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3 years ago

An object with a mass of 62.5 kg accelerates 12.3 m/s2 when an unknown force is applied to it. What is the amount of the force?

Physics

1 answer:

AlexFokin [52]3 years ago

4 0

Answer:768.75N

Explanation:

mass=62.5kg

acceleration=12.3m/s^2

Force=mass x acceleration

Force=62.5 x 12.3

Force=768.75N

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A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. A horizontal force of 2.5 N is applied o

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Answers is F=7.84 N

Friction force resists the effect of horizontal force and trying to approch to a limiting force.

we have formula for limiting friction force between block and floor

F=Ц N

where N=mg

putting values we get answer.

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3 years ago

What is the electric current measured in

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An ampere (AM-pir), or amp

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2 years ago

In 1993 Ileana Salvador of Italy walked 3.0km in under 12min. Suppose that during her walk Salvador is observed to steadily incr

padilas [110]

The distance covered is 115 m

Explanation:

The motion of Ileana is a uniformly accelerated motion (constant acceleration), therefore we can use the following suvat equation:

$s=(\frac{u+v}{2})t$

where

s is the distance covered

u is the initiaal velocity

v is the final velocity

t is the time elapsed

In this problem, we have:

u = 4.20 m/s

v = 5.00 m/s

t = 25.0 s

Therefore, we can re-arrange the equation to find the distance covered:

$s=(\frac{4.20+5.00}{2})(25.0)=115 m$

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3 years ago

Which object has the most kinetic energy? a bus a car a plane a bicycle

loris [4]

Answer:I would guess a plane

Assuming they all Thad the same velocity....

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3 years ago

A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C

Vinil7 [7]

Answer:

75 kmh⁻¹

Explanation:

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$d_{ab}$ = distance traveled from station A to station B

$t_{ab}$ = time of travel between station A to station B

we know that

$Time = \frac{distance}{speed}$

$t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }$

Given that :

$d_{ab} = 4 d_{bc}$

$v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}$

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$t_{ab}$ = time of travel between station A to station B

$d_{ab}$ = distance traveled from station A to station B

we know that

$distance = (speed) (time)$

$d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc}$ = time of travel for train from station B to station C

we know that

$distance = (speed) (time)$

$d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}$

Given that :

$t_{ab} = 3 t_{bc}$

$v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}$

4 0

3 years ago