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2 years ago

The speed of the 2 kg cart after 5 seconds is ? cm/s.

Physics

1 answer:

motikmotik2 years ago

3 0

The answer to this question is 1cm/s

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When a car drives over a speed bump and oscillates up and down in simple harmonic motion, at which position during the motion is

Oksana_A [137]

Using the second Law of Newton, F = m * a, you know that acceleration is maximum when the force is maximum.

Using Hooke's Law, F = K Δx, you know that the force is maximum when the displacement from the equilibrium (Δx) is maximum.

So the answer is that the acceleration is maximum at the maximum amplitude x = a.

3 0

3 years ago

A certain unbalanced force gives a 5kg object an acceleration of 15 m/s2. What would be the acceleration if the same force was a

stira [4]

Answer:

a2 = 2.5 m/s2

Explanation:

F1 = m1 a1 We use the same force so F1 = F2

= 5kg × 15m/s2 F2 = m2 a2

= 75N a2 is required

a2 = F2 / m2

= 75N / 30 kg

= 2.5 m/s2

7 0

3 years ago

During free fusion, the eyes _______ in order to view a stereogram without a stereoscope. A. use the pictorial depth cue B. are

Andru [333]

Answer:

Explanation:

During free fusion, the eyes converge or diverge in order to view a stereogram without a stereoscope.

7 0

2 years ago

The mass of a planet is 3.7 x 1024 kg. If the planet has a radius of 9.2 x 106 m what is the acceleration of gravity for a perso

nalin [4]

Explanation:

It s given that,

Mass of a planet, $M=3.7\times 10^{24}\ kg$

Radius of a planet, $R=9.2\times 10^{6}\ m$

(1) We need to find the acceleration due to gravity for a person on the surface of the planet. Its formula is given by :

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

$g=\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{(9.2\times 10^{6}\ m)^2}$

$g=2.91\ m/s^2$

(2) The escape velocity is given by :

$v=\sqrt{\dfrac{2GM}{R}}$

$v=\sqrt{{\dfrac{2\times 6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{9.2\times 10^{6}\ m}}$

v = 7324.61 m/s

Hence, this is the required solution.

3 0

3 years ago

9. A 50 kg physics i student trudges from the 1st floor to the 3rd floor, going up a total height of 13 m.

SVETLANKA909090 [29]

Answer:

6370 J

Explanation:

By the law of energy conservation, the work done by the student would be the change in potential enegy from 1st floor to 3rd floor, or a change of 13 m

$W = E_p = mgh$

where m = 50kg is the mass of the student, g = 9.8 m/s2 is the gravitational constant and h = 13 m is the height difference

$W = 50*9.8*13 = 6370 J$

3 0

2 years ago