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3 years ago

An object with little mass won't require a lot of force to move.

Physics

1 answer:

Anna35 [415]3 years ago

7 0

Answer:

True

Explanation:

Heavier objects (objects with more mass) are more difficult to move and stop. Heavier objects (greater mass) resist change more than lighter objects. Example: Pushing a bicycle or a Cadillac, or stopping them once moving. The more massive the object (more inertia) the harder it is to start or stop.

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Mr. Brewer drove his car at a speed of 60 miles/hour as he traveled from his house to Troy Prep. What additional information wou

Lubov Fominskaja [6]

Answer:how long it took him to get there

Explanation:

4 0

3 years ago

А masd<br>Of 500kg a raised to a height of 6m In 30s<br>Find (a) Workdone .

Mazyrski [523]

Answer:

Work done is 882000joule.

power is 29400watt.

Explanation:

given,

Mass(m)=500kg

Acceleration due to gravity(g)=9.8m/s²

Height(h)=6m

Time taken(t)=30s

Workdone=?

Power=?

now,

workdone=force*displaxement

= m*g*h

=500*9.8*6

=8,82,000joule

so, the work done by the man is 8,82,000joule.

then,

power=workdone/time taken

=8,82,000/30

=29,400watt

so, the required power to lift a load is 29,400watt.

5 0

3 years ago

I dont need answers, i just want to know what you guys think....just anything

Murljashka [212]

Answer:

I am not sure really.

Explanation:

I just don't know

7 0

3 years ago

A 60-kg swimmer suddenly dives horizontally from a 150-kg raft with a speed of 1.5 m/s. The raft is initially at rest. What is t

scZoUnD [109]

I’m almost positive it 60 m/s

7 0

3 years ago

A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with

Komok [63]

Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

$\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\ g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s$

$\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM$

Therefore the speed in RPM will be 62.64 RPM.

5 0

3 years ago