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4 years ago

An object with a mass of 2.0 kg is accelerated at 5.0 m/s/s. the net force acting on the mass is

2 answers:

7 0

First we need to know the equation:
F= Mass times acceleration
F = 2.0 kg times 5.0 m/s^2
multiply them to get the net force!
F= 10 N
N is newton
Hope this heps

Arlecino [84]4 years ago

3 0

Answer:

Fn = 10 N

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : Net force acting on the mass or algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

Data

m = 2.0 kg : mass of the object

a = 5.0 m/s² : acceleration of the object

Calculation of the net force (Fn) acting on the mass

We apply the formula (1)

∑F = m*a, ∑F = Fn

Fn = m*a

Fn = (2.0 kg)* (5.0 m/s²)

Fn = 10 N

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2 years ago

Two balls are dropped from rest and allowed to fall. If one ball is allowed to fall for 1 s and the other for 3 s compare the di

kirill115 [55]

The second ball traveled a greater distance when compared to the first ball because the second ball spent more time in motion.

The given parameters;

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time of fall of the second ball, t = 3 s

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The distance traveled by the first ball is calculated as follows;

$h = u_0t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\h = (0.5\times 9.8\times 1^2)\\\\h = 4.9 \ m$

The distance traveled by the second ball is calculated as follows;

$h = \frac{1}{2} gt^2\\\\h = (0.5\times 9.8\times 3^2)\\\\h = 44.1\ m$

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2 years ago

A police officer uses a radar gun to determine the speed of a car. A specialized radar gun uses ultraviolet light to determine t

Virty [35]

Answer:

A transverse and D electromagnetic

on edg2020

Explanation:

I got it right on the test

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3 years ago

Read 2 more answers

You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the

Sphinxa [80]

Answer:

Option (c) : 20°C

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3 years ago

A projectile has an initial horizontal velocity of 34.0 M/s at the edge of a roof top. Find the horizontal and vertical componen

Sveta_85 [38]

Answer:

$v_x=34 m/s$

$v_y=53.9\ m/s$

Explanation:

<u>Horizontal Launch</u>

When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled by gravity until it eventually hits the ground.

The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:

vx=v

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$v_y=g.t$

The horizontal component of the velocity is always the same:

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$v_y=9.8*5.5=53.9$

$v_y=53.9\ m/s$

8 0

3 years ago