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marta [7]
2 years ago
14

When a 9.9 kg object is in free fall, it feels a force of 97.02 N. what is its acceleration?

Physics
1 answer:
Free_Kalibri [48]2 years ago
6 0

The acceleration of an object in free fall is 9.8 m/s^2

Explanation:

We can solve the problem by using Newton's second law of motion, which states that:

F=ma

where

F is the net force on an object

m is the mass of the object

a is its acceleration

For the object in this problem, we have:

m = 9.9 kg is its mass

F = 97.02 N is the net force on the object

So, we can rearrange the equation to find its acceleration:

a=\frac{F}{m}=\frac{97.02}{9.9}=9.8 m/s^2

This value is actually known as acceleration of gravity (g), and it is the acceleration that any object in free fall has near the Earth's surface, regardless of its mass.

Learn more about Newton's second law here:

brainly.com/question/3820012

And about free fall:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

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A foot player runs 1.6m/s and has a KE of 790 J. What is his mass?
Mariana [72]
The equation for kinetic energy is,

Ke = (1/2)mv^2.

You're given a kinetic energy of 790 joules, and a speed of 1.6 m/s. Plugging these values into the equation, we get,

790 = (1/2)(1.6)^2(m).

Solving for m, we get,

m = (790)/(0.5(1.6)^2).

I'll let you crunch out those numbers for yourself :D

If you have any questions, feel free to ask. Hope this helps!
3 0
3 years ago
Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where l
goblinko [34]

Answer:

Δt = 5.29 x 10⁻⁴ s = 0.529 ms

Explanation:

The simple formula of the distance covered in uniform motion can be used to find the interval between when the sound arrives at the right ear and the sound arrives at the left ear.

\Delta s = v\Delta t\\\\\Delta t = \frac{\Delta s}{v}

where,

Δt = required time interval = ?

Δs = distance between ears = 18 cm = 0.18 m

v = speed of sound = 340 m/s

Therefore,

\Delta t = \frac{0.18\ m}{340\ m/s}

<u>Δt = 5.29 x 10⁻⁴ s = 0.529 ms</u>

4 0
3 years ago
A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/
hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is 9.1\cdot 10^{-4} J

C) The work done by the electric field is -2.4\cdot 10^{-3} J

Explanation:

A)

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the particle

\theta is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

  • The force is directed vertically upward (because the field is directed vertically upward)
  • The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

\theta=90^{\circ}

and cos 90^{\circ}=0: therefore, the work done on the charge by the electric field is zero.

B)

In this case, the charge move upward (same direction as the electric field), so

\theta=0^{\circ}

and

cos 0^{\circ}=1

Therefore, the work done by the electric force is

W=Fd

and we have:

F=qE is the magnitude of the electric force. Since

E=4.30\cdot 10^4 V/m is the magnitude of the electric field

q=32.0 nC = 32.0\cdot 10^{-9}C is the charge

The electric force is

F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J

C)

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

\theta=90^{\circ}+45^{\circ}=135^{\circ}

Moreover, we have:

F=1.38\cdot 10^{-3} N (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

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5 0
3 years ago
A girl delivering newspapers covers her route by traveling 3.00 blocks west, 4.00 blocks north, and then 6.00 blocks east. What
borishaifa [10]

Answer:

13 blocks

Explanation:

The total distance the student travels is 13 blocks.

 Distance is the length of path covered during the motion of a body.

 To find distance:

 Total distance  = Number of blocks to the west + number of blocks to the north + number of blocks to the east

 Total distance  = 3blocks + 4blocks + 6blocks  = 13blocks

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Anika [276]
I believe the answer is B. that's where the asetroid belt is.
4 0
3 years ago
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