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3 years ago

11

When serving a tennis ball, a player hits the ball when its velocity is zero (at the highest point of a vertical toss). The racq

uet exerts a force of 540 N on the ball for 5.00 ms, giving it a final velocity of 45.0 m/s. Using these data, find the mass of the ball

1 answer:

Lilit [14]3 years ago

8 0

Answer:60 gm

Explanation:

Given

initial velocity of ball $u=0$

Force exerted by racquet $F=540 N$

time period of force $t=5\ ms$

final velocity of ball $v=45\ m/s$

Racquet imparts an impulse to the ball which is given by

$J=F\Delta t=\Delta P$

$J=540\times \Delta t=m(45-0)$

$m=\frac{540\times 5\times 10^{-3}}{45}$

$m=60\ gm$

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two negative charges that are both -3.0 C push each other apart with a force of 19.2 N how far apart are the charges

Hoochie [10]

The electrostatic force between two charges q1 and q2 is given by
$F=k_e \frac{q_1 q_2}{r^2}$
where $k_e =8.99 \cdot 10^9 N m^2 C^{-2}$ is the Coulomb's constant and r is the distance between the two charges.

If we use F=19.2 N and q1=q2=-3.0 C, we can find the value of r, the distance between the two charges by re-arranging the previous formula:
$r= \sqrt{k_e \frac{q_1 q_2}{F} }= \sqrt{ 8.99 \cdot 10^9 N m^2 C^{-2} \frac{(-3.0C)^2}{19.2 N} } =6.49 \cdot 10^4 m=64.9 km$

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3 years ago

Which particles are heavy and which are light

VashaNatasha [74]

Answer:

Protons and neutrons are heavy, Electrons are extremely light

Explanation:

Protons and neutrons are heavier than electrons and reside in the nucleus at the center of the atom. Electrons are extremely lightweight and exist in a cloud orbiting the nucleus.

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3 years ago

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A car traveling at 30 m/s north slows to a stop in 15 seconds. What is the acceleration? PLEASE HELP

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The answer is C..........

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A resin tube and a PVC rod are both rubbed against styrofoam. Do

Ivan

The concepts of electrostatics allow us to find the result for the question about the electric forces in the two tubes is:

The tubes are attracted by having charges of different signs.

<h3>Electrostatics</h3>

Electrostatics studies the transfer of charge between bodies when they rub together, in generating the charges of the insulating bodies they are not mobile, so to transfer them from one body to another they must be in contact.

The force electric between charges is of two types:

Attractive. If the charges are of different signs.

Repulsive. If the charges are of the same sign.

The transferred charge depends on the bodies involved in the tables showns that the resin is positively charged when rubbed, the PVC acquires very little charge and polystyrene acquires a negative charge due to rubbing.

They indicate that the resin is rubbed with polystyrene, therefore the resin must acquire a positive charge and the polystyrene a negative charge.

Then the PVC is rubbed with the polystyrene, which initially we will assume neutral, the PVC does not acquire a charge and the polystyrenes acquires a negative charge, if the polystyrene is isolated from the ground, it shares this negative charge with the PVC, therefore the two materials remain with half of the negative charge.

Finally, we bring the resin, which is positively charged, closer to the PVC, which has a slight negative charge, therefore the two bodies must attract each other.

In conclusion using the concepts of electrostatics we can find the result for the question about the electric forces in the two tubes is:

The tubes are attracted by having charges of different signs.

Learn more about electrostatics here: brainly.com/question/17692887

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2 years ago

An engineer examining the oxidation of SO 2 in the manufacture of sulfuric acid determines that Kc = 1.7 x 108 at 600 K:

kozerog [31]

Answer: $p_{SO_2}=0.017atm$

Explanation:

We are given:

$K_c=1.7\times 10^8$

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

Where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $600K$

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$K_p=1.7\times 10^8\times (0.0821\times 600)^{-1}\\\\K_p=3.4\times 10^6$

The chemical reaction follows the equation:

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

The expression for $K_p$ for the given reaction follows:

$K_p=\frac{(p_{SO_3})^2}{ p_{O_2}\times {(p_{SO_2})^2}}$

We are given:

$p_{SO_3}=300atm$ $p_{O_2}=100atm$

Putting values in above equation, we get:

$3.4\times 10^6=\frac{(300)^2}{100\times {(p_{SO_2})^2}}$

$p_{SO_2}=0.017atm$

Hence, the partial pressure of the $SO_2$ at equilibrium is 0.017 atm.

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3 years ago