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3 years ago

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An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . Suppose that

the quarterback takes 0.30 s to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

1 answer:

RUDIKE [14]3 years ago

4 0

Answer:

0.0241875 m

Explanation:

$m_1$ = Mass of quarterback = 80 kg

$m_2$ = Mass of football = 0.43 kg

$v_1$ = Velocity of quarterback

$v_2$ = Velocity of football = 15 m/s

Time taken = 0.3 seconds

In this system as the linear momentum is conserved

$m_1v_1+m_2v_2=0\\\Rightarrow v_1=-\frac{m_2v_2}{m_1}\\\Rightarrow v_1=-\frac{0.43\times 15}{80}\\\Rightarrow v_1=0.080625\ m/s$

Assuming this velocity is constant

$Distance=Velocity\times Time\\\Rightarrow Distance=0.080625\times 0.3\\\Rightarrow Distance=0.0241875\ m$

The distance the quarterback will move in the horizontal direction is 0.0241875 m

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An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in

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Answer:

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Explanation:

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Assume that the density of water is $\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ . To the volume of water displaced from its mass, divide mass with density $\rho(\text{water})$ .

$\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}$ .

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$V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}$ .

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Average density is equal to mass over volume.

To find the mass of the sample from its weight, divide with $g$ .

$\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg$ .

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Divide mass with volume to find the average density.

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To obtain how rapidly is the area enclosed by the ripple increasing after 12 seconds, we calculate dA/dt

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$\frac{dA}{dt} = 2\pi (48)(4)$

dA/dt = 1206.528 ft²/sec

Therefore, after 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec

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