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sukhopar [10]
3 years ago
13

The gravitational acceleration on the surface of the moon is 1/6 what it is on earth. If a man could jump straight up 0.7 m (abo

ut 2 feet) on the earth, how high could he jump on the moon?
Physics
1 answer:
LenaWriter [7]3 years ago
5 0

Answer:

on moon he can jump 4.2 m high

Explanation:

given data

gravitational acceleration on moon a(m) = 1/6

jump = 0.7 m

to find out

how high could he jump on the moon

solution

we know gravitational acceleration on earth a =  g = 9.8 m/s²

so on moon am =  9.8 * \frac{1}{6}  = 1.633 m/s²

so if he jump on earth his speed will be for height 0.7 m i s

speed v = \sqrt{2gh}

v = \sqrt{2(9.8)0.7}

v = 3.7 m/s

so if he hump on moon

height will be

height = \frac{v^2}{2*a(m)}  

put here value

height =  \frac{3.7^2}{2*1.633)}  

height = 4.2 m

so on moon he can jump 4.2 m high

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A 1.0 kg copper rod rests on two horizontal rails 1.0 m apart and carries a current of 50 A from one rail to the other.
vagabundo [1.1K]

Answer

given,

mass of copper rod = 1 kg

horizontal rails = 1 m

Current (I) = 50 A

coefficient of static friction = 0.6

magnetic force acting on a current carrying wire is

           F = B i L

Rod is not necessarily vertical

F_x =i L B_d

F_y= i L B_w

the normal reaction N = mg-F y

static friction       f = μ_s (mg-F y )

horizontal acceleration is zero

F_x-f = 0

iLBd = \mu_s(mg-F_y )

 B_w = B sinθ

 B_d = B cosθ

iLB cosθ= μ_s (mg- iLB sinθ)

B = \dfrac{\mu_smg}{i(cos\theta +\mu_s sin\theta)}

\theta =tan{-1}{\mu_s}

\theta =tan{-1}{0.6}

\theta = 31^0

B = \dfrac{0.6\times 1 \times 9.8}{50(cos31^0 +0.6 sin31^0)}

       B = 0.1 T

4 0
3 years ago
Steam enters a one-inlet, two-exit control volume at location (1) at 360°C, 100 bar, with a mass flow rate of 2 kg/s. The inlet
yaroslaw [1]

Answer:

The inlet velocity is 21.9 m/s.

The mass flow rate at reach exit is 1.7 kg/s.

Explanation:

Given that,

Mass flow rate = 2 kg/s

Diameter of inlet pipe = 5.2 cm

Fifteen percent of the flow leaves through location (2)  and the remainder leaves at (3)

The mass flow rate is

m_{2}=0.15\times2

We need to calculate the mass flow rate at reach exit

Using formula of mass

m_{3}=m_{1}-m_{2}

m_{3}=2-0.15\times2

m_{3}=1.7\ kg/s

We need to calculate the inlet velocity

Using formula of velocity

v=\dfrac{m}{\rho A}

Put the value into the formula

v=\dfrac{2}{42.868\times\dfrac{\pi}{4}\times(5.2\times10^{-2})^2}

v=21.9\ m/s

Hence, The inlet velocity is 21.9 m/s.

The mass flow rate at reach exit is 1.7 kg/s.

7 0
3 years ago
Help me plaese on this
Crazy boy [7]

Answer:

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Explanation:

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5 0
2 years ago
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What work do you think is done when you carry a 20 N weight backpack for a 1000 m walk? will the work be positive, negative, or
dezoksy [38]

The work done is positive and is equal to 20000 J

<h3>What is work done?</h3>

Work done is defined as the product of force and the distance moved by the force.

Mathematically:

  • Work done = force * distance

The work done by the force = 20 * 1000 = 20000J

The work done is positive and is equal to 20000 J

Learn more about work done at: brainly.com/question/25923373

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8 0
2 years ago
Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci
Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W

\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})

h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}

2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99

6 0
3 years ago
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