A 38,500 kg sphere is located 2.55 m from a 15,400 kg sphere. What is the gravitational force, rounded to the nearest thousandth of a Newton, between the two spheres?

Answer: An electroscope is a device that detects static electricity by using thin metal or plastic leaves, which separate when charged. ... Electrical charges move to the metal and down to the foil leaves, which then repel each other. Since each leaf has the same charge (positive or negative), they repel each other.

Answer:

An object with a suspected static electric charge is brought near the metal plate or ball of the electroscope. Electrical charges move to the metal and down to the foil leaves, which then repel each other. Since each leaf has the same charge (positive or negative), they repel each other.

Answer:

The change in momentum of the car is  30,000 kg m/s                  

Explanation:

Given,

The force exerted on the car to slow down, F = 7500 N

The time period of force, t = 4 s

The rate of change of momentum of the object is equal to the force acting on it.

Therefore,

                                    (mv - mu) / t = F

Where v and u are the final and initial velocity of the car. The change in momentum of the car,

                                    mv - mu = F x t

                                                   = 7500 x 4

                                                   = 30,000 kg m/s

Hence the change in momentum of the car is  30,000 kg m/s                  

Answer:

The sound travels at [tex]v_{s}=11.4 \mathrm{m} / \mathrm{s}[/tex]

Option: c

Explanation:

Unknown source plays of middle C (fs) = 262 Hz

The sound wave from this source have to travel to raise the pitch to C sharp is (fd) = 272 Hz

[tex]\begin{array}{l}{velocity of sound in air(v)=343 \mathrm{m} / \mathrm{s}, f_{\mathrm{s}}=262 \mathrm{Hz}, f_{\mathrm{d}}=271 \mathrm{Hz}} \\ {velocity of receiver(v_{\mathrm{d}})=0 \mathrm{m} / \mathrm{s},velocity of source( v_{\mathrm{s}}) \text { is unknown }}\end{array}[/tex]

[tex]\text { Speed of sound } \mathrm{V}_{\mathrm{S}}=343 \mathrm{m} / \mathrm{s}[/tex]

[tex]f_{\mathrm{d}}=f_{\mathrm{s}}\left(\frac{v-v_{\mathrm{d}}}{v-v_{\mathrm{s}}}\right)[/tex]

[tex]\frac{f_{d}}{f_{s}}=\left(\frac{v-v_{d}}{v-v_{s}}\right)[/tex]

[tex]\left(v-v_{s}\right)=\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)[/tex]

[tex]v_{s}=v-\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)[/tex]

Substitute the given values in the formula,

[tex]v_{s}=343+\frac{262}{271}(343-0)[/tex]

[tex]v_{s}=343+0.966(343)[/tex]

[tex]v_{s}=343-331.33[/tex]

[tex]v_{s}=11.4 \mathrm{m} / \mathrm{s}[/tex]

Therefore, The sound travels at [tex]v_{s}=11.4 \mathrm{m} / \mathrm{s}[/tex]

Answer:

Temperature decreases because the number of collision of the molecules decreases as they escape or evaporate. Molecules are in constant motion. Increase in temperature leads to increase in average kinetic energy of the molecules.

The kinetic theory of matter states that all matter is made up of tiny particles that are constantly moving. The faster the particles are moving, the higher the temperature of the matter.

When a liquid evaporates, the fastest-moving particles are the ones that are most likely to escape from the surface of the liquid. This leaves behind the slower-moving particles, which means that the average temperature of the liquid decreases.

Another way to think about it is that evaporation is a cooling process because it removes the fastest-moving particles from the liquid. These particles have the most kinetic energy, so their removal lowers the average kinetic energy of the remaining particles and thus lowers the temperature of the liquid.

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1) Mass of the continent: [tex]2.2\cdot 10^{21} kg[/tex]

2) Kinetic energy: 1118 J

3) Speed of the jogger: 5.3 m/s

Explanation:

1)

First of all, we calculate the volume of the continent. It is a slab of side

[tex]L=5100 km = 5.1\cdot 10^6 m[/tex]

and thickness

[tex]t=30 km = 3.0\cdot 10^4 m[/tex]

So its volume is

[tex]V=tL^2=(3.0\cdot 10^4)(5.1\cdot 10^6)^2=7.8\cdot 10^{17} m^3[/tex]

The density of the slab is

[tex]\rho = 2850 kg/m^3[/tex]

Therefore, we can calculate the mass using the relationship

[tex]\rho = \frac{m}{V}[/tex]

where m is the mass. And solving for m,

[tex]m=\rho V=(2850)(7.8\cdot 10^{17})=2.2\cdot 10^{21} kg[/tex]

2)

The kinetic energy of the continent is given by

[tex]K=\frac{1}{2}mv^2[/tex]

where

[tex]m=2.2\cdot 10^{21} kg[/tex] is its mass

v = 3.8 cm/year is its speed

We have to convert the speed into m/s. Keeping in mind that

1 cm = 0.01 m

[tex]1 year = 365\cdot 24\cdot 60 \cdot 60 =3.15\cdot 10^7 s[/tex]

We find

[tex]v=3.18 \frac{cm}{y} \cdot \frac{0.01}{365\cdot 24 \cdot 60 \cdot 60}=1.0\cdot 10^{-9} m/s[/tex]

So now we can find the kinetic energy:

[tex]K=\frac{1}{2}(2.2\cdot 10^{21})(1.0\cdot 10^{-9})^2=1118 J[/tex]

3)

The kinetic energy of the jogger is given by

[tex]K=\frac{1}{2}m'v'^2[/tex]

where

m' = 80 kg is the mass of the jogger

v' is the speed of the jogger

Here we want the jogger to have the same kinetic energy of the continent, so

[tex]K=1118 J[/tex]

And by re-arranging the equation, we can find what speed the jogger must have:

[tex]v'=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1118)}{80}}=5.3 m/s[/tex]

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b) the net force on the car is zero.

Explanation:

Let's analyze each option one by one:

a) the force from the engine is greater than all the forces of friction.  --> FALSE. In fact, the car is moving at constant velocity: this means that its acceleration is zero,

a = 0

and so Newton's second law becomes

[tex]\sum F = ma = 0[/tex]

where [tex]\sum F[/tex] is the net force on the car and m is its mass. This means that the net force on the car is zero: so, the force from the engine cannot be greater than all the forces of friction, otherwise the net force cannot be zero.

b) the net force on the car is zero.  --> TRUE, for what we said at point A)

c) the inertia is changing.  --> FALSE. The inertia of an object just depend on the mass and the velocity of the object: as neither the mass nor the velocity are changing in this problem, then the inertia of the car is not changing.

d) the forces of friction are proportional to the acceleration of the car.  --> FALSE. Generally, the force of friction acting on an object moving on a flat surface is

[tex]F_f = \mu mg[/tex]

where [tex]\mu[/tex] is the coefficient of friction, m is the mass, and g the acceleration of gravity. Therefore, the force of friction does not depend on the acceleration of the car.

e) All of the above. --> FALSE

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a) The horizontal component of the velocity is 7.4 m/s

b) The vertical component of the velocity is 3.8 m/s

c) The balloon reaches the highest point after 0.39 s

d) The maximum height is 0.74 m

e) The total time of flight is 0.78 s

f) The range of the balloon is 5.77 m

Explanation:

a)

The motion of the balloon is the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

The horizontal component of the velocity (which is constant) is given by

[tex]v_x = u cos \theta[/tex]

where

u = 8.3 m/s is the initial velocity of the balloon

[tex]\theta=27^{\circ}[/tex] is the angle of projection

Substituting,

[tex]v_x = (8.3)(cos 27^{\circ})=7.4 m/s[/tex]

b)

The vertical component of the initial velocity of a projectile is given by

[tex]u_y = u sin \theta[/tex]

where

u is the initial velocity

[tex]\theta[/tex] is the angle of projection

Here we have

u = 8.3 m/s

[tex]\theta=27^{\circ}[/tex]

Substituting,

[tex]u_y = (8.3)(sin 27^{\circ})=3.8 m/s[/tex]

c)

The vertical component of the velocity of the balloon follows the suvat equation

[tex]v_y = u_y - gt[/tex]

where

[tex]v_y[/tex] is the vertical velocity at time t

[tex]u_y = 3.8 m/s[/tex] is the initial vertical velocity

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

The balloon reaches the maximum height when the vertical velocity becomes zero:

[tex]v_y = 0[/tex]

So we get:

[tex]0=u_y -gt\\t=\frac{u_y}{g}=\frac{3.8}{9.8}=0.39 s[/tex]

d)

The maximum height of the balloon can be calculated using the suvat equation:

[tex]s=u_y t - \frac{1}{2}gt^2[/tex]

where

[tex]u_y = 3.8 m/s[/tex] is the initial vertical velocity

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

t = 0.39 s is the time at which the highest point is reached

Substituting,

[tex]s=(3.8)(0.39)-\frac{1}{2}(9.8)(0.39)^2=0.74 m[/tex]

e)

The total time of flight of a projectile is twice the time needed to reach the maximum height, and it is given by

[tex]t=\frac{2u_y}{g}[/tex]

where

[tex]u_y[/tex] is the initial vertical velocity

[tex]g[/tex] is the acceleration of gravity

Here we have

[tex]u_y = 3.8 m/s[/tex]

[tex]g=9.8 m/s^2[/tex]

Substituting,

[tex]t=\frac{2(3.8)}{9.8}=0.78 s[/tex]

f)

The range of a projectile is the horizontal distance covered by the projectile, so it can be found by multiplying its horizontal velocity (which is constant) by the time of flight:

[tex]d=v_x t[/tex]

where

[tex]v_x[/tex] is the horizontal velocity

t is the time of flight

Here we have

[tex]v_x = 7.4 m/s[/tex]

[tex]t = 0.78 s[/tex]

Substituting,

[tex]d=(7.4)(0.78)=5.77 m[/tex]

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The maximum tension in the rope during the swing is the sum of the gravitational force and the centripetal force, which can be calculated with the given mass, rope length, and speed, resulting in a tension of 1602.48 N.

To calculate the maximum tension in the rope during the swing, we need to consider the forces acting on the man at the lowest point of the swing. These forces are the gravitational force (weight) and the centripetal force required to maintain the circular motion. The tension in the rope is the sum of these two forces.

The gravitational force (Fg) on the man can be calculated using the equation Fg = m  imes g, where m is the man's mass (88.00 kg) and g is the acceleration due to gravity (9.81 m/s2). The centripetal force (Fc) required for circular motion can be calculated using the equation Fc = m imes v2 / r, where v is the man's speed (10.20 m/s) at the bottom of the swing and r is the radius of the swing, which is equal to the length of the rope (12.0 meters).

Therefore, the maximum tension (T) in the rope would be:

T = Fg + Fc

T = (88.00 kg  imes 9.81 m/s2) + (88.00 kg  imes (10.20 m/s)2 / 12.0 m)

Calculating each part:

Fg = 863.28 N (2 decimal places)

Fc = 739.20 N (2 decimal places)

Thus, the maximum tension in the rope would be:

T = 863.28 N + 739.20 N

T = 1602.48 N (rounded to two decimal places)

Tension, centripetal force, and gravitational force are crucial components in determining the tension in the rope during circular motion.

The distance covered is 30.2 m

Explanation:

The motion of Cynthia is a uniformly accelerated motion (constant acceleration), so we can find the distance covered by using the following suvat equation:

[tex]s=ut+\frac{1}{2}at^2[/tex]

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem,

u = 0 (since Cynthia started from rest)

t = 3.2 s

[tex]a=5.9 m/s^2[/tex] is her acceleration

Substituting,

[tex]s=0+\frac{1}{2}(5.9)(3.2)^2=30.2 m[/tex]

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Using the equation of motion S = ut + ½at2, we calculated that Susan traveled 30.216 meters during the initial stage of her downhill luge, where she accelerated at 5.9 m/s2 for 3.2 seconds.

To calculate how far Susan traveled during the initial stage of her downhill luge, where she accelerated from rest at 5.9 m/s2 for 3.2 seconds, we use the equation of motion for constant acceleration:

S = ut + ½at2

Here, S represents the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.

Since Susan starts from rest, her initial velocity u = 0. We plug in the acceleration a = 5.9 m/s2 and time t = 3.2 s into the equation:

S = (0)(3.2) + ½(5.9)(3.2)2

S = 0 + ½(5.9)(10.24)

S = 30.216 m

Therefore, Susan traveled 30.216 meters during the initial stage of her downhill luge.

Answer:

The height of the hill is, h = 38.42 m

Explanation:

Given,

The horizontal velocity of the soccer ball, Vx = 15 m/s

The range of the soccer ball, s = 42 m

The projectile projected from a height is given by the formula

                             S = Vx [Vy + √(Vy² + 2gh)] / g

Therefore,

                            h = S²g/2Vx²                     (Since Vy = 0)

Substituting the values

                             h = 42² x 9.8/ (2 x 15²)

                                = 38.42 m

Hence, the height of the hill is, h = 38.42 m

Answer:

The work done by the force of gravity, W = 0 J

Explanation:

Given data,

The weight of the box, w = 1500 N

Therefore the gravitational force acting on the box is , F = 1500 N

The displacement of the box S = 0

The work done is defined as the product of force acting on the body to the displacement it caused.

The formula for work done is,

                                           W = F x S

                                                = 1500 N x 0

                                                = 0

Hence. the work done by the force of gravity is, W = 0 J

The work done by the force of gravity on a box weighing 1500 N depends upon the distance the box is moved. If we assume a free fall situation, the work done would be calculated using the formula: W = 1500N * h, where h is the height from which the box falls.

The calculation of the work that the force of gravity does on an object comes down to two important factors: the force of gravity acting on the object, represented as weight, and the distance the object is moved. In this case, the box weighs 1500 N, which is the force of gravity acting on it. If we assume that the box is in free fall, i.e., it's falling to the ground under gravity, we can calculate the work done by the force of gravity.

To calculate this, we need to know the height from which the box is falling (let's denote that as h). The work done, according to the work-energy principle, is equal to the force times the distance covered and the cosine of the angle between force and distance vectors. Considering the angle between the gravity force and the displacement is 0 degrees in a free fall, the work done by gravity is W = 1500N * h * cos(0) = 1500N * h. Here, cos(0) is equal to 1.

As you can see, the work done by gravity relies heavily on the distance/height textbox falls. If the box doesn't move, then no work is done by gravity even though the force of gravity exists, because work is dependent on displacement.

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In a collision between two 300 kg spaceships, one at rest and the other moving at 6 m/s, the final velocity after they stick together is 2 m/s.

When two objects with masses of 300 kg collide and stick together, their momentum is conserved. Utilizing the principle of conservation of momentum, we can calculate their final velocity after the collision.

In this case, since both spaceships have equal masses and one is at rest while the other moves with a known speed, the final velocity after the collision would be 2 m/s.

a) The frequency of the photon is [tex]7.16\cdot 10^{20}Hz[/tex]

b) The wavelength of the photon is [tex]4.19\cdot 10^{-13} m[/tex]

c) The wavelength of the photon is about 100 times larger than the nuclear radius

Explanation:

a)

The energy of a photon is given by

[tex]E=hf[/tex] (1)

where:

[tex]h=6.63\cdot 10^{-34} Js[/tex] is the Planck constant

f is the frequency of the photon

The photon in this problem has an energy of

[tex]E=2.5 MeV = 2.5\cdot 10^6 eV[/tex]

And keeping in mind that

[tex]1eV = 1.6\cdot 10^{-19} J[/tex]

we can convert to Joules:

[tex]E=(2.5\cdot 10^6)(1.9\cdot 10^{-19})=4.75\cdot 10^{-13} J[/tex]

And now we can use eq.(1) to find the frequency of the photon:

[tex]f=\frac{E}{h}=\frac{4.75\cdot 10^{-13}}{6.63\cdot 10^{-34}}=7.16\cdot 10^{20}Hz[/tex]

b)

The wavelength of a photon is related to its frequency by the equation

[tex]c=f\lambda[/tex]

where

[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light

f is the frequency

[tex]\lambda[/tex] is the wavelength

For the photon in this problem,

[tex]f=7.16\cdot 10^{20}Hz[/tex]

Re-arranging the equation, we find its wavelength:

[tex]\lambda=\frac{c}{f}=\frac{3\cdot 10^8}{7.16\cdot 10^{20}}=4.19\cdot 10^{-13} m[/tex]

c)

The size of the nuclear radius is approximately

[tex]d \sim 10^{-15} m[/tex]

While we see that the wavelength of this photon is

[tex]\lambda=4.19\cdot 10^{-13} m[/tex]

Therefore, the ratio between the wavelength of the photon and the nuclear radius is

[tex]\frac{\lambda}{d}=\frac{\sim 10^{-13}}{\sim \cdot 10^{-15}}=100[/tex]

So, the wavelength of the photon is approximately a factor 100 times larger than the nuclear radius.

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Answer:

The mass of the earth is, M = 5.98 x 10²⁴ kg

Explanation:

Given data,

The acceleration due to gravity, g = 9.8 m/s²

The radius of Earth, R = 6400 km

                                    =6.4 x 10⁶ m

The universal gravitational constant, G = 6.67 x 10 ⁻¹¹ Nm²kg⁻²

Using the equation,

                         GM = gR²

                            M = gR² / G

Substituting the values,

                            M = 9.8 x (6.4 x 10⁶)² / 6.67 x 10 ⁻¹¹

                                = 5.98 x 10²⁴ kg

Hence, the mass of the earth is, M = 5.98 x 10²⁴ kg

The acceleration of the racing car is 2 m/s² and the displacement over the 11 seconds is 605 m.

The given physics problem involves calculating the acceleration and the displacement of a racing car that has increased its velocity from 44 m/s to 66 m/s in a time interval of 11 seconds.

To find the acceleration, we use the formula for average acceleration, which is the change in velocity divided by the time taken. The change in velocity (Δv) is 66 m/s - 44 m/s = 22 m/s. So, the acceleration (a) is 22 m/s divided by 11 s, which equals 2 m/s².

For the displacement, we can use the kinematic equation s = ut + 0.5at², where u is the initial velocity, a is the acceleration, and t is the time. With u = 44 m/s, a = 2 m/s², and t = 11 s, we get the displacement s = 44 m/s * 11 s + 0.5 * 2 m/s² * (11 s)², resulting in 484 m + 121 m, or a total displacement of 605 m.

The acceleration of the racing car is [tex]\( 2 \text{ m/s}^2} \)[/tex] and the displacement of the racing car during this time is [tex]\( 605 \text{ m}} \)[/tex].

To find the acceleration and displacement of the racing car, we will use the following kinematic equations:

1. Acceleration [tex](\(a\))[/tex]:

[tex]\[ a = \frac{\Delta v}{\Delta t} \][/tex]

2. Displacement [tex](\(s\))[/tex]:

[tex]\[ s = v_i t + \frac{1}{2} a t^2 \][/tex]

Step 1: Calculate the Acceleration

Given:

Initial velocity [tex](\(v_i\))[/tex]: 44 m/s

Final velocity [tex](\(v_f\))[/tex]: 66 m/s

Time interval [tex](\(\Delta t\))[/tex]: 11 s

The change in velocity [tex](\(\Delta v\))[/tex] is:

[tex]\[ \Delta v = v_f - v_i = 66 \text{ m/s} - 44 \text{ m/s} = 22 \text{ m/s} \][/tex]

Using the acceleration formula:

[tex]\[ a = \frac{\Delta v}{\Delta t} = \frac{22 \text{ m/s}}{11 \text{ s}} = 2 \text{ m/s}^2 \][/tex]

Step 2: Calculate the Displacement

Using the kinematic equation for displacement:

[tex]\[ s = v_i t + \frac{1}{2} a t^2 \][/tex]

Given:

Initial velocity [tex](\(v_i\))[/tex]: 44 m/s

Time [tex](\(t\))[/tex]: 11 s

Acceleration [tex](\(a\))[/tex]: [tex]2 m/s^2[/tex]

Substitute the values into the equation:

[tex]\[ s = 44 \text{ m/s} \times 11 \text{ s} + \frac{1}{2} \times 2 \text{ m/s}^2 \times (11 \text{ s})^2 \][/tex]

[tex]\[ s = 484 \text{ m} + \frac{1}{2} \times 2 \text{ m/s}^2 \times 121 \text{ s}^2 \][/tex]

[tex]\[ s = 484 \text{ m} + 1 \times 121 \text{ m} \][/tex]

[tex]\[ s = 484 \text{ m} + 121 \text{ m} \][/tex]

[tex]\[ s = 605 \text{ m} \][/tex]

Answer:

iron

Explanation:

Increasing the number of coils of wire wrapped around the nail increases the strength of the electromagnet, as measured by the number of paper clips the magnet can pick up

iron

Explanation:

Increasing the number of coils of wire wrapped around the nail increases the strength of the electromagnet, as measured by the number of paper clips the magnet can pick up

Answer:

15N

Explanation:

According to Newton's Second Law of Motion

F = m*a

mass = m = 5Kg

acceleration = a = 3m/s^2

=> F = 5kg * 3m/s^2

=> F = 15 N

F = m a

The force is (5kg)x(3 m/s^2)= 15 Newtons, and it's still there.

If the force stops, the acceleration stops.

The acceleration of an object in circular motion is calculated by the formula ac = v²/r. Given that in case a, both the speed and radius are twice that of case b, the acceleration in case a is twice that of case b. This shows the squared relationship between speed and acceleration in circular motion.

In physics, the acceleration of an object moving in a circular path, also known as centripetal acceleration, is calculated by the formula ac = v²/r where v is the velocity or speed of the object and r is the radius of the circle.

In case a, the speed is twice that of case b and the circle's radius is also two times of case b. If we plug these values into our formula, we can see that the acceleration in case a (let's call it acA) could be represented as acA = (2v)² / 2r = 4v²/2r = 2v²/r. In case b, the acceleration (acB) would be represented as acB = v² / r.

Hence, comparing both, you can see the acceleration in case a is twice that in case b. Therefore, even if the speed and the radius are increased by the same factor, the acceleration increases by a factor of that increase due to the squared relationship between velocity and acceleration in circular motion.

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The acceleration in case A is twice that of case B.Given the conditions that the speed and radius of circle in case A are twice those in case B.

Let's analyze the relationship between acceleration, speed, and radius in circular motion. We know the acceleration of an object in uniform circular motion can be given by the equation:

|a| = |v|²/r

From the problem, we have two cases:

[tex]Case A: Speed (|v_A|) is \twice \ that \ of \ case \ B (|v_B|).\\\\Case A: Radius (|r_A|) \ is \ also \ two\ times \ that\ of \ case\ B (|r_B|).[/tex]

Let's denote the acceleration in case A as [tex]|a_A|[/tex] and in case B as [tex]|a_B|[/tex]

For case A:

[tex]|a_A| = |v_A|^2 / |r_A|[/tex]

For case B:

[tex]|a_B| = |v_B|^2 / |r_B|[/tex]

Given that [tex]|v_A| = 2|v_B|[/tex] and [tex]|r_A| = 2|r_B|[/tex], substituting these into the equations gives:

[tex]|a_A| = (2|v_B|)^2 / (2|r_B|)[/tex]

Simplifying this, we get:

[tex]|a_A| = 4|v_B|^2 / 2|r_B| = 2(|v_B|^2 / |r_B|) = 2|a_B|[/tex]

Therefore, the acceleration in case A is twice that of case B.

Answer:

180

Explanation:

The Moon's surface gravity is about 1/6th as powerful or about 1.6 meters per second per second. The Moon's surface gravity is weaker because it is far less massive than Earth. A body's surface gravity is proportional to its mass, but inversely proportional to the square of its radius.

Answer:

160kg

Explanation:

Given m= 95kg g= 1.6345m/s²

Unknown Fg= ?

Formula Fg=mg

Fg=(95)(1.6345)

Fg= 155.2775

Fg=160kg

Answer:

Answer: C

explanation:

They could be same or different

ie:

|-5|,|-5| = 5,5

|-5|,|5| = 5,5

Answer:

C

Explanation:

9.8 m/s/s as well

Answer:

look at the explanation

Explanation:

This pith-ball electroscope is used to detect the presence of a static electricity charge. The two lightweight “pith” balls suspended from the strings are attracted to objects with a static electric charge. The pith balls can also be charged by touching them to an object with a static electric charge.

Answer:

B

Explanation:

F = ma , a = F/m

a1 = F/10 and a2 = F/4

Since Force is constant, a2 will we greater than a1

Answer:

A

Explanation:

Answer:

a

Explanation:

a

Answer:

The acceleration of the train is 5 m/s².

Explanation:

Given:

let the initial velocity of a train = 5 m/s and

final velocity of a train = 45 m/s

time taken = 8 s

To find:

acceleration: ?

Solution:

We define acceleration as change in velocity per unit time that is the difference between the final velocity and initial velocity divided by time.

[tex]Acceleration = \frac{\textrm{final velocity} - \textrm{initial velocity}}{time} \\[/tex]

On substituting the above values we get the required acceleration

[tex]Acceleration = \frac{45 - 5}{8}\\ =\frac{40}{8}\\ =5\ m/s^{2}[/tex]

Therefore,the acceleration of the train is 5 m/s².

Answer:

B)velocity of object decreases

Explanation:

Consider the positive x axis as positive direction

Assume a body moving in negative x-axis direction

It's acceleration also alone negative x-axis direction

So according to our consideration

velocity and acceleration values are negative

That is both are towards negative x direction

But as both velocity and acceleration are in same direction, MAGNITUDE of velocity increases

But as magnitude increases in negative direction, velocity value decreases

But speed value increases(As speed is scalar and velocity is a vector)

The correct answer is a. potential energy. At the top of a hill, right before it plunges downward, a roller coaster car has potential energy due to its position relative to the ground. This potential energy is converted into kinetic energy as the car moves downward and gains speed.

The force of friction is 25 N

Explanation:

For this problem, we can apply Newton's second law of motion along the horizontal direction:

[tex]\sum F_x = ma_x[/tex]

where

[tex]\sum F_x[/tex] is the net force in the horizontal direction

m is the mass of the block

[tex]a_x[/tex] is the horizontal acceleration

Here  the block is moving at constant speed, so its acceleration is zero, therefore:

[tex]a=0 \rightarrow \sum F_x = 0[/tex] (1)

The net force in the horizontal direction can be written as:

[tex]\sum F_x = Fcos \theta -F_f[/tex] (2)

where

Combining (1) and (2), we find the magnitude of the force of friction:

[tex]Fcos \theta -F_f = 0\\F_f = F cos \theta =(50)(cos 60^{\circ})=25 N[/tex]

Learn more about friction:

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Answer:

SOUND ENERGY is the energy which gets released or formed.

Explanation:

Remember , Energy is ALWAYS conserved. It can neither be destroyed nor be created(made).

∴

Energy gets converted significantly to sound energy - from the following options.

Final answer:

The type of energy made by a door slamming is sound energy, which is a result of the door's mechanical energy being partially transformed into pressure waves in the air.

Explanation:

When a door slams, it produces sound energy as a result of the forceful closure that creates pressure waves in the air we perceive as sound. While the door itself experiences mechanical energy, which is necessary for the motion of slamming, the type of energy being asked about in this case is the one that is made as a consequence of that action, which is sound energy. Mechanical energy is the sum of kinetic and potential energy within a system and is responsible for physical movement, as when a person pushes a door causing it to slam. Upon impact, part of this mechanical energy is transformed into sound energy.

Answer:

(C) Plants use oxygen to make their own food

Answer:

B.

Explanation:

We can use process of elimination here.  A is not correct.  Animals do not make their own food.  C is not correct.  Plants use carbon dioxide, not oxygen to power photosynthesis.  D is not correct.  Decomposers break down dead animals.  Therefore, B is correct.  Plants are food for some consumers.