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3 years ago

Two football players each applied 100 n of force to sack the qb and move him 3 meters downfield. How much work was done?

1 answer:

4 0

<span>We know that the equation for work is W=force x distance, where work is in Joules, force is in Newtons and distance is in meters. In this situation, W= (100*2) x 3 which would result in W=600 J. The work done was 600 Joules.</span>

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The most abundant element in earth's continental crust (by weight) is____?

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if you place 0°c ice into 0°c water in an insulated container, what will happen? Will some ice melt, will more water freeze, or

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A uniform electric field exists in the region between two oppositely charged plane parallel plates. a proton is released from re

dedylja [7]

According to Newton's second law

E.e = a * mp ..... (1)

where

E is the magnitude of the electric field; e = 1.6 * 10^-19 is the elementary charge; mp = 1.67*10^-27 kg is the proton mass; a is the acceleration.

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Electric field

The physical field that surrounds electrically charged particles and exerts a force on all other charged particles in the field, either attracting or repelling them, is known as an electric field (also known as an E-field). It can also refer to a system of charged particles' physical field. Electric charges and time-varying electric currents are the building blocks of electric fields. The electromagnetic field, one of the four fundamental interactions (also known as forces) of nature, manifests itself in both electric and magnetic fields.

To learn more about an electric field refer here:

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2 years ago

What is it? Please help me.

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3 years ago

Water moves through a constricted pipe in steady, ideal flow. At the

Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: $1.8\cdot 10^{-3} m^3/s$

Explanation:

To solve the problem, we can use Bernoulli's equation, which states that

$p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2$

where

$p_1=1.75\cdot 10^4 Pa$ is the pressure in the lower section of the tube

$h_1 = 0$ is the heigth of the lower section

$\rho=1000 kg/m^3$ is the density of water

$g=9.8 m/s^2$ is the acceleration of gravity

$v_1$ is the speed of the water in the lower pipe

$p_2$ is the pressure in the higher section

$h_2 = 0.250 m$ is the height in the higher pipe

$v_2$ is hte speed in the higher section

We can re-write the equation as

$v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}$ (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-section of the lower pipe, with

$r_1 = 3.00 cm =0.03 m$ is the radius of the lower pipe (half the diameter)

$A_2 = \pi r_2^2$ is the cross-section of the higher pipe, with

$r_2 = 1.50 cm = 0.015 m$ (radius of the higher pipe)

So we get

$r_1^2 v_1 = r_2^2 v_2$

And so

$v_2 = \frac{r_1^2}{r_2^2}v_1$ (2)

Substituting into (1), we find the speed in the lower section:

$v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s$