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NeX [460]
3 years ago
6

A roller coaster car rapidly picks up speed as it rolls down a slope. As it starts down the slope, its speed is 4m/s. But 3 seco

nds later at the bottom of the slope its speed is 22m/s. What is its average acceleration?
Physics
1 answer:
slamgirl [31]3 years ago
7 0

Answer:

6 m/s^2

Explanation:

The average acceleration of the car is given by:

a=\frac{v-u}{t}

where

v is the final speed

u is the initial speed

t is the time elapsed

For the car in this problem,

v = 22 m/s

u = 4 m/s

t = 3 s

Therefore, the acceleration of the car is

a=\frac{22 m/s-4 m/s}{3 s}=6 m/s^2

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What is the energy of motion.
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Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with an
Alona [7]

Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

F=\dfrac{kQ^2}{r^2} ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}.....(2)

Dividing equation (1) and (2), we get :

\dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}

Hence, the correct option is (d) i.e. " 4F/9"

7 0
3 years ago
Please help and I don't mean to sound rude but, ONLY ANSWER IF YOUR GOING TO DO ALL 4 QUESTIONS
anzhelika [568]

Answer:

for first question is 2

for second question 1

for third question 2

for forth question 1

Explanation:

i hope i helped

6 0
3 years ago
A circular loop of wire with a radius of 20 cm lies in the xy plane and carries a current of 2 A, counterclockwise when viewed f
Zina [86]

To solve this problem we will apply the concept related to the magnetic dipole moment that is defined as the product between the current and the object area. In our case we have the radius so we will get the area, which would be

A=\pi r^2

A =\pi (0.2)^2

A =0.1256 m^2

Once the area is obtained, it is possible to calculate the magnetic dipole moment considering the previously given definition:

\mu=IA

\mu=2(0.1256)

\mu=0.25 A\cdot m^2

Therefore the magnetic dipole moment is 0.25A\cdot m^2

6 0
3 years ago
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