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Lorico [155]
1 year ago
11

A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio

n of Tmax = 7.5 N before it will break. What isthe maximum possible speed of the ball at the top of the loop, in meters per second?
Physics
1 answer:
natulia [17]1 year ago
8 0

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}

For the maximum velocity of the ball at the top of the vertical circular motion,

v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}

where g is the acceleration due to gravity,

Substituting the known values,

\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

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In the flow past a compression corner, the upstream Mach number and pressure are 3.5 and 1 atm, respectively. Downstream of the
yulyashka [42]

Answer:

\theta=23.7^{\circ}

Explanation:

The ratio of pressure 2 to 1 us 5.48/1= 5.48 rounded off as 5.5.

Referring to table A.2 of modern compressible flow then M_{\beta_1}=2.2

Also

M_{\beta_1}=M_1 sin \beta and making sin\beta the subject of the formula then

sin\beta=\frac {2.2}{3.5}\\\beta=38.94^{\circ}

Making reference to \theta-\beta-M diagram then

\theta=23.7^{\circ}

4 0
3 years ago
A person travelled in the -x direction at a constant speed of 2.5 m/s. The person's final position is -10.0 m, and the initial p
Nostrana [21]

Answer:

(c) time required to travel = 8 sec

Explanation:

We have given the final position = -10 m on x axis

And the initial position =10 m

So total distance = 10-(-10)=20 m

The speed is given as 2.5 m/sec

We have tof ind the time required by the person to travel

Time is given by t=\frac{distance}{speed}=\frac{20}{2.5}=8 sec

So the option (c) is correct option

4 0
3 years ago
You drive on Interstate 10 from San Antonio to Houston, half the time at 75 km/h and the other half at 106 km/h. On the way back
r-ruslan [8.4K]

Answer:

Explanation:

a ) from San Antonio to Houston let distance be d km .

Average speed = total distance / total time

time = distance / speed

Total time = (d / 2 x 75 ) +( d / 2 x 106 )

= .0067 d + .0047 d

= .0114 d

Average speed  = d / .0114 d = 87.72 km /h

b ) from Houston back to San Antonio

Total time = (d / 2 x 106 ) +( d / 2 x 75 )

= .0047 d + .0067 d

= .0114 d

Average speed  = d / .0114 d = 87.72 km /h

c )

For entire trip :

total distance = 2d

total time = 2 x .0114 d

Average speed  = 2 d / 2 x .0114 d

= 87.72 km /h .

4 0
2 years ago
If a 54 kg sprinter can accelerate from a standing start to a speed of 10 m/s in 3 s, what average power is generated?
lora16 [44]

Answer:

Power, P = 600 watts

Explanation:

It is given that,

Mass of sprinter, m = 54 kg

Speed, v = 10 m/s

Time taken, t = 3 s

We need to find the average power generated. The work done divided by time taken is called power generated by the sprinter i.e.

P=\dfrac{W}{t}

Work done is equal to the change in kinetic energy of the sprinter.

P=\dfrac{\dfrac{1}{2}mv^2}{t}

P=\dfrac{\dfrac{1}{2}\times 54\ kg\times (10\ m/s)^2}{3\ s}

P = 900 watts

So, the average power generated by the sprinter is 900 watts. Hence, this is the required solution.

3 0
3 years ago
The parallel plates in a capacitor, with a plate area of 8.00 cm2 and an air-filled separation of 2.70 mm, are charged by a 8.70
MrRa [10]

Answer:

a)  ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J,  d)  W = 150 10⁻¹² J

Explanation:

Let's find the capacitance of the capacitor

         C = \epsilon_o \frac{A}{d}

         C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³

         C = 2.62 10⁻¹² F

for the initial data let's look for the accumulated charge on the plates

          C = \frac{Q}{\Delta V}

          Q₀ = C ΔV

           Q₀ = 2.62 10⁻¹² 8.70

           Q₀ = 22.8 10⁻¹² C

a) we look for the capacity for the new distance

          C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³

          C₁ = 1.04 10⁻¹² F

       

          C₁ = Q₀ / ΔV₁

          ΔV₁ = Q₀ / C₁

          ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²

          ΔV₁ = 21.9 V

b) initial stored energy

          U₀ = \frac{Q_o}{ 2C}

          U₀ = (22.8 10⁻¹²)²/(2  2.62 10⁻¹²)

          U₀ = 99.2 10⁻¹² J

c) final stored energy

          U_f = (22.8 10⁻¹²) ² /(2  1.04 10⁻⁻¹²)

          U_f = 249.9 10⁻¹² J

d) the work of separating the plates

as energy is conserved work must be equal to energy change

          W = U_f - U₀

          W = (249.2 - 99.2) 10⁻¹²

          W = 150 10⁻¹² J

note that as the energy increases the work must be supplied to the system

6 0
3 years ago
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