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3 years ago

Consider the data collected in science class. Different masses were thrown with varied amounts of

Physics

2 answers:

hichkok12 [17]3 years ago

6 0

Answer:

a. in all cases, the acceleration was the same

Explanation:

Do the math - F = ma. In each case, the acceleration is 5 m/s2.

lesya [120]3 years ago

4 0

Answer:

A: In all cases, the acceleration was the same.

Explanation:

I know this because its a clear obvious answer not only that it was one of my USA TESTPREP questions and it was right.

All you mainly have to do is the math - F=ma , In each case , the acceleration is 5 m/s squared

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A fire is burning in a fireplace. Where is the radiation?

mrs_skeptik [129]

I think it’s A.
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3 years ago

G a magnetic field perpendicular to the plane of a wire loop is uniform in space but changes with time t in the region of the lo

Luden [163]

Answer:

e = Δφ / Δt induced emf is proportional to enclosed flux

Also φ = B * A flux is proportional to area and enclosed field

If the induced emf e increases with time than the flux and hence the magnetic field is increasing with time (replace B with G)

Since e = ΔG * A / Δt if e is linear then G must also be linear and be proportional to the time

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1 year ago

A bus moving along a level road increases its speed from 20 m/s to 35 m/s in 15.0s. What is the car's acceleration?

Ksju [112]

Explanation:

initial velocity U = 20m/s

Final velocity V = 35m/s

time = 15.0 secs

change in velocity = 35 - 15

= 20m/s

acceleration a = change in velocity/time V/t

a = (35-20)/15

a= 15/15

Hence, your acceleration is 1m/s^2

5 0

2 years ago

1. Describe what is happening at point A when these 3 vectors act on point A. 2. Describe what happen to the resultant vector at

zavuch27 [327]

Answer: Don't know sorry

Explanation:

4 0

3 years ago

In an engine governor, the two spheres (total mass of 1.0kg) are at 0.05m and rotating at 37rad/s If the engine increases the an

kirill115 [55]

Here we can say that there is no external torque on this system

So here we can say that angular momentum is conserved

so here we will have

$I_1\omega_1 = I_2\omega_2$

now we have

$I_1 = mr^2$

$I_1 = (1kg)(0.05^2)$

$I_1 = 25\times 10^{-4} kg m^2$

similarly let the final distance is "r"

so now we have

$I_2 = mr^2$

$I_2 = 1r^2$

now from above equation we have

$(25\times 10^{-4})37 = (r^2)(58)$

$r = 0.04 m$

so final distance is 0.04 m between them

8 0

3 years ago