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Komok [63]
2 years ago
6

If you know the amount of the unbalanced force acting upon an object and the mass of the object, using Newton's 2nd Law what cou

ld you calculate about the object? A) acceleration B) friction C) inertia D) velocity
Physics
2 answers:
densk [106]2 years ago
8 0
A) Acceleration is the answer
Zielflug [23.3K]2 years ago
7 0
Hey there!

<span>If you know the amount of the unbalanced force acting upon an object and the mass of the object, using Newton's 2nd Law what could you calculate about the object?

Answer: A) acceleration

Hope this helps
Have a great day (:
</span>
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Which statement correctly describes how a bar magnet should be placed on a globe to correctly align with Earth's magnetic field?
Marizza181 [45]

Answer:

The answer is B. When the magnet is placed on a globe to correctly align with Earth’s magnetic field, it is considered to be suspended freely. The Earth has geographical poles as well with North and South poles. Since unlike poles attract, the South Pole of the magnet will be attracted to the geographical North.

Explanation:

5 0
2 years ago
Read 2 more answers
If Pete ( mass=90.0kg) weights himself and finds that he weighs 30.0 pounds, how far away from the surface of the earth is he
shutvik [7]

Answer: 9938.8 km

Explanation:

1 pound-force = 4.48 N

30.0 pounds-force = 134.4 N

The force of gravitation between Earth and object on the surface of is given by:

F = \frac{GMm}{R^2} = mg

Where M is the mass of the Earth, m is the mass of the object, R (6371 km) is the radius of the Earth.

At height, h above the surface of the Earth, the weight of the object:

(mg)'= \frac{GMm}{(R+h)^2}

we need to find "h"

taking the ratio of two:

\frac{mg}{(mg)'}=\frac{(R+h)^2}{R^2}\\ \Rightarrow \frac{90kg \times 9.8 m/s^2}{134.4 N}=\frac{(R+h)^2}{R^2}\\ \Rightarrow 6.56 R^2= (R+h)^2 \Rightarrow h= (2.56-1)R\\ \Rightarrow h = 1.56 R = 1.56 \times 6371 km = 9938. 8 km

Hence, Pete would weigh 30 pounds at 9938.8 km above the surface of the Earth.

5 0
2 years ago
Two identical carts travel at the same speed toward each other, and then a collision occurs. The graphs show the momentum of eac
madam [21]

Explanation :

The interaction between two objects is termed as the collision. The collision can be of two types i.e. elastic collision and inelastic collision.

In this case, two identical carts travel at the same speed toward each other, and then a collision occurs. In an inelastic collision, the momentum before and after the collision remains the same but its kinetic energy gets lost.

After the collision, both the object sticks over each other and moves with one velocity.

Out of the given graph, the graph that shows a perfectly inelastic collision is attached. It shows that after the collision both the carts move with the same velocity.

5 0
2 years ago
A source charge generates an electric field of 1236 N/C at a distance of 4 m. What is the magnitude of the source charge?
guajiro [1.7K]

The correct answer to the question is-  2.2\ \mu C

CALCULATION:

As per the question, the electric field generated by the source charge is 1236  N/C at a distance of 4 m.

Hence , electric field  E =  1236 N/C.

The distance of the point R = 4m

We are asked to calculate the charge possessed by the source.

The electric field produced by a source charge of Q at a distance R is calculated as -

                    Electric field E = \frac{1}{4\pi \epsilon_{0}}\frac{Q}{R^2}

Here, \epsilon_{0} is called the absolute permittivity of the free space.

Hence, the charge of source is calculated as -

                                         Q = E\times 4\pi \epsilon_{0}\times R^2

                                            = 1236\times \frac{1}{9\times 10^9}\times (4)^2\ Coulomb

                                            = 2197.33\times 10^{-9}\ C

                                             = 2.19733\times 10^{-6}\ C

                                             = 2.2\ \mu C

Hence, the charge of source is 2.2\ \mu C

3 0
2 years ago
Read 2 more answers
Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea
madreJ [45]
In order to answer this exercise you need to use the formulas

 S = Vo*t + (1/2)*a*t^2

Vf = Vo + at

The data will be given as

Vf = final velocity = ?

Vo = initial velocity = 1.4 m/s

a = acceleration = 0.20 m/s^2

s = displacement = 100m

And now you do the following:

100 = 1.4t + (1/2)*0.2*t^2

t = 25.388s

and

Vf = 1.4 + 0.2(25.388)

Vf = 6.5 m/s

So the answer you are looking for is 6.5 m/s
7 0
2 years ago
Read 2 more answers
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