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3 years ago

If a 2 cm vector represents speed, and 1 cm equals 5 m/s, how fast is the represented

Physics

1 answer:

Ne4ueva [31]3 years ago

6 0

Answer: 10m/s

Explanation: Since a single vector with length of 1cm expresses 5m/s, the vector which has a length doubled from the original vector should have the speed which is also doubled.

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A window washer who does not want to change his position will want the forces acting on him to be ____________.

natali 33 [55]

My answer is a balanced

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3 years ago

How much do you have to increase the vertical distance to double the impact speed of an object?

sergey [27]

Twice the distance to double an impact

5 0

3 years ago

A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on

MAXImum [283]

Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

$Ec=\frac{1}{2} *m*v^{2}$

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

$W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}$

$W=\frac{1}{2} *m*(v2^{2}-v1^{2})$

In this case:

W=?
m= 2,145 kg
v2= 12 $\frac{m}{s}$

v1= 25 $\frac{m}{s}$

Replacing:

$W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})$

W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

3 0

3 years ago

The cylinder with piston locked in place is immersed in a mixture of ice and water and allowed to come to thermal equilibrium wi

lukranit [14]

Answer:

a. volume of gas: (decreases)

b. temperature of gas: (same)

c. internal energy of gas: (same)

d. pressure of gas: (increases)

Explanation:

We have a gas (let's suppose that is ideal) in a piston with a fixed volume V.

Then we put in a reservoir at 0°C (the mixture of water and ice)

remember that the state equation for an ideal gas is:

P*V = n*R*T

and:

U = c*n*R*T

where:

P = pressure

V = volume

n = number of mols

R = constant

c = constant

T = temperature.

Now, we have equilibrium at T = 0°C, then we can assume that T is also a constant.

Then in the equation:

P*V = n*R*T

all the terms in the left side are constants.

P*V = constant

And knowing that:

U = c*n*R*T

then:

n*R*T = U/c

We can replace it in the other equation to get:

P*V = U/c = constant.

Now, the piston is (slowly) moving inwards, then:

a) Volume of the gas: as the piston moves inwards, the volume where the gas can be is smaller, then the volume of the gas decreases.

b) temperature of the gas: we know that the gas is a thermal equilibrium with the mixture (this happens because we are in a slow process) then the temperature of the gas does not change.

c) Internal energy of the gas:

we have:

P*V = n*R*T = constant

and:

P*V = U/c = constant.

Then:

U = c*Constant

This means that the internal energy does not change.

d) Pressure of the gas:

Here we can use the relation:

P*V = constant

then:

P = (constant)/V

Now, if V decreases, the denominator in that equation will be smaller. We know that if we decrease the value of the denominator, the value of the quotient increases.

And the quotient is equal to P.

Then if the volume decreases, we will see that the pressure increases.

4 0

2 years ago

Pls help will give brainlest

Karo-lina-s [1.5K]

Respon

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llpop

io.

7 0

3 years ago

Read 2 more answers