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3 years ago

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In a skating stunt known as "crack-the-whip," a number of skaters hold hands and form a straight line. They try to skate so that

the line rotates about the skater at one end, who acts as the pivot. The skater farthest out has a mass of 80.0 kg and is 6.10 m from the pivot. He is skating at a speed of 6.80 m/s. Determine the magnitude of the centripetal force that acts on him.

1 answer:

4vir4ik [10]3 years ago

7 0

Answer:

The magnitude of the centripetal force that acts on him

Explanation:

Given that,

Mass = 80.0 kg

Distance = 6.10 m

Speed = 6.80 m/s

We need to calculate the magnitude of the centripetal force that acts on him

Using formula of the centripetal force

$F_{c}=\dfrac{mv^2}{r}$

Where, F = force

m = mass

v = speed

r = distance

Put the value into the formula

$F_{c}=\dfrac{80.0\times(6.80)^2}{6.10}$

$F_{c}=606.4\ N$

Hence, The magnitude of the centripetal force that acts on him

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In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue

Andru [333]

Answer:

a) p₀ = 1.2 kg m / s, b) p_f = 1.2 kg m / s, c) θ = 12.36, d) v_{2f} = 1.278 m/s

Explanation:

For this exercise we define a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved

a) the initial impulse is

p₀ = m v₁₀ + 0

p₀ = 0.6 2

p₀ = 1.2 kg m / s

b) as the system is isolated, the moment is conserved so

p_f = 1.2 kg m / s

we define a reference system where the x-axis coincides with the initial movement of the cue ball

we write the final moment for each axis

X axis

p₀ₓ = 1.2 kg m / s

p_{fx} = m v1f cos 20 + m v2f cos θ

p₀ = p_f

1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ

1.2482 = v_{2f} cos θ

Y axis

p_{oy} = 0

p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ

0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ

0.2736 = v_{2f} sin θ

we write our system of equations

0.2736 = v_{2f} sin θ

1.2482 = v_{2f} cos θ

divide to solve

0.219 = tan θ

θ = tan⁻¹ 0.21919

θ = 12.36

let's look for speed

0.2736 = v_{2f} sin θ

v_{2f} = 0.2736 / sin 12.36

v_{2f} = 1.278 m / s

4 0

3 years ago

A wire carries 3.7 A of current. A second wire is placed parallel to the first 8.0 cm away. What is the current flowing through

IgorC [24]

Answer:

The current in second wire is 5.0 A.

(B) is correct option.

Explanation:

Given that,

Current in first wire = 3.7 A

Distance = 8.0 cm

We need to calculate the magnetic field due to the current carrying wire

Using formula of magnetic field

$B=\dfrac{\mu_{0}I}{2\pi r}$

Where, I = current

r = distance

Put the value into the formula

For first wire

$B = \dfrac{\mu_{0}\times3.7}{2\pi \times3.4\times10^{-2}}$ ...(I)

For second wire,

The distance is 8-3.7 = 4.3 cm

$B' = \dfrac{\mu_{0}\times I'}{2\pi \times(8-3.4)\times10^{-2}}$ ...(II)

The magnetic field in both the wires,

From equation (I) and (II)

$\dfrac{\mu_{0}\times3.7}{2\pi \times3.4\times10^{-2}}= \dfrac{\mu_{0}\times I'}{2\pi \times(8-3.4)\times10^{-2}}$

$I'=\dfrac{3.7\times4.3\times10^{-2}}{3.4\times10^{-2}}$

$I'=4.68\ A\ approx = 5.0\ A$

Hence, The current in second wire is 5.0 A.

8 0

4 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 body accelerate uniformly from rest at 2m/s square.Calculate its velocity after traveling 9m

Thepotemich [5.8K]

The correct answer for this question is 6m/s. I hope this helps.

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3 years ago

Whoever answers correctly gets brainlist!

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Infared = used by police

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5 0

3 years ago

Read 2 more answers