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2 years ago

When an object experiences uniform circular motion the direction of the net force is?

1 answer:

3 0

In an uniform circular motion, the direction of the net force on the object is radially inward, passing through the center of the circle.

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A rock is thrown from a bridge at an angle 30∘ below horizontal.immediately after the rock is released, is the magnitude of its

Wittaler [7]

<span>The magnitude of the rock is equal to g. After the rock is released, there are no more forces acting on it, yet gravity remains. The initial inputs, on a bridge, at an angle of 30 deg below horizontal do not matter after the release.</span>

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3 years ago

only a relatively small part of the electromagnetic spectrum is visible. what determines which bands of the electromagnetic spec

swat32

Answer:

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Explanation:

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2 years ago

Only 5 questions plz answer.

Masja [62]

Question 18: a
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2 years ago

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A 3600 kg rocket traveling at 2900 m/s is moving freely through space on a journey to the moon. The ground controllers find that

Nana76 [90]

Answer:

m=417.24 kg

Explanation:

Given Data

Initial mass of rocket M = 3600 Kg

Initial velocity of rocket vi = 2900 m/s

velocity of gas vg = 4300 m/s

Θ = 11° angle in degrees

To find

m = mass of gas

Solution

Let m = mass of gas

first to find Initial speed with angle given

Vi=vi×tanΘ...............tan angle

Vi= 2900m/s × tan (11°)

Vi=563.7 m/s

Now to find mass

m = (M ×vi ×tanΘ)/( vg + vi tanΘ)

put the values as we have already solve vi ×tanΘ

m = (3600 kg ×563.7m/s)/(4300 m/s + 563.7 m/s)

m=417.24 kg

7 0

3 years ago

Two separate disks are connected by a belt traveling at 5m/s. Disk 1 has a mass of 10kg and radius of 35cm. Disk 2 has a mass of

pantera1 [17]

Explanation:

Given that,

Linear speed of both disks is 5 m/s

Mass of disk 1 is 10 kg

Radius of disk 1 is 35 cm or 0.35 m

Mass of disk 2 is 3 kg

Radius of disk 2 is 7 cm or 0.07 m

(a) The angular velocity of disk 1 is :

$v=r_1\omega_1\\\\\omega_1=\dfrac{v}{r_1}\\\\\omega_1=\dfrac{5}{0.35}\\\\\omega_1=14.28\ rad/s$

(b) The angular velocity of disk 2 is :

$v=r_2\omega_2\\\\\omega_2=\dfrac{v}{r_2}\\\\\omega_2=\dfrac{5}{0.07}\\\\\omega_2=71.42\ rad/s$

(c) The moment of inertia for the two disk system is given by :

$I=I_1+I_2\\\\I=\dfrac{1}{2}m_1r_1^2+\dfrac{1}{2}m_2r_2^2\\\\I=\dfrac{1}{2}(m_1r_1^2+m_2r_2^2)\\\\I=\dfrac{1}{2}\times (10\times (0.35)^2+3\times (0.07)^2)\\\\I=0.619\ kg-m^2$

Hence, this is the required solution.

6 0

2 years ago