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Umnica [9.8K]
2 years ago
14

An Olympic high jumper, with a mass of 82 kg, has a

Physics
1 answer:
Digiron [165]2 years ago
8 0

Answer:

I don't really know

Explanation:

I really wanted to help you, but then I realized i didnt know how to

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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
Read 2 more answers
Space-faring astronauts cannot use standard weight scales (since they are constantly in free fall) so instead they determine the
valentinak56 [21]

Answer:

ma = 48.48kg

Explanation:

To find the mass of the astronaut, you first calculate the mass of the chair by using the information about the period of oscillation of the empty chair and the spring constant. You use the following formula:

T=2\pi\sqrt{\frac{m_c}{k}}     (1)

mc: mass of the chair

k: spring constant = 600N/m

T: period of oscillation of the chair = 0.9s

You solve the equation (1) for mc, and then you replace the values of the other parameters:

m_c=\frac{T^2k}{4\pi^2}=\frac{(0.9s)^2(600N/m)}{4\pi^2}=12.31kg    (2)

Next, you calculate the mass of the chair and astronaut by using the information about the period of the chair when the astronaut is sitting on the chair:

T': period of chair when the astronaut is sitting = 2.0s

M: mass of the astronaut plus mass of the chair = ?

T'=2\pi\sqrt{\frac{M}{k}}\\\\M=\frac{T'^2k}{4\pi^2}=\frac{(2.0s)^2(600N/m)}{4\pi^2}\\\\M=60.79kg (3)

Finally, the mass of the astronaut is the difference between M and mc (results from (2) and (3)) :

m_a=M-m_c=60.79kg-12.31kg=48.48kg

The mass of the astronaut is 48.48 kg

3 0
3 years ago
Give an equation for kinetic energy<br>(supplement)​
Leya [2.2K]

Answer:

The equatiom of kinetic energy is 1/2×m×v² where m represents mass and v is velocity :

k.e =  \frac{1}{2}m {v}^{2}

The unit for kinetic energy is measured in Joules (J).

6 0
3 years ago
an asteroid whose mass is 2.0 x 10^-4 times the mass of earth revolves in a circular orbit around the sun at a distance that is
Xelga [282]

Answer:

0.0002

Explanation:

First you have to solve exponents first, so you solve 10 to the power of negative 4 which is 0.0001 and now you have 2.0x0.0001 and that equals 0.0002

5 0
3 years ago
A steam Rankine cycle operates between the pressure limits of 1500 psia in the boiler and 2 psia in the condenser. The turbine i
AlladinOne [14]

Answer:

a. Mass flow rate through the boiler = 5.462lbm/s

b. Power produced by the turbine = 2525.8kW

c. The rate of heat supply in the boiler = 6901.42Btu/s

d. Thermal efficiency of the cycle = 34.3%

Explanation:

In order to provide a solution, we must assume that ;

- The system is operating at a steady condition

- Kinetic and potential energy changes are negligible

Now from steam tables, we calculate specific volume v and enthalpy h as,

h_1 = 95.96Btu/lb (  h_1 = h_f at 2psia )

v_1 = 0.016238ft^3/lb ( v_1 = v_f at 2psia )

w_{p,in} = v_1(P_2-P_1) = 0.016238(1500-2) * \frac{1}{5.404} = 4.501 Btu/lb

w_p = h_2 - h_1\\h_2 = w_p+h_1=4.501+95.96=100.461Btu/lb

h_3 = 1364.0Btu/lb

s_3 = 1.5073Btu/lb.R

( at P_3 = 1500psia & T_3 = 800^0F )

P_4 = 2psia\\S_4 = S_3\\x_4S = \frac{S_4-S_f}{S_{fg}}=\frac{1.5073-0.1783}{1.7374}=0.765

( S_f & S_{fg} when pressure is 2psia)

h_4S = h_f+x_4S*h_{fg}=95.96+(0.765)(1021.0)=877.025Btu/lb

n_T= \frac{h_3-h_4}{h_3-h_4S}\\ h_4=h_3-n_T(h_3-h_4S)=1364.0-0.90(1364.0-877.025)=925.7Btu/lb

Therefore,

q_{in}=h_3-h_2=1364.0-100.461=1263.54Btu/lb\\q_{out}=h_4-h_1=925.7-95.96=829.74Btu/lb\\w_{net}=q_{in}-q_{out}=1263.54-829.74=433.8Btu/lb

To calculate the mass flow rate of steam in the cycle, we use the formula

W_{net}=mw_{net}\\m=\frac{W_{net}}{w_{net}} =\frac{2500}{433.8}=5.763*(\frac{0.94782Btu}{1Kj} )=5.462lb/s

where 1Kj = 0.947817 Btu

The power output and the rate of heat addition are calculated thus,

W_{T,out}=m(h_3-h_4)=(5.462lb/s)*(1364-925.7)Btu/lb*(\frac{1Kj}{0.94782Btu} )\\=5.462*438.3*1.055=2525.8KW

Q_{in}=mq_{in}=5.462(1263.54)=6901.46Btu/s

The thermal efficiency of the cycle can be found thus;

n_{th}=\frac{W_{net}}{Q_{in}} =\frac{2500}{6901.46}*(\frac{0.94782Btu}{1Kj} ) =0.343

= 34.3%

5 0
3 years ago
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