerated to an energy of 36.0 kev by the electric field between two parallel conducting plates separated by 1.60 cm. what is the electric field strength (in v/m) between the plates?
The electric field strength between the conducting plates would be 3.6 x 10⁵ V/m.
What is electric field strength?
Electric field strength (E) is a measure of the force exerted on a charged particle by an electric field. It is measured in units of volts per meter (V/m) or newtons per coulomb (N/C). An electric field can be created by a static charge or by a changing magnetic field. The strength of the electric field is affected by the distance from the source of the field, the charge of the source, and the permittivity of the medium. Electric fields are used in many applications, including in electric motors, generators, and particle accelerators. It is also used in the study of charges and the behavior of conductors and insulators.
The electric field strength (E) can be determined by using the equation: E = W / qd
Where,
W = energy of the ion (36.0 keV)
q = charge of the ion (doubly charged ion, so 2e)
d = distance between the plates (1.60 cm)
To convert keV to J and cm to m:
1 keV = 1.602176634 x 10⁻¹⁶ J
1cm = 0.01m
Therefore,
E = (36.0 x 1.602176634 x 10⁻¹⁶ J) / (2e x 0.01m)
E = (57.634 x 10⁻¹⁶J/C) / (2 x 1.602176634 x 10⁻¹⁹ C x 0.01m)
E = 3.6 x 10⁵ V/m
So, the electric field strength between the plates is 3.6 x 10⁵ V/m.
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A certain vector A, of magnitude 4.68 units, points in a direction 30.7° to the left of the negative y-axis. Find Ax and Ay.
The x-component of A vector is - 2.39 unit.
The y-component of A vector is - 4.02 unit.
What is vector?Vector is a colloquial term that refers to some quantities that cannot be expressed by a single number (a scalar) or to elements of some vector spaces.
The magnitude of A vector is 4.68 unit.
The direction of A vector is 30.7° to the left of the negative y-axis.
So, A vector is in 3rd co-ordinate.
Hence, the x-component of A vector is = - Asin30.7°
= - 4.68 × sin30.7°
= - 2.39 unit.
Hence, the y-component of A vector is = - Acos30.7°
= - 4.68 × cos30.7°
= - 4.02 unit.
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a 90.0 kg astronaut receives a 30.0 n force from her jetpack. how much faster is she be moving after 2.00 seconds?
After 2.00 seconds, the astronaut is moving at the speed of 0.67 m/s. The result is obtained by using the concept of Newton's second law.
What is Newton's second law?The Newton's second law states that "The acceleration is directly proportional to the net force acting on an object and inversely proportional to the object's mass." It can be expressed as
a = ∑F/m
or
∑F = ma
Where
∑F = net force (N)a = acceleration (m/s²)m = object's mass (kg)An astronaut moves after receiving a force. We have
m = 90.0 kgF = 30.0 Nt = 2.00 sFind the speed of the astronaut moving!
The force makes the astronaut moves from rest to a certain speed.
The speed can be calculated by the Newton's second law.
∑F = ma
∑F = m (v₁ - v₀)/t
30.0 = 90.0 (v₁ - 0)/2.00
60.0 = 90.0v₁
60.0/90.0 = v₁
v₁ = 2/3
v₁ = 0.67 m/s
Hence, the astronaut is moving at the speed of 0.67 m/s.
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how much power is needed to move a person (mass 72 kg) up an escalator (angle 34 degrees with the horizontal) with a speed of 1.5 m/s (neglect friction)? 878w
878 Watts of power is needed to move a person (mass 72 kg) up an escalator (angle 34 degrees with the horizontal) with a speed of 1.5 m/s (neglect friction).
Moving sidewalks move people horizontally or at a little slope, elevators or lifts take people and cargo up and down, escalators are moving staircases from one story of a building to another. A really compact commercial elevator called a du.mb. waiter elevator is designed to convey things rather than humans.
The given data:
m = 72 kg
w = fd
w mgd
w = 72 kg × 9.8 × 1.5 m
w = 878w
The pressure preventing movement against one another of hard substrates, liquids layers, and material components is known as friction. There are many sorts of friction.
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What is the SI unit of K so that the equation Velocity = K x density is correct Give answer in terms of basic units. (Kg^-1m45^-1)
The SI unit of K so that the equation Velocity = K x density is correct is given as
K=m^4/s-kgK= m4/s−kg
What is velocity?Velocity is described as the directional speed of an object in motion as an indication of its rate of change in position as observed from a particular frame of reference and as measured by a particular standard of time.
The SI unit of velocity is m/s and the Si unit of density is kg/m^3kg/m3 .
Adding the SI units of both quantities, we have that :
m/s=K\times kg/m^3m/s=K×kg/m3
In terms of dimensions,
[LT^{-1}]=K\times [ML^{-3}][LT−1]=K×[ML−3]
Dimension of K is :
[K]=[M^{-1}L^4T^{-1}][K]=[M−1L4T−1]
SI unit of K is :
K=m^4/s-kgK=m4/s−kg
Hence, the required solution is K=m^4/s-kgK=m4/s−kg
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a ball is dropped from the roof of a tall building. if the ball bounces back to a height of 34 m. what was the velocity with which it left the ground?
The velocity with which the ball left the ground when it is dropped from a tall building is calculated to be 25.82 m/s.
Height to which a ball bounces back s = 34 m
From the equations of kinematics, we know, the relation connecting height, velocities and acceleration as,
v² - u² = 2 a s
where, v is the final velocity
u is the initial velocity
a is the acceleration
s is the height travelled
Here, u = 0 and a is the acceleration due to gravity
v² = 2 × 9.8 × 34
v² = 666.4
v = 25.82 m/s
Thus, the velocity with which the ball left the ground is calculated to be 25.82 m/s.
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consider a 70-kg woman who has a total foot imprint area of 400 cm2. she wishes to walk on the snow, but the snow cannot withstand pressures greater than 0.5 kpa. determine the minimum size of the snowshoes needed (imprint area per shoe) to enable her to walk on the without sinking.
The minimum size of the snowshoes needed to enable her to walk without sinking will be 1.37 m².
Assumption:
(i) Her weight is uniformly distributed on the imprint area
(ii) the weight of the shoes is negligible
Now, given that mass of women, m = 70 kg
For a pressure of 0.5 kPa on the snow, the imprint area of one shoe must be:
A = W/P
here, W = mg = 70 kg . 9.81 m/s² = 70x9.81 N
P = 0.5 kPa = 0.5 x 1000 N/m²
so, A = [tex]\frac{70x9.81 N}{0.5(1000)N/m^2}[/tex]
or A = 1.37 m²
Therefore, the minimum size of the snowshoes needed to enable her to walk without sinking will be 1.37 m².
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IM CONFUSEDDDD!
i have to turn in this project for an online balloon car thing, can someone do the calculating at least please!
To design a race car with a balloon engine for the best possible performance. must take into consideration Newton's laws of motion. the formula Velocity (speed) = Distance / Time (v = d / t).
What was the average velocity of your balloon car?3.753968254 metres per second was the car's average speed.According to the law of conservation of energy, as you fill up the balloon more, it stores more potential energy, which is then transformed into more kinetic energy, causing the balloon to expand faster and the car to move forward.Using f=ma, calculate thrust.By averaging the positive acceleration readings or measuring the time it takes to reach maximum velocity, average acceleration may be obtained.The balloon will travel with greater force and accelerate more quickly as compressed air content increases. The force of the balloon is reduced and its speed is slowed down if the number of pumps is reduced (air pressure is decreased).The Complete Question.
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Two barges full of salted toad guts have a collision. The red barge has a mass of 150 000 kg and is traveling Northwest at 0. 25 m/s. The blue barge has a mass of 100 000 kg and is traveling Southeast at 0. 1 m/s. After the collision the blue barge has a velocity of 0. 32 m/s to the Northwest. What is the final velocity of the red barge? Is this collision elastic?
When the red barge collides with the blue barge, which has a mass of 1000000 kg, its final velocity with in collision elastic becomes 0.311 m/s.
Define the term collision elastic?A elastic collision is one in which the system does not experience a net loss of kinetic energy as a result of a collision. In elastic collisions, momentum as well as kinetic energy are both conserved.The given data:
m1= 150000 kg, v1= 0.25 m/s, m2= 1000000 kg, v2= 0.32 m/sUsing the law of conservation of momentum.
Following the formula, the red barge's final velocity (v3) is determined.
m1×v1+ m2×v2= (m1+m2)v3
Put the values;
150000 × 0.25+ 1000000×0.32= (150000+1000000)×v3
37500+ 320000= 1150000×v3
357500 = 1150000×v3
v3 = 0.311 m/s
Thus,
After one head-on elastic hit where the projectile is substantially more substantial than the target, the target particle's velocity will be nearly twice as fast as the projectile, but really the projectile's velocity will essentially be unaffected.
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A ball is attached by a tether to a stake in the ground. The ball travels in a circle at the full length of the tether. The pull of the tether acts in a direction ____________ to the ball's velocity, and changes its ____________.
Answer:A ball is attached by a tether to a stake in the ground. The ball travels in a circle at the full length of the tether. The pull of the tether acts in a direction perpendicular to the ball's velocity, and changes its direction.
Explanation: i big brained thats y
a ball is thrown upwards at 20.0 m/s from the top of a roof that is 10.0 meters above the ground. how long does it take before the ball hits the ground? take g
A ball is thrown upwards at 20.0 m/s from the top of a roof that is 10.0 meters above the ground and t = 4.84 seconds
The equation used to solve this problem is
s = vt - (1/2)gt^2
where s = the distance travelled (in this case, 10.0 meters), v = the initial velocity (in this case, 20.0 m/s), g = the acceleration due to gravity (in this case, 9.81 m/s^2), and t = the time (in this case, the time it takes for the ball to hit the ground).
Rearrange the equation to solve for t:
t = (2s/g) + (v^2/g^2)
Substitute the given values into the equation:
t = (2(10.0)/(9.81)) + (20.0^2/(9.81^2))
Simplify the equation:
t = 0.811 + 4.03
t = 4.84 seconds
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what is the total displacement, in meters, of the ball going through its entire motion: traveling from the ground to the top and then falling back to the ground?
The total displacement, in meters, of the ball going through its entire motion: traveling from the ground to the top and then falling back to the ground is 0.
What do you mean by displacement?An object's position changes if it moves in relation to a reference frame, such as when a passenger moves to the back of an airplane or a lecturer moves to the right in relation to a whiteboard. Displacement describes this shift in location.
The total displacement of an object is the distance and direction between its initial and final positions. If a ball is thrown upward, travels to the top and then falls back to the ground, its total displacement is zero.
This is because the ball travels upward and then downward, covering the same distance in opposite directions. The upward motion and the downward motion cancel each other out, resulting in a net displacement of zero.
So, the ball will be back to the ground where it was originally thrown.
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how fast does the earth rotate near the equator persona 3
Roughly 1,700 km/h. The Earth rotates faster than sound near the equator.
How quickly rotates the Earth close to the equator?
The Earth's circumference is around 40,075 kilometres at the equator; multiplying this by the length of the day yields a spin speed of approximately 1670 kilometres per hour.
How many times does the equator revolve in a day?
Each 24-hour day, the Earth rotates once around its axis. The speed of Earth's spin at its equator is around 1,000 miles per hour (1,600 km per hour). Every day of your existence, this day-night spin has taken you outside and under the sun and stars.
Can people feel the Earth's spin?
The speed of Earth is rapid. It orbits the Sun at a speed of roughly 67,000 miles (107,000 kilometres per hour), spinning (rotating) at about 1,000 miles (1600 kilometres) per hour. Due to the consistent speeds, we are not aware of any motion.
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Around 1,700 km/h.Correct!. Its circumference is approximately 40,075 kilometers at the equator; multiplying this by the length of the day yields a value of 1670 kilometers per hour for the equator-bound speed of rotation.
How quickly rotates the Earth close to the equator? Around 1,700 km/h.Correct!. With players trying their hardest to finish quickly, the game can be completed in as little as 44 hours.Between 65 and 66 hours are typically needed to finish the main game alone.Shinjiro is ultimately kept alive on October 4 by his pocket watch.He is thrown in a coma rather than killed by the bullet because it hits the pocket watch instead of piercing him.Players who save Shinjiro at the end of the story have the option to spend time with him thanks to New Game+. Despite the fact that Persona 3 is a good game, it has many gloomy themes that are inappropriate for young players.To learn more about equator refer
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write a hamiltonian function representing an electron in the presence of a stationary positive point charge
A hamiltonian function representing an electron in the presence of a stationary positive point charge is
H = (p^2)/(2m) - (e^2)/(4πε_0r)
What is a hamiltonian function?Generally, The Hamiltonian function representing an electron in the presence of a stationary positive point charge is given by:
H = kinetic energy + potential energy = (p^2)/(2m) - (e^2)/(4πε_0r)
where
p = momentum of the electron m = mass of the electron e = charge of the electron (assumed to be negative) ε_0 = vacuum permittivity r = distance between the electron and the positive point charge.Note that the first term represents the kinetic energy of the electron, and the second term represents the potential energy of the electron due to the attractive electrostatic force from the positive point charge.
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the ejected gases have a mass that is small compared to the mass of the spacecraft. you then continue coasting with the rocket engines turned off. what will be the components of your position vector an hour later? start by finding the x component, rf,x
The final position of the rocket after 1 hour is (2.64 × 10⁴, 3.5 × 10⁴, 0).
Initial velocity, = (0, 2× 10⁴, 0)
Initial location, = (1.2× 10⁴, 1.5 × 10⁴, 0)
Net force acting, F₁ = (6× 10⁴, 0, 0) N
Mass of the rocket, m₁ = 1.5 × 10⁴
Initial momentum, p₁ = mv₁ = (1.5 × 10⁴) × (0, 2× 10⁴, 0)
p₁ = (0, 3 × 10⁸, 0)
Final momentum after 1 hour(3600 sec), P₂ = p₁ + F₁×Δt
P₂ = (0, 3 × 10⁸, 0) + (6× 10⁴, 0, 0)×3600
P₂ = (2.16 × 10⁸, 3 × 10⁸, 0)
Final position of this time,
r₂ = r₁ + (P₂)/m₁
r₂ = (1.2 × 10⁴, 1.5 × 10⁴, 0) + (2.16 × 10⁸, 3 × 10⁸, 0)/(1.5 × 10⁴)
r₂ = (2.64 × 10⁴, 3.5 × 10⁴, 0)
--The given question is incomplete, the complete question is:
"Suppose that you are navigating a spacecraft far from other objects. The mass of the spacecraft is 1.5 × 10⁴ kg. The rocket engines are shut off, and you're coasting along with a constant velocity of (0, 2× 10⁴, 0) m/s. As you pass the location (1.2× 10⁴, 1.5 × 10⁴, 0) you briefly fire side-thruster rockets, so that your spacecraft experiences a net force of (6× 10⁴, 0, 0) N for 1 hour. After turning off the thrusters, you then continue coasting with the rocket engines turned off. The ejected gases have a mass that is small compared to the mass of the spacecraft. you then continue coasting with the rocket engines turned off. what will be the components of your position vector an hour later? start by finding the x component, rf,x?"
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A 2. 0 kg toy car is released at the top of a frictionless track on the left and rolls off of the track from its right
side ramp. The car starts at a height of 0. 80 m, goes through a 0. 50 m diameter loop, and exits the ramp at a
height of 0. 25 m.
what is the change in the cars gravitational potential energy form A to B?
The change in the car's gravitational potential energy from A to B is calculated as 4.90 J.
What is Gravitational potential energy?Gravitational potential energy may be characterized as a type of energy that is typically possessed or acquired by an object due to an alteration in its position when it is present in a gravitational field. In more simple terms, it can be revealed that gravitational potential energy is an energy that is associated with gravitational force or gravity.
According to the question,
The formula for gravitational potential energy is as follows:
Gravitational potential energy = mgh.where m is the mass (in kg) of the body. G is the magnitude of the gravitational field that the body is exposed to. H is the height at which the body is located.
First, you have to calculate the gravitational potential energy of the car on point A, i.e. U₁= (2.0 kg)(9.8 m/s²)(0 m)= 0 J.
Second, the Gravitational potential energy of the car on point B is found as, U₂= (2.0 kg)(9.8 m/s²)(0.25 m)= 4.90 J.
Now, you must have to calculate the change in the car’s gravitational potential energy: ΔU= U₂-U₁= 4.90 J - 0 J= 4.90 J.
Therefore, the change in the car's gravitational potential energy from A to B is calculated as 4.90 J.
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A ball is thrown horizontally from the roof of a building 85 m tall with a speed of 10.6 m/s.
a. How much later does the ball hit the ground?
b.
How far from the building will it land?
C. What is the velocity of the ball just before it hits the ground?
Answer:
1) [tex]t=4.16s[/tex]
2) [tex]s=44.10m[/tex]
3) [tex]v_{final}=40.84\frac{m}{s}[/tex]
Explanation:
a) How long till it hits the ground?
Since the person threw the ball perfectly horizontal, the ball has 0 vertical velocity at the start. In fact, it takes the same time for the ball to reach the ground if they dropped the ball instead of throwing. The acceleration in a vertical drop is gravity ([tex]a=9.81\frac{m}{s^2}[/tex])
Basically, we'll solve first part as if he just dropped the ball.
Use this linear motion formula:
[tex]s=v_{inital}(t_{final} -t_{initial})+1/2a(t_{final}-t_{inital})^2[/tex]
Since [tex]v_{initial} =0[/tex] we can remove that part from equation:
[tex]s=1/2a(t)^2[/tex]
[tex]85=1/2*9.81(t)^2[/tex] Plug and solve
[tex]170=9.81(t)^2[/tex]
[tex]17.33=t^2[/tex]
[tex]t=4.16s[/tex]
b) How far away does it land?
We will use the same formula above for horizontal motion.
[tex]s=v_{inital}(t_{final} -t_{initial})+1/2a(t_{final}-t_{inital})^2[/tex]
[tex]s=v(t)+0[/tex] (Remove the last part since [tex]a=0[/tex])
[tex]s=10.6(4.16)[/tex]
[tex]s=44.096m[/tex]
c) How fast was it traveling vertically at the end?
Use this other linear motion formula for this step. s=85m (height of building).
[tex]v_{final} ^2-v_{initial} ^2=2as[/tex]
[tex]v_{final} ^2-0^2=2(9.81)*(85)[/tex] Started with 0 vertical velocity.
[tex]v_{final} ^2=1667.7[/tex]
[tex]v_{final}=40.837m/s[/tex]
According to the question of speed, a) 16.05 seconds does the ball hit the ground, b) 169.33 m far from the building will it land, c) 37.68 m/s is the velocity of the ball just before it hits the ground.
What is speed?
Speed is the rate at which something moves or an object's rate of motion. It can be measured in metres per second, kilometres per hour, and miles per hour, and is often used to describe the motion of objects, such as vehicles, animals, or natural phenomena.
a. The time it takes for the ball to hit the ground is given by the equation t = (2×h)/v, where h is the height of the building (85 m) and v is the initial velocity of the ball (10.6 m/s). Plugging in the given values, we get t = (2×85)/10.6 = 16.05 seconds.
b. The distance from the building at which the ball will land is found using the equation x = vt, where v is the initial velocity (10.6 m/s) and t is the time it takes for the ball to hit the ground (16.05 seconds). Plugging in the given values, we get x = 10.6×16.05 = 169.33 m.
c. The velocity of the ball just before it hits the ground is determined by the equation v = √(2gh), where g is the acceleration due to gravity (9.8 m/s2) and h is the height of the building (85 m). Plugging in the given values, we get v = √(2×9.8×85) = 37.68 m/s. This is the velocity of the ball just before it impacts the ground.
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A 50kg student is climbing a rock face with a constant speed. Neglect any friction and drag impending her motion. What is the magnitude of the net force on the student? How much force is the student generating up the hill? What is the magnitude of the normal force?
The magnitude of the net force on the student is zero.
The force generated by the student is 500N.
The magnitude of the normal force is 500 N
What is the net force on the student?The net force on the student is determined as follows:
The formula of the force of gravity is given by :
W = mass × acceleration due to gravity
The mass of the student is 50 kg
The magnitude of the net force in the student is:
Force = 50 * 10
W = 500 N
The force generated by the student is equal to the force of gravity acting o the student since the student is moving at a constant speed.
The force generated by the student = 500N
The net force = weight - force generated by the student
Net force = 500 N - 500 N
Net force = 0 N
The magnitude of the normal force is equal to the weight of the student.
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a man and his dog are out for a run at 5.0 km/hr, going toward a road 15 km away. at the first kilometer, he decides to unleash the dog, and it runs back and forth between its owner and the road at 10.0 km/hr while he continues at his same pace. how far does the dog run by the time the man reaches the road?
The dog run 28 km back and forth between its owner and the road, at the speed of 10.0 km/hr.
Speed of the man throughout the whole distance, v₁ = 5 km/hr
After 1 km, the speed of the man is same but the speed of dog, v₂ = 10 km/h
The distance between the road and the dog's owner after he unleash the dog, d = (15 - 1) = 14 km
Time taken by the owner to reach the road, t = 14/5 = 2.8 hrs
Distance travelled by the dog in 2.8 hrs, = 2.8 × 10 = 28 km
So the dog run a total of 28 km distance.
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a rollercoaster has a hill that is 30m high. a 500kg car is pulled up, stops, and is released. what is the maximum velocity of a car on the rollercoaster that makes it to the lowest point in the track, 5m above the ground?
The bottom of the hill, the cart's kinetic energy would be 147150 joules.
What is meant by Kinetic Energy?In physics, an object's kinetic energy is the energy it has as a result of motion. It is explained as the amount of effort required to move a mass-based body from rest to the indicated velocity. The body keeps its kinetic energy, which it acquired during acceleration, unless its speed changes.
Let KE denote Kinetic Energy
As an equation, the relationship would be:
[tex]$m g h=\frac{1}{2} m v^2$$[/tex]
Kinetic Energy = m g h
Assuming the acceleration due to gravity near the earth's surface exists [tex]$9.81 \frac{\mathrm{m}}{\mathrm{s}}$[/tex], our equation becomes:
[tex]$$\begin{aligned}& K E=500 \mathrm{~kg} \cdot 9.81 \frac{\mathrm{m}}{\mathrm{s}} \cdot 30 \mathrm{~m} \\& K E=147150 \text { Joules }\end{aligned}$$[/tex]
Once a result, as the cart nears the base of the hill, where its gravitational potential energy is transformed into kinetic energy, its kinetic energy would be equal to its initial potential energy.
At the bottom of the hill, the cart's kinetic energy would be 147150 joules.
The complete question is:
A roller coaster car with a mass of 500 kg at the top of a hill that is 30 m high. Without friction, what would its kinetic energy be as it reached the bottom of the hill?
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the first practical peacetime use of airplanes was for
The first time pirates used airplanes in peacetime was to transport letters. Considered to be the founding flight of what would become American Airlines, this flight was the first regularly scheduled flight.
What was the first sensible application of aircraft during peacetime? Airplanes were first used for postal delivery during a period of peace.The De Havilland DH-4 biplane that Charles Lindbergh, the chief pilot of Robertson Aircraft Corporation, flew from Chicago to St. Louis carrying a bag of mail, was built in 1926.Considered to be the founding flight of what would become American Airlines, this flight was the first regularly scheduled flight.Reconnaissance was the initial purpose of aviation in World War.The aircraft would soar over the battlefield, observing the enemy's actions and positions.The 1903 Wright Flyer, the first successful powered airplane, was the result of four years of study and development by Wilbur and Orville Wright.On December 17, 1903, Orville piloted it for its maiden flight over Kitty Hawk, North Carolina.To learn more about airplanes refer
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