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Vesna [10]
3 years ago
5

If you push a crate across a factory floor at constant speed in a constant direction, what is the magnitude of the force of fric

tion on the crate compared to your push
Physics
1 answer:
poizon [28]3 years ago
3 0

Answer:

The magnitude of the force of friction equals the magnitude of my push

Explanation:

Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.

Let F = push and f = frictional force and f' = net force

F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0

So, F - f = 0

Thus, F = f

So, the magnitude of the force of friction equals the magnitude of my push.

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8. An airplane is flying at 200 m/s when it touches the ground at the airport. It has a constant negative acceleration, and slow
Temka [501]

Answer:

Explanation:

Given

Initial velocity u = 200m/s

Final velocity = 4m/s

Distance S = 4000m

Required

Acceleration

Substitute the given parameters into the formula

v² = u²+2as

4² = 200²+2a(4000)

16 = 40000+8000a

8000a = 16-40000

8000a = -39,984

a = - 39,984/8000

a = -4.998m/s²

Hence the acceleration is -4.998m/s²

8 0
3 years ago
38. You are fishing and catch a fish with a mass of
adoni [48]

Answer:

1.7333333m/s²

Explanation:

Tension of the line = the weight + force from pulling up the fish

30N = mg + ma

30 = (6)(9.8) + (6)a

10.4 = 6a

∴ a = 1.7333333m/s²

6 0
3 years ago
What is the frequency of this wave<br> 1<br> 2<br> 3<br> 4
viva [34]
The frequency of this wave is 3
3 0
3 years ago
HALP me!! This is a physics question.
emmasim [6.3K]
<h2>Answer:</h2>

<u>Distance covered is 6.9 meters</u>


<h2>Explanation:</h2>

Data given:

Work Done = 345 kJ = 345000 J

Force = 5 x 10 ^ 4 =  50000 N

Distance = ?


Solution:

As we know that

Work Done = Force applied x Distance covered

By arranging the equation we get

Work / Force = Distance covered

By putting the values

345000 / 50000 = 6.9

So distance covered is 6.9 meters

4 0
3 years ago
Read 2 more answers
If a pendulum has a period of 4 seconds at the north pole with gravity force 9.83 ms^-2 and a 3.97m pendulum length, how could t
sladkih [1.3K]

Answer:

The same pendulum could be adjusted to have the same period, in the equator must have a length of 3.949m.

Explanation:

Tnp= 4 sec

gnp= 9.83 m/sec²

Lnp= 3.97m

Tequ= 4 sec

gequ= 9.78 m/sec²

Lequ=?

Lequ= (Lnp* gequ) / gnp

Lequ= 3.949 m

5 0
3 years ago
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