Answer:
The value is [tex]F = 1.568 *10^{-9} \ N[/tex]
Explanation:
From the question we are told that
The mass of the first lead sphere is [tex]m = 1.60 \ kg[/tex]
The mass of the second lead sphere is [tex]M = 16 \ g = 0.016 \ kg[/tex]
The separation between masses is [tex]r = 3.30 \ cm = 0.033 \ m[/tex]
Generally the gravitational force between each sphere is mathematically represented as
[tex]F = \frac{G * m * M }{r^2 }[/tex]
Here G is the gravitational constant with value [tex]G = 6.67 *10^{-11 } \ m^3 \cdot kg^{-1} \cdot s^{-2}[/tex]
[tex]F = \frac{6.67 *10^{-11 } * 1.60 * 0.016 }{0.033^2 }[/tex]
=> [tex]F = 1.568 *10^{-9} \ N[/tex]
PLEASE HELP!!!!
Suppose a tractor was moving forward at 6.0 m/s. It runs into a pile of hay and is
brought to a stop in 1.1 s. If the tractor had a mass of 2550 kg, with what force did
the pile of hay stop the tractor?
A. -5.5N
B. -14000N
C. 15000N
D. -15000N
E. -0.0021N
F. -0.000072
Answer:
B )-14000N
Explanation:
F=mv-mu
t
F =2550(0)-2550(6)
1.1
F = -13909.09
approximately -14000N
Suppose a tractor was moving forward at 6.0 m/s. It runs into a pile of hay and is brought to a stop in 1.1 s. If the tractor had a mass of 2550 kg, the force did the pile of hay stop the tractor 14000N
What are the types of force ?Force can be a unit of pushing or pulling of any object which result from the object’s interaction or movement, without applying force the objects can not be moved, can be stopped or change the direction.
Force is a quantitative parameter between two physical bodies, means an object and its environment, there are various types of forces in nature.
If an object present in its moving state will be either static or motion, the position of the object will be changed if it is pushed or pulled and The external push or pull upon the object is mainly called as Force.
The contact force that occurs when we apply some effort on an object such as Spring Force, Applied Force, Air Resistance Force, Normal Force, Tension Force, Frictional Force
Non-Contact forces are the type of forces that occur from a distance such as Electromagnetic Force, Gravitational Force, Nuclear Force
F=mv-mu
t
F =2550(0)-2550(6)
1.1
F = -13909.09
approximately -14000N
For more details Force, visit
brainly.com/question/13691251
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A child blows a leaf from rest straight up in the air. the leaf has a constant upward acceleration of magnitude 1.0 m by s square. how much time does it take the leaf to displace 1.0m upwards?
Answer:
√2
Explanation:
From the question, we're given that the
Acceleration of the leaf is 1 m/s²
Change in displacement of the leaf is 1 m/s.
Again, from the question, we can tell that the initial velocity u = 0, since the object starts at rest
Now, to solve this, we don't the equation of motion to ur
S = ut + 1/2at², substituting the whole parameters, we then have
1 = 0 * t + 1/2 * 1 * t²
1 = 1/2 * t²
t²/2 = 1
t² = 2
t = √2 seconds
Therefore the time it takes the leaf to dislodge is 2 seconds
BERE
Which describes the positions on a horizontal number line?
0
O All points to the left of one are positive.
O All points to the right of one are positive.
O All points to the left of zero are negative.
O All points to the right of zero are negative.
Mark this and return
Save and Exit
Next
Submit
Answer:
All points to the left of zero are negative
Explanation:
Answer:
C
Explanation:
on edge
A basketball is rolling rightward onto the court with a speed of 4.0 m/s and slows down with a consta leftward acceleration of magnitude 0.50 m/s^2 over 14 m.
What is the velocity of the basketball after rolling for 14 m?
a. 2.0 m/s
b. 5.5 m/s
c. 7.5 m/s
d. 1.4 m/s
Answer:
d. 1.4 m/s
Explanation:
Given;
initial velocity of the `basketball, u = 4.0 m/s
acceleration of the basketball, a = -0.5 m/s² (leftward)
distance the basketball traveled, d = 14 m
the final velocity of the basketball, v = ?
(this final velocity should be less than the initial velocity since the ball is slowing down at a constant rate)
Apply the following the kinematic equation;
v² = u² + 2ad
v² = 4² + 2(-0.5)14
v² = 16 - 14
v² = 2
v = √2
v = 1.41 m/s
Therefore, the final velocity of the basketball is 1.4 m/s.
A circuit with a 12 V battery and lamp has a current of 3 A. What is the resistance of the lamp?
A 15.0kg block is dragged over a rough, horizontal surface by a 70.0-N force acting at 20.0 degrees above the horizontal. The block is displaced 5.0 m, and the coefficient of kinetic friction is 0.3.
a. Draw a free body diagram. Draw your coordinate system and label the axes.
b. Calculate the work done on the block by the 70 N force.
c. Calculate the work done on the block by the normal force.
d. Calculate the work done on the block by the gravitational force.
e. Calculate the work done on the block by the force of friction.
Answer:
[tex]W=70 * 5cos20 = 328.89 J[/tex]
[tex]W_n = 0[/tex]
[tex]W_g=0[/tex]
[tex]W_f= -184.59J[/tex]
Work done is 0
Explanation:
From the question we are told that
Weight of block =15.0kg
Force acting on the block = 70.0N
At an angle of 20 degree
Displacement of block is 5m
Coefficient of kinetic friction 0.3
b) Generally work done by force is give by [tex]W=fdcos \theta[/tex]
therefore
[tex]W=70 * 5cos20 = 328.89 J[/tex]
c) there is no work done by the normal force in this scenario because
normal force in this case is perpendicular to the displacement of the motion
[tex]W_n = 0[/tex]
d) The displacement in the vertical direction is 0
Therefore the gravitational work done is 0 [tex]W_g=0[/tex]
e)Generally in finding work done by friction we first find frictional force
Mathematically the equation for frictional force is given [tex]f = \alpha N[/tex]
Given that
[tex]N=mg-Fsin20[/tex]
[tex]N= 15.0*9.8 - 70 sin20[/tex]
[tex]N=123 N[/tex]
[tex]f=0.3* 123.06 = 36.92N[/tex]
Mathematically solving to get work done by frictional force [tex]W_f[/tex]
[tex]W_f= -fd\\W_f = -36.92 * 5[/tex]
[tex]W_f= -184.59J[/tex]
the frictional force work done is [tex]W_f= -184.59J[/tex]
A hockey player strikes a puck that is initially at rest. The force exerted by the stick on the puck is 975 N, and the stick is in contact with the puck for 0.0049 s.
(a) What is the impulse imparted by the stick to the puck.
___________ kg m/s
(b) What is the speed of the puck (m= 1.67 kg)just after it leaves the hockey stick?
____________ m/s
Explanation:
Given that,
The force exerted by the stick on the puck is 975 N
The stick is in contact with the puck for 0.0049 s
Initial speed of the puck, u = 0 (at rest)
(a) We need to find the impulse imparted by the stick to the puck.
Impulse = Force × time
J = 4.7775 kg-m/s
(b) Mass of the puck, m = 1.76 kg
We need to find the speed of the puck just after it leaves the hockey stick.
Let the speed be v.
As impulse is equal to the change in momentum.
[tex]J=m(v-u)\\\\4.7775=1.67(v-0)\\\\v=\dfrac{4.7775}{1.67}\\\\v=2.86\ m/s[/tex]
So, when the puck leaves the hockey stick its speed is 2.86 m/s.
A truck covers 40.0 m in 9.50 s while uniformly slowing down to a final velocity of 2.75 m/s.
a. Find its original speed.
b. Find its acceleration.
Explanation:
Given that,
Distance covered, d = 40 m
Time, t = 9.5 s
Final velocity, v = 2.75 m/s
(a) Let u be the original speed of the truck. We can find it using first equation of motion.
[tex]v=u+at\\\\2.75=u+2.75\times 9.5\\\\2.75-26.125=u\\\\u=-23.375\ m/s[/tex]
(b) Acceleration = rate of change of velocity
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{2.75-(-23.375)}{9.5}\\\\=2.75\ m/s^2[/tex]
So, the original speed is -23.375 and acceleration is 2.75 m/s².
For the following types of electromagnetic radiation, how do the wavelength, frequency, and photon energy change as one goes from the top of the list to the bottom?
a. radio waves
b. infared radiation
c. visible light
d. ultraviolet radiation
e. gamma radiation
Answer:
Wavelength, frequency and the photon energy changes as the one goes across the ranges of the electro-magnetic radiations.
Explanation:
Electro-magnetic radiations may be defined as the form of energy that is radiated or given by the electro-magnetic radiations. The visible light that we can see is the one of the electro-magnetic radiations. Other forms are the radio waves, gamma waves, UV rays, infrared radiations, etc.
The wavelength of the radiations decreases as we go from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.
The frequency of the radiations increases when we move from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.
The photon energy of the radiations increases when we move from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.
In schematic diagrams, currents are indicated using arrows. What do the arrows indicate? a) the direction of motion of the electrons b) the direction of the current vector c) the direction of motion of the charge carriers d) the direction that positive charge carriers would move e) nothing; they are just a convenient drawing tool
Answer:
D
Explanation:
The direction that positive charge would move
Calculate the escape velocity
the moon's surface given that a man on the moon has 1/6 his weight on earth
Answer:
v = 2.38 × 10³ m/s
Explanation:
Escape velocity, v = √(2gR) where g = acceleration due to gravity on planet and R = radius of planet.
Since it is given that the weight of the man on the moon is 1/6 his weight on earth, and g' = acceleration due to gravity on moon and g = acceleration due to gravity on earth and m = mass of man,
mg' = mg/6
g' = g/6
Since g = 9.8 m/s²,
g'= 9.8 m/s² ÷ 6
g' = 1.63 m/s²
The escape velocity of the moon is thus v = √(2g'R) where R = radius of moon = 1.737 × 10⁶ m.
Substituting these into v, we have
v = √(2g'R)
v = √(2 × 1.63 m/s² × 1.737 × 10⁶ m)
v = √[5.663 × 10⁶ (m/s)²]
v = 2.38 × 10³ m/s
A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surface. At a certain time, both engines turn on. The one on the left gives the spacecraft an acceleration component in the x direction of ax = 5.10 m/s2, while the one on the back gives an acceleration component in the ydirection of ay = 7.30 m/s2. The engines turn off after firing for 675 s, at which point the spacecraft has velocity components of vx = 3630 m/s and vy = 4276 m/s.What was the magnitude and direction of the spacecraft's initial velocity, before the engines were turned on?Express the direction as an angle measured counterclockwise from the x -axis.
Answer:
v₀ = 677.94 m / s , θ = 286º
Explanation:
We can solve this exercise using the kinematic expressions, let's work on each axis separately.
X axis
has a relation of aₓ = 5.10 m / s², the motor is on for a time of t = 675 s, reaching the speed vₓ = 3630 m / s, let's use the relation
vₓ = v₀ₓ + aₓ t
v₀ₓ = vₓ - aₓ t
let's calculate
v₀ₓ = 3630 - 5.10 675
v₀ₓ = 187.5 m / s
Y Axis
[tex]v_{y}[/tex] = v_{oy} - a_{y} t
v_{oy} = v_{y} - a_{y} t
let's calculate
v_{oy} = 4276 - 7.30 675
v_{oy} = -651.5 m / s
we can give the speed starts in two ways
a) v₀ = (187.5 i ^ - 651.5 j ^) m / s
b) in the form of module and angle
Let's use the Pythagorean theorem
v₀ = [tex]\sqrt{v_{ox}^{2} + v_{oy}^{2} }[/tex]
v₀ = [tex]\sqrt{187.5^{2} +651.5^{2} }[/tex]
v₀ = 677.94 m / s
we use trigonometry
tan θ = [tex]\frac{v_{oy} }{v_{ox} }[/tex]
θ = tan⁻¹ \frac{v_{oy} }{v_{ox} }
θ = tan⁻¹ ([tex]\frac{-651.5}{187.5}[/tex])
θ = -73.94º
This angle measured from the positive side of the x-axis is
θ‘ = 360 - 73.94
θ = 286º
derive an expression for torque experiend by an electric dipole placed in a uniform electric field
Answer:
The torque τ on an electric dipole with dipole moment p in a uniform electric field E is given by τ = p × E where the "X" refers to the vector cross product. Ref: Wikipedia article on electric dipole moment.
Explanation:
define alpha and beta
alpha is the excess return on an investment after adjusting for market related volatility and random fluctuations.
beta is a measure of volatility relative to a benchmark ,such as the S&P 500.
Explanation:
alpha and beta are two different parts of an equation used to explain the performance of stocks and investments funds. But in maths alpha and beta is the Greek alphabet
What acceleration will you give a 22.4 kg box if you push it with a force of 83.1N
Answer:
mass =22.4kg
force=83.1N
a=?
f=ma
a=f/m
a=83.1/22.4
a=3.70m/s^2
Mental processes refers to
overt actions and reactions.
only animal behavior.
internal, covert processes.
outward behavior.
Which of the following is an instantaneous speed?
A: All of the above
B: 80 ft/s
C. 80 yds./min
D. 80 km/hr
Answer:
A: All of the above
Explanation:
The instantaneous speed of an object is simply the current seed of the object at any given time. The SI unit is m/S and it is a vector quantity.
Therefore, according to the given options, they all have SI units that are consistent with distance and time which makes them all an example of instantaneous speed.
Two particles are separated by 0.38 m and have charges of -6.25 x 10-°C and 2.91 x 10-°C. Use Coulomb's law to predict the force between the particles if the distance is cut in half. The equation for Coulomb's law is F = kqi 42, and the constant, k, equals 9.00 x 109 Nm2/C2 2
Answer:
-4.35 × 10^-6 N
Explanation:
i just answered it on ap3x :)
An object has an average acceleration of + 6.24 m/s ^ 2 for 0.300 s . At the end of this time the object's velocity is + 9.31 m/s . What was the object's initial velocity?
Answer:
initial velocity = 7.44 m/s (3 s.f.)
Explanation:
a = 6.24 m/s² t = 0.300 s v = 9.31 m/s u = ?
v = u + at
9.31 = u + (6.24 x 0.300)
9.31 = u + 1.872
u = 7.438
= 7.44 m/s (3 s.f.)
Hope this helps!
A student is performing an experiment that involves the charge on a metal sphere that is attached to a charged electroscope. A charged rod is brought near the sphere without touching it. As a result the leaves of the electroscope separate more. The rod is then removed, and the leaves return to their initial separated position. The student repeats the procedure, but this time the electroscope is grounded and the ground is removed before the rod is removed from near the sphere. The leaves again end up separated. What can be concluded about the charge on the separated leaves of the electroscope
Answer:
The leaves have a charge in each experiment, but the sign of the charge cannot be determined.
Explanation:
In the first experiment, A charged rod is brought near the sphere without touching it. As a result the leaves of the electroscope separate more.
Thus indicates that there are charges involved. Now, like charges would repel like what is happening here but we don't know if they are both positive or negative because in both cases, they will still repel.
Now for the second experiment, electroscope is grounded and the ground is removed before the rod is removed from near the sphere. The leaves end up being separated again.
Similar to the first time, it's clear there are charges but the charges repel. Thus, they are the same sign charges but we don't know if they are both positive or negative.
Thus, in both cases we can conclude that the leaves have charges but we don't know their signs.
You work at a garden store for the summer. You lift a bag of fertilizer with a force of 112 N, and it moves upward with an acceleration of 0.790 m/s^2.
a. What is the mass of the fertilizer bag?
b. How much does the fertilizer bag weigh?
Given :
Force provided, F = 112 N.
Acceleration of the bag, a = 0.79 m/s².
To Find :
a. What is the mass of the fertilizer bag?
b. How much does the fertilizer bag weigh?
Solution :
We know, force is given by :
F = ma
m = F/a
m = 112/0.79 kg
m = 141.77 kg
Now, weight is given by :
W = mg
W = 141.77 × 9.8 N
W = 1389.35 N
Therefore, the mass of fertilizer bag is 141.77 kg and weight us 1389.35 N.
A thin-walled vessel of volume V contains N particles which slowly leak out of a small hole of area A. No particles enter the volume through the hole. Find the time required for the number of particles to decrease to N/2. Express your answer in terms of A, V, and v.
Answer:
[tex]\frac{V}{2av}[/tex]
Explanation:
From the question we are told that
Volume V
Contains N particles
Leaks from a small hole of area A
Generally the equation for Flow rate is given as
Volume Flow Rate [tex]V_r = A * v[/tex]
Mathematically we find the time taken to flow half way which is given by
[tex]\frac{(V/2)}{A*v}[/tex]
Therefore the time taken is
[tex]\frac{V}{2av}[/tex]
The energy of a photon is ________ proportional to its wavelength.
Answer:
The amount of energy is directly proportional to the photon's electromagnetic frequency and thus, equivalently, is inversely proportional to the wavelength. The higher the photon's frequency, the higher its energy. Equivalently, the longer the photon's wavelength, the lower its energy.
Explanation:
Plz mark brainliest thanks
For an object like a planet, with a typical temperature of a few hundred kelvin, what kind of blackbody radiation would it principally emit
Answer:
Low-temperature blackbody
Explanation:
There are 3 types of blackbody temperatures.
Low-temperature blackbody
High temperature extended area blackbody
High-temperature cavity blackbody
A Low-temperature blackbody is a type of black body radiation that has the range of -40° C to 175° C, typically between 233 K and 448 K. A perfect fit for the temperature range mentioned in the question, "a few hundred Kelvin". Therefore, it's the kind of blackbody temperature that the object would emit.
If a net horizontal force of 0.8 N is applied to a toy whose mass is 1.2 kg, acceleration is?
Hello!
[tex]\large\boxed{a = \frac{2}{3}m/s^{2}}[/tex]
Use the equation F = m · a to solve. We are given the force (N) and mass (kg), so we can solve for the acceleration by plugging in the given values:
0.8 = 1.2a
0.8 / 1.2 = a
a = 2/3 m/s²
Sugar crystals enter a dryer at the rate of 1000 kg h-1 and at 20% w.b. moisture content. They leave the dryer at 3% w.b. moisture content. If the drying process requires 3000 kJ kg-1 of water removed, estimate the amount of heat required per hour and the rate of dry crystals out of the dryer.
Answer:
The dryer evaporate 200 - 24.74 kg of water per hour
To remove 1kg of water it need 3000 K J
So to remove, 175.26 Kg
it need 5.257x [tex]10^{5}[/tex] KJ of heat per hour.
Explanation:
In one hour, the amount of sugar entering = 1000 kg
w.b moisture content is defined as,
weight of water / weight of water + weight of dry
[tex]W_{w}[/tex]/[tex]W_{w}[/tex] + [tex]W_{d}[/tex] x 100
[tex]W_{w}[/tex] + [tex]W_{d}[/tex] = 1000 kg when entering
it has 20% moisture content when entering
[tex]W_{w}[/tex] = 0.2 x 1000 = 200 kg
when leaving it has 3% moisture content then weight of dry material
[tex]W_{d}[/tex] = 1000 - 200 = 800 Kg
[tex]\frac{W_{w}^{'} }{W_{w}^{'} + W_{d}^{'} }[/tex] = 0.03
[tex]\frac{W_{w}^{'} }{W_{w}^{'} + 800 }[/tex] = 0.03
[tex]W_{w} ^{'}[/tex] = 0.03 x [tex]W_{w} ^{'}[/tex] + 0.03 x 800
[tex]W_{w} ^{'}[/tex] = 24.74 kg
When leaving the dryer, the crystals has total weight of the water = 24.74 kg per hour.
The dryer evaporate 200 - 24.74 kg of water per hour
To remove 1kg of water it need 3000 K J
So to remove, 175.26 Kg
it need 5.257x [tex]10^{5}[/tex] KJ of heat per hour.
In a test run, a certain car accelerates uniformly from zero to 20.4 m/s in 2.60 s.
Required:
a. What is the magnitude of the cars acceleration?
b. How long does it take the car to change speed from 10.0 m/s to 20 m/s.
c. Will doubling the time always double the change in speed? why?
Answer:
(a) The acceleration is 7.85 m/s²
(b) It takes the car to change speed from 10.0 m / s to 20 m / s in a time of 1.27 seconds.
(c) Doubling the time will double the change in velocity if the acceleration is kept constant.
Explanation:
(a) Acceleration is the physical quantity that measures the rate of change of velocity with time. That is, acceleration relates changes in speed with the time in which they occur, that is, it measures how fast the changes in speed are.
The average acceleration is calculated using the following expression:
[tex]a=\frac{vf-vi}{t}[/tex]
where a is the acceleration, vf is the final velocity, vi is the initial velocity and t is the time.
In this case:
vf= 20.4 m/svi=0 m/st= 2.60 sReplacing:
[tex]a=\frac{20.4 \frac{m}{s} - 0\frac{m}{s} }{2.60 s}[/tex]
a= 7.85 m/s²
The acceleration is 7.85 m/s²
(b) In this case you know:
a= 7.85 m/s²vf= 20 m/svi= 10 m/sReplacing:
[tex]7.85 \frac{m}{s^{2} } =\frac{20 \frac{m}{s} - 10\frac{m}{s} }{t}[/tex]
and solving you get:
[tex]t=\frac{20 \frac{m}{s} - 10\frac{m}{s} }{7.85 \frac{m}{s^{2} } }[/tex]
t=1.27 s
It takes the car to change speed from 10.0 m / s to 20 m / s in a time of 1.27 seconds.
(c) Being:
[tex]a=\frac{vf-vi}{t}[/tex]
Then:
a*t= vf - vi
vf - vi represents the change in velocity. You can see that, if a (acceleration) is constant, then (vf - vi) is directly proportional to the time t: therefore, if t doubles, the change in velocity doubles as well.
In other words, doubling the time will double the change in velocity if the acceleration is kept constant.
An astronaut weighing 190 lbs on Earth is on a mission to the Moon and Mars.
Required:
a. What would he weigh in newtons when he is on the Moon?
b. How much would he weigh in newtons when he is on Mars, where the acceleration due to gravity is 0.38 times that on Earth?
Answer:
The weight is defined as:
W = m*g
where:
m = mass
g = gravitational acceleration.
We know that in Earth the astronaut weights 190 lb-f (this is force, not mass, the correct unit here is 190 lb*m/s^2)
then:
190 lb*m/s^2 = m*9.8m/s^2
(190 lb*m/s^2)/(9.8m/s^2) = 19.39 lb
Now we know the mass of the astronaut.
a) wieght on the moon in Newtons.
Newtons uses kilograms as the units of mass, then we need to rewrite the mass of the astronaut in kg.
we know that 1lb = 0.454 kg
Then 19.39 lb is equal to: 19.39*0.454 kg = 8.8 kg
We know that the acceleration due to gravity on the Moon is one-sixth that on Earth.
then: g = (9.8m/s^2)/6
And the weight of the astronaut in the moon will be:
W = 8.8 kg*(9.8m/s^2)/6 = 14.37 N
b) The weight on mars, where the acceleration due to gravity is 0.38 times that on Earth, we have:
g = (9.8m/s^2)*0.38
then the weight will be:
W = 8.8kg*(9.8m/s^2)*0.38 = 32.77 N
a 150kg roller coaster SITTING ON THE TOP OF A 200M HILL HAS HOW MUCH POTETNTIAL ENERGY
Answer:
Epot = 294300 [J]
Explanation:
Potential energy is defined as the product of mass by height by gravitational acceleration. The height is measured with respect to the reference level. At this reference level the potential energy is equal to zero.
[tex]E_{pot}=m*g*h\\[/tex]
where:
m = mass = 150 [kg]
g = gravity acceleration [m/s²]
h = elevation = 200 [m]
[tex]E_{pot}=150*9.81*200\\E_{pot}=294300 [J][/tex]
A solid is 5 cm tall, 3 cm wide, and 2 cm thick. It has a mass of 129 g. What is its
density?
Answer:
4.3 g/cm³ or 4.3g/cc
Explanation:
Volume(V) = Height × Length × Width
= 5cm × 3cm × 2cm
= 30cm³
Mass(m) = 129gram
So,
Density = m/V
= 129g/30cm³
= 4.3g/cc or 4.3g/cm³