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2 years ago

What force is required to accelerate a body with a mass of 15 kilograms at a rate of 8 m/s²?

2 answers:

mihalych1998 [28]2 years ago

6 0

Newton's Second law of motion:

Force = (mass) x (acceleration)

Force = (15kg) x (8m/s²) = 120 kg-m/s² = 120 newtons

USPshnik [31]2 years ago

4 0

The equation for force is force is equal to mass times acceleration or $F = ma$ where:

- Force or F is measured in Newtons or N (which is also kilogram-meter per second squared),
- Mass or m is measured in kilograms or kg, and acceleration is measured in meters per second or m/s2.

This leads to an answer in kilogram-meter per second squared or kg- m/s2<span>So if we multiply a body with a mass of 15 kilograms to a rate of 8 meters per second, we get 120 kilogram-meter per second squared. </span>

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The position of an object in simple harmonic motion is defined by the function y = (0.50 m) sin (πt/2). Determine the maximum sp

Gnoma [55]

The maximum speed of the object under simple harmonic motion is 0.786 m/s.

The given parameters:

Position of the particle, y = 0.5m sin(πt/2)

<h3>Wave equation for simple harmonic motion;</h3>

y = A sin(ωt + Ф)

where;

A is the amplitude = 0.5 m
ω is the angular speed = π/2

The maximum speed of the object is calculated as follows;

$V_{max} = A \omega\\\\V_{max} = 0.5 \times \frac{\pi}{2} = \frac{\pi}{4} \ m/s = 0.786 \ m/s$

Thus, the maximum speed of the object under simple harmonic motion is 0.786 m/s.

Learn more about simple harmonic motion here: brainly.com/question/17315536

5 0

2 years ago

An object of mass m = 5.0 kg hangs from a cord around a light pulley: The length of the cord between the oscillator and the pull

puteri [66]

Answer:

$\mu=0.0049Kg/m$

Explanation:

When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:

$f_n=\frac{nv}{2L}$

Where $L$ is the length of the string and $v$ the velocity of propagation. Use this expression to find the value of $v$ .

$f_6=\frac{6v}{2L}\\(150)=\frac{6v}{2(2)} \\150=\frac{3v}{2} \\3v=150(2)\\ v=\frac{300}{3} \\v=100m/s$

The velocity of propagation is given by the expression:

$v=\sqrt{\frac{T}{\mu }$

Where $\mu$ is the desirable variable of the problem, the linear mass density, and $T$ is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:

$T=W=mg=(5)(9.81)=49.05N$

With the value of the tension and the velocity you can find the mass density:

$v=\sqrt{\frac{T}{\mu}$

$v^2=\frac{T}{\mu}\\ \mu=\frac{T}{v^2} =\frac{49.05}{(100)^2} =\frac{49.05}{10000} =0.0049Kg/m$

6 0

2 years ago

All Physicsts over here plz help in these questions!!!!!!!!!

mrs_skeptik [129]

Hello!

First one we can use that PE=mgh so we have

4.37*10^5J/(9.12*10^3kg*9.80m/s^2)= 4.89m

Second one we can use Newton’s Second Law
F=ma and in this case F=mg so we have

g= 3.28*10^-2N/6*10^-3kg = 5.47m/s^2

Hope this helps. Any questions please ask. Thank you.

6 0

3 years ago

Read 2 more answers

The position of a particle as it moves along an y axis is given by y = (2.0cm)sin(πt/4), with t in second and y in centimeters.

irina [24]

Part a)

At t = 0 the position of the object is given as

$x = 0$