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3 years ago

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A 0.1 kg bouncy ball moves toward a brick wall with a speed of 11 m/s. After colliding with the wall, the ball travels at a spee

d of 8.8 m/s in the OPPOSITE DIRECTION. What is the magnitude of the impulse experience by the ball?

1 answer:

mr_godi [17]3 years ago

4 0

Impulse is change in momentum. The initial momentum was
0.1kg*11m/s=1.1kg m/s.
The second momentum is 0.1*(-8.8m/s)=-0.88 (since it's moving backward now)
The difference is P1-P2 = 1.1-(-0.88)=1.98kg m/s

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A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

3 years ago

During a compression at a constant pressure of 290 Pa, the volume of an ideal gas decreases from 0.62 m3 to 0.21 m3. The initial

Aloiza [94]

Answer:

a) -41.1 Joule

b) 108.38 Kelvin

Explanation:

Pressure = P = 290 Pa

Initial volume of gas = V₁ = 0.62 m³

Final volume of gas = V₂ = 0.21 m³

Initial temperature of gas = T₁ = 320 K

Heat loss = Q = -160 J

Work done = PΔV

⇒Work done = 290×(0.21-0.62)

⇒Work done = -118.9 J

a) Change in internal energy = Heat - Work

ΔU = -160 -(-118.9)

⇒ΔU = -41.1 J

∴ Change in internal energy is -41.1 J

b) V₁/V₂ = T₁/T₂

⇒T₂ = T₁V₂/V₁

⇒T₂ = 320×0.21/0.62

⇒T₂ = 108.38 K

∴ Final temperature of the gas is 108.38 Kelvin

5 0

3 years ago

If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2

kaheart [24]

Answer:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

So then the difference of temperature across the material would be $\Delta T = 375 K$

Explanation:

For this case we can use the Fourier Law of heat conduction given by the following equation:

$Q = -kA \frac{\Delta T}{\Delta x}$ (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 represent the cross sectional area

Q= 3KW represent the rate of heat transfer

$\Delta T$ is the temperature of difference that we want to find

$\Delta x=2.5 cm =0.025 m$ represent the thickness of the material

If we solve $\Delta T$ in absolute value from the equation (1) we got:

$\Delta T =\frac{Q \Delta x}{Ak}$

First we convert 3KW to W and we got:

$Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W$

And we have everything to replace and we got:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

So then the difference of temperature across the material would be $\Delta T = 375 K$

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3 years ago

Electric kettle with the resistance of 20-02.is used-to-heat:4 kg of water. If

jeka94

The guy that answered this is fake and he’s doing it for free points so
Just repost it and hope he doesn’t find it

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3 years ago

What is the formula for velovity?

GarryVolchara [31]

Answer:

if it's velocity u talking of.....

Explanation:

then it's displacement/ time

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3 years ago