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2 years ago

How long must a 400w electrical engine work in order to produce 300kj of work

Physics

1 answer:

vladimir1956 [14]2 years ago

8 0

Answer:

300000 J / 400 J/s = 750 s = 12.5 minutes

Explanation:

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A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc

Harman [31]

Answer:

$E=88200\ J$

Explanation:

Given:

mass of car, $m=1500\ kg$

distance of skidding after the application of brakes, $d=15\ m$

coefficient of kinetic friction, $\mu_k=0.4$

So, the energy dissipated during the skidding of car:

Frictional force:

$f=\mu_k.N$

where N = normal reaction by ground on the car

$f=0.4\ties 1500\times 9.8$

$f=5880\ N$

Now from the work-energy equivalence:

$E=f.d$

$E=5880\times 15$

$E=88200\ J$ is the dissipated energy.

3 0

3 years ago

A submarine is stranded on the bottom of the ocean with its hatch 21.0 m below the surface. calculate the force (in n) needed to

Tanzania [10]

Answer:

28,400 N

Explanation:

Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:

$p_{top} = p_{atm} + \rho g h=1.013\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(21.0 m)=3.071 \cdot 10^5 Pa$

On the lower part of the hatch, there is a pressure equal to

$p_{bot}=p_{atm}=1.013\cdot 10^5 Pa$

So, the net pressure acting on the hatch is

$p=p_{top}-p_{bot}=3.071 \cdot 10^5 Pa - 1.013\cdot 10^5 Pa=2.058 \cdot 10^5 Pa$

which acts from above.

The area of the hatch is given by:

$A=\pi r^2 = \pi (\frac{0.420 m}{2})^2=0.138 m^2$

So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:

$F=pA=(2.058\cdot 10^5 Pa)(0.138 m^2)=28,400 N$

8 0

3 years ago

You do 45.0 joules of work and 3.00 seconds how much power do you use

Arada [10]

Answer:

15 watt

Explanation:

Power is the rate at which work is done.

This means you divide the work done with the amount of time used to perform the work.

The formula for Power is : P = W/t where;

W= work done in J = 45

t= time in seconds = 3 sec

P= 45/ 3 = 15 watt

8 0

2 years ago

A _____ space zone is one that is obstructed in some way

telo118 [61]

A closed space zone is one that is obstrutced in some way.

Hope this helps :)

5 0

2 years ago

Read 2 more answers

10. Solve the following numerical problems

frosja888 [35]

Answer:

$\boxed {\boxed {\sf 120 \ Joules}}$

Explanation:

Work is equal to the product of force and distance.

$W=F*d$

The force is 8 Newtons and the distance is 15 meters.

$F= 8 \ N \\d= 15 \ m$

Substitute the values into the formula.

$W= 8 \ N * 15 \ m$

Multiply.

$W= 120 \ N*m$

1 Newton meter is equal to 1 Joule
Our answer of 120 N*m equals 120 J

$W= 120 \ J$

The work done is 120 Joules

3 0

2 years ago