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3 years ago

11

A 5 kg rock is lifted 3m find amount of work done a= 98 J

1 answer:

Levart [38]3 years ago

6 0

Answer:

How long would it take a machine to do 13,000 joules of work if the power rating ... An object gains 15 joules of potential energy as it is lifted vertically 5.0 meters. ... As the time required to do a given quantity of work decreases, the power ... How much work is being done to the system by the person of the sled moves 10 m?

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What is the similarities of a furnace and the sun?

Svetlanka [38]

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They are both extremely hot, they both produce a form of light, they both have/use fire(typically)

Explanation:

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A building is being knocked down with a wrecking ball, which is a big metal sphere that swings on a 12-m-long cable. You are (un

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Because you know that gravity is in m/s^2 so, period will be measured in seconds. You know the cable is 12m long and gravity is 9.81 solve for T (period) 2π12sqrt(9.81)=6.94922

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What is the ultimate source of energy for an ecosystem?

Romashka-Z-Leto [24]

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3 years ago

A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas

Radda [10]

Answer:

change in momentum, $\Delta p=7.65 \,kg.m.s^{-1}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

Average Force, $F=144.3396\,N$

$F_x=125.0018\,N$

$F_y=72.1698\,N$

Explanation:

Given:

angle of kicking from the horizon, $\theta= 30^{\circ}$

velocity of the ball after being kicked, $v=18 m.s^{-1}$

mass of the ball, $m=0.425\, kg$

time of application of force, $t=5.3\times 10^{-2}\,s$

We know, since body is starting from the rest

$\Delta p=m.v$ .....................(1)

$\Delta p=0.425\times 18$

$\Delta p=7.65 \,kg.m.s^{-1}$

Now the components:

$\Delta p_x= 7.65\times cos 30^{\circ}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

similarly

$\Delta p_y= 7.65\times sin 30^{\circ}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

also, impulse

$I=F\times t$ .........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

$F\times t=m.v$

$F\times 5.3\times 10^{-2}= 7.65$

$F=144.3396\,N$

Now, the components

$F_x=144.3396\times cos 30^{\circ}$

$F_x=125.0018\,N$

&

$F_y=144.3396\times sin 30^{\circ}$

$F_y=72.1698\,N$

6 0

3 years ago

A woman stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.64

Galina-37 [17]

Answer:

The downwards acceleration is 3.53 m/s2.

Explanation:

Let the true weight is m g.

The reading of the balance, R = 0.64 mg

Let the acceleration is a.

As the apparent weight is less than the true weight so the elevator goes down wards with some acceleration.

Use Newton's second law

m g - R = m a

m g - 0.64 m g = m a

0.36 g = a

a = 3.53 m/s2

6 0

3 years ago