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Rashid [163]
3 years ago
14

A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 45.0kg . Initially open and at re

st, the door is struck at its center by a handful of sticky mud with mass 0.700 kg, traveling perpendicular to the door at 12.0m/s just before impact
A) Find the final angular speed of the door.
answer in rad/s
B) Does the mud make a significant contribution to the moment of inertia?
Yes or No
Physics
1 answer:
aniked [119]3 years ago
5 0

Answer:

0.19rad/s and Yes

Explanation:

From the principle of conservation of momentum it means momentum before and after collision is the same.

Momentum before collision is 0.700 kg×12 = 8.4Ns

Momentum of the door = mass of door × velocity of door

8.4Ns = mass of door × velocity of door

Velocity of door = 8.4Ns/45 =0.19m/s

But velocity V= w×r ;

w-angular velocity

r- raduis = width

w= 0.19/1m = 0.19rad/s

2. Yes it did because it resisted The moment of inertia and ensued the locking of the door.

You might be interested in
A 0.0780 kg lemming runs off a
kotegsom [21]

Answer:

5.01 J

Explanation:

Info given:

mass (m) = 0.0780kg

height (h) = 5.36m

velocity (v) = 4.84 m/s

gravity (g) = 9.81m/s^2

1. First, solve for Kinetic energy (KE)

KE = 1/2mv^2

1/2(0.0780kg)(4.84m/s)^2 = 0.91 J

so KE = 0.91 J

2. Next, solve for Potential energy (PE)

PE = mgh

(0.0780kg)(9.81m/s^2)(5.36m) = 4.10 J

so PE = 4.10 J

3. Mechanical Energy , E = KE + PE

Plug in values for KE and PE

KE + PE = 0.91J + 4.10 J = 5.01 J

4 0
3 years ago
three condensers are connected in series across a 150 volt supply. The voltages across them are 40,50 and 60 volts respectively,
ioda

Explanation:

Given that,

The voltages across them are 40,50 and 60 volts respectively, and the charge on each condenser is 6×10⁻⁸ C.

(a) Capacitance of capacitor 1,

C_1=\dfrac{Q}{V_1}\\\\C_1=\dfrac{6\times 10^{-8}}{40}\\\\C_1=1.5\times 10^{-9}\ F\\\\C_1=1.5\ nF

Capacitance of capacitor 2,

C_2=\dfrac{Q}{V_2}\\\\C_2=\dfrac{6\times 10^{-8}}{50}\\\\C_2=1.2\times 10^{-9}\ F\\\\C_2=1.2\ nF

Capacitance of capacitor 3,

C_3=\dfrac{Q}{V_3}\\\\C_3=\dfrac{6\times 10^{-8}}{60}\\\\C_3=1\times 10^{-9}\ F\\\\C_3=1\ nF

(b) The equivalent capacitance in series combination is :

\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}\\\\\dfrac{1}{C}=\dfrac{1}{1.5}+\dfrac{1}{1.2}+\dfrac{1}{1}\\\\C=0.4\ nF

Hence, this is the required solution.

5 0
3 years ago
Nonrenewable energy resources do not include which of the following?
Tems11 [23]
Hydrogen fuel cells is the answer
8 0
3 years ago
Read 2 more answers
How has the farming industry been affected by the use of fossil fuels?
Mashutka [201]

Fossil fuels have led to an increase in food costs.

Explanation:

Buried organic material over million of years form combustible geologic deposits of organic materials called fossil fuels that can be converted crude oil, natural gas, or coal.

Fossil fuel are currently used as a source of energy in machine and equipment applied in the farming industry. Mechanization has been made possible due to this type of fuel availability thus larger farms have been prepared for cultivation.In addition, other processes such as planting,weeding and harvesting are currently performed by farming machines which are propelled by fossil energy. So in general, fossil fuels in the farming industry has led to increase in amount of crop harvested because more farms are under cultivation and farming process is now mechanized. However, due to use of machines and energy, the production cost has increased, thus to cover for this increased cost, the prices for food products increases, making the cost of food commodities to rise.

Learn More

Fossil fuels: brainly.com/question/13660400

Keyword: Farming, industry, fossil fuel

#LearnwithBrainly

3 0
3 years ago
These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the neg
Kay [80]

Answer:

Following are the solution to the given question:

Explanation:

For charging plates that are connected in a similar manner:

Calculating the total charge:

\to q =q_1 + q_2 = C_1V_1 +C_2V_2 =1320 + 2714 = 4034 \mu C

Calculating the common potential:

\to V = \frac{q}{C}= \frac{q}{(C_1 + C_2)} =\frac{4034}{6.8} = 593 \ V\\\\

Calculating the charge after redistribution:

When: \\\\q = q_{1}' + q_{2}' = q_1 + q_2        

\to q_{1}' = C_1V = 2.2 \times 593 = 1305\ \mu C\\  \\  \to               q_{2}' = C_2V = 4.6 \times 593 = 2729 \ \mu C

6 0
3 years ago
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