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Vitek1552 [10]
3 years ago
5

A 3000 n force acts on a 200 kg object what is the acceleration of the object

Physics
2 answers:
lesantik [10]3 years ago
7 0
F=ma. a= F/m = 3000/200 = 15 m/s
Virty [35]3 years ago
3 0

Explanation:

As we know that relation between force, mass and acceleration is as follows.

                              F = ma

where,    F = force

              m = mass

               a = acceleration

So, it is given that force is 3000 N and mass is 200 kg. Hence, calculate the acceleration as follows.

               F = ma

              3000 N = 200 \times a

                  a = \frac{3000}{200}

                     = 15 m/s^{2}

Thus, we can conclude that acceleration of the given object is 15 m/s^{2}.

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A bus travels a distance of 120 km with a speed of 40km per hour and returns with a speed of 30km per hour calculate the average
Vsevolod [243]

Answer:

35 km/hr

Explanation:

Average speed = (total of the speed)/(the sets of speeds given)

Direction does not matter in this instance since speed is only magnitude,

Average speed = (30 + 40)/2

Average speed = 70 ÷ 2

= 35 km/hr

3 0
3 years ago
What is the wave speed of a wave that has a frequency of 250 Hz and a wavelength of 0.35 m?
azamat

Answer:

87.5 m/s

Explanation:

The speed of a wave is given by

v=\lambda f

where

v is the wave speed

\lambda is the wavelength

f is the frequency

In this problem, we have

f=250 Hz is the frequency

\lambda=0.35 m is the wavelength

Substituting into the equation, we find

v=(0.35 m)(250 Hz)=87.5 m/s

6 0
3 years ago
A person hears a siren as a fire truck approaches and passes by. The frequency varies from 480Hz on approach to 400Hz going away
alekssr [168]

Answer:

31.2 m/s

Explanation:

f_{app} = Frequency of approach = 480 Hz

f_{aw} = Frequency of going away = 400 Hz

V = Speed of sound in air = 343 m/s

v = Speed of truck

Frequency of approach is given as

f_{app} = \frac{Vf}{V - v}                           eq-1

Frequency of moving awayy is given as

f_{aw} = \frac{Vf}{V + v}                          eq-2

Dividing eq-1 by eq-2

\frac{f_{app}}{f_{aw}} = \frac{V + v}{V - v}

\frac{480}{400} = \frac{343 + v}{343 - v}

v = 31.2 m/s

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7 0
2 years ago
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