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liubo4ka [24]
3 years ago
14

Obviously, we can make rockets to go very fast, but what is a reasonable top speed? assume that a rocket is fired from rest at a

space station in deep space, where gravity is negligible. assume that c = 3.00 108 m/s.
Physics
1 answer:
steposvetlana [31]3 years ago
3 0

 

I’ve answered this problem before and there were 2 parts in this problem.

The solution would be like this for this specific problem:

<span>A.    </span><span>Vf = Vi + Vex*ln(Mi / Mf) </span><span>
<span>0.002 * 3e8m/s = 0 + 2000m/s * ln(Mi / Mf) </span>
<span>300 = ln(Mi / Mf) </span>
<span>1.9e130 = Mi / Mf </span></span>

 

<span>B.    </span><span>4000m/s = 2000m/s * ln(Mi / Mf) </span><span>
<span>2 = ln(Mi / Mf) </span>
<span>7.389 = Mi / Mf </span>
<span>Mf = Mi / 7.389 = 0.135*Mi<span> </span></span></span>

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baseball is hit into the air at an initial speed of 37.2 m/s and an angle of 49.3 ° above the horizontal. At the same time, the
Agata [3.3K]

Answer:

The average speed of the fielder is 5.24 m/s

Explanation:

The position vector of the ball after it was hit can be calculated using the following equation:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem.

When the ball is caught, its position vector will be (see r1 in the figure):

r1 = (r1x, 0.873 m)

Then, using the equation of the position vector written above:

r1x = x0 + v0 · t · cos α

0.873 m = y0 + v0 · t · sin α + 1/2 · g · t²

Since the frame of reference is located at the point where the ball was hit, x0 and y0 = 0. Then:

r1x = v0 · t · cos α

0.873 m = v0 · t · sin α + 1/2 · g · t²

Let´s use the equation of the y-component of r1 to obtain the time of flight of the ball:

0.873 m = 37.2 m/s · t · sin 49.3° - 1/2 · 9.8 m/s² · t²

0 = -0.873 m + 37.2 m/s · t · sin 49.3° - 4.9 m/s² · t²

Solving the quadratic equation:

t = 0.03 s and t = 5.72 s.

It would be impossible to catch the ball immediately after it is hit at t = 0.03 s. Besides, the problem says that the ball was caught on its way down. Then, the time of flight of the ball is 5.72 s.

With this time, we can calculate r1x which is the horizontal distance traveled by the ball from home:

r1x = v0 · t · cos α

r1x = 37.2 m/s · 5.72 s · cos 49.3°

r1x = 1.39 × 10² m

The distance traveled by the fielder is (1.39 × 10² m - 1.09 × 10² m) 30.0 m.

The average velocity is calculated as the traveled distance over time, then:

average velocity = treveled distance / elapsed time

average velocity = 30.0 m / 5.72 s = 5.24 m/s

8 0
3 years ago
A slice of pizza has 500 kcal. If we could burn the pizza and use all the heat to warm a 50-L container of cold water, what woul
topjm [15]

Answer:

increase in temperature of water is 10° C

Explanation:

Given data

pizza = 500 kcal = 500000 calories

cold water = 50 L

to find out

increase in temperature of water

solution

we know heat formula that is

heat = m × specific heat × Δt

here m is mass = 50000 gram and Δt is change in temperature

and specific heat = 1 cal / gram C

so put here all value and find Δt

500000 = 50000 × 1  × Δt

Δt = 10° C

so increase in temperature of water is 10° C

4 0
3 years ago
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NEED HELP!!! ANSWER THESE 5 QUESTIONS FOR 25 POINTS!!!! PLEASE ANSWER!! I WILL GIVE YOU BRAINLIEST
ladessa [460]
1.A
2. C
3. Not Sure
4. Not Sure
5. Biometrics can help to identify
who's at risk for injuries and when
they're able to safely return, and
they can gauge athlete readiness to
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4 0
3 years ago
Read 2 more answers
What does the diagram represent?
solong [7]

Answer:

Atoms in a element on the periodic table I dont know which one tho so try google that sorry I am not much of a help

Explanation:

4 0
3 years ago
In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m
ikadub [295]

(a) -39.4^{\circ}

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

p_y =0

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2

where

m is the mass of each ball

v_1= 4.60 m/s is the velocity of the cue ball after the collision

v_2 = 3.40 m/s is the velocity of the second ball after the collision

\phi_1=28.0^{\circ} is the angle of the cue ball with the x-axis

\phi_2 is the angle of the second ball

Solving for \phi_2, we find the angle between the direction of motion of the second ball and the original direction of motion:

sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m

The initial total momentum along the x-direction as

p_x = m u

where

m is the mass of the cue ball

u is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

mu = 6.69 m\\u = 6.69 m/s

7 0
3 years ago
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