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Vika [28.1K]
3 years ago
13

A car is towing a boat on a trailer. The driver starts from rest and accelerates to a velocity of +15.2 m/s in a time of 22.3 s.

The combined mass of the boat and trailer is 573 kg. The frictional force acting on the trailer can be ignored. What is the tension in the hitch that connects the trailer to the car?
Physics
1 answer:
scZoUnD [109]3 years ago
4 0

Answer:

The tension is 384 N.

Explanation:

To calculate the acceleration, we know the velocity and the time, so:

a=Δv/t

a=\frac{15\frac{m}{s} }{22.3s}

a=0.67\frac{m}{s^{2} }

To calculate the tension we need to apply Newton's second law:

∑F=m*a

F=573Kg *0.67\frac{m}{s^{2} }\\F=384N

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A gas undergoes a process in a piston–cylinder assembly during which the pressure-specific volume relation is pv1.3 = constant.
Galina-37 [17]

Answer:

Change in specific internal Energy=250\ \rm Btu/lb

Explanation:

Given:

  • Mass of the gas, m=0.4 lb
  • Initial pressure and volume are p_1=160\ \rm lbf/in^2\ and\ v_1=1\ \rm ft^3\\
  • Final pressure and temperature are p_1=480\ \rm lbf/in^2
  • Heat transfer from the gas is 2.1 Btu

Since the process is isotropic we have

p_1v_1^{1.3}=p_2v_2^{1.3}\\160\times 1^{1.3}=480\times v_2^{1.3}\\v_2=0.43\ \rm ft^3\\

So the final volume of the gas is calculated.

Work in any isotropic is given by w

w=\dfrac{p_1v_1-P_2v_2}{n-1}\\\\w=\dfrac{160\times1-480\times0.43}{1.3-1}\\w=-154.67\ \rm Btu\\

According to the first law of thermodynamics we have

Q=\Delta U+w\\-2.1=\Delta U-154.67\\\Delta U=152.56\ \rm Btu\\

So the Specific Internal Change is given by

\Delta u=\dfrac{\Delta U}{m}\\\Delta u=\dfrac{152.56}{0.4}\\\Delta u=250\ \rm Btu/lb

So the specific Change in Internal energy is calculated.

6 0
3 years ago
1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position
antiseptic1488 [7]

Answer:

The value of the distance is \bf{14.52~cm}.

Explanation:

The velocity of a particle(v) executing SHM is

v = \omega \sqrt{A^{2} - x^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)

where, \omega is the angular frequency, A is the amplitude of the oscillation and x is the displacement of the particle at any instant of time.

The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e., x = 0.

The maximum velocity(\bf{v_{m}}) is

v_{m} = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)

Divide equation (1) by equation(2).

\dfrac{v}{v_{m}} = \dfrac{\sqrt{A^{2} - x^{2}}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)

Given, v = 0.25 v_{m} and A = 15~cm. Substitute these values in equation (3).

&& \dfrac{1}{4} = \dfrac{\sqrt{15^{2} - x^{2}}}{15}\\&or,& A = 14.52~cm

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Answer:

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A 0.0450-kg golf ball initially at rest is given a speed of 25.2 m/s when a club strikes. part a part complete if the club and b
Ksenya-84 [330]
We are given information:
m = 0.0450 kg
Δv = 25.2 m/s
Δt = 1.95 ms = 0.00195s

To find force we use formula:
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a is acceleration. To find it we use formula:
a = Δv / Δt 
a = 25.2 / 0.00195
a = 12923.1 m/s^2

Now we can find force:
F = 0.0450 * 12923.1
F = 581.5 N 


To check the effect of the ball's weight on this movement we need to calculate it and then compare it to this force.
W = m * g
W = 0.0450 * 9.81
W = 0.44145 N 

We can see that weight is much smaller than the applied force so it's influence in negligible.
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