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patriot [66]
3 years ago
9

A 1460-kg submarine rises straight up towards the surface. Seawater exerts both an upward buoyant force of 16670 N on the submar

ine, as well as a downward resistive force of 1150 N. What is the submarine's acceleration?
Physics
1 answer:
Amiraneli [1.4K]3 years ago
5 0

Answer:

a = 0.63 m/s²

Explanation:

given,

mass of submarine = 1460-kg

upward buoyant force = 16670 N

downward resistive force = 1150 N

submarine acceleration = ?

assuming g = 10 m/s²

now,

B - (R + mg) = ma

16670 - 1150 - 1460 × 10 = 1460 × a

1460× a = 920

a = 0.63 m/s²

hence, the acceleration of submarine is equal to a = 0.63 m/s²

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Answer:

Explanation:

See the attached figure . See the forces acting on man pulling up the box .

Man is stationary so net force acting on man is zero .

T + R = Wman

R is the reaction force of the ground of second floor  .

R = Wman - T

3 0
3 years ago
Toy car in a science experiment covers 1.6 meters in half a second. If a the car travels at a steady speed, how far will it go i
Tanzania [10]
The answer is D. 32 m.

The simple equation that connects speed (v), time (t), and distance (d) can be expressed as:
v= \frac{d}{t}         ⇒ d=v*t

It is given:
v =  \frac{1.6m}{0.5s} = \frac{1.6m*2}{0.5s*2}= \frac{3.2m}{1s}  = 3.2 m/s
t = 10 s
d = ?

So:
d= v*t=3.2m/s*10s = 32m
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3 years ago
What happens to the mechanical advantage of a machine if the output force is less than the input force? What must happen to outp
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A machine that put in force is a Mechanical
5 0
3 years ago
An empty can of soda is yeeted (thrown) into a crowd. Then, a full can of soda is yeeted into a crowd. Throwing the full can too
4vir4ik [10]

Answer:

Newton's third law of motion.

Explanation:

We are told the force needed to throw the full soda can was more than that needed to throw the empty can.

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3 0
2 years ago
a bullet with a mass of 4.0g and a speed of 650m/s is fired at a block of wood with a mass of 0.095kg. the block rests on a fric
n200080 [17]

Part a)

Here in this we can use momentum conservation as there is no external force on it

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} +m_2v_{2f}

here we know that

m_1 = 0.004 kg

v_{1i} = 650 m/s

m_2 = 0.095

v_{2i} = 0

v_{2f} = 23 m/s

now by above equation

0.004*650 + 0.095* 0 = 0.004*v + 0.095*23

2.6 + 0 = 0.004*v + 2.185

v = 103.75 m/s

Part b)

Final kinetic energy of the system

KE_f = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2

KE_f = \frac{1}{2}*0.004*(103.75^2) + \frac{1}{2}*(0.095)*23^2

KE_f = 46.65 J

Initial Kinetic energy of the system will be

KE_f = \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2

KE_f = \frac{1}{2}*0.004*(650^2) + \frac{1}{2}*(0.095)*0^2

KE_f = 845 J

So here kinetic energy is decreased for this system

final energy is less than initial energy

6 0
3 years ago
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