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Alla [95]
2 years ago
11

Who will give me answer I will give you brainliest..... ​

Physics
1 answer:
kondaur [170]2 years ago
7 0
1 . W=mass times acceleration due to gravity
60kg times 9.8m/s2
= 588N

2. W=mg
1176N=m times 9.8
m=120kg


3. 1 hour=3600s
24 hours=?
24 times 3600
= 86400 seconds


4. 1000g=1kg
25000g=?
25000 times 1 divide by 1000
=25kg



5. 1000000mg=1kg
123000000=?
123000000 times 1 divide by 1000000
=123 kg
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What is the frequency of a wave that has a speed of 5 m/s and a wavelength of 0.5meter?
Debora [2.8K]

Explanation:

frequency =speed/wavelength

=5/0.5=10Hz

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3 years ago
Define average velocity.Immersive Reader
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The slope of a graph of position vs time

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9. ¿Por qué no es lo mismo distancia que desplazamiento?
masya89 [10]

Answer:

Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.

Explanation:

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The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete
oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\

make shear stress (τ) the subject of the formula

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of  0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2

Part (b) shear stress at a distance of  0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0}{2*12} =0

3 0
3 years ago
It is found that the most probable speed of molecules in a gas at equilibrium temperature
kaheart [24]

Answer:

\frac{T_2}{T_1} = 1

Explanation:

The root mean square velocity of the gas at an equilibrium temperature is given by the following formula:

v = \sqrt{\frac{3RT}{M} }

where,

v = root mean square velocity of molecules:

R = Universal Gas Constant

T = Equilibrium Temperature

M = Molecular Mass of the Gas

Therefore,

For T = T₁ :

v = \sqrt{\frac{3RT_1}{M} }

For T = T₂ :

v = \sqrt{\frac{3RT_2}{M} }

Since both speeds are given to be equal. Therefore, comparing both equations, we get:

\sqrt{\frac{3RT_1}{M} }=\sqrt{\frac{3RT_2}{M} }\\\\\frac{T_2}{T_1} = 1

8 0
3 years ago
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