a student pushes a box with a total mass of 50kg. what is the net force on the box if it accelerates 1.5 m/s^2?

snow_tiger [21]

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8 0

4 years ago

What derived unit is used to measure the slope of the line in this graph?

Alexxx [7]

Answer:

g/cm³

Explanation:

From the question given above,

The y-axis is representing mass (g)

The x-axis is representing volume (cm³)

Unit of slope =?

Slope of a graph is simply defined as the change in y-axis divided by the change in x-axis. Mathematically it is expressed as:

Slope = change in y-axis (Δy)/change in x-axis (Δx)

Slope = Δy/Δx

Thus, with the above formula, we can obtain the unit used for measuring the slope as follow:

y-axis = mass (g)

x-axis = volume (cm³)

Slope =.?

Slope = Δy/Δx

Slope = mass (g) /volume (cm³)

Slope = g/cm³

Therefore, the derive unit used for measuring the slope is g/cm³

5 0

3 years ago

Two separate masses on two separate springs undergo simple harmonic motion indefinitely (the surface is frictionless). In CASE 1

Artemon [7]

Answer:

$\frac{F_1}{F_2} =\frac{1}{2}$

Explanation:

Assuming the we have to find ratio maximum forces on the mass in each case

we know that in a spring mass system

F= Kx

K= spring constant

x= spring displacement

Case 1:

$F_1=2k\times d$

case 2:

$F_2=2k\times 2d$

therefore, $\frac{F_1}{F_2} = \frac{2K\times d}{2K\times 2d}$

$\frac{F_1}{F_2} =\frac{1}{2}$

4 0

3 years ago

Solve for work when

BlackZzzverrR [31]

So, <u>the value of the work is approximately 84.65 J</u>.

<h2>Introduction</h2>

Hi ! Here I will help you to discuss the subject about work that caused by force in amount value of angle. Work is affected by the force and displacement.

If related to the magnitude of the force, the amount of work will be proportional to the magnitude of the applied force. Thats mean, if the value of the force that applied on it is greater, then the value of the work will be greater.
If related to the magnitude of shift, the amount of work will be proportional to the magnitude of shift of object. Thats mean, if the value of the shift on it is greater, then the value of the work will be greater.

<h3>Formula Used</h3>

The work done by a moving object can be expressed in the equation:

If the Angle Is Ignored

$\boxed{\sf{\bold{W = F \times s}}}$

If the Angle Effect on Work

$\boxed{\sf{\bold{W = F \times s \times \cos(\theta)}}}$

With the following condition:

W = work that done by object (J)
F = force that applied (N)
s = shift or distance (m)
$\sf{\theta}$ = angle of elevation (°)

<h3>Solution</h3>

We know that :

F = force that applied = $\sf{1.41 \times 10^4}$ N
s = shift or distance = 84.9 m
$\sf{\theta}$ = angle of elevation = 45°

What was asked ?

W = work that done by object = ... J

Step by step :

$\sf{W = F \times s \times \cos(\theta)}$

$\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \cos(45^o)}$

$\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \frac{\sqrt{2}}{2}}$

$\sf{W = 119.709 \times \frac{\sqrt{2}}{2}}$

$\sf{W = 59.8545 \sqrt{2}}$

$\boxed{\sf{W \approx 84.65 \: J}}$

<h3>Conclusion</h3>

So, the value of the work is approximately 84.65 J.

3 0

1 year ago

A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t

notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is $1.37 m/s^2$

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, $F_f$ , backward

The frictional force can be written as

$F_f = \mu mg$

where

$\mu=0.03$ is the coefficient of kinetic friction

$m=150 kg$ is the mass of the motorbike

$g=9.8 m/s^2$ is the acceleration of gravity

Therefore the net force is given by

$\sum F = F - F_f = F - \mu mg$

And substituting, we find

$\sum F=250 - (0.03)(150)(9.8)=205.9 N$

2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

$\sum F = ma$