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kvasek [131]
3 years ago
7

A small rocket to gather weather data is launched straight up. Several seconds into the flight, its velocity is 140 m/s and it i

s accelerating at 20 m/s2 . At this instant, the rocket's mass is 51 kg and it is losing mass at the rate of 0.50 kg/s as it burns fuel. What is the net force on the rocket?
Physics
1 answer:
Free_Kalibri [48]3 years ago
7 0

Explanation:

The force will be 98 N....

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Which of the following are known to exist on the moon? select all that apply.
denis-greek [22]
B. ice and d basalt............
3 0
3 years ago
Read 2 more answers
What can you say about the speed of light in diamond?
Inessa05 [86]
So n=c/v, n= index, c=speed of light and v= speed of light in diamond. 2.42=c/v so v=c/2.42, c≈<span>3x108 m/sec</span><span> so v=</span><span>1.24x108 m/sec</span>.
<span>Hope this helps.</span>
7 0
3 years ago
The current in a lamp is 0.5 ampere when it is plugged into a standard wall outlet. What is the resistance of the lamp when it i
lidiya [134]

The resistance of the lamp plugged in to a standard wall outlet with a current of 0.5 amps is 240 Ω (ohms)

Explanation:

In the United States Of America the standard voltage is 120 v and their frequency is 60 Hz

Standard wall outlet voltage is 120 V

The current in the lamp is 0.5 ampere

Resistance (R) = V/ I

      = 120/0.5

      = 240Ω (ohms)

Thus the resistance of the lamp plugged in to a standard wall outlet with a current of 0.5 amps is 240 Ω (ohms).

8 0
3 years ago
2 m3 of an ideal gas are compressed from 100 kPa to 200 kPa. As a result of the process, the internal energy of the gas increase
rosijanka [135]

Answer:

work done is -150 kJ

Explanation:

given data

volume v1 = 2 m³

pressure p1 = 100 kPa

pressure p2 = 200 kPa

internal energy = 10 kJ

heat is transferred  = 150 kJ

solution

we know from 1st law of thermodynamic is

Q = du +W    ............1

put here value and we get

-140 = 10 + W

W = -150 kJ

as here work done is -ve so we can say work is being done on system

3 0
3 years ago
A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a
il63 [147K]

Answer:

0.0000109261200583 s

0.0109261200583

Explanation:

d_2 = Distance from right ear = 3 m

s = Distance between ears = 15 cm

v = Speed of sound in air = 343 m/s

Distance between the left ear and the bird

d_1=\sqrt{s^2+d_2^2}\\\Rightarrow d_1=\sqrt{0.15^2+3^2}\\\Rightarrow d_1=3.00374765918\ m=3.004\ m

Time

t=\dfrac{Distance}{Speed}

Time difference would be

\Delta T=\dfrac{d_1}{v}-\dfrac{d_2}{v}\\\Rightarrow \Delta T=\dfrac{3.00374765918}{343}-\dfrac{3}{343}\\\Rightarrow \Delta T=0.0000109261200583\ s

The time difference is 0.0000109261200583 s

Time period is given by

T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{1000}\\\Rightarrow T=10^{-3}\ s

The ratio is

\dfrac{\Delta T}{T}=\dfrac{0.0000109261200583}{10^{-3}}\\\Rightarrow \dfrac{\Delta T}{T}=0.0109261200583

The ratio is 0.0109261200583

7 0
3 years ago
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