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Feliz [49]
3 years ago
11

A 160.-kilogram space vehicle is traveling along a straight line at a constant speed of 800. Meters per second. The magnitude of

the net force on the space vehicle is
Physics
2 answers:
Alisiya [41]3 years ago
5 0

<u>Answer:</u> The magnitude of the force on the space vehicle is 0 N

<u>Explanation:</u>

Force is defined as the push or pull on an object with some mass that causes change in its velocity.

It is also defined as the mass multiplied by the acceleration of the object.

Mathematically,

F=m\times a

where,

F = force exerted on the space vehicle

m = mass of the space vehicle = 160 kg

a = acceleration of the space vehicle = 0m/s^2    (Speed is constant)

Putting values in above equation, we get:

F=160kg\times 0m/s^2\\\\F=0N

Hence, the magnitude of the force on the space vehicle is 0 N

natulia [17]3 years ago
4 0

Answer:

Zero

Explanation:

As force acting on the body is equal to the product of mass and acceleration.

Acceleration is equal to rate of change in velocity.

Here velocity is constant so acceleration is zero.

It means the net force acting on the vehicle is zero.

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When a carousel is in motion, the movement of the carousel horse can best be described as?
tatiyna
Supposing the carousel is rotating with constant speed, the movement is uniform angular motion.
5 0
3 years ago
A motorboat heads due east at 16 m/s across a river that flows due south at 9.0 m/s. a.) What is the resultant velocity of the b
alukav5142 [94]

Answer:

a)V=18.35 m/s (South -East)

b) t =7.41 m/s

c)D= 66.70 m

Explanation:

Given that

Velocity of boat in east direction = 16 m/s

Velocity of river = 9 m/s

a)The resultant velocity V

V=\sqrt{16^2+9^2}\ m/s

V=18.35 m/s (South -East)

b)

We know that

Distance = Velocity x time

Lets t time takes to cross the river

136 = 18.35 x t

t =7.41 m/s

c)

   The distance covered downstream  

We know that

Distance = Velocity x time

t= 7.41 s

D= 7.41 x 9 m

D= 66.70 m

3 0
2 years ago
A 75.0-kg ice skater moving at 10.0 m/s crashes into a stationary skater of equal mass. After the collision, the two skaters mov
Ksenya-84 [330]

Answer:

Explanation:

Momentum change for either skater is mΔv = 75.0(5.0) = 375 kg•m/s

As a change in momentum is equal to an impulse

375 = FΔt

F = 375/0.100 = 3750 N

As 3750 N < 4500 N no bones are broken.

4 0
3 years ago
Lindsay is late to class and is running down the hallway at a speed of 2 m/s the length of the hallway is 150 meters . How long
Mamont248 [21]

Explanation:

a=v-u\t

2=150-0/t

2t=150

t=75sec

7 0
3 years ago
Suppose you are standing on top of a hemisphere of radius r and you kick a soccer ball horizontally such that it has velocity v.
Ksivusya [100]

|v| =\sqrt{ G \cdot M / r}, where

  • M the mass of the planet, and
  • G the universal gravitation constant.

Explanation:

Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.

<em>Equation 1</em>  (see below) relates net force the object experiences, \Sigma F to its orbit velocity v and its mass m required for it to stay in orbit :

\Sigma F = m \cdot v^{2} / r <em>(equation 1)</em>

The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force \Sigma F acting on the soccer ball shall equal to its weight, W = m \cdot g where g the gravitational acceleration constant. Thus

\Sigma F = W = m \cdot g <em>(equation 2)</em>

Substitute equation 2 to the left hand side of <em>equation 1</em> and solve for v; note how the mass of the soccer ball, m, cancels out:

m \cdot g = \Sigma F = m \cdot v^{2} / r \\ v^{2} = g \cdot r \\ |v| = \sqrt{g \cdot r} \; (|v| \ge 0) <em>(equation 3)</em>

<em>Equation 4 </em> gives the value of gravitational acceleration, g, a point of negligible mass experiences at a distance r from a planet of mass M (assuming no other stellar object were present)

g = G \cdot M/ r^{2} <em>(equation 4)</em>

where the universal gravitation <em>constant</em> G = 6.67408 \times 10^{-11} \cdot \text{m}^{3} \cdot \text{kg}^{-1} \cdot \text{s}^{-2}

Thus

\begin{array}{lll}|v| &=& \sqrt{g \cdot r}\\ & =&\sqrt{ G \cdot M / r}\end{array}

3 0
3 years ago
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