Question 11 of 22
A horse of mass 180 kg gallops at a speed of 8 m/s. What is the momentum
of the horse?
Answers
1440
22.5
845
1955
Momentum = (mass) x (speed)
If you work the problem in the same units as the given data, then you get the momentum in units of kilogram-meters per second, and your horse has 1,440 of them.
Answer:
A
Explanation:
1440 kg*m/s
A red apple reflects ______
light and absorbs all other
colours.
A red apple reflects Red light and absorbs all other colours.
g You decide to play fetch with your dog, who is sitting nextto you, so you throw a ball down a narrow hallway. The ballcomes to a stop 3.9 m down the hallway. The dog, startingfrom rest, runs after the ball with a constant acceleration of0.70m/s2until she reaches the ball. She grabs the ball whilestill running down the hallway uniformly accelerating(slowingdown) for 4.7 more seconds until she comes to a stop. What isthe total distance the dog travels to grab the ball and come toa final stop, starting from rest
You decide to play fetch with your dog, who is sitting nextto you, so you throw a ball down a narrow hallway. The ballcomes to a stop 3.9 m down the hallway. The dog, startingfrom rest, runs after the ball with a constant acceleration of0.70m/s2until she reaches the ball. She grabs the ball whilestill running down the hallway uniformly accelerating(slowingdown) for 4.7 more seconds until she comes to a stop. What isthe total distance the dog travels to grab the ball and come toa final stop, starting from rest
How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W
Complete question:
How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W. Assume the average energy per photon in 78.0 keV.
Answer:
The number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec
Explanation:
Given;
power of the flux produced, P = 1 W = 1 J/s
energy per photon, E = 78 keV
Convert the energy per photon to J
E = 78 x 10³ x 1.6 x 10⁻¹⁹ = 1.248 x 10⁻¹⁴ J / photon
let the number of photons = n
n(1.248 x 10⁻¹⁴ J / photon) = 1 J/s
[tex]n = \frac{1 \ J/s}{1.248 \times 10^{-14}\ J/photon } = 8.01 \times 10^{13} \ photons/s[/tex]
Therefore, the number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec
Please helppppppp I need it todayyyyyt!!!!!
the 120-lb woman jogs up the flight of stairsThe 120-lb woman jogs up the flight of stairs in 5 seconds. Determine her average power output.
Answer:
Power = 24.41Watts
Explanation:
Find the diagram attached
Power output = Force * distance/Time
Given
Force = 120lb
Distance = 9inches
Time = 5sec
Since
1lb = 4.4482216153 N
120lb = 120 * 4.4482216153
120lb = 533.787N
9in to meters
9in = 0.0254*9
9in = 0.2286N
Power = 533.787*0.2286/5
Power = 24.41Watts
If a fisherman applies a horizontal force with magnitude 47.0 NN to the box and produces an acceleration of magnitude 3.20 m/s2m/s2, what is the mass of the box
Answer:
The correct solution is "14.6875 kg".
Explanation:
Given values:
Force,
F = 47.0 N
Acceleration,
a = 3.20 m/s²
Now,
⇒ [tex]Force=Mass\times Acceleration[/tex]
or,
⇒ [tex]F=ma[/tex]
⇒ [tex]47.0=m\times 3.20[/tex]
⇒ [tex]m=\frac{47.0}{3.20}[/tex]
⇒ [tex]=14.6875 \ kg[/tex]
A box attached to a spring is being pulled across a flat frictionless surface. The spring constant is 45 N/m, the box is accelerating at
1.3 m/s2, and the spring is stretched by 0.88 m. What is the mass of the box?
26 kg
O 38 kg
30 kg
O 33 kg
Plzzzzz help
Answer:
Mass = 30.46 kg
Explanation:
Given the following data;
Spring constant = 45 N/m
Acceleration = 1.3 m/s²
Extension = 0.88 m
To find the mass of the box;
First of all, we would determine the force acting on the spring.
Force = spring constant * extension
Force = 45 * 0.88
Force = 39.6 N
Next, we find the mass using Newton's second equation of motion.
Force = mass * acceleration
39.6 = mass * 1.3
Mass = 39.6/1.3
Mass = 30.46 kg
Probability of a woman wearing green to walk into a restaurant
A 20-g bullet is shot vertically into an 2.8-kg block. The block lifts upward 9 mm. The bullet penetrates the block and comes to rest in it in a time interval of 5 ms. Assume the force on the bullet is constant during penetration and that air resistance is negligible. What is the speed of the bullet just before the impact
Answer:
The speed of the bullet just before the impact is 701 m/s
Explanation:
Given;
mass of the bullet, m₁ = 20 g = 0.02 kg
mass of the block, m₂ = 2.8 kg
displacement of the block, d = 9 mm = 9 x 10⁻³ m
duration of motion of the bullet, t = 5 ms = 5 x 10⁻³ s
Apply the principle of conservation of energy;
The final kinetic energy of the bullet = maximum potential energy of the block
[tex]\frac{1}{2} m_1v^2 = m_2gh\\\\v^2 = \frac{2m_2gh}{m_1} \\\\v= \sqrt{\frac{2m_2gh}{m_1} } \\\\v = \sqrt{\frac{2 \times 2.8 \times 9.8 \times (9\times 10^{-3})}{0.02} } \\\\v = 4.97 \ m/s[/tex]
Apply the principle of conservation of linear momentum, to determine the initial velocity of the bullet before the impact.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
u₁ is the initial velocity of the bullet
u₂ is the initial velocity of the block = 0
m₁u₁ + 0 = v(m₁ + m₂)
m₁u₁ = v(m₁ + m₂)
0.02u₁ = 4.97(2.8 + 0.02)
0.02u₁ = 14.02
u₁ = 14.02 / 0.02
u₁ = 701 m/s
Therefore, the speed of the bullet just before the impact is 701 m/s
One of the earliest vertebrate animal groups that evolved in the early Paleozoic Era
are
A circular loop of wire with radius 10.0 cm is located in the xy-plane in a region of uniform magnetic field. A field of 2 T is directed in the z-direction, which is upward. (a) What is the magnetic flux through the loop
Answer:
[tex]\phi=628.3[/tex]
Explanation:
From the question we are told that
Radius [tex]r=10.0 cm[/tex]
Magnetic field[tex]B=2T[/tex]
Generally the equation for area of circular path is mathematically given by
[tex]Area=\pi r^2[/tex]
[tex]A=\pi 10^2[/tex]
[tex]A=314.15m^2[/tex]
Generally the equation for Magnetic flux is mathematically given by
[tex]\phi=BA[/tex]
[tex]\phi=2*314.15[/tex]
[tex]\phi=628.3[/tex]
A cart weighing 40 pounds is placed on a ramp incline 15 degrees to the horizon. The cart is held in place by a rope inclined 60 degrees to the horizontal. find the force that the rope must exert on the cart to keep it from rolling down the ramp.
Answer: [tex]14.64\ N[/tex]
Explanation:
Given
Inclination of ramp is [tex]\theta=15^{\circ}[/tex]
Rope is inclined [tex]\phi=60^{\circ}[/tex] to the horizontal
Weight of cart [tex]W=40\ lb[/tex]
from the diagram, rope is at angle of [tex]45^{\circ}[/tex] w.r.t ramp
Sine component of weight pulls down the cart Cosine component of force applied through rope held it at the position
[tex]\Rightarrow 40\sin 15^{\circ}=F\cos 45^{\circ}\\\\\Rightarrow F=40\cdot \dfrac{\sin 15^{\circ}}{\cos 45^{\circ}}\\\\\Rightarrow F=40\times 0.366\\\Rightarrow F=14.64\ N[/tex]
a force in the opposite direction
of the motion
forces: forces applied to an object in opposite directions that are not equal in size. Unbalanced forces result in a change in motion. friction: the force that opposes the motion or tendency toward motion of two objects that are in contact.
You throw a glob of putty straight up toward the ceiling, which is 3.50 mm above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.50 m/sm/s. Part A What is the speed of the putty just before it strikes the ceiling
Answer: [tex]4.65\ m/s[/tex]
Explanation:
Given
Distance putty has to travel is 3.5 m
The initial speed of putty is 9.50 m/s
Using equation of motion to determine the velocity of putty just before it hits ceiling
[tex]v^2-u^2=2as[/tex]
[tex]\Rightarrow v^2-(9.5)^2=2(-9.8)(3.5)\\\\\Rightarrow v^2=9.5^2-68.6\\\Rightarrow v=\sqrt{90.25-68.6}\\\Rightarrow v=4.65\ m/s[/tex]
So, the velocity of putty just before hitting is [tex]4.65\ m/s[/tex]
What is the overall reaction potential for reaction below
Answer:
potential energy is a form of energy that a body possess to its position.Potential energy is equal to potential energy is equal to mass times acceleration due to gravity times height.
what's the difference between mass and inertia in a tabular form
Answer:
to be honest i dont know
Explanation:
^^
Suppose an astronomer observes a binary star system where the stars are separated by 2.0 AU , and they have an orbital period of 7.0 years . Using Newton's version of Kepler's Third Law, find the combined mass of the stars.
Answer:
4.408 [tex]\mathsf{M_{sun}}[/tex]
Explanation:
According to Kelper's Third Law, the equation of the combined mass (m₁+m₂) can be expressed as:
[tex](m_1 + m_2) = \dfrac{\text{(distance between stars)}^3}{\text{(orbital period)}^2}[/tex]
[tex]\text{combined mass}(m_1+m_2)} =\dfrac{(6.0)^3}{(7)^2} \ M_{sun}[/tex]
[tex]\text{combined mass}(m_1+m_2)} =\dfrac{216}{49} \ M_{sun}[/tex]
combined mass (m₁+m₂) = 4.408 [tex]\mathsf{M_{sun}}[/tex]
Two workers are sliding 450 kg kg crate across the floor. One worker pushes forward on the crate with a force of 380 NN while the other pulls in the same direction with a force of 230 NN using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor
Answer:
The coefficient of kinetic friction on the floor is 0.138
Explanation:
Given;
mass of the crate, m = 450 kg
force applied by the first worker, F₁ = 380 N
force applied by the second worker in the same direction as the first worker, F₁ = 230 N
frictional force opposing the motion of the box = -[tex]F_k[/tex]
Apply Newton's second law of motion;
∑F = ma
[tex]F_1 + F_2 - F_k = ma[/tex]
If the crate slides with constant speed, acceleration is zero (0).
[tex]F_1 + F_2 - F_k = ma = 0\\\\F_1 + F_2 - F_k = 0\\\\F_k = F_1 + F_2\\\\\mu _kmg= F_1 + F_2\\\\\mu _k = \frac{F_1 + F_2}{mg} \\\\\mu _k = \frac{380 + 230}{450 \times 9.8} \\\\\mu _k = 0.138[/tex]
Therefore, the coefficient of kinetic friction on the floor is 0.138
А bus has started to move from
the rest with an acceleration of
0.25 m/s². find its final velocity
An athlete training for an event does 6.53 104 J of work during a workout and gives off 5.97 105 J of heat. Consider the athlete to be like a heat engine. (a) Determine the magnitude of the change in internal energy of the athlete. J (b) What is the efficiency of the athlete
Answer:
(a) The magnitude of the change in internal energy is 6.623 x 10⁵ J
(b) the efficiency of the athlete is 10.94 %
Explanation:
Given;
work done by the athlete (system), W = 6.53 x 10⁴ J
the heat given off by the athlete (system), Q = 5.97 x 10⁵ J
The simple diagram below will be used to illustrate the direction of the energy flow assuming a heat engine.
Q← ⊕ →W
The work, W, points away from the system since the system does the work
The heat, Q, points away from the system since heat is given off
Apply first law of thermodynamic;
ΔU = Q + W
where;
q is the heat flowing into or out of the system
(+q if the heat is flowing into the system
(-q if the heat is leaving the system
w is the work done by or on the system
(+w if the work is done on the system by the surrounding
(-w if the work is done by the system to the surrounding
Thus, from the above explanation, the change in internal energy of the system is calculated as;
ΔU = -Q - W
ΔU = - 5.97 x 10⁵ J - 6.53 x 10⁴ J
ΔU = -6.623 x 10⁵ J
The magnitude of the change in internal energy = 6.623 x 10⁵ J
(b) the efficiency of the athlete;
[tex]Efficiency = \frac{W}{Q} \times 100\%\\\\Efficiency = \frac{6.53 \times 10^4}{5.97 \times 10^5} \times 100\%\\\\Efficiency = 10.94 \ \%[/tex]
different between pressure and force
Force is mass into acceleration
and pressure is force applied per unit area.
The pair of forces described by Newton third law must be
Answer:
The answer is Newton's third law of motion states that every action has an equal and opposite reaction. This means that force always act in pairs
The pair of forces described by Newton third law must be in opposite direction.
What is Newton's third law of motion ?Every action have equal and opposite reaction. for example when we fire bullet from a gun, the gun will recoil back and bullet moves forward. In case of rocket, rocket is fired, thrust is reaction of force applied by the gas on the floor.
The motion of lift from an airfoil in which the air is diverted downward by the airfoil's action and the wing is pushed upward in response.
When a spinning ball moves, the air is deflected to one side, and the ball responds by travelling in the other direction.
A jet engine's motion generates thrust, and hot exhaust gases rush out the back of the engine, producing thrust in the opposite direction.
To know more about motion :
https://brainly.com/question/26603017
#SPJ3.
Chuck and Jackie stand on separate carts, both of which can slide without friction. The combined mass of Chuck and his car, mcartmcart, is identical to the combined mass of Jackie and her cart. Initially, Chuck and Jackie and their carts are at rest.
Chuch then picks up a ball of mass mballmball and throws it to Jackie, who catches it. Assume that the ball travels in a straight line parallel to the ground (ignore the effect of gravity). After Chuck throws the ball, his speed relative to the ground is vc. The speed of the thrown ball relative to the ground is vb. Jackie catches the ball when it reaches her, and she and her cart begin to move. Jackie's speed relative to the ground after she catches the ball is vj.
When answering the questions in this problem, keep the following in mind:
1. The original mass mcartmcart of Chuck and his cart does not include the mass of the ball.
2. The speed of an object is the magnitude of its velocity. An object's speed will always be a non-negative quantity.
Required:
a. Find the relative speed u between Chuck and the ball after Chuck has thrown the ball.
b. What is the speed vb of the ball (relative to the ground) while it is in the air?
c. What is Chuck's speed vc (relative to the ground) after he throws the ball?
d. Find Jackie's speed vj (relative to the ground) after she catches the ball in terms of vb.
e. Find Jackie's speed vj (relative to the ground) after she catches the ball in terms of u.
Explanation:
a.)
We find the relative speed
u = vb + vc
b.)
chuck and the cart are at a rest position
mcartvc = mballvb
from part a above,
vc = u - vb
mcartu = vb(mcart + mball)
make vb the subject of the equation
[tex]vb=\frac{mcartu}{mcart +mball}[/tex]
c.)
from anser a,
vb = u - vc
then mcart vc = mball(u-vc)
[tex]vc= \frac{Mballu}{Mcart+Mball}[/tex]
d.
Mballvb = (Mcart + Mball)vj
we make vj the subject to get her relative speed
[tex]vj=\frac{MballVb}{Mcart+Mball}[/tex]
e.
given the solution in part d above,
we have
[tex]vj=\frac{MballVb}{Mcart+Mball}[/tex]
remember,
vb = u - vc
such that
[tex]Vj = \frac{Mball(u-v)}{Mball +Mcart}[/tex]
thank you!
Two long, straight wires are fixed parallel to one another a distance do apart. The wires carry equal constant currents 1, in the same direction. The attractive magnetic force per unit length between them if f = F/L. What is the force per unit length between the wires if their separation is 2d, and each carries current 2I0?
A. f/4
B. f/2
C. 3f/2
D.) 2f
Answer:
Option D
Explanation:
From the question we are told that:
The attractive magnetic force per unit length as
[tex]f = F/L[/tex]
Separation Distance [tex]x=2d[/tex]
Generally the equation for Magnetic force between two current carrying wire is mathematically given by
[tex]\frac{F}{\triangle l}=\frac{\mu_0I_1I_2}{\mu \pi x}[/tex]
[tex]\frac{F}{\triangle l }=\frac{I_1I_2}{ x}[/tex]
Where
[tex]x=2r[/tex]
And
[tex]I_1=I_2=>2I[/tex]
Then
[tex]\frac{F}{\triangle l}=>\frac{2*2}{2}*f[/tex]
[tex]\frac{F}{\triangle l}=>2f[/tex]
Therefore s the force per unit length between the wires if their separation is 2d
[tex]\frac{F}{\triangle l}=>2f[/tex]
Option D
A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is the magnitude
Answer:
Centripetal acceleration = 0.79 m/s²
Explanation:
Given the following data;
Radius, r = 2.6 km
Time = 360 seconds
Conversion:
2.6 km to meters = 2.6 * 1000 = 2600 meters
To find the magnitude of centripetal acceleration;
First of all, we would determine the circular speed of the car using the formula;
[tex] Circular \; speed (V) = \frac {2 \pi r}{t}[/tex]
Where;
r represents the radius and t is the time.Substituting into the formula, we have;
[tex] Circular \; speed (V) = \frac {2*3.142*2600}{360} [/tex]
[tex] Circular \; speed (V) = \frac {16338.4}{360} [/tex]
Circular speed, V = 45.38 m/s
Next, we find the centripetal acceleration;
Mathematically, centripetal acceleration is given by the formula;
[tex] Centripetal \; acceleration = \frac {V^{2}}{r}[/tex]
Where;
V is the circular speed (velocity) of an object.r is the radius of circular path.Substituting into the formula, we have;
[tex] Centripetal \; acceleration = \frac {45.38^{2}}{2.6}[/tex]
[tex] Centripetal \; acceleration = \frac {2059.34}{2600}[/tex]
Centripetal acceleration = 0.79 m/s²
Matter's resistance to a change in motion is called _____ and is directly proportional to the mass of an object. For an object to change its state of motion, a force must be applied to it.
A. Velocity
B. Inertia
C. Distance
D. Area
Answer:
B) Inertia is the resistance of any physical object
Work is done when you lift an object to a certain height. If the force exerted is greater than the weight of the object, input work is greater than the output work. Where does the extra energy go?
Work is done when you lift an object to a certain height. If the force exerted is greater than the weight of the object, input work is greater than the output work. Then the extra energy goes in overcoming the gravitational acceleration and heating up of body etc
A train with mass 3.3 x 107 kg starts from rest and accelerates to a speed of 42
m/s. What is the initial kinetic energy of the train?
Answer:
kinetic energy of the train = 2,910.6 x 10⁷ joule
Explanation:
Given:
Mass of train = 3.3 x 10⁷ kg
Speed of train = 42 m/s
Find:
kinetic energy of the train
Computation:
kinetic energy = (1/2)(m)(v²)
kinetic energy of the train = (1/2)(3.3 x 10⁷)(42²)
kinetic energy of the train = (1/2)(3.3 x 10⁷)(1,764)
kinetic energy of the train = (3.3 x 10⁷)(882)
kinetic energy of the train = 2,910.6 x 10⁷ joule
Answer: The initial kinetic energy of the train is [tex]2910.6 \times 10^{7} J[/tex].
Explanation:
Given: Mass = [tex]3.3 \times 10^{7} kg[/tex]
Speed = 42 m/s
Kinetic energy is the energy acquired by an object due to its motion.
Formula to calculate kinetic energy is as follows.
[tex]K.E = \frac{1}{2}mv^{2}[/tex]
where,
m = mass of object
v = speed of object
Substitute the values into above formula as follows.
[tex]K.E = \frac{1}{2}mv^{2}\\= \frac{1}{2} \times 3.3 \times 10^{7} kg \times (42 m/s)^{2}\\= 2910.6 \times 10^{7} kg m^{2}/s^{2} (1 J = 1 kg m^{2}/s^{2})\\= 2910.6 \times 10^{7} J[/tex]
Thus, we can conclude that the initial kinetic energy of the train is [tex]2910.6 \times 10^{7} J[/tex].
is anyone online??just asking
Answer:
me...:(
Explanation:
Answer:
hello I'm online here thanks for the points (◔‿◔)