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frozen [14]
3 years ago
14

slader A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of

25 m/s and it bounces back with this same speed. The ball is in contact with the wall 0.050 s. What is the magnitude of the average force exerted on the wall by the ball?
Physics
1 answer:
Ivan3 years ago
6 0

Answer: 800N

Explanation:

Given :

Mass of ball =0.8kg

Contact time = 0.05 sec

Final velocity = initial velocity = 25m/s

Magnitude of the average force exerted on the wall by the ball is can be calculated using the relation;

Force(F) = mass(m) * average acceleration(a)

a= (initial velocity(u) + final velocity(v))/t

m = 0.8kg

u = v = 25m/s

t = contact time of the ball = 0.05s

Therefore,

a = (25 + 25) ÷ 0.05 = 1000m/s^2

Therefore,

Magnitude of average force (F)

F=ma

m = mass of ball = 0.8

a = 1000m/s^2

F = 0.8 * 1000

F = 800N

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Harrizon [31]

Answer:

mass of the fish is 8.11 kg

Explanation:

As we know that the frequency of oscillation of spring block system is given as

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

here we know that the reading of scale varies from 0 to 155 N from length varies from x = 0 to x = 10 cm

Now we have

k = \frac{155}{0.10} N/m

k = 1550 N/m

so now we have

2.20 = \frac{1}{2\pi}\sqrt{\frac{1550}{m}}

m = 8.11 kg

so mass of the fish is 8.11 kg

4 0
3 years ago
The speed of light in air is 3 x 108 m/s. The speed of light in ice is 2.29 x 108 m/s. What is the refractive index from air to
Studentka2010 [4]

Answer:

η = 1.31

Explanation:

The formula for the refractive index of from air to some other medium is given by the following formula:

\eta = \frac{c}{v}\\

where,

η = refractive index = ?

c = speed of light in air = 3 x 10⁸ m/s

v = speed of light in ice = 2.29 x 10⁸ m/s

Therefore, using these values in the equation we get:

\eta = \frac{3\ x\ 10^8\ m/s}{2.29\ x\ 10^8\ m/s} \\

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4 0
2 years ago
Two protons are maintained at a separation of nm. Calculate the electric potential due to the two particles at the midpoint betw
Liono4ka [1.6K]

Answer:

The electric potential is approximately 5.8 V

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero

Explanation:

The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:

\displaystyle{U=\frac{q}{4\pi \epsilon_0r}} (1)

where q is the charge of the particle, \epsilon_0 the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and r is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:

\displaystyle{U_{midpoint}=\frac{q}{4\pi \epsilon_0r}}+\frac{q}{4\pi \epsilon_0r}}=\frac{q}{2\pi \epsilon_0r}}}

Substituting the values q=1.602 \cdot10^{-19}\ C, \displaystyle{\frac{1}{4\pi\epsilon_0}=8.99\cdot 10^9 N\cdot m^2\cdot C^{-2}} and r=0.5 \cdot 10^{-9} m we obtain:

\displaystyle{U_{midpoint}=\frac{q}{2\pi \epsilon_0r}}=5.759 \approx 5.8 V}

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.

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F=ma
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