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nexus9112 [7]
3 years ago
14

NEED HELP ASAP! DUE NOW!!

Physics
1 answer:
docker41 [41]3 years ago
4 0
Oh okay the first person who answered is incorrect but


0—> PE=0J

1—>PE=1J

20—>PE=20J

50—>PE=50J

100—>PE=100J
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If an area receives a large amount of insulation is it likely to become warm or cold?
Levart [38]

Answer: heat

Insulation Traps Heat. Keeping the cold air out

Explanation:

5 0
3 years ago
a 70 kg desk sits on the floor motionless If gravity is pulling it with an acceleration of 9.8 m/s/s how much force is gravity e
adell [148]

Since force is mass*acceleration,

F = 70kg * 9.8 m/s

3 0
3 years ago
AM radio signals have frequencies between 550 kHz and 1600 kHz (kilohertz) and travel with a speed of 3.0Ã108m/s. What are the w
Westkost [7]

Answer:

The wavelength of these signals is as follow:

  • Wavelength of 550 kHz is 545.45 m
  • Wavelength of 1600 kHz is 187.5 m

Explanation:

Given that:

Frequency = 550 kHz & 1600 kHz

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As we know that frequency is expressed by the following equation:

  • Frequency = Velocity / Wavelength ---- (1)

For 550 kHz:

The equation can be rearranged as

Wavelength = Velocity / Frequency

Wavelength = (3.0 x 10⁸ m/s) / (550 x 1000 Hz)

Wavelength = 545.45 m

For 1600 kHz:

Wavelength = Velocity / Frequency

Wavelength = (3.0 x 10⁸ m/s) / (1600 x 1000 Hz)

Wavelength = 187.5 m

5 0
3 years ago
If x2 = 60, what is the value of x? plus or minus square root 30 plus or minus square root 60 ±30 ±120
Zielflug [23.3K]
X2 = 60
/ 2 / 2
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4 0
3 years ago
Computers A and B implement the same ISA. Computer A has a clock cycle time of 200 ps and an effective CPI of 1.5 for some progr
lesya692 [45]

Answer:

Computer A is 1.41 times faster than the Computer B

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Assume that number of instruction in the program is 1

Clock time  of computer A is CT_{A} =200 ps

Clock time  of computer B is CT_{B} =250 ps

Effective CPI of computer A is CPI_{A} =1.5

Effective CPI of computer B isCPI_{B} =1.7

CPU time of A is

CPU_{time}=instructions \times CPA_{A} \times CT_{A}\\CPU_{time}=1 \times 1.5 \times 200=300 sec

CPU time of B is

CPU_{time}=instructions \times CPA_{B} \times CT_{B}\\CPU_{time}=1 \times 1.7 \times 250=425 sec

Hence Computer A is Faster by \frac{425}{300} =1.41

Computer A is 1.41 times faster than the Computer B

4 0
3 years ago
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