All categories

3 years ago

If an object, initially at rest, accelerates at the rate of 25m/s2, what will the magnitude of the displacement be after 50s

Physics

1 answer:

Lady bird [3.3K]3 years ago

5 0

Answer:

31250 meters

Explanation:

Given data

Intitially at rest, the velocity will be

u= 0m/s

acceleration a= 25m/s^2

Time= 50s

We know that the expression for the displacement is given as

S=U+ 1/2at^2

S= 0+ 1/2*25*50^2

S= 12.5*2500

S=31250 meters

Hence the displacement is 31250 meters

You might be interested in

A solid sphere of radius R carries a fixed, uniformly distributed charge q. Obtain an expression for the magnitude of the electr

NISA [10]

Answer:

The electric field outside the sphere will be $\dfrac{qr}{4\pi\epsilon_{0}R^3}$ .

Explanation:

Given that,

Radius of solid sphere = R

Charge = q

According to figure,

Suppose r is the distance between the point P and center of sphere.

If $\rho$ be the volume charge density,

Then, the charge will be,

$q=\rho\times\dfrac{4}{3}\pi R^3$ .....(I)

Consider a Gaussian surface of radius r.

We need to calculate the electric field outside the sphere

Using formula of electric field

$\oint{\vec{E}\cdot \vec{dA}}=\dfrac{Q}{\epsilon_{0}}$

$E\times4\pi r^2=\dfrac{\rho\dotc \dfrac{4}{3}\pi r^3}{\epsilon_{0}}$

Put the value from equation (I)

$E\times4\pi r^2=\dfrac{qr^3}{\epsilon_{0}R^3}$

$E=\dfrac{qr}{4\pi\epsilon_{0}R^3}$

Hence, The electric field outside the sphere will be $\dfrac{qr}{4\pi\epsilon_{0}R^3}$ .

4 0

3 years ago

Technically, how would you know if any work was done on an object?

kherson [118]

Answer:

through reading the question carefully and using yourself as an exampke

8 0

3 years ago

12. Suppose a person uses a mechanical iack to lift half the weight of a car with a

Likurg_2 [28]

a) The force that must be applied is 73.5 N

b) The actual efficiency is 82 %

Explanation:

Since the jack is 100% efficient, all the input work is converted into output work. So we can write:

$W_{in} = W_{out}\\F_{in} d_{in} = F_{out} d_{out}$

where:

$F_{in}$ is the input force applied on the jack

$d_{in}$ is the arm of the input force

$F_{out}$ is the output force applied on the jack

$d_{out}$ is the arm of the output force

Here we have:

$F_{out}=\frac{mg}{2}= \frac{(1200 kg)(9.8 m/s^2)}{2}=5880 N$ is the output force (half the weight of the car)

$d_{in} = 40.0 cm = 0.40 m$ is the arm of the input force

$d_{out} = 5.0 mm = 0.005 m$ is the arm of the output force

Solving for $F_{in}$ , we find the force that must be applied in input to lift the car:

$F_{in} = \frac{F_{out}d_{out}}{d_{in}}=\frac{(5880)(0.005)}{0.40}=73.5 N$

The efficiency of the jack is given by the ratio between the output work and the input work:

$\eta = \frac{W_{out}}{W_{in}}=\frac{F_{out}d_{out}}{F_{in}d_{in}}$

where we have:

$F_{out}=5880 N$ is the output force (half the weight of the car)

$d_{in} = 40.0 cm = 0.40 m$ is the arm of the input force

$d_{out} = 5.0 mm = 0.005 m$ is the arm of the output force

Here we are told that the input force this time is

$F_{in}=90.0N$

Substituting into the equation, we find the new efficiency of the jack:

$\eta = \frac{(5880)(0.005)}{(90.0)(0.40)}=0.82$

Which means an efficiency of 82%.

Learn more about levers:

brainly.com/question/5352966

#LearnwithBrainly

8 0

4 years ago

A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is

maksim [4K]

D = v^2 / 2ug

d= 3.5^2 / 0,15 x 9.8 m/s^2

the answer should be around 4.2m

hope this helps

8 0

3 years ago

lesya [120]

Answer:

Terrorists:scary,ruthless,morbid. Pretty:beautiful,sweet. Gay:happy,sweet,nice

Explanation:

Reeee

5 0

3 years ago