Explian the construction of electric motor?

Gayle runs at a speed of 3.85 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. After s

enot [183]

Answer:

Final velocity at the bottom of hill is 15.56 m/s.

Explanation:

The given problem can be divided into four parts:

1. Use conservation of momentum to determine the speed of the combined mass (Gayle and sled)

From the law of conservation of momentum (perfectly inelastic collision), the combined velocity is given as:

$p_i = p_f$

$m_1u_1 + m_2v_2 = (m_1 + m_2)v$

$v = \frac{(m_1u_1 + m_2v_2)}{(m_1 + m_2)}$

$v=\frac{[50.0\ kg)(3.85\ m/s) + 0]}{(50.0\ kg + 5.00\ kg)}= 3.5\ m/s$

2. Use conservation of energy to determine the speed after traveling a vertical height of 5 m.

The velocity of Gayle and sled at the instant her brother jumps on is found from the law of conservation of energy:

$E(i) = E(f)$

$KE(i) + PE(i) = KE(f) + PE(f)$

$0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)$

$v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]}$

Here, initial velocity is the final velocity from the first stage. Therefore:

$v(f) = \sqrt{[(3.5)^2+2(9.8)(5.00-0)]}= 10.5\ m/s$

3. Use conservation of momentum to find the combined speed of Gayle and her brother.

Given:

Initial velocity of Gayle and sled is, $u_1(i)=10.5$ m/s

Initial velocity of her brother is, $u_2(i)=0$ m/s

Mass of Gayle and sled is, $m_1=55.0$ kg

Mass of her brother is, $m_2=30.0$ kg

Final combined velocity is given as:

$v(f) = \frac{[m_1u_1(i) + m_2u_2(i)]}{(m_1 + m_2)}$

$v(f)=\frac{[(55.0)(10.5) + 0]}{(55.0+30.0)}= 6.79$ m/s

4. Finally, use conservation of energy to determine the final speed at the bottom of the hill.

Using conservation of energy, the final velocity at the bottom of the hill is:

$E(i) = E(f)$

$KE(i) + PE(i) = KE(f) + PE(f)$

$0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)$

$v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]} \\v(f)=\sqrt{[(6.79)^2 + 2(9.8)(15 - 5.00)]}\\v(f)= 15.56\ m/s$

6 0

3 years ago

A rod of length L is hinged at one end. The moment of inertia as the rod rotates around that hinge is ML2 /3. Suppose a 2.00-m r

Stels [109]

To solve this problem we will apply the concepts related to the moment of inertia and Torque, the latter both its translational and rotational expression.

According to the information given the moment of inertia of the body would be

$I = \frac{1}{3} mL^2$

Replacing we have

$I = \frac{1}{3} (3kg)(2m)^2$

$I = 4 kg * m^2$

Now the translational torque would be the product between the force applied (Its own Weight) and the distance (Its center of mass at the middle)

$\tau = F*r$

$\tau = mg (\frac{L}{2})$

$\tau = (3)(9.8)(\frac{2}{2})$

$\tau = 29.4N\cdot m$

Now the rotational torque is defined as the product between the moment of inertia and the angular acceleration, then,

$\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}$

Replacing,

$\alpha = \frac{29.4}{4}$

$\alpha =7.35 rad / s^2$

Therefore the angular acceleration is $7.35rad/s^2$

4 0

3 years ago

Density Questions

Katen [24]

Explanation:

let's take the volume of each first:

ice cream: 10×10×4=400cm^3

freezer: 40×40×20=32,000cm^3

now we just divide: 32000/400=80

5 0

3 years ago

How would the mathematical model change if the direction that the object traveled was reversed

victus00 [196]

A line that is falling towards the x axis represents an object that is negatively accelerating, or slowing down. When the line hits the x axis, the object has stopped moving. If the graph continues below the x axis, the object has changed direction and is moving backwards at increasing velocity.

7 0

3 years ago

Derive the dimension of Power

stellarik [79]

Answer:

The dimension of power is energy divided by the time or $[ML^2T^-3]$