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Nesterboy [21]
2 years ago
8

What tangential speed, v, must the bob have so that it moves in a horizontal circle with the string always making an angle θ fro

m the vertical? Express your answer in terms of some or all of the variables m , L, and θ, as well as the free-fall acceleration g.

Physics
1 answer:
Rainbow [258]2 years ago
4 0

Answer:

v = √(Lsinθ tanθ)

Explanation:

From the diagram attached,

v = the tangential speed.

r = the radius of the horizontal circle.

T = tension in the string.

θ =  the angle that the string makes with the vertical

m =  Bob's mass  (mg = the weight)

F =  centripetal force

L = the length of the string

From geometry,

r = Lsin θ

Thus, the centripetal acceleration is given as;

a = v²/r = v²/Lsin θ

Force = mass x acceleration

Thus,

Centripetal force = mv²/Lsin θ

Let's balance the vertical forces to obtain,

T cosθ - mg = 0

Thus, T cosθ = mg - - - - (eq1)

Similarly, let's balance the horizontal forces to obtain;

T sinθ = F = mv²/Lsin θ

So, T sinθ = mv²/Lsin θ - - - - (eq2)

Let's divide eq 2 by eq 1 to get;

Tsinθ/Tcosθ = (mv²/Lsin θ)/mg

tanθ = gv²/Lsinθ

Thus, v² =  Lsinθ tanθ

v = √(Lsinθ tanθ)

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A helicopter, starting from rest, accelerates straight up from the roof of a hospital. The lifting force does work in raising th
kobusy [5.1K]

Answer:

24,267.6 watts

Explanation:

from the question we are given the following:

mass (m) = 810 kg

final velocity (v) = 7 m/s

initial velocity (u) = 0 m/s

time (t) = 3.5 s

final height (h₁) = 8.2 m

initial height (h₀) = 0 m

acceleration due to gravity (g) = 9.8 m/s^{2}

find the power

power = \frac{work done}[time}

and

work done = change in kinetic energy (K.E) + change in potential energy (P.E)

work done = (0.5 mv^{2} - 0.5 mu^{2} ) + ( mgh₁ - mgh₀)

since u and h₀ are zero the work done now becomes

work done = (0.5 mv^{2}) + ( mgh₁ )                    

work done = (0.5 x 810 x 7^{2}) + ( 810 x 9.8 x 8.2)

work done = 84, 936.6 joules

recall that power = \frac{work done}[time}

power = \frac{84,936.6}[3.5}

power = 24,267.6 watts

7 0
2 years ago
An ice hockey puck is tied by a string to a stake in the ice. the puck is then swung in a circle. what force is producing the ce
Taya2010 [7]
In a circular motion scenario, the force that pulls the revolving object towards the centre is the force that produces the centripetal acceleration. So, in this case, the tension on the string is the force that pulls the puck towards the centre.

Therefore, it is the tension in the string that causes the centripetal acceleration of the puck

Hope I helped!! xx
8 0
2 years ago
Read 2 more answers
A motion sensor is used to create the graph of a student’s horizontal velocity as a function of time as the student moves toward
Marrrta [24]

Answer:

Position xf is farther away from the sensor than x0, and ax is negative

Explanation:

                     Area of trapezoidal are=

                            =\frac{1}{2} *(1.5+0.75)+\frac{1}{2} (1+0.75)(-0.5)

                            =\frac{1}{2} *(2.25-1.75*0.5)

                            =0.6875 m

As the area is positive therefore displacement from xo is positive

                                  ax=(change in velocity)/(Time)

                                  ax=\frac{-0.5-0}{3} =-\frac{1}{6} ms2

3 0
2 years ago
A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth
BaLLatris [955]
The answer would be 2.8m height on earth takes 
2.8=1/2*9.8*t^2 => <span>s = ut +1/2at^2 </span>
8 0
2 years ago
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin
Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 14,000 + 10,000x − 26,000x^{2}, where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

     W = \int_{0}^{0.54} F dx

         = \int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx

         = [14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}

         = 7560 + 1458 - 1364.69

         = 7653.31 J

or,      = 7.65 kJ       (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0
2 years ago
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