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Fed [463]
1 year ago
8

A rocket takes off from Earth's surface, accelerating straight up at 47.2 m/s2. Calculate the normal force (in N) acting on an a

stronaut of mass 80.9 kg, including her space suit. (Assume the rocket's initial motion parallel to the +y-direction. Indicate the direction with the sign of your answer.)
HINT
Physics
1 answer:
lions [1.4K]1 year ago
3 0

Answer:

Approximately 4.61\times 10^{3}\; {\rm N} upwards (assuming that g = 9.81\; {\rm m\cdot s^{-2}}.)

Explanation:

External forces on this astronaut:

  • Weight (gravitational attraction) from the earth (downwards,) and
  • Normal force from the floor (upwards.)

Let (\text{normal force}) denote the magnitude of the normal force on this astronaut from the floor. Since the direction of the normal force is opposite to the direction of the gravitational attraction, the magnitude of the net force on this astronaut would be:

\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}.

Let m denote the mass of this astronaut. The magnitude of the gravitational attraction on this astronaut would be (\text{weight}) = m\, g.

Let a denote the acceleration of this astronaut. The magnitude of the net force on this astronaut would be (\text{net force}) = m\, a.

Rearrange \begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned} to obtain an expression for the magnitude of the normal force on this astronaut:

\begin{aligned}(\text{normal force}) &= (\text{net force}) + (\text{weight}) \\ &= m\, a + m\, g \\ &= m\, (a + g) \\ &= 80.9\; {\rm kg} \times (47.2\; {\rm m\cdot s^{-2}} + 9.81\; {\rm m\cdot s^{-2}}) \\ &\approx 4.61 \times 10^{3}\; {\rm N}\end{aligned}.

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Answer:

The total time is \bf{52850~Yr}.

Explanation:

Given:

The distance between Galactic Capitol and Earth is, d = 2.5 \times 10^{20}~m

The speed of light is, c = 3 \times 10^{8}~m/s

The time taken by the message to reach and come back to Earth is given by

t &=& \dfrac{2d}{c}\\~~&=& \dfrac{2(2.5 \times 10^{20}~m)}{3 \times 10^{8}~m/s}\\~~&=& 1.6 \times 10^{12}~s\\~~&=& \dfrac{1.6 \times 10^{12}~s}{(86400 \times 365)~s}~Yr\\~~&\approx& 52849~Yr

In addition they take 1 year to decide to reply.

Thus, the total time is 25850~Yr.

7 0
3 years ago
you are given two circuits with two batteries of emf e and internal resistance r1 each. circuit a has the batteries connected in
jeka57 [31]

Answer:

In which direction does the current in circuit A flow?

counterclockwise

<h2>What is the power dissipated by the resistor of resistance R2 for circuit A, given that E=10 V, R1=300ohms, and R2=5000ohms? </h2><h2>Calculate the power to two significant figures.</h2><h2>0.064W</h2><h2 /><h2>For what ratio of R1 and R2 would power dissipated by the resistor of resistance R2 be the same for circuit A and circuit B?</h2><h2>R1/R2 = 1 </h2><h2 /><h2>Under which of the following conditions would power dissipated by the resistance R2 in circuit A be bigger than that of circuit B? </h2><h2>Some answer choices overlap; choose the most restrictive answer.</h2><h2>R2>R1</h2><h2> </h2>

Explanation:

5 0
2 years ago
A car is traveling at 40 m/s for 20 seconds. How far did it travel in this time?
barxatty [35]
<h3>Answer:</h3>

800 meters

<h3>Explanation;</h3>

<u>We are given;</u>

  • Speed as 40 m/s
  • Time as 20 seconds

We are required to determine the distance traveled

  • Speed refers to the rate of change in distance.
  • It is given by;

Speed = Distance ÷ time

Rearranging the formula;

Distance = speed × time

In this case;

Distance = 40 m/s × 20 sec

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Thus, the distance traveled by the car is 800 m

7 0
2 years ago
Where would you expect to have more touch receptors: on the palm of your hand or on the back of your hand? Explain your reasonin
anygoal [31]

Answer:

ive answered this

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3 years ago
Point charge μC is located at x =, y = , point charge is located at x = 0m. What are (a)the magnitude and (b)direction of the to
pishuonlain [190]

Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0

Explanation:

  • a) First of all find the distance between the two charges;
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  • r = √( 0.4² + 0.3²)
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hence, the force F = 2Kq1q2cosθ /r²...............equation 1

but cosθ = y/r = 0.3/0.5

cosθ = 0.6

plugging back to equation 1;

F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2

F = 540 x 10^-3

Magnitude of Force = 0.54N

b) Direction is at angle 90

6 0
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