Ask question

Ask question

All categories

3 years ago

11

You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the oars) toward you, it moves a dis

tance of 1.3 m in a time of 0.96 s. The readout on the display indicates that the average power you are producing is 77 W. What is the magnitude of the force that you exert on the handle?

1 answer:

tangare [24]3 years ago

5 0

Answer:

56.86153 N

Explanation:

t =Time taken

F = Force

Power

$P=\frac{W}{t}\\\Rightarrow W=P\times t\\\Rightarrow W=77\times 0.96\\\Rightarrow W=73.92\ Joules$

Work done

$W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow F=\frac{73.92}{1.3}\\\Rightarrow F=56.86153\ N$

The magnitude of the force that is exerted on the handle is 56.86153 N

You might be interested in

All of the members of a community belong to the same species tf

Anna35 [415]

That is false

I hope this helps!

3 0

3 years ago

Why is water always used in energy generation (water -> steam -> turbine) instead of a liquid with a lower specific heat c

frez [133]

There are multiple reasons for this. First of all, water is available in almost every place on the Earth. It doesn't pollute the air, doesn't cause health use and is easily handle.

Other factor is the fact that water has a really high specific heat. This means that water, and more specifically steam, can aborb and transport more energy. A lower heat capacity would imply the need to boil more of the liquid to obtain the same amount of energy. This combine with the fact that water expands at a large rate when boiling, combine with everything mentioned previously, and you get a liquid with all the characteristics that a efficient turbine requires to work.

8 0

3 years ago

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

$F_d=mg$

Now, we determine the mass of the raindrop using the following formula:

$m=\rho_{water}V$

Where:

$\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}$

The volume is the volume of a sphere, therefore:

$m=\rho_{water}(\frac{4}{3}\pi r^3)$

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

$m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)$

Solving the operations:

$m=1.39\times10^{-5}kg$

Now, we substitute the values in the formula for the drag force:

$F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})$

Solving the operations:

$F_d=1.36\times10^{-4}N$

Now, we substitute in the formula:

$1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2$

Now, we solve for the velocity:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2$

Now, we substitute the values. We will use the area of a circle:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2$

Substituting the radius:

$\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2$

Solving the operations:

$70.67\frac{m^2}{s^2}=v^2$

Now, we take the square root to both sides:

$\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\ \\ 8.4\frac{m}{s}=v \\ \end{gathered}$

Therefore, the velocity is 8.4 m/s

7 0

1 year ago

What factors determine the presence of electrical energy?

luda_lava [24]

One of them would be power source.

8 0

4 years ago

A watt is a measure of power (the rate of energy change) equal to 1 j>s. (a) calculate the number of joules in a kilowatt- ho

GalinKa [24]

Before solving this, you must know the definition of these units of measurement. Watt is a measurement of power which is the amount of energy per unit time in seconds. Energy, on the other hand, is expressed through the SI unit Joules. Thus, power is the amount of energy in Joules per second.

From here, you can use the dimensional analysis technique. Also, you should know that 1 kilowatt is equal to 1000 watts, and 24 hours is equal to 86,400 seconds. Then,

100 = Energy/86,400
Energy = 8,640,000 Joules

4 0

4 years ago