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3 years ago

Describe the transformation of mechanical energy in a ball when it is thrown straight up and then comes back down

Physics

1 answer:

zhenek [66]3 years ago

6 0

I believe that the mechanical energy would transform from starting out as kinetic, the reaching the top it would be potential, then go back to kinetic as it is falling back down.

I'm not 100% sure that this is right but if I had to take a guess this is what I would say.

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A small ball is attached to the lower end of a 0.800-m-long string, and the other end of the string is tied to a horizontal rod.

mestny [16]

Answer:

a = 17.68 m/s²

Explanation:

given,

length of the string, L = 0.8 m

angle made with vertical, θ = 61°

time to complete 1 rev, t = 1.25 s

radial acceleration = ?

first we have to calculate the radius of the circle

R = L sin θ

R = 0.8 x sin 61°

R = 0.7 m

now, calculating at the angular velocity

$\omega =\dfrac{2\pi}{T}$

$\omega =\dfrac{2\pi}{1.25}$

ω = 5.026 rad/s

now, radial acceleration

a = r ω²

a = 0.7 x 5.026²

a = 17.68 m/s²

hence, the radial acceleration of the ball is equal to 17.68 rad/s²

7 0

4 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago

Two flywheels of negligible mass and different radii are bonded together and rotate about a common axis (see below). The smaller

jekas [21]

Answer:

$F_2 = 29.54 N$

Explanation:

As we know that the combination is maintained at rest position

So we will take net torque on the system to be ZERO

so we know that

$\tau = \vec r \times \vec F$

here we will have

$\vec r_1 \times F_1 = \vec r_2 \times F_2$

so we have

$13 \times 50 = 22 \times F_2$

so we have

$F_2 = \frac{13 \times 50}{22}$

$F_2 = 29.54 N$

8 0

3 years ago

Whats the answer to this question show in the picture 2 questions

astraxan [27]

1) D

2) I would say A, but not 100%, its the only one that makes sense tho

3 0

3 years ago

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At what minimum speed must a roller coaster be traveling when upside down at the top of a circle so that the passengers do not f

Basile [38]

Jjjjbhubhhbvfvgvgvggb. v vvbbhhhhb

7 0

3 years ago