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Fynjy0 [20]
2 years ago
10

You have a ball with a mass 1.3 kg tied to a rope, and you spin it in a circle of radius 1.8m. If the ball is moving at a speed

of 3 m/s, what is the centripetal force action on the ball?
Physics
1 answer:
BlackZzzverrR [31]2 years ago
7 0

Answer:

6.5\; {\rm N}.

Explanation:

When an object travel at a speed of v in a circle of radius r, the (centripetal) acceleration of that object would be a = (v^{2} / r).

In this question, the ball is travelling at v = 3\; {\rm m\cdot s^{-1}} in a circle of radius r = 1.8\; {\rm m}. The (centripetal) acceleration of this ball would be:

\begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(3\; {\rm m\cdot s^{-1}})^{2}}{1.8\; {\rm m}} \\ &= 5\; {\rm m\cdot s^{-2}}\end{aligned}.

By Newton's Laws of Motion, for an object of mass m, if the acceleration of that object is a, the net force on that object would be m\, a. Since the acceleration of this ball is a = 5\; {\rm m\cdot s^{-2}}, the net force on this ball would be:

\begin{aligned} F &= m\, a \\ &= 1.3\; {\rm kg} \times 5\; {\rm m\cdot s^{-2}} \\ &= 6.5\; {\rm N} \end{aligned}.

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A soccer player kicks a 0.44 kg ball with a force of 57.6 N, what is the ball’s acceleration
noname [10]

THE BALL'S ACCELERATION IS 130.90\frac{m}{s}

Explanation:

According to Newton's second law "the force (F) acting on an object is equal to the mass (m) of an object times its acceleration (a)", meaning that when the soccer player kicks a ball, a force is acting on the ball, therefore increasing it's acceleration

SO FOR CALCULATING THIS WE WILL USE NEWTON LAW,

   FORCE = MASS × ACCELERATION

WE ARE GIVEN

FORCE = 57.6N

MASS= .44KG

SO HERE WE APPLYING FORMULA ,WE WILL GET

ACCELERATION = \frac{FORCE}{MASS}

ACCELERATION = \frac{57.6N}{.44KG}

ACCELERATION =130.90\frac{m}{s}

8 0
3 years ago
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.20×10^6 N, one at an angle 14.0∘ west of north,
laila [671]

Answer:

1.45544 J

Explanation:

See attachment

5 0
3 years ago
Two 0.20-kg balls, moving at 4 m/s east, strike a wall. Ball A bounces backwards at the same speed. Ball B stops. Which statemen
muminat

Answer:

Option A

Explanation:

From the question we are told that:

Mass m=0.20kg

Velocity v=4m/s

Generally the equation for momentum for Ball A is mathematically given by

Initial Momentum

 M_{a1}=mV

 M_{a1}=0.2*4

 M_{a1}=0.8

Final Momentum

 M_{a2}=-0.8kgm/s

Therefore

 \triangle M_a=-1.6kgm/s

Generally the equation for momentum for Ball B is mathematically given by

Initial Momentum

 M_{b1}=mV

 M_{b1}=0.2*4

 M_{b1}=0.8

Final Momentum

 M_{b2}=-0 kgm/s

Therefore

 |\triangle M_a|>|\triangle Mb|

Option A

4 0
3 years ago
A car slows down uniformly from a speed of 30.0 m/s to rest in 7.20 s
Ede4ka [16]

When acceleration is constant, the average velocity is given by

\bar v=\dfrac{v+v_0}2

where v and v_0 are the final and initial velocities, respectively. By definition, we also have that the average velocity is given by

\bar v=\dfrac{\Delta x}{\Delta t}=\dfrac{x-x_0}{t-t_0}

where x,x_0 are the final/initial displacements, and t,t_0 are the final/initial times, respectively.

Take the car's starting position to be at t_0=0\,\mathrm s. Then

\dfrac{v+v_0}2=\dfrac{x-x_0}t\implies x=x_0+\dfrac12(v+v_0)t

So we have

x=0\,\mathrm m+\dfrac12\left(0\,\dfrac{\mathrm m}{\mathrm s}+30.0\,\dfrac{\mathrm m}{\mathrm s}\right)(7.20\,\mathrm s)=108\,\mathrm m

You also could have first found the acceleration using the equation

v=v_0+at

then solve for x via

x=x_0+v_0t+\dfrac12at^2

but that would have involved a bit more work, and it turns out we didn't need to know the precise value of a anyway.

7 0
3 years ago
How does Mercury's close proximity to the sun and thin atmosphere affect its ability to maintain liquid water
alekssr [168]

Answer

A thin atmosphere does not supply much oxygen, and the heat from the sun would evaporate it, because mercury is close to the sun.

5 0
4 years ago
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