All categories

2 years ago

How much total surface of the moon is illuminated by the sun when it is at quarter phase?

1 answer:

6 0

50% of the moon is always illuminated, however during it's quarter phase means that we only see a quarter of what's really lit up. So it LOOKS like the moon is only 25% lit and 75% dark, it's truly 50/50. We only see that 25% since we can see it from one angle.

You might be interested in

Two tiny particles having charges 20.0 μC and 8.00 μC are separated by a distance of 20.0 cm What are the magnitude and directio

Alecsey [184]

Answer:

The magnitude and direction of electric field midway between these two charges is $10.8\times10^{5}\ N/C$ along AB.

Explanation:

Given that,

First charge $q_{1}= 20\mu C$

second charge $q_{2}= 8\mu C$

Distance = 20 cm

We need to calculate the electric field

For first charge,

Using formula of electric field

$E_{1}= \dfrac{kq_{1}}{r^2}$

Put the valueinto the formula

$E_{1}=\dfrac{9\times10^{9}\times20\times10^{-6}}{10\times10^{-2}}$

$E_{1}=18\times10^{5}\ N/C$

Direction of electric field along AB

We need to calculate the electric field

For second charge,

Using formula of electric field

$E_{2}= \dfrac{kq_{2}}{r^2}$

Put the valueinto the formula

$E_{2}=\dfrac{9\times10^{9}\times8\times10^{-6}}{10\times10^{-2}}$

$E_{2}=7.2\times10^{5}\ N/C$

Direction of electric field along AO

We need to calculate the net electric field at midpoint

$E_{net}=E_{1}-E_{2}$

$E_{net}=(18-7.2)\times10^{5}\ N/C$

$E_{net}=10.8\times10^{5}\ N/C$

Direction of net electric field along AB

Hence, The magnitude and direction of electric field midway between these two charges is $10.8\times10^{5}\ N/C$ along AB.

8 0

2 years ago

At NASA's Zero Gravity Research Facility in Cleveland, Ohio, experimental payloads fall freely from rest in an evacuated vertica

Diano4ka-milaya [45]

Answer:

(a). Energy is 64,680 J

(b) velocity is 51.43m/s

Explanation:

(a).

The potential energy $P$ of the payload of mass $m$ is at a vertical distance $h$ is

$P =mgh$ .

Therefore, for the payload of mass $m = 50kg$ at a vertical distance of $h = 132 m$ , the potential energy is

$P = (50kg)(9.8m/s^2)(132m)$

$\boxed{P = 64,680J}$

(b).

When the payload reaches the bottom of the shaft, all of its potential energy is converted into its kinetic energy; therefore,

$mgh= \dfrac{1}{2}mv^2$

$v= \sqrt{2gh}$

$v = \sqrt{2*9.8*135}$

$\boxed{v = 51.43m/s}$

(c).

The velocity in mph is

$\dfrac{51.43m}{s} * \dfrac{3600s}{hr} * \dfrac{1mile}{1609.34m}$

$\boxed{v= 115.0mph}$

5 0

3 years ago

1. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bott

Kobotan [32]

0.4 x 18 = 7.2 kg m/s

The momentum of the bottle after being hit is 0.2 x 25 = 5 kg m/s

7.2 - 5 = 2.2 kg m/s is the motmentum of the ball now

the velocity is 2.2/0.4 = 5.5 m/s

7 0

3 years ago

Calculation using Ohm's Law: If a circuit has a voltage of 500 V and a resistance of 250 ohms, what is the current?

konstantin123 [22]

I = V / R

Current = (voltage) / (resistance)

Current = (500 V) / (250 ohms)

Current = (500/250) Amperes

<em>Current = 2 Amperes</em>

6 0

2 years ago

Read 2 more answers

Diffrence between:- movable pulley and fixed pully

kotykmax [81]

Answer:

fixed pulley: A pulley system in which the pulley is attached to a fixed point and the rope is attached to the object. ... movable pulley: A pulley system in which the pulley is attached to the object; one end of the rope is attached to a fixed point and the other end of the rope is free.

Explanation:

8 0

2 years ago