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dedylja [7]
3 years ago
8

Two runners start at the same point and jog at a constant speed along a straight path. Runner A starts at time t = 0 s, and Runn

er B starts at time t = 2.5 s. The runners both reach a distance 64 m from the starting point at time t = 25 s. If the runners continue at the same speeds, how far from the starting point will each be at time t = 45 s?
Physics
1 answer:
Radda [10]3 years ago
8 0

Answer:

See explanation below

Explanation:

First thing to do here, is to calculate the speed of each runner. The problem states that both runners jog at a constant speed. We are given data for both runners, the starting time and the time where both of them reach the same distance.

With these data we can calculate the speed of both runners. Let's do it with runner 1:

If by t  = 25 s it reach a 64 m of distance, then his speed:

V = d/t   (1)

Replacing we have:

V₁ = 64 / 25 = 2.56 m/s

With this speed we can calculate the distance at t = 45 s so:

d = V * t    (2)

d₁ = 2.56 * 45

<h2>d₁ = 115.2 m</h2>

So Runner 1 will be at 115.2 from the starting point at 45 seconds.

For runner 2, he began at t = 2.5 s, but he reach the 64 m at 25 s, so his speed would be:

V₂ = 64 / (25-2.5) = 2.84 m/s

And the distance at t = 45 s

d₂ = 2.84 * 45

<h2>d₂ = 127.8 m</h2>

Runner 2 is 127.8 m from the starting point at 45 seconds.

Hope this helps

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Answer:

A)0.00966 N/C

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Explanation:

We are given;

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Now, as a result of change in magnetic field, an emf will be induced in it. Thus, , electric field is induced and given by the formula :

∫E•dr = d/dt∫B.A •dA

This gives;

E(2πr) = dB/dt(πr²)

Gives;. 2E = dB/dt(r)

E = dB/dt × 2r

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E = 0.23 × 2(0.021)

E = 0.00966 N/C

The magnitude of the electric field induced in the ring has a magnitude of 0.00966 N/C

B) The direction of electric field will be in a counterclock wise direction when viewed by someone on the south pole of the magnet

6 0
2 years ago
while playing her guitar , karen plucks one string with increasin levels of force. what effect does this have on the sound produ
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A uniform solid disk with a mass of 24.3 kg and a radius of 0.364 m is free to rotate about a frictionless axle. Forces of 90.0
a_sh-v [17]

Answer:

a. -12.7 Nm

b. -7.9 rad/s^2

Explanation:

I have attached an illustration of a solid disk with the respective forces applied, as stated in this question.

Forces applied to the solid disk include:

F_1 = 90.0N\\F_2 = 125N

Other parameters given include:

Mass of solid disk, M = 24.3kg

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b.)  The angular acceleration of the disk can be found thus:

using the formula for the Moment of Inertia of a solid disk;

I_{disk} = {\frac{1}{2}}Mr^2

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We then relate the torque and angular acceleration (\alpha) with the formula:

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4 0
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Suppose a rock is dropped off a cliff with an initial speed of 0m/s. What is the rocks speed after 5 secounds, in m/s, if it enc
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Answer:

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