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3 years ago

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A 52.3 kg student is standing at rest on roller skates when another student throws a bag of 3.5 kg bag of sand. After the studen

t catches the bag they are pushed a distance of 2 meters in 1 second. What was the initial velocity of the sand bag being thrown?

1 answer:

IRINA_888 [86]3 years ago

4 0

PLEASE HELP, I need help too !! Please !

Explanation:

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Give an example of a normal force. type below.

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Normal force is the force exerted when an object is on an surface. So an example could be a pile of books on top of a table.

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3 years ago

A 70 kg human sprinter can accelerate from rest to 10 m/s in 3.0 s. During the same time interval, a 30 kg greyhound can go from

ladessa [460]

Answer:

$P_1 = 1166.7 Watt$

$P_2 = 2000 Watt$

Explanation:

Average power for the human sprinter is given as

$Power = \frac{\Delta E}{\Delta t}$

so we have

$P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}$

$P = \frac{\frac{1}{2}(70)(10^2) - 0}{3}$

$P_1 = 1166.7 Watt$

Average power for greyhound is given as

$P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}$

$P = \frac{\frac{1}{2}(30)(20^2) - 0}{3}$

$P_2 = 2000 Watt$

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3 years ago

Two engineering students, John with a weight of 96 kg and Mary with a weight of 48 kg, are 30 m apart. Suppose each has a 0.04%

djyliett [7]

Answer:

$6.8370869499\times 10^{20}\ N$

Explanation:

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

e = Charge of electron = $1.6\times 10^{-19}\ C$

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Z = Atomic number of water = 18

M = Molar mass of water = 0.018 kg/mol

m = Mass of person

The charge is given by

$q=imbalance\times n\times e$

Total number of protons and electrons in each sphere

$n=\dfrac{mN_AZe}{M}$

$q=imbalance\times \dfrac{mN_AZe}{M}$

$q_1=0.0004\times \dfrac{96\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=3699916.8\ C$

$q_2=0.0004\times \dfrac{48\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=1849958.4\ C$

Electrical force is given by

$F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times 3699916.8\times 1849958.4}{30^2}\\\Rightarrow F=6.8370869499\times 10^{20}\ N$

The electrostatic force of attraction between them is $6.8370869499\times 10^{20}\ N$

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