If you apply a 500N force to a car 200N car for 10 seconds but it doesn't move. How much work was done on the car?

PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer these correctly!!!! (40pts)

denpristay [2]

Answer:

False, true

Explanation:

3 0

2 years ago

Read 2 more answers

The owner of a company that manufactures drinking cups decides it would be impressive to build an inground swimming pool that is

garri49 [273]

Answer:

The depth is 5.15 m.

Explanation:

Lets take the depth of the pool = h m

The atmospheric pressure ,P = 101235 N/m²

The area of the top = A m²

The area of the bottom = a m²

Given that A= 1.5 a

The force on the top of the pool = P A

The total pressure on the bottom = P + ρ g h

ρ =Density of the water = 1000 kg/m³

The total pressure at the bottom of the pool = (P + ρ g h) a

The bottom and the top force is same

(P + ρ g h) a = P A

P a +ρ g h a = P A

ρ g h a = P A - P a

$h=\dfrac{P ( A-a)}{\rho g a}$

$h=\dfrac{P ( 1.5 a-a)}{\rho g a}$

$h=\dfrac{P ( 1.5- 1)}{\rho g}$

$h=\dfrac{101235 ( 1.5- 1)}{1000\times 9.81}\ m$

h=5.15 m

The depth is 5.15 m.

7 0

2 years ago

30. How do you make a conclusion?

FromTheMoon [43]

B. You compare the data

6 0

2 years ago

What is the average speed of a car that travels 60 meters in 2<br> seconds?

blondinia [14]

Answer:

30 m/s

Explanation:

Speed is distance over time. 60 meters / 2 seconds, = 30 m/s.

6 0

2 years ago

Aloop of wire of area 71 cm^2 is placed with its plane parallel to a 16 mt magnetic field. the loop is then rotated so that its

kkurt [141]

Answer:

Approximately 1.62 × 10⁻⁴ V.

Explanation:

The average EMF in the coil is equal to

$\displaystyle \frac{\text{Final Magnetic Flux} - \text{Initial Magnetic Flux}}{2}$ ,

Why does this formula work?

By Faraday's Law of Induction, the EMF $\epsilon$ induced in a coil (one loop) is equal to the rate of change in the magnetic flux $\Phi$ through the coil.

$\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t))$ .

Finding the average EMF in the coil is similar to finding the average velocity.

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt$ .

However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

$\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0)$ .

Hence the equation

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}$ .

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in $\Phi(t)$ won't matter.

Apply this formula to this question. Note that $\Phi$ , the magnetic flux through the coil, can be calculated with the equation

$\Phi = B \cdot A \cdot N \; \sin{\theta}$ .

For this question,

$B = \rm 16\; mT = 16\times 10^{-3}\; T$ is the strength of the magnetic field.
$A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2}$ is the area of the coil.
$N = 1$ is the number of loops in the coil.
$\theta$ is the angle between the field lines and the coil.
At $\rm 0\;s$ , the field lines are parallel to the coil, $\theta = 0^{\circ}$ .
At $\rm 0.7\; s$ , the field lines are perpendicular to the coil, $\displaystyle \theta = 90^{\circ}$ .