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svp [43]
3 years ago
13

A block of mass 5 kg is descending with a constant velocity on an inclined plane at 30°. What is the value of the frictional for

ce?
Physics
1 answer:
sattari [20]3 years ago
4 0

Answer:28.9N

Explanation:

F=UR R=mg R=5×10 R=50N

U=tan© U=Tan30 U=1/√(3)

F=UR F=1/√(3) ×50

F=28.9N

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In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these s
riadik2000 [5.3K]

Answer:

Part a)

\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}

Part b)

\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}

Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

\tau = I\alpha

here we will have

\tau = (m_1g - m_2g)(\frac{L}{2})

now we will have inertia of two masses given as

I = (m_1 + m_2)(\frac{L}{2})^2

now we have

I = (m_1 + m_2)\frac{L^2}{4}

now the angular acceleration is given as

\alpha = \frac{\tau}{I}

so we have

\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}

Part b)

Now if the rod is not massles then we will have total inertia given as

I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}

so we will have

I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}

now the acceleration is given as

\alpha = \frac{\tau}{I}

\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}

7 0
3 years ago
Help me please, please !
steposvetlana [31]

Answer:

b and d

a, c, e, and f

Explanation:

Ideal gas law:

PV = nRT

Solving for temperature:

T = PV / (nR)

Therefore, temperature is directly proportional to pressure and volume, and inversely proportional to the number of molecules.

T = k PV / N

Let's say that T₀ is the temperature when P = 100 kPa, V = 4 L, and N = 6×10²³.

a) T = k PV / N = T₀

b) T = k (2P) V / N = 2T₀

c) T = k (P/2) (2V) / N = T₀

d) T = k PV / (N/2) = 2T₀

e) T = k P (V/2) / (N/2) = T₀

f) T = k (P/2) V / (N/2) = T₀

b and d have the highest temperature,

a, c, e, and f have the lowest temperature.

8 0
3 years ago
Read 2 more answers
A circular loop ( radius = 0.5 m) carries a current of 3.0 A and has unit normal vector of (2i - j +2k)/3 what is x component of
Roman55 [17]

Answer:

T=9.42Nm

Explanation:

From the question we are told that:

Radius r= 0.5 m

Current I= 3.0 A

Normal vector n=\frac{(2i - j +2k)}{3}

Magnetic field B= (2i-6j) T

Generally the equation for Area is mathematically given by

 A=\pi r^2

 A=3.1415 *0.5^2

 A=0.7853 m^2

Generally the equation for Torque is mathematically given by

 T=A(i'*B)

Where

 i'*B= \begin{bmatrix}2&-1&2\\2&-6&0\end{bmatrix}

 X\ component\ of\ i'*B= [(-1 * 0)-(2*-6)]

 X\ component\ of\ i'*B=12

Therefore

 T=0.7853*12

 T=9.42Nm

7 0
3 years ago
The orbital period of an object is 2 x 10^7 and its total radius is 4 x 10^4 m.
SVETLANKA909090 [29]

Answer:

12566

Explanation:

Use formula: vt=2pir/t

7 0
3 years ago
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Rebecca was at the playground and found a rusted nail. What happened to the nail? A) It went through a physical change because t
BaLLatris [955]
Buff dude skid did di d I’d
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