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3 years ago

Erica throws a tennis ball against a wall, and it bounces back. Which force is responsible for sending the ball back to Erica?

Physics

2 answers:

balandron [24]3 years ago

8 0

Answer:

The normal force that the wall exerts on the ball.

Explanation:

Erica throws a tennis ball against a wall, and it bounces back. Here Newton's third law of motion follows. When the ball hits the wall, the wall will also exert an equal and opposite force on the ball as per third law. The opposite force that wall exert on ball is normal force which is responsible for sending the ball back to Erica.

Hence, the force is normal force exerted by the wall on the ball.

matrenka [14]3 years ago

6 0

Kinetic energy is responsible for this

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According to Archimedes’ principle, the mass of a floating object equals the mass of the fluid displaced by the object. Use this

Andrew [12]

Answer:

Part a)

$\rho = 0.55 g/cm^3$

Part b)

$\rho_L = 1.49 g/cm^3$

Part c)

Since we know that the base area will remain same always

so here the length and width of the object is not necessary to obtain the above data in such type of questions

Explanation:

Part a)

As we know that when cylinder float in the water then weight of the cylinder is counter balanced by the buoyancy force

So here we know

buoyancy force is given as

$F_b = \rho_w V_{sub} g$

$F_b = (1 g/cm^3) (30 - 13.5) Ag$

$F_b = 16.5 Ag$

Now we know that the weight of the cylinder is given as

$W = \rho (30 cm)A g$

now we have

$\rho (30 cm) A g = 16.5 A g$

$\rho = 0.55 g/cm^3$

Part b)

When the same cylinder is floating in other liquid then we will have

$F_b = \rho_L (30 - 18.9 )A g$

so we have

$\rho_L (11.1) Ag = 0.55(30) Ag$

$\rho_L = 1.49 g/cm^3$

Part c)

Since we know that the base area will remain same always

so here the length and width of the object is not necessary to obtain the above data in such type of questions

3 0

2 years ago

To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T

Fantom [35]

Answer:

$V_1= 3.4*10^7m/s$

Explanation:

From the question we are told that

Nucleus diameter $d=5.50-fm$

a 12C nucleus

Required kinetic energy $K=2.30 MeV$

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

$K_2 +U_2=K_1+U_1$

where

$K_1$ =initial kinetic energy

$K_2$ =final kinetic energy

$U_1$ =initial electric potential

$U_2$ =final electric potential

mathematically

$U_2 = \frac{Kq_pq_c}{r_2}$

where

$r_f$ =distance b/w charges

$q_c$ =nucleus charge $=6(1.6*10^-^1^9C)$

$K$ =constant

$q_p$ =proton charge

Generally kinetic energy is know as

$K=\frac{1}{2} mv^2$

Therefore

$U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2} mv_1^2 +U_1$

Generally equation for radius is $d/2$

Mathematically solving for radius of nucleus

$R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})$

$R=2.75*10^-^1^5m$

Generally we can easily solving mathematically substitute into v_1

$q_p$ $=6(1.6*10^-^1^9C)$

$K_1=9.0*10^9 N-m^2/C^2$

$U_1= 0$

$R=2.75*10^-^1^5m$

$K=2.30 MeV$

$m= 1.67*10^-^2^7kg$

$V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }$

$V_1= 3.4*10^7m/s$

Therefore the proton must be fired out with a speed of $V_1= 3.4*10^7m/s$

8 0

3 years ago

A dogsled team is shown pulling a man on a sled. Below the picture is a free body diagram with 4 force vectors. The first vector

Scorpion4ik [409]

Answer:

yes

Explanation:

its on edg.

8 0

3 years ago

Read 2 more answers

A car accelerates uniformly from rest to a speed of 10m/s. Calculate its acceleration

Maurinko [17]

Answer:

Give four examples of landforms where both the water and the land around it are flat.

Explanation:

8 0

2 years ago

If the resistance of dry skin is 200 times larger than the resistance of wet skin, how do the maximum voltages without shock com

Lelu [443]

Answer:

B) The voltage on dry skin needs to be 200 times larger than the voltage on wet skin.

Explanation:

This is the complete question

A person will feel a shock when a current of greater than approximately 100μ A flows between his index finger and thumb. If the resistance of dry skin is 200 times larger than the resistance of wet skin, how do the maximum voltages without shock compare in each scenario?

A) The voltage on dry skin needs to be 200 times smaller than the voltage on wet skin.

B) The voltage on dry skin needs to be 200 times larger than the voltage on wet skin.

C) The voltage on dry skin is the same as the voltage on wet skin.

D) The voltage on dry skin needs to be 40,000 times larger than the voltage on wet skin.

Ohm's law states that electric current is proportional to voltage and inversely proportional to resistance.

the equation is written as

V = IR

Where V is the voltage

I is the current

R is the resistance

for this case, the current I is 100μ A = 100 x 10^16 A

resistance of wet skin = R

resistance of dry skin = 200R

for the wet skin, voltage will be

V = IR = $100*10^{-6} R$

for dry skin, voltage will be

V = IR = $100*10^{-6}*200R = 0.02R$

Comparing both voltages

$0.02R$ ÷ $100*10^{-6} R$ = 200

this means that the voltage on the wet skin should be 200 times lesser than the voltage on the dry skin or the voltage on the dry skin should be 200 times more than the voltage on the wet skin.

4 0

3 years ago