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S_A_V [24]
2 years ago
9

A 540 kg satellite moves through deep space with a speed of 27 m/s. A booster rocket on the satellite fires for 1.4 s, giving a

force to the satellite. The speed of the rocket increases to 31 m/s. Calculate the force given by the booster rocket to the satellite. (Assume that the mass of the satellite remains constant.)
A)1543 N
B)2160 N
C)11,957 N
D)3024 N
Physics
2 answers:
prohojiy [21]2 years ago
6 0

Answer: a

Explanation: because the answer is 1.4444444 and that's the closest

Tema [17]2 years ago
5 0
Answer: A. Hope this helped you
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6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri
mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0
3 years ago
Look at the Grand Canyon cross section. What is the 4th event to take place in the Grand Canyon's geologic history?
mart [117]

Answer:

bass limestone

Explanation:

if I'm right bass limestone is the fourth one because 3 are under it

8 0
2 years ago
An electron is released form rest in a region of space where a uniform electric field is present. Joanna claims that its kinetic
djyliett [7]

Answer:

Neither.

Explanation:

When an electron is released from rest, in an uniform electric field, it will accelerate moving in a direction opposite to the field (as the field has the direction that it would take a positive test charge, and the electron carries a negative charge).

It will move towards a point  with a higher potential, so its kinetic energy will increase, while its potential energy will decrease:

⇒ ΔK + ΔU = 0 ⇒ ΔK = -ΔU = - (-e*ΔV)

As ΔV>0, we conclude that the electric potential energy decreases while the kinetic energy increases in the same proportion, in order to energy be conserved, in absence of non-conservative forces.

4 0
3 years ago
What are examples of chemical energy
klio [65]

-- the big flash of light and heat coming out of the head
of a match when it gets hot enough;

-- the explosion of a tiny bit of gunpowder that can send
a bullet many miles;

-- the energy captured from a few drops of burning gasoline
that moves a car;

-- the energy in the carbohydrates you eat that is used
to move you around;

3 0
3 years ago
determine the loudness (in decibels) of the sound at a rock concert if the intensity of the sound is 1 x 10–1 w/m2. remember, th
EleoNora [17]

The loudness of the sound at the rock concert, where the intensity of the sound is1 x 10⁻¹ Wm⁻² is  110 dB.

Here we are dealing with loudness which is the perception of the Intensity of the sound.

The formula  to refer to in order  to  find the value of the loudness of a sound is ,

  db= 10log(I/I₀)

As we are provided with the current intensity which is  1 x 10⁻¹ Wm⁻². and the initial intensity which is  1 x 10⁻¹² Wm⁻².

So, by substituting the required values in the formula we get

db= 10 * log( 1 x 10⁻¹ /1 x 10⁻¹²)

 = 10 * 11 log(10)

 = 110

So, the result is 110 dB.

To know more about the intensity of sound refer to the link brainly.com/question/9323731?referrer=searchResults.

To know more about questions related to loudness refer to the link brainly.com/question/21094511?referrer=searchResults.

#SPJ4

4 0
1 year ago
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