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crimeas [40]
3 years ago
7

What two forces act on falling bodies?

Physics
1 answer:
monitta3 years ago
6 0
The two forces acting on the object would be 1. Weight because of gravity pulling the object toward the ground and 2. Drag resisting this motion
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A large magnetic flux change through a coil must induce a greater emf in the coil than a small flux change. A) True B) False
bixtya [17]

Answer:

False

Explanation:

Faraday's law gives the relationship between the induced emf and the rate of change of magnetic flux i.e.

\epsilon=\dfrac{-d\phi}{dt}

The given statement "A large magnetic flux change through a coil must induce a greater emf in the coil than a small flux change" is false. The reason is that if the rate of change of magnetic flux is greater, then its will induce more emf. It would mean it does not say about emf.

Hence, it is false.

3 0
4 years ago
Starting at 1.0 m/s, a cheetah runs with a constant acceleration for 4.8 s reaching a speed of 28 m/s. What is the acceleration
LenaWriter [7]

Answer:

c.5.6m/s^2

Explanation:

Initial velocity of cheetah,u=1 m/s

Time taken by cheetah =4.8 s

Final velocity of cheetah,v=28 m/s

We have to find the acceleration of this cheetah.

We know that

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

Where v=Final velocity of object

u=Initial velocity of object

t=Time taken by object

Using the formula

Then, we get

Acceleration, a=\frac{28-1}{4.8}=\frac{27}{4.8} m/s^2

Acceleration=a=5.6 m/s^2

Hence, the acceleration of cheetah=5.6m/s^2

5 0
3 years ago
Name 2 things centripetal force acts on.
ira [324]

The centripetal force acts upon an object moving in a circle at constant speed.  The centripetal force acts perpendicular to the direction of motion , the speed of object will remain constant.

3 0
3 years ago
PLEASE HELP FAST!!!
ELEN [110]

Answer:

1.It improved by using the Mars survey probe, that took a portrait of the planet that helped us see the climate of Mars.

Explanation:

I watched the video.

7 0
3 years ago
A 4.5-kg, three legged stool supports a 89-kg person. If each leg of the stool has a cross-sectional diameter of 2.8 cm and the
kkurt [141]

Answer:

4.96 × 10⁵ Pa

Explanation:

F = mg

F = (m_{person}+m_{stool})g\\\\F =  (4.5 + 89)*9.8\\\\F = 916.3 N

This force is evenly distributed on the three leg

radius, r = d/2

= 2.8 / 2

= 1.4 cm = 0.014 m

total cross sectional area of the three legs, A = 3*pi*r^2

= 3\times\pi \times0.014^2\\\\= 1.847\times10^-^3m^2

Pressure due to weight,

P = Weight/A

P = / 1.847 × 10⁻³\\P = \frac{ 916.3N}{1.847\times10^-^3} \\\\P= 496032.9Pa

P = 4.96 × 10⁵ Pa

8 0
3 years ago
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