How much work is done when 100 N of force is applied to a rock to move it 20 m

find the gravitational force this shell exerts on a 1.60 kg point mass placed at the distance 5.01 m from the center of the shel

RideAnS [48]

The gravitational force of the shell exerts is 4.25m x 10¯¹² N.

We need to know about gravitational force to solve this problem. The gravitational force is the force caused by two masses of objects. The magnitude of gravitational force can be determined as

F = G.m1.m2 / R²

where F is the gravitational force, G is the gravitational constant (6.674 × 10¯¹¹ Nm²/kg²), m1 and m2 are the mass of the object and R is the radius.

From the question above, we know that

m1 = 1.6 kg

m2 = m

R = 5.01 m

By substituting the following parameters, we get

F = G.m1.m2 / R²

F = 6.674 × 10¯¹¹ . 1.6 . m / 5.01²

F = 4.25m x 10¯¹² N

where m is the mass of the shell

For more on gravitational force at: brainly.com/question/19050897

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7 0

1 year ago

What is the velocity of the car, in meters per second, just after it hits a 135-kg deer initially running at 10.5 m/s in the sam

Gnom [1K]

Answer:

V = 27.46 m/s

Explanation:

given,

mass of the deer(m) = 135 Kg

speed of the deer (u) = 10.5 m/s

assuming,

mass of the car(M) = 900 Kg

initial velocity of car (v) = 30 m/s

using conservation of momentum

m u + M v = (M + m )V

V is the velocity of the car as deer is on the car

135 x 10.5 + 900 x 30 = (900 + 135 ) V

28417.5 = 1035 V

V = 27.46 m/s

so, the velocity of car is equal to V = 27.46 m/s

5 0

4 years ago

A car with a mass of 1,000 kg moving with an initial velocity of 8 m/s is brought to rest by colliding with a truck. The collisi

ruslelena [56]

Answer:

-40,000 N

Explanation:

First, use the kinematics equation v(f) = v(i) + at. Final velocity is 0, initial is 8, and time is 0.2 seconds. Solving for a, you get -40 m/s^2. Then, use Newton’s second law, F=ma, to find the force. F = (1000)(-40) = -40,000 N.

4 0

3 years ago

53. A recently discovered planet has a mass four times as great as Earth's and a radius twice as large as Earth's. What will be

marusya05 [52]

Answer:

$g' = g = 9.81 m/s^2$

so gravity will be same as that of surface of earth

Explanation:

As we know that the acceleration due to gravity is given as

$g = \frac{GM}{R^2}$

here we have

$M = 4M_e$

$R = 2R_e$

we know that for earth we have

$g = 9.81 = \frac{GM_e}{R_e^2}$

now if the radius and mass is given as above

$g' = \frac{G(4M_e)}{(2R_e)^2}$

$g' = \frac{GM_e}{R_e^2}$

$g' = g = 9.81 m/s^2$

so gravity will be same as that of surface of earth

6 0

3 years ago

The titanium shell of an SR-71 airplane would expand when flying at a speed exceeding 3 times the speed of sound. If the skin of

Fed [463]

Answer:

The 10-meter long rod of an SR-71 airplane expands 0.02 meters (2 centimeters) when plane flies at 3 times the speed of sound.

Explanation:

From Physics we get that expansion of the rod portion is found by this formula:

$\Delta l = \alpha\cdot l_{o}\cdot (T_{f}-T_{o})$ (Eq. 1)

Where:

$\Delta l$ - Expansion of the rod portion, measured in meters.

$\alpha$ - Linear coefficient of expansion for titanium, measured in $\frac{1}{^{\circ}C}$ .

$l_{o}$ - Initial length of the rod portion, measured in meters.

$T_{o}$ - Initial temperature of the rod portion, measured in Celsius.

$T_{f}$ - Final temperature of the rod portion, measured in Celsius.

If we know that $\alpha = 5\times 10^{-6}\,\frac{1}{^{\circ}C}$ , $l_{o} = 10\,m$ , $T_{o} = 0\,^{\circ}C$ and $T_{f} = 400\,^{\circ}C$ , the expansion experimented by the rod portion is:

$\Delta l = \left(5\times 10^{-6}\,\frac{1}{^{\circ}C} \right)\cdot (10\,m)\cdot (400\,^{\circ}C-0\,^{\circ}C)$

$\Delta l = 0.02\,m$

The 10-meter long rod of an SR-71 airplane expands 0.02 meters (2 centimeters) when plane flies at 3 times the speed of sound.

4 0

3 years ago