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2 years ago

An object accelerates 13.0 m/s2 when a force of 4.4 newtons is applied to it. What is the

Physics

1 answer:

velikii [3]2 years ago

6 0

Answer:

0.3384 N

Explanation:

Acceleration = 13 m/s^2

Force = 4.4 N

Force = mass * acceleration

mass = force / acceleration

mass = 4.4 / 13

mass = 0.3384 N

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b.Law of conservation of mass

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The wavelength of red helium-neon laser light in air is 632.8 nm.(a) What is its frequency?(b) What is its wavelength in glass t

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(a) $4.74 \cdot 10^{14}Hz$

The frequency of a wave is given by:

$f=\frac{v}{\lambda}$

where

v is the wave's speed

$\lambda$ is the wavelength

For the red laser light in this problem, we have

$v=c=3\cdot 10^8 m/s$ (speed of light)

$\lambda=632.8 nm=632.8\cdot 10^{-9} m$

Substituting,

$f=\frac{3\cdot 10^8 m/s}{632.8 \cdot 10^{-9} m}=4.74 \cdot 10^{14}Hz$

(b) 427.6 nm

The wavelength of the wave in the glass is given by

$\lambda=\frac{\lambda_0}{n}$

where

$\lambda_0 = 632.8\cdot 10^{-9} m$ is the original wavelength of the wave in air

n = 1.48 is the refractive index of glass

Substituting into the formula,

$\lambda=\frac{632.8\cdot 10^{-9}m}{1.48}=427.6\cdot 10^{-9}m=427.6 nm$

(c) $2.02\cdot 10^8 m/s$

The speed of the wave in the glass is given by

$v=\frac{c}{n}$

where

$c = 3\cdot 10^8 m/s$ is the original speed of the wave in air

n = 1.48 is the refractive index of glass

Substituting into the formula,

$v=\frac{3\cdot 10^8 m/s}{1.48}=2.02\cdot 10^8 m/s$

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3 years ago

The student makes a single pile of the 500 sheets of paper. Which a metre rule, she measured the height of the pile. The height

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Answer: please see attached work.

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3 years ago

Who water rocket starts from rest and roses straight up with an acceleration of 5 m/s until it runs out of water 2.5 seconds lat

Kitty [74]

Answer:

23. 4375 m

Explanation:

There are two parts of the rocket's motion

1 ) accelerating (assume it goes upto h1 height )

using motion equations upwards

$s = ut+\frac{1}{2}*a*t^{2} \\h_1=0+\frac{1}{2}*5*2.5^{2} \\=15.625 m$

Lets find the velocity after 2.5 seconds (V1)

V = U +at

V1 = 0 +5*2.5 = 12.5 m/s

2) motion under gravity (assume it goes upto h2 height )

now there no acceleration from the rocket. it is now subjected to the gravity

using motion equations upwards (assuming g= 10m/s² downwards)

V²= U² +2as

0 = 12.5²+2*(-10)*h2

h2 = 7.8125 m

maximum height = h1 + h2

= 15.625 + 7.8125

= 23. 4375 m

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3 years ago