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Mumz [18]
2 years ago
14

The2 archer uses a force of 120 N. The force acts on an area of 0.5 cm2 on the archer's fingers. . Calculate the pressure on the

archer's fingers.
Physics
1 answer:
Mkey [24]2 years ago
4 0

Answer:

Pressure = 24000 N/m²

Explanation:

Given the following data;

Force = 120N

Area = 0.5cm² to meters = 0.5/100 = 0.005 m²

To find the pressure;

Pressure= force/area

Pressure= 120/0.005

Pressure = 24000 N/m²

Therefore, the pressure on the archer's fingers is 24000 Newton per meters square.

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For a machine with 35-cm -diameter wheels, what rotational frequency (in rpm) do the wheels need to pitch a 85 mph fastball?
Inessa05 [86]

Answer:

The rotational frequency must be 2073.56 rpm

Explanation:

Notice that we need to obtain a rotational frequency in "rpm" (revolutions per minute), so we better start by converting all the given information into the appropriate units:

The magnitude of the velocity for the pitch is given in miles per hour, while the diameter of the machine's wheels is given in cm. Let's reduce all units of length into meters(using the metric system), and the units of time into minutes.

Conversion of the 85 mph  speed into meters per minute:

Recall that 1 mile equals 1609.34 meters, and that 1 hour equals 60 minutes, so we write:

85\,\frac{miles}{hour} = 85\,\frac{1609.34\,m}{60\,min} =2279.898\,\frac{m}{min}

which can be rounded to approximately 2280 m/min.

We also convert the 35 cm diameter into meters:

diameter = 0.35 m

Now we use the equation that relates angular velocity (w) and the radius (R) of the circular movement, with tangential velocity (v_t), in order to obtain the angular velocity of the wheel:

v_t=w*R\\w=\frac{v_t}{R}

but recall that this angular velocity is given in radians per unit of time. So first find the radius of the wheel (half its diameter). R = 0.175 m

So we have:

w=\frac{2280}{0.175}\frac{radians}{min} \\w=13028.57\,\frac{radians}{min}

And now, recalling that 2\pi radians equal one revolution, we convert the angular velocity ot revolutions per minute by dividing the "w" we found by 2\pi :

rotational frequency = \frac{13028.57}{2\pi} \frac{rev}{min} = 2073.56 \frac{rev}{min}

6 0
2 years ago
One game at the amusement park has you push a puck up a long, frictionless ramp. You win a stuffed animal if the puck, at its hi
Aneli [31]

Answer:

Explanation:

Given

initial speed(u)=5 m/s

Final speed(v)=4 m/s

Distance traveled=3 m

using equation of motion

v^2-u^2=2as

4^2-5^2=2(a)(3)

a=\frac{-3}{2}=-1.5 m/s^2

after this its final velocity will be zero

v^2-u^2=2as

0^2-4^2=2\times (-1.5)\times s

s=5.33 m

Total distance=3+5.33=8.33 m

Thus he will not be able to win the game

4 0
2 years ago
Suppose the foreman had released the box from rest at a height of 0.25 m above the ground. What would the crate's speed be when
Arturiano [62]

Answer:

v = 2.21 m/s

Explanation:

The foreman had released the box from rest at a height of 0.25 m above the ground.

We need to find the speed of the crate when it reaches the bottom of the ramp. Let v is the velocity at the bottom of the ramp. It can be calculated using conservation of energy as follows :

mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 0.25} \\\\v=2.21\ m/s

So, its velocity at the bottom of the ramp is 2.21 m/s.

4 0
1 year ago
THE RELATIVE ANGLE AT THE KNEE CHANGES FROM 0O TO 85O DURING THE KNEE FLEXION PHASE OF A SQUAT EXERCISE. IF 10 COMPLETE SQUATS A
Mice21 [21]

Answer: angular distance = 1700° and 29.7 rad

      also the angular displacement = 0

Explanation:

To explain this, i will give a breakdown of this works.

we are asked to find both the angular distance and displacement the knee undergo.

Ok to get the distance of the knee, we would first take note that for one to squat down and get back up, the knee would travel through 85° of flexion to gpo down, and also through another 85° of extension to return standing (upright). So, the actual angular distance of the squat is 170°.

taking ten squats, the knee would have to go through 170° motion times 10 i.e;

10 * 170° = 1700°

Therefore the angulaar distance is 1700°

now converting this distance to radians since we will be required to have our answer in both degree and rad.

Given that 2pi = 360°, it means that one degree will give 57.3°;

∅ (rad) = ∅ (deg) * 2π/360°

∅ (rad) = 1170° * 2π/360°  = 29.7 rad

∅ (rad) = 29.7 rad

b. For the other part, let us remember that angular displacement is equal to angular distance divided by time, so the angular displacement displacement of the knee will be zero, because the knee's position at the final third will be the same as the initial position.

cheers i hope this helps!!!!

π

5 0
2 years ago
There are lots of examples of ideal gases in the universe, and they exist in many different conditions. In this problem we will
elena-14-01-66 [18.8K]

Answer:

P = ρRT/M

Explanation:

Ideal gas equation is given as follows generally:

PV = nRT (1)

P = pressure in the containing vessel

V = volume of the containing vessel

n = number of moles

R = gas constant

T = temperature in K

n = m/M

m = mass of the gas contained in the vessel in g

M = molar mass in g/mol

ρ = m/V

Density of the gas = ρ

Substituting for n in (1)

PV = mRT/M. (2)

Dividing equation (2) through by V

P = m/V ×RT/M

P = ρRT/M

5 0
1 year ago
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