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AlekseyPX
3 years ago
11

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

e displacement at 2 seconds.
Physics
1 answer:
dalvyx [7]3 years ago
4 0

Answer:

Explanation:

a )  V = 3 cos(0.5t)

differentiating with respect to t

dv /dt = -3 x .5 sin0.5t

= -1.5 sin0.5t.

acceleration = - 1.5 sin 0.5t

when t = 3 s

acceleration = - 1.5 sin 1.5

= - 1.496 ms⁻²

v = 3 cos.5t

b )  dx/dt = 3 cos 0.5 t

dx = 3 cos 0.5 t dt

integrating on both sides

x = 3 sin .5t / .5

x = 6 sin0.5t

At t = 2 s

x = 6 sin 1

x = 5.05 m

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~D. Very dusty irregular galaxy.

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4 0
3 years ago
A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.79 rad/s . Its total moment of inertia is 1790
Nezavi [6.7K]

Answer:

w_2=0.467rad/s

Explanation:

Four people standing on the ground each of mass and usually this questions have to find the final angular velocity

m_t=4*70kg=280kg

The radius r=4.2/2=2.1m

Angular velocity  w_1=0.79rad/s

The moment of inertia total is I_t=1790 kg/m^2

Momento if inertia

I_1=m_t*r^2

I_1=280kg*(2.1m)^2=1234.8kg*m^2

Angular momentum

I_1*w_1=I_t*w_2

Solve to w2

w_2=\frac{I_1*w_1}{I_t}

w_2=\frac{1790kg*m^2*0.79rad/s}{3024.8kg*m^2}

w_2=0.467rad/s

8 0
3 years ago
distant galaxy emits light that has a wavelength of 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm.
a_sh-v [17]

Answer:

The speed of the galaxy relative to the Earth is 3.09\times 10^6\ m/s.

Explanation:

We have,

(a) Wavelength emitted by light at distant galaxy is 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm. It can be seen that the wavelength of light reduces as it reaches Earth. It is called Red shift. As per Doppler's effect, we can say that the galaxy is receding from the Earth.

(b) Let v is the speed of the galaxy relative to the Earth. It can be given by :

v=c(\dfrac{\lambda'}{\lambda}-1)\\\\v=3\times 10^8\times (\dfrac{438.6 }{434.1 }-1)\\\\v=3\times 10^8\times (\dfrac{438.6}{434.1}-1)\\\\v=0.0103\cdot3\cdot10^{8}\\\\v=3.09\times 10^6\ m/s

So, the speed of the galaxy relative to the Earth is 3.09\times 10^6\ m/s.

4 0
3 years ago
Because of interstellar dust, astronomers can see at most about 5 kpc into the disk of the galaxy at visual wavelengths. What pe
NNADVOKAT [17]

Answer:

96%

Explanation

Let A the total area of the galaxy, is modeled as a disc:

A = πR^2 = π (25 kpc)^2

And let a be the area that astronomers are able to see:

a = πr^2 = π(5 kpc)^2

The percentage that can be seen is equal to 100 times the ratio of the areas, of the galaxy and the "visible" part:

P = 100 a/A = (5/25)^2 = 100/25 = 4%

Therefore, the percentage of the galaxy not included, i.e. not seen is:

(100-4)% = 96%

5 0
3 years ago
while working out a man performed 2525j of work in 19seconds . what was his power A:132.9w. B:241.5w C 47.975w. D100.5w
Olenka [21]
A: 132.9w because 2525\19 is how much energy transferred per second which is also known as the power
7 0
3 years ago
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