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slamgirl [31]
3 years ago
5

A car’s momentum is p when it is traveling with a velocity of v. If the velocity of that car doubles, what is the new momentum o

f the car? 1/2p p 2p 4p
Physics
2 answers:
Andreas93 [3]3 years ago
6 0

Answer:

C. 2p

Explanation:

On Edgenuity 2020

mamaluj [8]3 years ago
5 0
The answer C. 2p. I took the quiz and I got it correct.
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32. Only a small percentage of the energy
nordsb [41]
Only a small Percentage of the energy emitted by the sun strikes earth because, since the earth is going in circles round and round, the sun only hits part of the earth and not fully. So for example if the earth is going around and the sun hits Africa then in a couple minutes it will go to the next country and that country will have sunlight. Let me know if you need anything else or if this isn't partially correct. GLAD TO HELP! :)
5 0
3 years ago
A war wolf is a device used during the middle ages to assault fortifications with large rocks. A simple trebuchet is constructed
Katarina [22]

Answer:v=41.23 m/s

Explanation:

Given

mass of heavy object m_1=52 kg

distance of m_1 from the axle r_1=14 cm

mass of rock m_2=123 gm

Length of rod =4.1 m

distance of m_2 from axle r_2=4.1-0.14=3.96 m

Net torque acting is

T_{net}=m_1gr_1-m_2gr_2

T_{net}=52\times 0.14\times g-0.123\times 3.96\times g

T_{net}=6.793\times 9.8

T_{net}=66.57 N-m

Work done by T_{net} is converted to rock kinetic Energy

thus

T_{net}\times \theta =\frac{mv^2}{2}

Where \theta =angle\ turned =\frac{\pi }{2}

v= velocity\ at\ launch

66.57\times \frac{\pi }{2}=\frac{0.123\times v^2}{2}

v^2=66.57\times \pi

v=\sqrt{1700.511}

v=41.23 m/s

3 0
4 years ago
For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is
ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

where A(t)=A₀e^{\frac{-bt}{2m}}

 A₀ is the amplitude at t=0 and

w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

⇒e^{\frac{bt}{2m}}=8

applying logarithm on both sides

⇒\frac{bt}{2m}=ln(8)

⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

b) w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}

T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

⇒T'=\frac{2\pi}{2.86}=2.2s

let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

8 0
3 years ago
What is the kinetic energy of a 120-cm thin uniform rod with a mass of 450 g that is rotating about its center at 3.60 rad/s?
goldfiish [28.3K]

Answer:

1.05 J.

Explanation:

Kinetic Energy: This is the energy possessed by a body due to its motion. The S.I unit of kinetic energy is Joules (J). The formula of kinetic energy is given as

Ek = 1/2mv²................. Equation 1

Where Ek = kinetic energy, m = mass of the uniform rod, v = liner velocity of the rod.

But,

v = αr .......................... Equation 2

Where α = angular velocity of the rod, r = radius of the circle.

Given: α = 3.6 red/s, r = 120/2 = 60 cm = 0.6 m.

Substitute into equation 2

v = 3.6(0.6)

v = 2.16 m/s.

Also given: m = 450 g = 0.45 kg.

Substitute into equation 1

Ek = 1/2(0.45)(2.16²)

Ek = 1.05 J.

4 0
3 years ago
(EARTH SPACE SCIENCE QUESTION)
Sergeeva-Olga [200]

*may be egg shaped

*has no new stars being formed

*has almost no gas or dust between stars

I literally just took notes on elliptical galaxies a week ago wow

8 0
3 years ago
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