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anastassius [24]
3 years ago
7

A 900 kg car travelling at 12 m/s due east collides with a 600 kg car travelling at 24 m/s due north. As a result of the collisi

on, the two cars lock together and move in what final direction?
45.0° N of E

53.1° N of E

63.3° N of E

69.5° N of E
Physics
1 answer:
katrin [286]3 years ago
8 0

Calculate force of each car:

P1 = 900 kg x 12m/s = 10,800

P2= 600 x 24 = 14,400

Degree of travel = arctan(14,300/10800)

Degree of travel = 53.1 N of E

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A cord is attached to the box and run through a pulley directly above the box, so that the cord is vertical. The free end of the
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Answer:

The answer is given here would be a simplified equation, seeing as there are some missing variables in the question.

<u>F1 = T- 46, 674.656 gm/s² </u>

Explanation:

<em>Note: Once we have the mass of the second object and/or acceleration of the cord, we can solve for the force of the ground acting on the box.</em>

To calculate the force caused by gravity on the basic pulley system we use the following equation:

F2 = M2 x g; where g= gravitational acceleration (a constant equal to 9.8 m/s²). The mass M2 = 10.5 lb = 4762.72g

∴ F2 = 4762.72g x 9.8 m/s²

= 46, 674.656 gm/s² or 46, 674.656 N

But since this F2 is acting in a downlowrd direction, it would be negative.

Tension of the cord, T = Mass, x × acceleration. ( x is in the pulley diagram)

⇒ F1 = T - F2

<u>F1 = T- 46, 674.656 gm/s² </u>

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3 years ago
A wave has a period of 2s and a wavelength of 5m. What is the speed of the wave?
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A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the surface of the earth.(a)Determine the magnitude of the gravitational
a_sh-v [17]

Answer:

a). Determine the magnitude of the gravitational force exerted on each by the earth.

Rock: F = 49.06N

Pebble: F = 29.44N

(b)Calculate the magnitude of the acceleration of each object when released.

Rock: a =9.8m/s^{2}

Pebble:  a =9.8m/s^{2}

Explanation:

The universal law of gravitation is defined as:

F = G\frac{m1m2}{r^{2}}  (1)

Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.

<em>Case for the rock </em>m = 5.0 Kg<em>:</em>

m1 will be equal to the mass of the Earth m1 = 5.972×10^{24} Kg and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth r = 6371000m.

F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}

F = 49.06N

Newton's second law can be used to know the acceleration.

F = ma

a =\frac{F}{m} (2)

a =\frac{(49.06 Kg.m/s^{2})}{(5.0 Kg)}

a =9.8m/s^{2}

<em>Case for the pebble </em>m = 3.0 Kg<em>:</em>

F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}

F = 29.44N

a =\frac{F}{m}

a =\frac{(29.44 Kg.m/s^{2})}{(3.0 Kg)}

a =9.8m/s^{2}

3 0
3 years ago
Read 2 more answers
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