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kompoz [17]
3 years ago
12

An object is allowed to fall freely near the surface of an unknown planet. The object falls 72 meters from rest in 4.0 seconds.

The acceleration due to gravity on that planet is
Physics
2 answers:
dimaraw [331]3 years ago
7 0
<span>it fairly is going to attain a speed of 24 m/s in a 2d, yet between t = 0 and t = a million, it fairly is not any longer vacationing at that speed, yet at slower speeds. it fairly is 12 meters. ?D = [ ( a?T^2 + 2?Tv_i ) ] / 2 the place: ?D = displacement a = acceleration ?T = elapsed time v_i = preliminary speed ?D = [ ( 24m/s^2 • 1s • 1s + 2 • 1s • 0m/s ) ] / 2 ?D = 24 / 2 ?D = 12m</span>
Zanzabum3 years ago
4 0

Answer:

Acceleration due to gravity on that planet, a = 9 m/s²

Explanation:

Initial velocity, u =  0 m/s

Acceleration due to gravity , a = ?

Displacement, s = 72 m

Time , t = 4 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

    72 = 0 x 4 + 0.5 x a x 4²

   a = 9 m/s²

Acceleration due to gravity on that planet, a = 9 m/s²

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) a 1.0 kilogram laboratory cart moving with a velocity of 0.50 meter per second due east collides with and sticks to a similar
ZanzabumX [31]
Momentum would be the same before and after the collision
 Before the collision:
 Momentum of the single cart: 1 * 0.50 = 0.50
 After the collision
 velocity = 0.25m / s
 1 * 0.25 + 1 * 0.25 =
 0.25 * (1 + 1) =
 0.25 * 2 =
 0.50
 Now new momentum will be 0.5
 answer
 the same before and after the collision
4 0
3 years ago
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The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas
Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 m,/sec^2

(c) Escape velocity on earth is 3.563\times 10^4m/sec

Explanation:

We have given radius of mars R_{mars}=6.9\times 10^3km=6.9\times 10^6m and radius of earth R_{E}=1.3\times 10^4km=1.3\times 10^7m

Mass of earth M_E=5.972\times 10^{24}kg

So mass of mars M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg

Volume of mars V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3

So density of mars d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3

Volume of earth  V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3

So density of earth d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3

(A) So the ratio of mean density \frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735

(B) Value of g on mars

g is given by g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2

(c) Escape velocity is given by

v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec

5 0
3 years ago
Read 2 more answers
To move a heavy object, like a refrigerator what could be used to help decrease the frictional force?
Olenka [21]

Answer:

A. Remove everything in the refrigerator to lighten the load.

B. Put a lubricant between the surface of the object and the floor

C. Use round objects, like pencils , to decrease the friction and push the refrigerator over the pencils more easily

Explanation:

Force of friction is a resistance force which acts between two surfaces which are in relative motion. Friction is both boon and bane. Due to friction, we are able to sit, walk etc but also, due to friction there is dissipation of energy. Friction can be reduced by applying lubricants, reducing contact area, reducing the load.

F = μN where N is the normal force which depends on the mass.

Thus, by reducing the load, force of friction can be reduced. Round objects like wheels can also be used. By this the contact area reduces.

8 0
3 years ago
What is the relationship between increase speed of a car and energy in a<br> collision?
sweet [91]
The relationship between a car and energy is that the car uses gas to produce speed within energy needs to be powered
6 0
3 years ago
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A box slides downwards at a constant velocity on an inclined surface that has a coefficient of friction uK = .58 The angle of th
mojhsa [17]

Answer: The angle of inclination is nearly 30°

Explanation:

For a body on an inclined plane, the coefficient of friction between the body and the plane is equal to the ratio of the moving force applied to the body to the frictional force acting on the body.

If uK coefficient of friction;

Fm is the moving force

R is the normal reaction on the body

Mathematically uK = Fm/R

Fm = WSin(theta)

R = Wcos(theta)

uK = Wsin(theta)/Wcos(theta)

uK = tan(theta)

theta = arctan(uK)

If uK is 0.58

theta = arctan0.58

theta = 30°

The angle of the inclined will be 30°

7 0
3 years ago
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