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inna [77]
2 years ago
10

5. Find the velocity of a train that traveled 75 km in 35 minutes. (answer

Physics
1 answer:
iogann1982 [59]2 years ago
3 0

75km=75000m

35min=2100 seconds

Velocity=Displacement/Time

=75000/2100

=750/21

=35.71 m/s

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A 0.155 kg arrow is shot upward
Soloha48 [4]

Answer: 244.05 J

Explanation:

To find speed  at 30 m above the ground use equation:

V²=Vo²-2Gs

V0=31.4m/s

s=30m

G=9.81m/s²

-----------------------

V²=31.4²-2*9.81*30

V²=985.96+588.6

V²=1574.56

V=39.68m/s ---speed of arrow on 30 m obove the ground

Use equation for kinetic enrgy:

Ke=mV²/2

m=0.155kg

V=39.68m/s

-------------------------

Ke=0.155kg*(39.68m/s)²/2

Ke=0.155*1574.5/2

Ke=244.05J

6 0
3 years ago
Trình bày về Thí nghiệm tán xạ Rutherford, phát hiện proton.
Setler [38]

Answer:

Any other language I don't know this language

6 0
3 years ago
A tube is sealed at both ends and contains a 0.0100-m long portion of liquid. The length of the tube is large compared to 0.0100
Ahat [919]

Answer:

31.321 rad/s

Explanation:

L = Tube length

A = Area of tube

\rho = Density of fluid

v = Fluid velocity

m = Mass = \rho Al

Centripetal force is given by

F=\dfrac{mv^2}{L}\\ F=\dfrac{m(\omega L)^2}{L}\\ F=m\omega^2\\ F= 0.01A\rho\omega^2L

Pressure is given by

P=\dfrac{F}{A}=\rho gL\\\Rightarrow \dfrac{0.01A\rho\omega^2L}{A}=\rho gL\\\Rightarrow 0.01\omega^2=g\\\Rightarrow \omega^2=\dfrac{g}{0.01}\\\Rightarrow \omega=\sqrt{\dfrac{g}{0.01}}\\\Rightarrow \omega=\sqrt{\dfrac{9.81}{0.01}}\\\Rightarrow \omega=31.321\ rad/s

The angular speed of the tube is 31.321 rad/s

5 0
3 years ago
Need asap !
Tanya [424]

D. 51 N. The minimum applied force that will cause the television slide is 51 N.

In order to solve this problem we have to use the force of static friction equation Fs = μs*n, where μs is the coefficient of static friction, and n is the normal force m*g.

With μs = 0.35, and n = 15kg*9.8m/s² = 147 N

Fs = (0.35)(147 N)

Fs = 51.45 N

Fs ≅ 51 N

3 0
3 years ago
Read 2 more answers
A boy with a mouse is 80 kg stands on a scale in an ascending elevator with an acceleration of 3 m/s What is the resultant upwar
Rus_ich [418]

Answer:

Upward force on the boy will be 1024 N      

Explanation:

We have given mass of the boy m = 80 kg

Acceleration due to gravity g=9.8m/sec^2

It is given that elevator is ascending with acceleration of 3m/sec^2

So net acceleration on the boy = 9.8+3 = 12.8 m/sec^2 ( As the car elevator is moving ascending )

So upward force on the boy will be equal to F = m(g+a)

So F=80\times 12.8=1024N

So upward force on the boy will be 1024 N

3 0
3 years ago
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