Ask question

Ask question

All categories

3 years ago

9

A man pushes a 35.2 kg box across a frictionless floor with a force of 128 N. What is the acceleration of the box

2 answers:

e-lub [12.9K]3 years ago

6 0

The formula for Force is F = MA

F = 128N.
M = 35.2kg.128
= 35.2A

Divide both sides by 35.2

A = ~3.636
So the answer is 3.636 m/s².

vodka [1.7K]3 years ago

4 0

The formula for Force is F = MA, or Force is equivalent to the product of Mass and Acceleration.

F = 128N.
M = 35.2kg.

128 = 35.2A
Divide both sides by 35.2 to solve for the acceleration.
A = ~3.636

The acceleration is 3.636 m/s^2.

I hope this helps!

You might be interested in

_____ have played a major role in altering Earth's atmospheric composition over time.A. TsunamisB. VolcanoesC. HurricanesD. Eart

sammy [17]

I believe the answer is valcanos because they release sulfur in to the air and other chemicals so it willl affect the anispher

4 0

3 years ago

Read 2 more answers

A car is moving at 10 m/s to the right. It accelerates for 10 s after which it is moving at 5 m/s to the left. What was the car'

NNADVOKAT [17]

Answer:

Acceleration, $a=-1.5\ m/s^2$

Explanation:

It is given that,

Initial velocity of the car, u = 10 m/s (in right)

Final velocity of the car, v = -5 m/s (in left)

Time taken, t = 10 s

Let a is the acceleration of the car. It can be calculated using the equation of kinematics. The equation is as :

$v=u+at$

$a=\dfrac{v-u}{t}$

$a=\dfrac{-5-10}{10}$

$a=-1.5\ m/s^2$

So, the acceleration of the car is $-1.5\ m/s^2$ . Hence, this is the required solution.

6 0

3 years ago

Which object has the larger magnitude of its momentum?

Charra [1.4K]

The object that had the most 1000 ton weight has the most momentum

5 0

3 years ago

Does coefficient of thermal expansion vary with temperature?

N76 [4]

Answer:

yes

Explanation:

8 0

2 years ago

Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic

Nuetrik [128]

Answer:

(a) A = $3.90 \AA$

(b) $A = 4.50 \AA$

(c) $A = 5.51 \AA$

(d) $A = 9.02 \AA$

Solution:

As per the question:

Radius of atom, r = 1.95 $\AA = 1.95\times 10^{- 10} m$

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = $2\times 1.95 = 3.90 \AA$

(b) For body centered cubic lattice:

$A = \frac{4}{\sqrt{3}}r$

$A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA$

(c) For face centered cubic lattice:

$A = 2{\sqrt{2}}r$

$A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA$

(d) For diamond lattice:

$A = 2\times \frac{4}{\sqrt{3}}r$

$A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA$

6 0

2 years ago