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Sergio039 [100]
3 years ago
12

The available source of charge that pushes a charge through a circuit is

Physics
1 answer:
sergey [27]3 years ago
8 0
     Voltage, because the voltage applied to a resistor generates a current, which is the load flow. 

Letter C

If you notice any mistake in my english, please let me know, because i am not native.
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Baseball reporters say that long fly balls that would have carried for home runs in July "die" in the cool air of Octo- ber and
anastassius [24]

Answer:

Consider the followig calculation

Explanation:

a)  use deal equation:

PV = nRT                      

ρ = m/V,= ==> V = m/ρ

therefore,

ρ = Pm/RT

convert 95 oF in degree

95 oF = 308.15 K

1 atm = 1.013 * 105 pascal

ρ = 1.013*105 * 29 * /8.314 * 308.15

                   = 1.146 kg/m3

b)  again use ideal gas equation:

ρ = Pm/RT

T = 50 oF = 283.15 K

1 atm = 1.013 * 105 pascal

molar mass will be same

ρ = 1.013 * 105 * 29 / 8.314 * 283.15

ρ = 1.248 kg / m3

So,

c) . more than density of the hot, dry air computed in part (a)

7 0
3 years ago
How do we use energy transformation in our daily lives?
olga55 [171]

Answer:hat are some examples of energy transformation?

The Sun transforms nuclear energy into heat and light energy.

Our bodies convert chemical energy in our food into mechanical energy for us to move.

An electric fan transforms electrical energy into kinetic energy.

Explanation:

4 0
3 years ago
Read 2 more answers
A truck is traveling 70.0 kph away from you. The driver is blowing the horn, which has a frequency of 400 Hz. The air temperatur
mr_godi [17]
There are several information's already given in the question. Based on those information's the answer can be easily deduced.
Speed of the truck = 70 kph
                              = 19.44 m/s
Speed of sound = 343 m/s
Then
Observed frequency = 400 Hz ∗ 343 m/s/<span>19.44m/s + 343 m/<span>s
                                 = 378.54
                                 = 379 Hz</span></span>
4 0
4 years ago
A vertical spring (spring constant =160 N/m) is mounted on the floor. A 0.340-kg block is placed on top of the spring and pushed
AleksandrR [38]

(a) 3.5 Hz

The angular frequency in a spring-mass system is given by

\omega=\sqrt{\frac{k}{m}}

where

k is the spring constant

m is the mass

Here in this problem we have

k = 160 N/m

m = 0.340 kg

So the angular frequency is

\omega=\sqrt{\frac{160 N/m}{0.340 kg}}=21.7 rad/s

And the frequency of the motion instead is given by:

f=\frac{\omega}{2\pi}=\frac{21.7 rad/s}{2\pi}=3.5 Hz

(b) 0.021 m

The block is oscillating up and down together with the upper end of the spring. The block will lose contact with the spring when the direction of motion of the spring changes: this occurs when the spring is at maximum displacement, so at

x = A

where A is the amplitude of the motion.

The maximum displacement is given by Hook's law:

F=kA

where

F is the force applied initially to the spring, so it is equal to the weight of the block:

F=mg=(0.340 kg)(9.81 m/s^2)=3.34 N

k = 160 N/m is the spring constant

Solving for A, we find

A=\frac{F}{k}=\frac{3.34 N}{160 N/m}=0.021 m

3 0
3 years ago
Complete the following table. Be sure to include units in your answer.
mel-nik [20]

Answer:

Complete the following table. Be sure to include units in your answer.

Explanation:

all you have to do is mautiply them

6 0
3 years ago
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