A 125g steel ball with a kinetic energy of .25j rolls along a horizontal track how high up an inclined track will the ball roll

A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the bea

serg [3582]

Answer:

$v = 3369.2 m/s$

Explanation:

The beacon is rotating at an angular speed of

$f = 10 rev/min$

so we have

$\omega = 2\pi f$

$\omega = 2\pi(\frac{10}{60})$

$\omega = 1.047 rad/s$

We know that

$v = r \omega$

At this point we have

$r = 2 miles = 2(1609 m)$

$r = 3218 m$

So we can conclude with

$v = 3218(1.047)$

$v = 3369.2 m/s$

6 0

2 months ago

A conducting sphere 45 cm in diameter carries an excess of charge, and no other charges are present. You measure the potential o

Maru [3345]

Answer:

The excess charge is $Q = 3.5 *10^{-7} \ C$

Explanation:

According to the question, we are informed that

The diameter is $d = 45 \ cm = 0.45 \ m$

The potential of the surface is $V = 14 \ kV = 14 *10^{3} \ V$

The radius of the sphere is

$r = \frac{d}{2}$

by plugging in given values

$r = \frac{0.45}{2}$

$r = 0.225 \ m$

The potential at the surface is mathematically expressed as

$V = \frac{k * Q }{r }$

Where k is Coulomb's constant with a value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

Based on the question stating there are no other charges, Q represents the excess charge

Therefore

$Q = \frac{V* r}{ k}$

inserting the numerical values

$Q = \frac{14 *10^{3} 0.225}{ 9*10^9}$

$Q = 3.5 *10^{-7} \ C$

7 0

1 month ago

A merry-go-round with a a radius of R = 1.63 m and moment of inertia I = 196 kg-m2 is spinning with an initial angular speed of

serg [3582]

Answer:

1) L = 299.88 kg-m²/s

2) L = 613.2 kg-m²/s

3) L = 499.758 kg-m²/s

4) ω₁ = 0.769 rad/s

5) Fc = 70.3686 N

6) v = 1.2535 m/s

7) ω₀ = 1.53 rad/s

Explanation:

Given

R = 1.63 m

I₀ = 196 kg-m²

ω₀ = 1.53 rad/s

m = 73 kg

v = 4.2 m/s

1) What is the magnitude of the initial angular momentum of the merry-go-round?

We utilize the formula

L = I₀*ω₀ = 196 kg-m²*1.53 rad/s = 299.88 kg-m²/s

2) What is the angular momentum magnitude of the person 2 meters prior to jumping onto the merry-go-round?

The equation we apply is

L = m*v*Rp = 73 kg*4.2 m/s*2.00 m = 613.2 kg-m²/s

3) What is the angular momentum of the person just before she hops onto the merry-go-round?

We utilize the formula

L = m*v*R = 73 kg*4.2 m/s*1.63 m = 499.758 kg-m²/s

4) What is the angular velocity of the merry-go-round after the individual jumps on?

We can apply the Principle of Conservation of Angular Momentum

L in = L fin

⇒ I₀*ω₀ = I₁*ω₁

where

I₁ = I₀ + m*R²

⇒ I₀*ω₀ = (I₀ + m*R²)*ω₁

At this point, we can determine ω₁

⇒ ω₁ = I₀*ω₀ / (I₀ + m*R²)

⇒ ω₁ = 196 kg-m²*1.53 rad/s / (196 kg-m² + 73 kg*(1.63 m)²)

⇒ ω₁ = 0.769 rad/s

5) Once the merry-go-round moves at this new angular speed, what force must the person exert to hold on?

We must calculate the centripetal force as follows

Fc = m*ω²*R

⇒ Fc = 73 kg*(0.769 rad/s)²*1.63 m = 70.3686 N

6) When the person is halfway around, they choose to simply release their hold on the merry-go-round to exit the ride.

What is the linear speed of the person the moment they exit the merry-go-round?

we can apply the equation

v = ω₁*R = 0.769 rad/s*1.63 m = 1.2535 m/s

7) What is the angular velocity of the merry-go-round after the individual releases their hold?

ω₀ = 1.53 rad/s

It returns to its original angular speed

8 0

1 month ago

You are at a train station, standing next to the train at the front of the first car. The train starts moving with constant acce

serg [3582]

The duration required for the seventh car to pass amounts to 13.2 seconds. The train's movement is characterized by uniform acceleration, enabling the application of suvat equations. Initially, we analyze the movement of the first car, utilizing the equation for distance s covered in time t, which corresponds to the length of one car, with u = 0 as the initial velocity and a representing acceleration, over t = 5.0 s. We can rearrange the equation reflecting L as the length of one car. This is similarly applicable for the initial seven cars, accounting for the distance of 7L and the required time t'. With constant acceleration retained, we can derive t' through substitution in the equation, leading to fundamental conclusions regarding the relationship exhibited in the graph of distance against time in uniformly accelerated motion.

8 0

2 months ago

Four wires are made of the same highly resistive material, cut to the same length, and connected in series. Wire 1 has resistanc

ValentinkaMS [3465]

Answer:

$V_2 = \frac{R_2 V_0}{R_1+R_2+R_3+R_4}$

Explanation:

The four wires are arranged in series: this setup indicates that the same current passes through them, while the total voltage from the battery, V0, equals the combined voltages across each resistor:

$V_0=V_1+V_2+V_3+V_4$