All categories

3 years ago

If a runner has a power output of 150 W over 10.0 s, then how much work does she do?

1 answer:

5 0

The amount of work done by a certain body is the product of the power given and the time. Mathematically,
Work = Power x time
Substituting the known values,
Work = (150 W) x (10 s)
= 1500 N.m
= 1500 J

You might be interested in

How much would it cost to cover the entire land area of the U.S. in dollar bills?

agasfer [191]

Answer:

$900 trillion

Explanation:

If Alaska is 20% of the contiguous US, then the approximate area of interest is ...

1200 miles × 3000 miles = 3.6×10^6 square miles.

The size of a dollar bill is about ...

(6.5 cm)·(15.5 cm) = 100.75 cm^2

One mile is 160,934.4 cm, so 1 square mile is about ...

1 mi^2 = (160,934.4 cm)^2 ≈ 2.59·10^10 cm^2

The number of dollars of interest is then ...

(3.6 · 10^6 mi^2)(2.59 · 10^10 cm^2)/(100.75 cm^2) ≈ 9.3·10^14

≈ 930 × 10^12 . . . dollars

It would cost about 900 trillion dollars to cover the land area of the US in $1 bills.

7 0

3 years ago

Compared with a force, a torque involves

ycow [4]

A rotational movement.

6 0

3 years ago

A luge run is straight for the first 10 m. If the velocity of the luger when he passes the start line is +25 m/s and his acceler

dangina [55]

V^2 =U^2 +2AS
V^2 = 25 ^2 + 2x9.5x10
V^2 = 625 + 1900 = 2525
V = 50.25

3 0

3 years ago

What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to e⃗

AleksandrR [38]

Electric potential energy of a dipole is given as

$U = - P.E$

$U = - PE cos\theta$

so change in the potential energy is given by

$U_f - U_i = (-PEcos\theta_2)- (-PEcos\theta_1)$

$\Delta U = PE (cos\theta_1 - cos\theta_2)$

here initially it was parallel to electric field while finally it is perpendicular to the electric field

so we have

$\Delta U = PE(cos 0 - cos 90)$

$\Delta U = PE$

so above is the change in potential energy of dipole

8 0

3 years ago

A jet airliner moving initially at 548 mph

sasho [114]

Let's choose the "east" direction as positive x-direction. The new velocity of the jet is the vector sum of two velocities: the initial velocity of the jet, which is

$v_1 =548 mph$ along the x-direction