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Lubov Fominskaja [6]
3 years ago
11

Which of the following definitions BEST defines a satellite?

Physics
1 answer:
Temka [501]3 years ago
5 0

Answer:

D, any object that revolves around another object.

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Which quote best reveals why sowers was successful in achieving her goal?
BabaBlast [244]
Sowers has spent her career fighting for gender equality and it passionate about it .
5 0
2 years ago
What is the difference between a 3 v and 12 v battery
Lorico [155]
V=IR
The more V, the more I
6 0
3 years ago
An alpha particle (a helium nucleus, consisting of two protons and two neutrons) has a radius of approximately 1.6 × 10-15 m. A
Snezhnost [94]

Answer:

9.96x10^-20 kg-m/s

Explanation:

Momentum p is the product of mass and velocity, i.e

P = mv

Alpha particles, like helium nuclei, have a net spin of zero. Due to the mechanism of their production in standard alpha radioactive decay, alpha particles generally have a kinetic energy of about 5 MeV, and a velocity in the vicinity of 5% the speed of light.

From this we calculate the speed as

v = 5% 0f 3x10^8 m/s (speed of light)

v = 1.5x10^7 m/s

The mass of an alpha particle is approximately 6.64×10−27 kg

Therefore,

P = 1.5x10^7 x 6.64×10^−27

P = 9.96x10^-20 kg-m/s

8 0
3 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
Margaret [11]

Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

8 0
2 years ago
Suppose a 2.0kg bird is flying at a speed of of 1m/s. It’s kinetic energy would be what?
Ganezh [65]

As we know that the formula of kinetic energy will be

KE = \frac{1}{2} mv^2

now here we know that

m = 2 kg

v = 1 m/s

so from the above equation we have

KE = \frac{1}{2}(2)(1^2)

KE = 1 J

7 0
3 years ago
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