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svetoff [14.1K]
3 years ago
9

It is 2058 and you are taking your grandchildren to Mars. At an elevation of 34.7 km above the surface of Mars, your spacecraft

is dropping vertically at a speed of 293 m/s. The spacecraft is to make a soft landing -- that is, at the instant it reaches the surface of Mars, its velocity is zero. Assume the spacecraft undergoes constant acceleration from the elevation of 34.7 km until it reaches the surface of Mars. What is the magnitude of the acceleration
Physics
1 answer:
Paul [167]3 years ago
8 0

Answer: 1.23\ m/s^2

Explanation:

Given

At an elevation of y=34.7\ km, spacecraft is dropping vertically at a speed of u=293\ m/s

Final velocity of the spacecraft is v=0

using equation of motion i.e. v^2-u^2=2as

Insert the values

\Rightarrow 0-(293)^2=2\times a\times (34.7\times 10^3)\\\\\Rightarrow a=-\dfrac{293^2}{2\times 34.7\times 10^3}\\\\\Rightarrow a=-1.23\ m/s^2

Therefore, magnitude of acceleration is 1.23\ m/s^2.

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If at 10m above the ground an object has 50J of Kinetic Energy, with 50J of Potential Energy. How high can the object travel?
steposvetlana [31]

Hi there!

\large\boxed{\text{B) 20 meters}}

We know that:

E_T = U + K

U = Potential Energy (J)

K = Kinetic Energy (J)

E = Total Energy (J)

At 10m, the total amount of energy is equivalent to:

U + K = 50 + 50 = 100 J

To find the highest point the object can travel, K = 0 J and U is at a maximum of 100 J, so:

100J = mgh

We know at 10m U = 50J, so we can solve for mass. Let g = 10 m/s².

50J = 10(10)m

m = 1/2 kg

Now, solve for height given that E = 100 J:

100J = 1/2(10)h

100J = 5h

<u>h = 20 meters</u>

3 0
3 years ago
What is the number of electrons that move past a point in a wire carrying 500 A of current in 4.0 minutes
mr Goodwill [35]
The current is defined as the amount of charge Q that passes through a given point of a wire in a time \Delta t:
I= \frac{Q}{\Delta t}
Since I=500 A and the time interval is
\Delta t=4.0 min=240 s
the charge is
Q=I \Delta t=(500 A)(240 s)=1.2 \cdot 10^5 C

One electron has a charge of q=1.6 \cdot 10^{-19}C, therefore the number of electrons that pass a point in the wire during 4 minutes is
N= \frac{Q}{q}= \frac{1.2 \cdot 10^5 C}{1.6 \cdot 10^{-19}C}=7.5 \cdot 10^{23} electrons
3 0
3 years ago
Define kinetic energy and derive its relation.
IRINA_888 [86]

Answer:

The kinetic energy of a body is the energy that it possessed due to its motion. Kinetic energy can be defined as the work needed to accelerate an object of a given mass from rest to its stated velocity. Kinetic energy depends upon the velocity and the mass of the body.

3 0
3 years ago
How would you go about measuring the speed of a vehicle? What measurements would you have to take? What calculations would you h
White raven [17]

Answer:

For a body moving at a uniform velocity you can calculate the speed by dividing the distance traveled by the amount of time it took, for example one mile in 1/2 hour would give you 2 miles per hour. If the velocity is non-uniform all you can say is what the average speed is.

3 0
3 years ago
A 72.9-kg base runner begins his slide into second base when moving at a speed of 4.02 m/s. The coefficient of friction between
elena-14-01-66 [18.8K]

Answer:

-589.05 J

Explanation:

Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner

So, W = ΔK

W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)

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

W = 1/2m(v₁² - v₀²)

W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)

W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)

W = 1/2 × 72.9 kg(-16.1604 m²/s²)

W = 1/2 × (-1178.09316 kgm²/s²)

W = -589.04658 kgm²/s²

W = -589.047 J

W ≅ -589.05 J

4 0
3 years ago
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