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3 years ago

@alexrobin13 A 62 kg box is lifted 12 meters off the ground. How much work was done?

Physics

2 answers:

34kurt3 years ago

6 0

7291.2! I'm for sure this is the right answer.

nevsk [136]3 years ago

3 0

Given:
weight=62 kgs
height= 12 m
gravity= 9.8
To find:
work done= mass*gravity*height
solution:
w=62*12*9.8
workdone=7291.2 J

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A baseball with a mass of 0.145 kilograms collides with a bat at a velocity of 44 meters/second. The ball bounces off the bat wi

dangina [55]

The answer is 13.5 N*S

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3 years ago

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A bowler throws a bowling ball of radius R = 11 cm along a lane. The ball slides on the lane with initial speed <img src="https:

s344n2d4d5 [400]

Answer:

a) v_com= Rω

b) -2.254 m/ $s^{2}$

c) 51.2 rad/ $s^{2}$

d) t=1.08 seconds

e) x=7.865m

f) v_roll=6.07m

Explanation:

Initially, the ball is travelling with v_com=v_0

Wen not rotating, at the initial stage the ball must be sliding along the surface.

This motion therefore generates a frictional force F_r at the point of contact.

Let the velocity at the point of contact be v_bottom

v_bottom=v_com-Rω

Therefore when ω=0, v_bottom=v_com

So when the ball begins rolling

v_com= Rω

F_r=μ_rmg

〖-F〗_r=ma_com

a_com=(〖-μ〗_r mg)/m

a_com=-μ_rg

a_com=-(0.23)(9.8)

a_com=-2.254m/s^2

Te negative sow decrearse

$\alpha$ =(μ_r mgR)/I = (〖5μ〗_r mgR)/2mRR

=(〖5μ〗_r g)/2R

=(5*(0.23)*(9.8))/(2*0.11)

=51.2 rad/s^2

t=v_0/(〖-a〗_com+Rα)

=8.5/(2.255+0.11*(51.2))

=8.5/7.886

=1.08 seconds

X=v_0 t+1/2 a_com t^2

X=8.5*(2.254) - 1/2 (2.254)*〖1.08〗^2

=7.865m

v_roll=v_0+a_com t_r

=8.5-(2.254)(1.08)

=6.07m/sec

Initially, the ball is travelling with v_com=v_0

Wen not rotating, at the initial stage the ball must be sliding along the surface.

This motion therefore generates a frictional force F_r at the point of contact.

a) Let the velocity at the point of contact be v_bottom

v_bottom=v_com-Rω

Therefore when ω=0, v_bottom=v_com

So when the ball begins rolling

v_com= Rω

b) F_r=μ_rmg

〖-F〗_r=ma_com

a_com=(〖-μ〗_r mg)/m

a_com=-μ_rg

a_com=-(0.23)(9.8)

a_com=-2.254m/s^2

Te negative sow decrearse

c) α=(μ_r mgR)/I = (〖5μ〗_r mgR)/2mRR

=(〖5μ〗_r g)/2R

=(5*(0.23)*(9.8))/(2*0.11)

=51.2 rad/s^2

d) t=v_0/(〖-a〗_com+Rα)

=8.5/(2.255+0.11*(51.2))

=8.5/7.886

=1.08 seconds

e) X=v_0 t+1/2 a_com t^2

X=8.5*(2.254) - 1/2 (2.254)*〖1.08〗^2

=7.865m

f) v_roll=v_0+a_com t_r

=8.5-(2.254)(1.08)

=6.07m/sec

7 0

3 years ago

A 3-kg object is attached to a spring and moving in simple harmonic motion. Its angular frequency is 20 rads/sec. When the mass-

Trava [24]

Answer:19.5 J

Explanation:

Given

mass of block=3 kg

angular frequency=20 rad/sec

spring constant $k=\omega _n^2m=1200 N/m$

we know total energy remain conserved

$E_T at x=0.1 m$

$E_T=E_P+E_K$

Where $E_K$ =kinetic energy

$E_P$ =potential Energy

$E_P=\frac{1}{2}kx^2$

$E_P=600\times 0.01=6 J$

$E_K=\frac{1}{2}mv^2$

$E_K=\frac{1}{2}\times 3\times 3^2=13.5 J$

$E_T=13.5+6=19.5 J$

When mass reaches amplitude its velocity becomes zero

there is only potential energy which is equal to Total energy

$E_T=E_P=19.5 J$

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3 years ago

Identify the smallest unit of an element

marysya [2.9K]

atom ................................................... sorry for the periods

3 0

3 years ago

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A bullet with a mass m b = 11.9 mb=11.9 g is fired into a block of wood at velocity v b = 261 m/s. vb=261 m/s. The block is atta

fredd [130]

Answer:

0.372 kg

Explanation:

The collision between the bullet and the block is inelastic, so only the total momentum of the system is conserved. So we can write:

$mu=(M+m)v$ (1)

where

$m=11.9 g = 11.9\cdot 10^{-3}kg$ is the mass of the bullet

$u=261 m/s$ is the initial velocity of the bullet

$M$ is the mass of the block

$v$ is the velocity at which the bullet and the block travels after the collision

We also know that the block is attached to a spring, and that the surface over which the block slides after the collision is frictionless. This means that the energy is conserved: so, the total kinetic energy of the block+bullet system just after the collision will entirely convert into elastic potential energy of the spring when the system comes to rest. So we can write

$\frac{1}{2}(M+m)v^2 = \frac{1}{2}kx^2$ (2)

where

k = 205 N/m is the spring constant

x = 35.0 cm = 0.35 m is the compression of the spring

From eq(1) we get

$v=\frac{mu}{M+m}$

And substituting into eq(2), we can solve to find the mass of the block:

$(M+m) \frac{(mu)^2}{(M+m)^2}=kx^2\\\frac{(mu)^2}{M+m}=kx^2\\M+m=\frac{(mu)^2}{kx^2}\\M=\frac{(mu)^2}{kx^2}-m=\frac{(11.9\cdot 10^{-3}\cdot 261)^2}{(205)(0.35)^2}-11.9\cdot 10^{-3}=0.372 kg$

4 0

3 years ago