If a 5kg ball is traveling at 20 m/s and it’s stopped in 4s, what’s the impulse on the ball?

Satellite A has an orbital radius 3.00 times greater than that of satellite B. Satellite B's orbital period around Earth is 120

igomit [66]

Answer:

To find the circumference (orbit) of an object, you use Pi x Diameter.

As you have the circumference of B, you divide it by Pi to get the Diameter.

So 120 divided by 3.141592654 = 38.2 minutes for the Diameter.

As' radius and Diameter will be 3x greater than B.

38.2 x 3 = 114.6

To get back to the orbital period, times 114.6 by Pi, and you will get 360 minutes

HOPE THIS HELPS AND PLS MARK AS BRAINLIEST

THNXX :)

7 0

3 years ago

Imagine that you move a substance from one container to another and its volume changes. What state of matter is that substance?

Naddika [18.5K]

If it's volume changes when you move it to the new container it would be a solid

4 0

3 years ago

Read 2 more answers

A silver wire with resistivity 1.59 × 10-8 Ω-m carries a current density of 4.0 A/mm2, What is the magnitude of the electric fie

Nat2105 [25]

Answer:

Electric field, E = 0.064 V/m

Explanation:

It is given that,

Resistivity of silver wire, $\rho=1.59\times 10^{-8}\ \Omega-m$

Current density of the wire, $J=4\ A/mm^2=4\times 10^6\ A/m^2$

We need to find the magnitude of the electric field inside the wire. The relationship between electric field and the current density is given by :

$E=J\times \rho$

$E=4\times 10^6\times 1.59\times 10^{-8}$

E = 0.0636 V/m

or

E = 0.064 V/m

So, the magnitude of electric field inside the wire is 0.064 V/m. Hence, this is the required solution.

6 0

3 years ago

A vessel with an unknown volume is filled with 10 kg of water at 90oC. Inspection of the vessel at equilibrium shows that 8 kg o

damaskus [11]

Hey there!

In this case, it is possible to solve this problem by using the widely-known steam tables which show that at 90 °C, the pressure that produces a vapor-liquid mixture at equilibrium is about 70.183 kPa (Cengel, Thermodynamics 5th edition).

Moreover, for the calculation of the volume, it is necessary to calculate the volume of the vapor-liquid mixture, given the quality (x) it has:

$x=\frac{m_{steam}}{m_{total}}$

Thus, since 8 kg correspond to liquid water, 2 kg must correspond to steam, so that the quality turns out:

$x=\frac{2kg}{10kg} =0.20$

Now, at this temperature and pressure, the volume of a saturated vapor is 2.3593 m³/kg whereas that of the saturated liquid is 0.001036 m³/kg and therefore, the volume of the mixture is:

$v=0.001036m^3/kg+0.2(2.3593-0.001036 )m^3/kg=0.4727m^3/kg$