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3 years ago

Kinetic Energy Assignment: Lab Report

Physics

1 answer:

kompoz [17]3 years ago

4 0

Answer:

so what do we gotta do?

Explanation:

is there any book or homework?

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Find out the kinetic energy of a body of mass 2kg moving with the speed of 3m/s

Varvara68 [4.7K]

Kinetic energy = 1/2mv2

= 1/2 x 2 x 3^2
= 9J

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4 years ago

Sheila did 110 J of work to move a chair 2 m to the right. How much force did Sheila use to move the chair?

Licemer1 [7]

Take 110 and divide it by 2 and there's your answer

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How do the magnetic poles of earth serve to protect life on earth?

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4 years ago

Read 2 more answers

To test a slide at an amusement park, a block of wood with mass 3.00 kgkg is released at the top of the slide and slides down to

dem82 [27]

Answer:

511.1 J

Explanation:

We are given that

Mass of wood block=m=3 kg

Vertical distance,h=23 m

Horizontal distance =x=30 m

Distance traveled in downward direction y=40 m

Initial velocity,u=0

$y=ut+\frac{1}{2} gt^2$

Where $g=9.8 m/s^2$

$40=0+\frac{1}{2}(9.8)t^2=4.9t^2$

$t^2=\frac{40}{4.9}$

$t=\sqrt{\frac{40}{4.9}}=2.86 s$

$x=v_x\times t$

$30=v_x(2.86)$

$v=v_x=\frac{30}{2.86}=10.49 m/s$

By work energy theorem

Change in kinetic energy=Work done= mgh-W

$\frac{1}{2}mv^2-\frac{1}{2}mu^2=3\times 9.8\times 23-W$

$\frac{1}{2}(3)(10.49)^2-0=676.2-W$

$W=676.2-\frac{1}{2}(3)(10.49)^2=511.1 J$

Hence, the work done due to friction on the block as it slides down the ramp=511.1 J

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3 years ago

a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate <br /&

Katen [24]

Answer:

a) T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

sin 30 = $\frac{T_x}{T}$

cos 30 = $\frac{T_y}{T}$

Tₓ = T sin 30

T_y = T cos 30

Y axis

T_y -W = 0

T cos 30 = mg (1)

X axis

Tₓ = m a

they relate it is centripetal

a = v² / r

we substitute

T sin 30 = m $\frac{v^2}{r}$ (2)

a) we substitute in 1

T = $\frac{mg }{cos 30}$

T = $\frac{ 0.2 \ 9.8}{cos \ 30}$

T = 2.26 N

b) from equation 2

v² = $\frac{T \ sin 30 \ r}{m}$

If we know the length of the string

sin 30 = r / L

r = L sin 30

we substitute

v² = $\frac{ T \ sin 30 \ L \ sin 30}{m}$

v² = $\frac{TL \ sin^2 30}{m}$

For the problem let us take L = 1 m

let's calculate

v = $\sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }$

v = 1.68 m / s

8 0

3 years ago