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2 years ago

Calculate the force needed to give a car of mass 800 kg an acceleration of 2.0 ms−2. plss quick

Physics

1 answer:

erik [133]2 years ago

4 0

The force needed to give a car of mass 800 kg an acceleration of 2.0 ms-² is 1600N.

<h3>How to calculate force?</h3>

The force needed to push an object can be calculated by multiplying the mass of the object by its acceleration as follows:

Force = mass × acceleration

According to this question, a car of mass 800 kg has an acceleration of 2.0 ms−². The force is calculated as follows:

Force = 800kg × 2m/s²

Force = 1600N

Therefore, the force needed to give a car of mass 800 kg an acceleration of 2.0 ms-² is 1600N.

Learn more about force at: brainly.com/question/13191643

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navik [9.2K]

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4 years ago

During exercise, the supply of oxygen and __________ to muscles increases. What one word completes the sentence?

kondaur [170]

Answer:

Glucose

Explanation:

During exercise, muscles need more energy. They get this energy by aerobic respiration, which uses oxygen and glucose to release energy.

Note: If the body fails to supply oxygen to the muscles fast enough, the muscles will produce energy by anaerobic respiration, which is not as effective as aerobic respiration in addition to producing lactic acid.

7 0

3 years ago

Read 2 more answers

World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude

Karo-lina-s [1.5K]

Answer:

A. The force exerted by the sprinter must be 9.6 × 10² N.

B. The force that propels the sprinter is exerted by the blocks.

Explanation:

Hi there!

Let´s begin with part B:

The sprinter exerts a force on the blocks and, as a reaction, the blocks exert a force on the sprinter that is of equal magnitude but opposite direction (Newton´s third law). This reaction of the blocks causes the acceleration of the sprinter.

Part A

The force exerted by the blocks can be calculated using Newton´s second law:

F = m · a

Where:

F = exerted force.

m = mass of the object being accelerated.

a = acceleration of the object after applying the force on the object.

F = m · a

F = 64 kg · 15 m/s²

F = 9.6 × 10² N

8 0

3 years ago

An open 1-m-diameter tank contains water at a depth of 0.7 m when at rest. As the tank is rotated about its vertical axis the ce

Mamont248 [21]

Answer:

Explanation:

To find the angular velocity of the tank at which the bottom of the tank is exposed

From the information given:

At rest, the initial volume of the tank is:

$V_i = \pi R^2 h_i --- (1)$

where;

height h which is the height for the free surface in a rotating tank is expressed as:

$h = \dfrac{\omega^2 r^2}{2g} + C$

at the bottom surface of the tank;

r = 0, h = 0

∴

$h = \dfrac{\omega^2 r^2}{2g} + C$

0 = 0 + C

C = 0

Thus; the free surface height in a rotating tank is:

$h=\dfrac{\omega^2 r^2}{2g} --- (2)$

Now; the volume of the water when the tank is rotating is:

dV = 2π × r × h × dr

Taking the integral on both sides;

$\int \limits ^{V_f}_{0} \ dV = \int \limits ^R_0 \times 2 \pi \times r \times h \ dr$

replacing the value of h in equation (2); we have:

$V_f} = \int \limits ^R_0 \times 2 \pi \times r \times ( \dfrac{\omega ^2 r^2}{2g} ) \ dr$

$V_f = \dfrac{ \pi \omega ^2}{g} \int \limits ^R_0 \ r^3 \ dr$

$V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{r^4}{4} \Big]^R_0$

$V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{R^4}{4} \Big] --- (3)$

Since the volume of the water when it is at rest and when the angular speed rotates at an angular speed is equal.

Then $V_f = V_i$

Replacing equation (1) and (3)

$\dfrac{\pi \omega^2}{g}( \dfrac{R^4}{4}) = \pi R^2 h_i$

$\omega^2 = \dfrac{4g \times h_i }{R^2}$

$\omega =\sqrt{ \dfrac{4g \times h_i }{R^2}}$

$\omega = \sqrt{\dfrac{4 \times 9.81 \ m/s^2 \times 0.7 \ m}{(0.5)^2} }$

$\omega = \sqrt{109.87 }$

$\mathbf{\omega = 10.48 \ rad/s}$

Finally, the angular velocity of the tank at which the bottom of the tank is exposed = 10.48 rad/s

6 0

3 years ago

A car traveling in a straight line has a velocity of 5.0 m/s. After an acceleration of 0.75 m/s/s, the cars velocity is 8.0. In

Bogdan [553]

Vs - velocity on beginning
ve - velocity on ending. You've got:
$v_s = 5 \frac{m}{s} \\ v_e=8 \frac{m}{s} \\ \hbox{Then:} \\ \Delta v=v_e - v_s = 8 \frac{m}{s} - 5\frac{m}{s}=3 \frac{m}{s} \\ a=0,75 \frac{m}{s^2} \\ \hbox{And from formula:} \\ a=\frac{\Delta v}{\Delta t} \qquad \Rightarrow \qquad \Delta t= \frac{\Delta v}{a} \\ \hbox{Substitute:} \\ \Delta t=\frac{3\frac{m}{s}}{0,75 \frac{m}{s^2}}= \frac{3}{\frac{3}{4}} s= 3 \cdot \frac{4}{3} s= 4 s$
So he needed 4 second.

3 0

3 years ago