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3 years ago

A wave transfers _____ from the source outward in the direction of wave motion.

1 answer:

5 0

A wave transfers energy from the source outward in the direction of wave motion.

To add, in physics, a wave is an oscillation accompanied by a transfer of energy that travels through a medium (space or mass). Frequency refers to the addition of time.

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Since you've determined that the power supply is a 700W dual rail, what does that make the maximum output power?

inysia [295]

700 makes the maximum output power.

Explanation:

In physics, power is the rate of doing work or of transferring heat, i.e. the amount of energy transferred or converted per unit time. The output power of a motor is the product of the torque that the motor generates and the angular velocity of its output shaft.

A joule is equal to one Newton-meter, which is the amount of work needed to move a 1 Newton force a distance of 1 meter. When you divide work by time, you get power, measured in units of joules per second. This is also called a Watt. 1 Watt = 1 Joule Sec. This is the formula to calculate output power.

6 0

3 years ago

An electric grinder uses a grinding wheel

luda_lava [24]

(1500 rev/min)(min / 60 s) / (3.0 s) = 8.33 rev/s²

(B) 
(1/2)(8.33 rev/s²)(3.0 s)² = 37.5 rev 

(C) 
(1500 rev/min)(min / 60 s)[2π(0.12 m) / rev] = 18.8 m/s

3 0

3 years ago

Read 2 more answers

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

4 years ago

A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.

aliina [53]

Answer:

9.43 m/s

Explanation:

First of all, we calculate the final kinetic energy of the car.

According to the work-energy theorem, the work done on the car is equal to its change in kinetic energy:

$W=K_f - K_i$

where

W = -36.733 J is the work done on the car (negative because the car is slowing down, so the work is done in the direction opposite to the motion of the car)

$K_f$ is the final kinetic energy

$K_i = 66,120 J$ is the initial kinetic energy

Solving,

$K_f = K_i + W = 66,120 + (-36,733)=29,387 J$

Now we can find the final speed of the car by using the formula for kinetic energy

$K_f = \frac{1}{2}mv^2$

where

m = 661 kg is the mass of the car

v is its final speed

Solving for v, we find

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.43 m/s$

3 0

3 years ago

How do you know that forces are balanced when static friction acts on an object?

lyudmila [28]

By looking at the acceleration of the object.
In fact, Netwon's second law states that the resultant of the forces acting on an object is equal to the product between the mass m of the object and its acceleration:
$\sum F = ma$

So, when static friction is acting on the object, if the object is still not moving we know that all the forces are balanced: in fact, since the object is stationary, its acceleration is zero, and so the resultant of the forces (left term in the formula) must be zero as well (i.e. the forces are balanced).

6 0

3 years ago