All categories

3 years ago

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:

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

17 copper wires of length l and diameter d are connected in parallel to form a single composite conductor of resistance R. What

Lubov Fominskaja [6]

Answer:

$\frac{D}{d} = 4.12$

Explanation:

As we know that resistance of one copper wire is given as

$r = \rho \frac{L}{a}$

here we know that

$a = \pi (\frac{d}{2})^2$

now we have

$r = \rho \frac{L}{\pi (\frac{d^2}{4})}$

$r = \rho \frac{4L}{\pi d^2}$

now we know that such 17 resistors are connected in parallel so we have

$R = \frac{r}{17}$

$R = \rho \frac{4L}{17 \pi d^2}$

Now if a single copper wire has same resistance then its diameter is D and it is given as

$R = \rho \frac{4L}{\pi D^2}$

now from above two equations we have

$\rho \frac{4L}{\pi D^2} = \rho \frac{4L}{17 \pi d^2}$

$D^2 = 17 d^2$

now we have

$\frac{D}{d} = 4.12$

3 0

3 years ago

Which part of the electromagnetic spectrum is nearest to radio waves?

Sav [38]

Answer:

i believe it is the microwaves that is closest to the radio waves.

Explanation:

8 0

3 years ago

Read 2 more answers

Un dulap are laturile de jos egale cu 60cm si1,1m,iar masa lui este de 25kg.Ce presiune exercita dulapul asupra podelei? URGENT!

Tcecarenko [31]

Answer:

okkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

7 0

3 years ago

Consider the sections of two circuits illustrated above. Select True or False for all statements. After connecting c and d to a

Basile [38]

Answer:

Follows are the solution to the given question:

Explanation:

In this question the missing file of the circuit is not be which is defined in attached file please find it.

In Option 1, this statement is true because the current is on $R_3 \ and \ R_4$ , that is the same.

In option 2, this statement is false because $Rcd=R_3+R_4$ therefore it implies that Rcd is always larger then $R_3$ .

In option 3, this statement is true because the voltage of $R_1 \ and \ R_2$ is always equal.

In option 4, this statement is true because $Rab= \frac{1}{(\frac{1}{R1}+\frac{1}{R2})}=\frac{R1R2}{(R1+R2)}. \frac{R2}{(R1+R2)}$ is always smaller then 1 therefore, $R_1 \times (\frac{R2}{(R1+R2)})$ is always equal to R1.