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2 years ago

Only 5 questions plz answer

Physics

2 answers:

Viktor [21]2 years ago

5 0

Convergent boundary for the 4th picture

Romashka [77]2 years ago

5 0

Answer:

Q,12-c

Q.15-d

Q.17-a

Q.14-a

Q.16-c

Explanation:

sorry if I'm wrong

Q means question

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A body of mass 3.0kg moves with a velocity 10 m/s. calculate the momentum of the body

Aleksandr-060686 [28]

P=30 kg•m/s. This will be your correct answer.

4 0

2 years ago

The angular velocity of a 755-g wheel 15.0 cm in diameter is given by the equation ω(t) = (2.00 rad/s2)t + (1.00 rad/s4)t 3. (a)

Lunna [17]

Answer:

The number of radians turned by the wheel in 2s is $\theta= 8\ radians$

The angular acceleration is $\alpha =14 rad/s^2$

Explanation:

The angular velocity is given as

$w(t) = (2.00 \ rda/s^2)t + (1.00 rad /s^4)t^3$

Now generally the integral of angular velocity gives angular displacement

So integrating the equation of angular velocity through the limit 0 to 2 will gives us the angular displacement for 2 sec

This is mathematically evaluated as

$\theta(t ) = \int\limits^2_0 {2t + t^3} \, dt$

$= [\frac{2t^2}{2} + \frac{t^4}{4}] \left\{ 2} \atop {0}} \right.$

$= [\frac{2(2^2)}{2} + \frac{2^4}{4}] - 0$

$= 4 +4$

$\theta= 8\ radians$

Now generally the derivative of angular velocity gives angular acceleration

So the value of the derivative of angular velocity equation at t= 2 gives us the angular acceleration

This is mathematically evaluated as

$\frac{dw}{dt} = \alpha (t) = 2 + 3t^2$

so at t=2

$\alpha (2) = 2 +3(2)^2$

$\alpha =14 rad/s^2$

7 0

2 years ago

An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the

garik1379 [7]

(a) 6.04 rev/s

The speed of the ball is given by:

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the ball from the centre of the circle

In situation 1), we have

$\omega=8.13 rev/s \cdot 2\pi = 51.0 rad/s$

r = 0.600 m

So the speed of the ball is

$v=(51.0 rad/s)(0.600 m)=30.6 m/s$

In situation 2), we have

$\omega=6.04 rev/s \cdot 2\pi = 37.9 rad/s$

r = 0.900 m

So the speed of the ball is

$v=(37.9 rad/s)(0.900 m)=34.1 m/s$

So, the ball has greater speed when rotating at 6.04 rev/s.

(b) $1561 m/s^2$

The centripetal acceleration of the ball is given by

$a=\frac{v^2}{r}$

where

v is the speed

r is the distance of the ball from the centre of the trajectory

For situation 1),

v = 30.6 m/s

r = 0.600 m

So the centripetal acceleration is

$a=\frac{(30.6 m/s)^2}{0.600 m}=1561 m/s^2$

(c) $1292 m/s^2$

For situation 2 we have

v = 34.1 m/s

r = 0.900 m

So the centripetal acceleration is

$a=\frac{v^2}{r}=\frac{(34.1 m/s)^2}{0.900 m}=1292 m/s^2$

5 0

2 years ago

Please Help!!!!!

Oliga [24]

Answer: Molecules of H202, H20 and 02 are still forming. ( A P E X )

Explanation: I know this is late but for anyone looking at this later

3 0

3 years ago

According to the law of refraction, light passing from air into a piece of glass at an angle of 30 degrees will cause the light

pantera1 [17]

Answer:

bend toward the normal line

Explanation:

When light passes from a less dense to a more dense substance, (for example passing from air into water), the light is refracted (or bent) towards the normal. In your question the light is moving from rarer to denser medium

4 0

3 years ago