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joja [24]
2 years ago
12

What is centripetal forcehello​

Physics
2 answers:
Gwar [14]2 years ago
8 0

Answer:

A centripetal force is a net force that acts on an object to keep it moving along a circular path

Bond [772]2 years ago
5 0

Answer:

A force that acts on a body moving in a circular path and directed towards the centre around which the body is moving is called Centripetal force.

Explanation:

When an object travels around a circular path with a constant speed, it experiences an accelerating centripetal force towards the centre.

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Why is there a difference between potential and kinetic energy?
inn [45]

Potential energy is stored energy because it has the potential to do something which laters turns into kinetic energy which is the moving energy.

3 0
2 years ago
In 2007, michael carter (u.s.) set a world record in the shot put with a throw of 24.77 m. what was the initial speed of the sho
Serga [27]

The initial speed of the shot is 15.02 m/s.

The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.

Pl refer to the attached diagram.

Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.

Write an expression for R.

R=u_xt=(ucos \theta)t

Therefore,

t=\frac{R}{ucos\theta} .......(1)

In the time t, the net displacement of the shotput is y in the downward direction.

Use the equation of motion,

y=u_yt-\frac{1}{2}gt^2=(usin\theta) t-\frac{1}{2}gt^2

Substitute the value of t from equation (1).

y=(ucos\theta)(\frac{R}{ucos\theta} )-\frac{1}{2} g(\frac{R}{ucos\theta} )^2\\ =Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

y=Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )\\ (-2.10m)=(24.77 m)(tan38.0^o)-\frac{(9.8 m/s^2)(24.77m)^2}{2u^2(cos38.0^o)^2} \\ u^2=225.71(m/s)^2\\ u=15.02m/s

The shot put was thrown with a speed 15.02 m/s.




7 0
2 years ago
Suppose a particle is accelerated through space (no gravity) by a 10 N force. Suddenly the particle encounters a second force of
iren2701 [21]

please dont mind me just looking for points

3 0
3 years ago
) A satellite of mass m has an orbital period T when it is in a circular orbit of radius R around the earth. If the satellite in
Mrrafil [7]

Answer:

A) T.

Explanation:

Kepler's third law states that the orbital period (T) of a satellite is related with the radius (R) and the mass of the object (M) it orbits:

T=\frac{2\pi R^{\frac{3}{2}}}{\sqrt{GM}}  

So the orbital period is independent of the mass of the satellite, that means no matter the mass every satellite at a radius R around the earth have an orbital period A.

4 0
3 years ago
Energy from the Sun arrives at the top of the Earth’s atmosphere with an intensity of 1.30kW/m21.30⁢kW/m2. How long does it take
vivado [14]

Answer:

T=1.384×10⁶seconds

Explanation:

Given data

p (Intensity)=1.30 kw/m²

E (Energy)=1.8×10⁹ J

A (Area)=1.00 m²

T (Time required)=?

Solution

E=PT   ................eq(i)

where E is energy

P is radiation power

T is time

Radiating Power is given as

P=pA

Where p is intensity

A is Area

Put P=pA in eq(i) we get

E=pAT

T=E/pA

T=\frac{(1.8*10^{9}}{(1.30*10^{3} )*(1.00)}  )\\T=1.38*10^{6} seconds

3 0
3 years ago
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