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3 years ago

Pendulum earth air lololo

Physics

2 answers:

zhannawk [14.2K]3 years ago

4 0

Is this a question?

Likurg_2 [28]3 years ago

4 0

Answer:

ooooookkkkkk????

Explanation:

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On earth, how can you tell up from down

bearhunter [10]

Answer:

I think is 3 I am not sure

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3 years ago

Water vapor is an example of a(n)

murzikaleks [220]

Answer:

Water vapor is steam. An example of water vapor is the floating mist above a pot of boiling water. Water in the form of a gas; steam.

Explanation:

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3 years ago

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

aniked [119]

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.

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3 years ago

What does hypothesis mean

k0ka [10]

Hypothesis means <span>a supposition or proposed explanation made on the basis of limited evidence as a starting point for further investigation</span>

7 0

3 years ago

Read 2 more answers

A 1-kilogram mass is attached to a spring whose constant is 21 N/m, and the entire system is then submerged in a liquid that imp

amm1812

Answer:

the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Explanation:

Given that,

mass, m = 1kg

spring constant k = 21N/M

damping force = $-\beta\frac{dx}{dt} = \frac{-10dx}{dt}$

$\beta = 10$

By Newtons second law ,

The diffrential equation of motion with damping is given by

$m\frac{d^2x}{dt^2} = -kx-\beta\frac{dx}{dt}$

substitute the value of m =1kg, k = 21N/M, and $\beta = 10$

$1\frac{d^2x}{dt^2} = -21x=10\frac{dx}{dt}$

$\frac{d^2x}{dt^2} + 10\frac{dx}{dt} + 21x = 0$

suppose the equation of the form $x =e^m^t$ ,

and the auxilliary equation is given by

$m^2 + 10m + 21 = 0\\\\m^2 + 7m+3m+21=0\\\\m(m+7)+3(m+7)=0\\\\(m+7)(m+3)=0\\\\m=-7\\m=-3$

The general solution for the above differential equation is

$x(t) =C_1e^{-3t}+C_2e^{-7t}$

Derivate with respect to t

$x'(t)=-3C_1e^{-3t}-7C_2e^{-7t}$

(a)

since time is 0 then mass is one meter below

so x(0) = 1

Also it start from rest , that implies , velocity is 0 and time is 0

$x'(0) = 0$

substitute the initial condition

$C_1 +C_2 = 1$

$-3C_1-7C_2=0$

Solve the above equation to get C₁ and C₂

$C_1 =\frac{7}{4} and C_2 = -\frac{3}{4}$

substitute for C₁ and C₂ in general solution

$x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Thus the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

3 0

3 years ago

Read 2 more answers