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Juliette [100K]
3 years ago
14

A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a

nd the blocks suddenly collide and couple. After the collision, the blocks have a common speed of .90 m/s. what was the speed v the lighter block before the collision'
Physics
1 answer:
sammy [17]3 years ago
5 0

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

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Ostrovityanka [42]

It's answer number four

8 0
3 years ago
An Earth satellite is orbiting at a distance from the Earth’s surface equal to one Earth radius (4 000 miles). At this location,
trasher [3.6K]

Answer:

The acceleration due to gravity is \dfrac{1}{4} times the value of g at the Earth’s surface.

(D) is correct option.

Explanation:

Given that,

Radius = 4000 miles

We need to calculate the gravitational force at surface

Gravitational force on the mass m on the surface of the earth

At r = R

F=mg

mg=\dfrac{GmM}{R^2}....(I)

We need to calculate the gravitational force at height

Gravitational force on a mass m from the center of the earth,

At r = R + R = 2 R

F'=mg'

mg'=\dfrac{GmM}{4R^2}....(II)

Dividing equation (II) by equation (I)

\dfrac{mg'}{mg}=\dfrac{\dfrac{GmM}{4R^2}}{\dfrac{GmM}{R^2}}

\dfrac{g'}{g}=\dfrac{1}{4}

Hence, The acceleration due to gravity is \dfrac{1}{4} times the value of g at the Earth’s surface.

4 0
3 years ago
A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police
Nastasia [14]

Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

1. x=x_{0}+vt

2. x=x_{0}+v_{0}t+\frac{at^{2}}{2}

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

vt = v_{0}t+\frac{at^2}{2}\\\\   16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}

We simplify the fraction present and rearrange for our formula so that it equals 0:

0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0

In the very last step we factored a common factor t. There is two possible solutions to the equation at t=0 and:

0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\  0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at t=0 s (when the SUV passed the police car) and t=15.56s(when the police car catches up to the SUV)

8 0
3 years ago
1.which of the following are true.
Ksivusya [100]

B is the correct answer

3 0
3 years ago
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An Olympic high jumper, with a mass of 82 kg, has a
Digiron [165]

Answer:

I don't really know

Explanation:

I really wanted to help you, but then I realized i didnt know how to

8 0
2 years ago
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