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malfutka [58]
3 years ago
9

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! I NEED THE ANSWER NOW

Physics
1 answer:
FrozenT [24]3 years ago
5 0
90 F = 43 OR 0.9F = 0.43
(F = 43 / 90 OR 0.43 / 0.9 =) 0.48 N

upwards force = downwards force
(R =) 1.2 N
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Suppose two children push horizontally, but in exactly opposite directions, on a third child in a sled. The first child exerts a
anyanavicka [17]

Answer:

Please, read the anser below

Explanation:

1. In order to calculate the acceleration of the children you use the Newton second law for the summation of the implied forces:

F_2-F_1-F_f=Ma          (1)

Where is has been used that the motion is in the direction of the applied force by the second child

F2: force of the second child = 92N

F1: force of the first child = 79N

Ff: friction force = 5.5N

M: mass of the third child = 24kg

a: acceleration of the third child = ?

You solve the equation (1) for a, and you replace the values of the other parameters:

a=\frac{F_2-F_1.F_f}{M}=\frac{96N-79N-5.5N}{24kg}=0.48\frac{m}{s^2}

The acceleration is 0.48m/s^2

2. The system of interest is the same as before, the acceleration calculated is about the motion of the third child.

3. An image with the diagram forces is attached below.

4. If the friction would be 150N, the acceleration would be zero, because the friction force is higher than the higher force between children, which is 92N.

Then, the acceleration is zero

3 0
3 years ago
Which clouds are often associated with thunder and lightning?
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Your answer is cumulonimbus clouds
3 0
3 years ago
A 15.0 cm object is 12.0 cm from a convex mirror that has a focal length of -6.0 cm. What is the height of the image produced by
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Read 2 more answers
A hot air balloon is traveling vertically upward at a constant speed of 4.5 m/s. When it is 28 m above the ground, a package is
ella [17]

Answer:

1.97 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow 21=4.5t+\frac{1}{2}\times -9.8\times t^2\\\Rightarrow 21=-4.5t-4.9t^2\\\Rightarrow 4.9t^2+4.5t-28=0\\\Rightarrow 49t^2+45t-280=0

Solving the above equation we get

t=\frac{-45+\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}, \frac{-45-\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}\\\Rightarrow t=1.97, -2.89

So, the time the package was in the air is 1.97 seconds

3 0
3 years ago
A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.
Advocard [28]

The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as

  • t=0.476v
  • t=1.967v
  • V2=4.323v

<h3>What is the potential across the capacitor?</h3>

Question Parameters:

A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.

at

  • t = 1.0 seconds
  • 5.0 seconds
  • 20.0 seconds.

Generally, the equation for the Voltage is mathematically given as

v(t)=Vmax=(i-e^{-t/t})

Therefore

For t=1

V=5(i-e^{-1/10})

t=0.476v

For t=5s

V2=5(i-e^{-5/10})

t=1.967

For t=20s

V2=5(i-e^{-20/10})

V2=4.323v

Therefore, the values of voltages at the various times are

  • t=0.476v
  • t=1.967v
  • V2=4.323v

Read more about  Voltage

brainly.com/question/14883923

Complete Question

A 1.0 μF capacitor is being charged by a 5.0 V battery through a 10 MΩ resistor.

Determine the potential across the capacitor when t = 1.0 seconds, 5.0 seconds, 20.0 seconds.

7 0
2 years ago
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