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sesenic [268]
2 years ago
14

Given a force of 88n and an acceleration of 4m/s2 what is the mass?

Physics
2 answers:
Hitman42 [59]2 years ago
5 0
F=ma
m=F/a = 88/4 = 22 kg

Leni [432]2 years ago
3 0

Answer:

Mass, m = 22 kg

Explanation:

It is given that,

Force, F = 88 N

Acceleration, a=4\ m/s^2

We need to find its mass. It can be calculated using the formula of force as per second law of Newton i.e.

F = m a

m=\dfrac{F}{a}

m=\dfrac{88\ N}{4\ m/s^2}  

m = 22 kg

So, the mass of the object is 22 kg. Hence, this is the required solution.

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rotational velocity initial = 0.90 rev/s 
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lo =  0.8*4.1 / 0.90 = 3.6 kg m2
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Throughout the problem, take the speed of sound in air to be 343 m/s . Part A Consider a pipe of length 80.0 cm open at both end
White raven [17]

Answer:

the lowest frequency f of the sound wave is 214.375 Hz

Explanation:

The computation of the lowest frequency f of the sound wave is shown below;

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= 0.8 m

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V1 = 214.375 Hz

Hence, the lowest frequency f of the sound wave is 214.375 Hz

We simply applied the above formula so that the correct value could come

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An object is placed 18 cm in front of spherical mirror.if the image is formed at 4cm to the right of the mirror, calculate it's
ivolga24 [154]
1) Focal length

We can find the focal length of the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i} (1)
where 
f is the focal length
d_o is the distance of the object from the mirror
d_i is the distance of the image from the mirror

In this case, d_o = 18 cm, while d_i=-4 cm (the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
\frac{1}{f}= \frac{1}{18 cm}- \frac{1}{4 cm}=- \frac{7}{36 cm}
from which we find
f=- \frac{36}{7} cm=-5.1 cm

2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.

3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
r=2f=2 \cdot 5.1 cm=10.2 cm
3 0
3 years ago
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