All categories

3 years ago

a shot putter accelerates a 7.3kg shot from rest to 14m/s in 1.5r seconds. what average power was developed?

Physics

1 answer:

Sedaia [141]3 years ago

3 0

Explanation:

$power = \frac{energy \: expended}{time}$

m = 7.3kg

u = 0

v = 14m/s

t = 1.5sec

P = (0.5×7.3×14²) ÷ 1.5

P = 476.93

P = 477 watt

You might be interested in

How long does it take light from the sun to reach earth?

storchak [24]

It takes approximately 8 minutes and 20 seconds.

5 0

4 years ago

Read 2 more answers

HELP ME!!! I don't know if you can see the picture on the right or not..

gizmo_the_mogwai [7]

It’s an assortment of compound molecules

7 0

3 years ago

Read 2 more answers

The average annual rainfall in Naples, Italy, is 39.6 inches, and the average annual rainfall in Fort Collins, Colorado, is 15 i

Nezavi [6.7K]

The simple answer would be; closeness to the equator combined with general climate.

7 0

4 years ago

Read 2 more answers

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp

slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

$Q=Av$

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

$Q=1.20 m^3/s$

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

$A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2$

So we can re-arrange the equation to find the speed of the water:

$v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s$

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

$Q_1 = Q_2\\A_1 v_1 = A_2 v_2$

where we have

$A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s$

and where $A_2$ is the cross-sectional area of the pipe at the second point.

Solving for A2,

$A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2$

And finally we can find the radius of the pipe at that point:

$A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m$

6 0

3 years ago

The eyes of some reptiles are sensitive to 850 nm light. If the minimum energy to trigger the receptor at this wavelength is 3.1

mr_godi [17]

Answer:

Minimum number of photons required is 1.35 x 10⁵

Explanation:

Given:

Wavelength of the light, λ = 850 nm = 850 x 10⁻⁹ m

Energy of one photon is given by the relation :

$E=\frac{hc}{\lambda}$ ....(1)

Here h is Planck's constant and c is speed of light.

Let N be the minimum number of photons needed for triggering receptor.

Minimum energy required for triggering receptor, E₁ = 3.15 x 10⁻¹⁴ J

According to the problem, energy of N number of photons is equal to the energy required for triggering, that is,

E₁ = N x E

Put equation (1) in the above equation.

$E_{1}=N\times\frac{hc}{\lambda}$

Substitute 3.15 x 10⁻¹⁴ J for E₁, 850 x 10⁻⁹ m for λ, 6.6 x 10⁻³⁴ J s for h and 3 x 10⁸ m/s for c in the above equation.

$3.15\times10^{-14} =N\times\frac{6.6\times10^{-34}\times3\times10^{8}}{850\times10^{-9}}$

N = 1.35 x 10⁵

8 0

3 years ago