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3 years ago

I’m still lost on this please help me on this I know it’s not D I got it wrong

Physics

1 answer:

RSB [31]3 years ago

5 0

Before you even look at the questions, look over the graph, so you know what kind of information is there.

The x-axis is "time". OK. You know that as the graph moves from left to right, it shows what's happening as time goes on.

The y-axis is "speed" of something. OK. When the graph is high, the thing is moving fast. When the graph is low, the thing is moving slow. When the graph slopes up, the thing is gaining speed. When the graph slopes down, the thing is slowing down. When the graph is flat, the speed isn't changing, so the thing is moving at a constant speed.

NOW you can look at the questions.

OMG ! It's only ONE question: What's happening from 'c' to 'd' ? Well I don't know. Perhaps we can figure it out if we LOOK AT THE GRAPH !

-- Between c and d, the graph is flat. The speed is not changing. It's the same speed at d as it was back at c .

What speed is it ?

-- Look back at the y-axis. The speed at the height of c and d is 'zero' .

-- The 2nd and 4th choices are both correct. From c to d, the speed is constant. The constant speed is zero. The car is not moving.

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a bullet moving with a velocity of 100m/s pierce a block of wood and moves out with a velocityof 10 m/s.if the thickness of the

erma4kov [3.2K]

The emerging velocity of the bullet is 71 m/s.

The bullet of mass m moving with a velocity u has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed v when it emerges from the block.

If the block exerts a resistive force F on the bullet and the thickness of the block is x then, the work done by the resistive force is given by,

$W=Fx$

This is equal to the change in the bullet's kinetic energy.

$W=Fx=\frac{1}{2} m(u^2-v^2)......(1)$

If the thickness of the block is reduced by one-half, the bullet emerges out with a velocity v₁.

Assuming the same resistive forces to act on the bullet,

$F(\frac{x}{2} )=\frac{1}{2} m(u^2-v_1^2)......(2)$

Divide equation (2) by equation (1) and simplify for v₁.

$\frac{\frac{Fx}{2} }{Fx} =\frac{(u^2-v_1^2)}{(u^2-v^2)} \\\frac{100^2-v_1^2}{100^2-10^2} =\frac{1}{2} \\v_1^2=5050\\v_1=71.06 m/s$

Thus the speed of the bullet is 71 m/s

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3 years ago

A closed circuit is formed by a solid conductor in contact with two parallel conducting rails that are closed on their left end,

castortr0y [4]

this is an equation that you need to solve for motional emf. motional emf=vBL, where v is velocity in meters/second, B is magnetic field in Teslas and L is length or distance the rails are apart from each other. when we plug everything into the formula given above, we get: motional emf=5m/s*0.80T*0.20m. solving all this we get 0.8 volts. pretty sure that since they are giving you the direction of the field, they want to know which way the current will flow . since the conductor is moving from left to right the area of the field is increasing which means magnetic flux is increasing as Ф(magnetic flux)=B(magnetic field)*A(area)*cosФ(little phi is the angle to the normal. in this case little fee is 0 degrees so the cosФ doesn't matter). so ↑Ф=B↑A. if magnetic flux is increasing, the induced magnetic field is in the opposite direction as the original magnetic field meaning the induced magnetic field will be out of the page. using the right hand rule which says that if the field is in to the page, the current should go clockwise and if the field is out of the page, the current is counterclockwise so that means that the current should be going counter clockwise since the induced field is going out of the screen. the top of the conducting wire will have its current go to the left and the bottom of the conducting wire will have the current go to the right.

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3 years ago

Calculate the angular velocity of the neptune (orbital period of 165 earth period) in its orbit around the sun.

ludmilkaskok [199]

The correct answer is: Angular velocity = $1.208 * 10^{-9}$ rad/s

Explanation:
The angular velocity is given as:
ω = $\frac{2\pi}{T}$ --- (1)

Where T = 165 * (365 days) * (24 hours/day) * (60 minutes/hour) * (60 seconds/minute) = 5203440000 s

Plug in the value in (1):
ω = $\frac{2\pi}{5203440000} = 1.208 * 10^{-9}$ rad/s

7 0

3 years ago

An object has an average acceleration of +6.18 m/s2 for 0.266 s . At the end of this time the object's velocity is +9.90 m/s .

Archy [21]

At the start of the 0.266 s, the object's speed was 8.26 m/s.

The question can only be talking about speed, not velocity.

4 0

3 years ago

One side of the moon always faces Earth because the time it takes the moon to spin on its axis is blank the time it takes the mo

Kamila [148]

The word to fill in the blank is "equal". Because the time taken to rotate (spin on its axis) is equal to the time of revolution (going around the earth), this means that both have the same rate of angular rotation. So for every bit that the moon goes around its orbit around earth, the moon itself rotates accordingly to present the exact same side to earth.

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3 years ago