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3 years ago

What is the force needed to give a .25 kg arrow an acceleration of 196m/s2?

2 answers:

5 0

<span>49N is the force needed to give a .25 kg arrow an acceleration of 196m/s2. F =ma â‡’ =( 0.25kg)(196m/s2) = 49N if the arrow is shot horizontally where the applied force is entirely in the x-direction.</span>

Lyrx [107]3 years ago

4 0

F=MA
F is force- we are going to solve for F
M is mass- quantity of matter that takes up an object
A is acceleration- change in velocity
F=MA
F=(0.25 kg)(196 m/s^2)
F= 49 N (newton-SI unit for force)
Hope this helps!

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2 years ago

A student is standing 8 m from a roaring truck engine that is measured at 20 dB. The student moves 4 m closer to

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Answer:80 dB

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3 years ago

A box of mass 50 kg is pushed hard enough to set it in motion across a flat surface. Then a 99-N pushing force is needed to keep

vitfil [10]

Answer:

0.20

Explanation:

The box is moving at constant velocity, which means that its acceleration is zero; so, the net force acting on the box is zero as well.

There are two forces acting in the horizontal direction:

- The pushing force: F = 99 N, forward

- The frictional force: $F_f=\mu mg$ , backward, with

$\mu$ coefficient of kinetic friction

m = 50 kg mass of the box

g = 9.8 m/s^2 gravitational acceleration

The net force must be zero, so we have

$F-F_f = 0$

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3 years ago

The burner on an electric stove has a power output of 2.0 kW. A 710 g stainless steel tea kettle is filled with 20∘C water and p

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Answer:

The volume of water that was in the kettle is 1170 $cm^{3}$

Explanation:

Given:

Power, P = 2.0 kW = 2000 W, Mass of stainless steel, $m_{s}$ = 710 g = 0.71 kg at temperature of $20^{0} C$

Part A:

If it takes time, t = 3.5 minutes to reach boiling point of water $100^{0} C$ , then from conservation of energy,

Total energy supplied by the burner = Total heat gained by the water and the stainless steel to rise from $20^{0} C$ to $100^{0} C$

i.e. Pt = $m_{s}$ $c_{s}$ (100 - 20 ) + $m_{w}$ $c_{w}$ (100 - 20 )

$m_{w}$ = $\frac{pt - 80m_{s} C_{s} }{80c_{w} } = \frac{2000*3.5*60 - 80*0.71* 450}{80*4200}$

$m_{w}$ = 1.17 kg

where $c_{w}$ = 4200 J/Kgk (specific heat capacity of water), $c_{s}$ = 450 J/Kgk (specific heat capacity of steel)

But volume of water in the the kettle, v = $\frac{mass}{density} = \frac{1.17}{1000}= 1.17 *10^{-3}$ $m^{3}$

∴ v = 1170 $cm^{3}$

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3 years ago