If a refrigerator draws about 0.5 amps when plugged into a 120 V wall outlet how much power (watts) does it use?

What is one way to increase the amplitude of a wave in a medium?

kobusy [5.1K]

Answer:

A

Explanation:

Vibrations will always increase amplitude

7 0

4 years ago

Loghan and Kyle wanted to see if the type of ice cream in the freezer affects how hard or soft it is. At three different times,

sweet-ann [11.9K]

Loghan and Kyle should have use the same type of ice cream for all three ice cream containers to get a more accurate result and data

4 0

3 years ago

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Could anyone help me with this?

tatiyna

Answer:

Answer is option C..

Mark as brainlist

3 0

3 years ago

A Huge water tank is 2m above the ground if the water level on it is 4.9m high and a small opening is there at the bottom then t

lapo4ka [179]

Answer:

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

Explanation:

Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ( $P_{atm}$ ), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:

$P_{1} + \rho\cdot \frac{v_{1}^{2}}{2} + \rho\cdot g \cdot z_{1} = P_{2} + \rho\cdot \frac{v_{2}^{2}}{2} + \rho\cdot g \cdot z_{2}$

Where:

$P_{1}$ , $P_{2}$ - Water total pressures inside the tank and at ground level, measured in pascals.

$\rho$ - Water density, measured in kilograms per cubic meter.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{1}$ , $v_{2}$ - Water speeds inside the tank and at the ground level, measured in meters per second.

$z_{1}$ , $z_{2}$ - Heights of the tank and ground level, measured in meters.

Given that $P_{1} = P_{2} = P_{atm}$ , $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{1} = 0\,\frac{m}{s}$ , $z_{1} = 6.9\,m$ and $z_{2} = 4.9\,m$ , the expression is reduced to this:

$\left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m) = \frac{v_{2}^{2}}{2} + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.9\,m)$

And final speed is now calculated after clearing it:

$v_{2} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m-4.9\,m)}$

$v_{2} \approx 6.263\,\frac{m}{s}$

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

5 0

4 years ago

What speed should a satellite with a mass of 1500 kg at 8,500 km above the center of the earth be traveling at in order to stay

hodyreva [135]

Answer:

6844.5 m/s.

Explanation:

To get the speed of the satellite, the centripetal force on it must be enough to change its direction. This therefore means that the centripetal force must be equal to the gravitational force.

Formula for centripetal force is;

F_c = mv²/r

Formula for gravitational force is:

F_g = GmM/r²

Thus;

mv²/r = GmM/r²

m is the mass of the satellite and M is mass of the earth.

Making v the subject, we have;

v = √(GM/r)

We are given;

G = 6.67 × 10^(-11) m/kg²

M = 5.97 × 10^(24) kg

r = 8500 km = 8500000

Thus;

v = √((6.67 × 10^(-11) × (5.97 × 10^(24)) /8500000) = 6844.5 m/s.

7 0

3 years ago

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