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Norma-Jean [14]
3 years ago
9

You throw a small toy ball that is covered with suction cups at the center of your glass patio door. When it hits it sticks to t

he glass, and because the door was not latched, it causes the door to swing open with an angular velocity of 0.19 rad/s. If the ball has a mass of 120 g and the patio door can be treated as a uniform box that is 2.3 m high, 1.0 m wide, and 0.030 m thick with a mass of 7.5 kg what speed did you throw the ball at
Physics
1 answer:
Natali [406]3 years ago
6 0

Answer:

Explanation:

moment of inertia of the door M = 1 /3 m ( l² + b² + d² )

= 1 / 3 x 7.5 x ( 2.3² + 1² + .03² )

= 1 / 3 x 7.5 x 6.2909

I = 15.7272.

moment of inertia of the door + small toy

15.7272 + .12 x .5²

= 15.7272 + .03

= 15.7572

Applying conservation of angular momentum law

mvr = M ω

m is mass of the toy thrown with velocity v at distance r from the axis , M is moment of inertia of door + toy and ω is angular velocity ith which door opens.

.12 x v x .5 =  15.7572 x .19

v = 49.89 m /s .

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A roller coaster travels around a vertical 8-m radius loop. Determine the speed at the top of the loop if the normal force exert
strojnjashka [21]

Answer:

v=10m/sec

Explanation:

From the question we are told that

Radius of vertical r= 8m

Force exerted by passengers is 1/4 of weight

Generally the net force acting on top of the roller coaster is give to be

F_N+Fg

where

F_N =forceof the normal

Fg= force due to gravity

Generally the net force is given to be FC(force towards center)

F_C =F_N + Fg

F_N =F_C -Fg

F_N=F_C-F_g

F_N=\frac{mv^2}{R} -mg

Mathematical we can now derive V

m_g + \frac{8m}{4}= \frac{mv^2}{8}

\frac{5mg}{4} =\frac{mv^2}{8}

v^2 =\frac{40*10}{4}

v=10m/sec

Therefore the speed of the roller coaster is given ton be v=10m/sec

5 0
3 years ago
What is the driving force of charge around a circuit
Virty [35]
A force of charge that drive around a circuit is call electeons
5 0
3 years ago
An electron moves at a speed of 1.0 x 104 m/s in a circular path of radius 2 cm inside a solenoid. The magnetic field of the sol
iogann1982 [59]

Answer:

(a) B = 2.85 × 10^{-6} Tesla

(b) I =  I = 0.285 A

Explanation:

a. The strength of magnetic field, B, in a solenoid is determined by;

r = \frac{mv}{qB}

⇒ B = \frac{mv}{qr}

Where: r is the radius, m is the mass of the electron, v is its velocity, q is the charge on the electron and B is the magnetic field

B = \frac{9.11*10^{-31*1.0*10^{4} } }{1.6*10^{-19}*0.02 }

  = \frac{9.11*10^{-27} }{3.2*10^{-21} }

B = 2.85 × 10^{-6} Tesla

b. Given that; N/L = 25 turns per centimetre, then the current, I, can be determined by;

B = μ I N/L

⇒    I = B ÷ μN/L

where B is the magnetic field,  μ is the permeability of free space = 4.0 ×10^{-7}Tm/A, N/L is the number of turns per length.

I = B ÷ μN/L

 = \frac{2.85*10^{-6} }{4*10^{-7} *25}

I = 0.285 A

5 0
3 years ago
For a mass oscillating on a spring at what positions are (a) velocity and (b) acceleration of the mass have maximum valeus?
EleoNora [17]

Answer:

a)At the mean position

b)At the extremes positions

Explanation:

Given that mass is having oscillation motion.

We know that

1. At the mean position -The velocity of the mass is maximum and the acceleration of the mass is minimum.The net force on the mass will be zero.

2. At the extreme position-The velocity of the mass is minimum and the acceleration of the mass is maximum.The net force on the mass will not be zero.

Therefore

a)At the mean position

b)At the extremes positions

3 0
3 years ago
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