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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, <em>the speed is constant</em>. The constant speed is zero. <em>The car is not moving</em>.

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A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

bekas [8.4K]

Answer:

Explanation:

Given that,

Radius r = 15cm = 0.15m

Area of the circular loop can be determined using the formula for area of a circle

A = π r²

A = π × 0.15²

A = 0.0708 m²

Magnetic field B = 1.2T in positive z direction

B = 1.2 •k T.

If loop is remove from the field in the time interval

∆t = 2.3ms = 2.3×10^-3s

We want to find the average EMF and it is given as

ε = —∆Φ/∆t

The final flux is zero

Φf = 0

Where magnetic flux is given as

Φi = BA Cosθ

Where θ=0 since the area and the magnetic field point in the same direction.

Φi = BA Cos0

Φi = BA

Φi = 1.2 × 0.0708

Φi = 0.0848 Vs

Then, ε = —∆Φ/∆t

ε = —(Φf — Φi) / ∆t

ε = —(0-0.0848) / (2.3×10^-3)

ε = 0.0848 / (2.3×10^-3)

ε = 36.88 V

The EMF is 36.88 Volts

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3 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin

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Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
$=16000x+10000 \frac{x^2}{2} - 26000 \frac{x^3}{3}$
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
$=8733 J=8.73 kJ$

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
$=12280 J=12.28 kJ$

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