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NeX [460]
3 years ago
14

When is kinetic energy transferred from one object to another? Question 24 options:

Physics
2 answers:
postnew [5]3 years ago
7 0
When two objects collide is the answer...Hope this helps :)
Alex3 years ago
3 0
Its B when two objects collide

Hope this helps :)
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An astronaut on a distant planet wants to determine its acceleration due to gravity. the astronaut throws a rock straight up wit
Tamiku [17]
Define
u = 16 m/s, the vertical launch velocity
g = acceleration due to gravity, measured positive downward
s = vertical distance traveled
t  = 21.2 s, total time of travel.

The vertical motion obeys the equation
s = ut - (1/2)gt²

When the rock is at ground level, s = 0.
Therefore
(16 m/s)(21.2 s) - 0.5*(g m/s²)*(21.2 s)² = 0
339.2 - 224.72g = 0
g = 1.5094 m/s²

Answer:
The acceleration due to gravity is 1.509 m/s² measured positive downward.


7 0
3 years ago
a car accelerates along a straight road from rest to 75km/h in 5.0s. What is the magnitude of its average acceleration? show up
kykrilka [37]

Answer:

4.186 m/s^2

Explanation:

First, convert km/hr to m/s:

(75 km/hr)(1000m/1km)(1hr/60min)(1min/60s) = 20.83 m/s

Then, divide 20.93 m/s by 5.0s

(20.93 m/s) / (5.0s) = 4.186 m/s^2

5 0
2 years ago
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
An unlabeled hierarchical diagram of various astronomical bodies is shown below. The labels A, B, C, and D can be used to repres
Paha777 [63]
A) The biggest astronomical object is the Universe, which contains billions of galaxies among which there is the Milky Way.
The Milky Way contains thousands of planetary systems, among which the Solar System.
The Solar System contains many <span>planets <span>(but only one star, the Sun)</span>,</span> among which there is Earth.
Therefore you can label:
A = Universe, B = Milky Way, C = Solar system, D = Earth

b) Given what we said before, you could label D also any other planet in the Solar System, therefore you can choose among Mercury, Venus, Mars, Jupiter, Saturn, Uranus, and Neptune.
8 0
3 years ago
What is the net charge of a metal ball if there are 21,749 extra electrons in it?
pickupchik [31]

Answer:

Q=3.47\times 10^{-15}\ C

Explanation:

Given that,

Number of extra electrons, n = 21749

We need to find the net charge on the metal ball. Let Q is the net charge.

We know that the charge on an electron is q=1.6\times 10^{-19}\ C

To find the net charge if there are n number of extra electrons is :

Q = n × q

Q=21749\times 1.6\times 10^{-19}\ C

Q=3.47\times 10^{-15}\ C

So, the net charge on the metal ball is 3.47\times 10^{-15}\ C. Hence, this is the required solution.

6 0
3 years ago
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