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notsponge [240]
3 years ago
5

A 70-kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.3 m/s. Neglect air resistan

ce, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.
Physics
1 answer:
yKpoI14uk [10]3 years ago
5 0

Answer:

h_{B} = 6.083\,m

Explanation:

Let assume that pole vaulter begins running at a height of zero. The pole vaulter is modelled after the Principle of Energy Conservation:

K_{A} = K_{B} + U_{g,B}

\frac{1}{2}\cdot m \cdot v_{A}^{2} = \frac{1}{2}\cdot m \cdot v_{B}^{2} + m\cdot g \cdot h_{B}

The expression is simplified and final height is cleared within the equation:

\frac{1}{2}\cdot (v_{A}^{2} - v_{B}^{2}) = g\cdot h_{B}

h_{B} = \frac{(v_{A}^{2}-v_{B}^{2})}{2\cdot g}

h_{B} = \frac{[(11\,\frac{m}{s} )^{2}-(1.3\,\frac{m}{s} )^{2}]}{2\cdot (9.807\,\frac{m}{s^{2}} )}

h_{B} = 6.083\,m

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Which of the following is same dimension quantity?
hram777 [196]

Answer:

Find the dimension of each and every quantity in all the options to check whether they are the same or not. We can use any one formula of each identity to find its dimension.

Complete step by step solution:

To find the dimension of a quantity, we can use any formula related to that quantity but we will use the easiest ones to save time.

Force-

from Newton’s law of motion,

F=maF=ma

Dimension of force =[M][LT−2]=[MLT−2]=[M][LT−2]=[MLT−2]

Work done-

W=F×sW=F×s

Dimension of work=[MLT−2][L]=[ML2T−2]=[MLT−2][L]=[ML2T−2]

Momentum-

p=mvp=mv

Dimension of momentum=[M][LT1]=[MLT−1]=[M][LT1]=[MLT−1]

Impulse-

I=F×tI=F×t

Dimension of impulse=[MLT−2][T]=[

3 0
2 years ago
Light is incident on the left face of an isosceles prism; with an apex angle of 49o, such that the light exiting the right face
Sunny_sXe [5.5K]

Answer:

\mu = 1.645

Explanation:

By Snell's law we know at the left surface

\theta_i = 19^o

\theta_r = ?

\mu_1 = 1

\mu_2 = \mu

now we have

1 sin19 = \mu sin\theta_r

0.33 = \mu sin\theta_r

now on the other surface we know that

angle of incidence = \theta_r'

\theta_e = 90

so again we have

\mu sin\theta_r' = 1 sin90

so we have

\theta_r = sin^{-1}\frac{0.33}{\mu}

\theta_r' = sin^{-1}\frac{1}{\mu}

also we know that

\theta_r + \theta_r' = 49

sin^{-1}\frac{0.33}{\mu} + sin^{-1}\frac{1}{\mu} = 49

By solving above equation we have

\mu = 1.645

3 0
3 years ago
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A mass of 2 kg traveling at 3 m/s undergoes a one-dimensional elastic collision with a group of four 1kg masses that are at rest
alukav5142 [94]
2kg + 3 kg = 5 kg as result of the conclusion
6 0
3 years ago
A dry cell gives static electricity true or false?
Rashid [163]

Answer:

False

Explanation:

6 0
3 years ago
A solid circular shaft and a tubular shaft, both with the same outer radius of c=co = 0.550 in , are being considered for a part
Norma-Jean [14]

Answer:

The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp

Explanation:

<u>Polar moment of Inertia</u>

(I_p)s = \frac{\pi(0.55)4}2

      = 0.14374 in 4

<u>Maximum sustainable torque on the solid circular shaft</u>

T_{max} = T_{allow} \frac{I_p}{r}

         =(14 \times 10^3) \times (\frac{0.14374}{0.55})

         = 3658.836 lb.in

         = \frac{3658.836}{12} lb.ft

        = 304.9 lb.ft

<u>Maximum sustainable torque on the tubular shaft</u>

T_{max} = T_{allow}( \frac{Ip}{r})

          = (14 \times10^3) \times ( \frac{0.13101}{0.55})

          = 3334.8 lb.in

          = (\frac{3334.8}{12} ) lb.ft

          = 277.9 lb.ft

<u>Maximum sustainable power in the solid circular shaft</u>

P_{max} = 2 \pi f_T

          = 2\pi(2.1) \times 304.9

          = 4023.061 lb. ft/s

          = (\frac{4023.061}{550}) hp

          = 7.315 hp

<u>Maximum sustainable power in the tubular shaft</u>

P _{max,t} = 2\pi f_T

            = 2\pi(2.1) \times 277.9

            = 3666.804 lb.ft /s

            = (\frac{3666.804}{550})hp

            = 6.667 hp

7 0
3 years ago
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