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photoshop1234 [79]
1 year ago
12

You slide a coin across the floor, and observe that it slows down and eventually stops. A sensitive thermometer shows that its t

emperature increased. What can we conclude? (Select all that apply.)
Physics
1 answer:
dolphi86 [110]1 year ago
6 0

Answer:

Explanation:

Coin slide across the floor and finally comes to stop with an increase in temperature.

We can conclude that the coin interacted with one or more object which reduces its velocity.

  • It is also possible that coin stop due to friction which elevates the coin temperature by converting some part of its energy into thermal energy  
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If Pete ( mass=90.0kg) weights himself and finds that he weighs 30.0 pounds, how far away from the surface of the earth is he
shutvik [7]

Answer: 9938.8 km

Explanation:

1 pound-force = 4.48 N

30.0 pounds-force = 134.4 N

The force of gravitation between Earth and object on the surface of is given by:

F = \frac{GMm}{R^2} = mg

Where M is the mass of the Earth, m is the mass of the object, R (6371 km) is the radius of the Earth.

At height, h above the surface of the Earth, the weight of the object:

(mg)'= \frac{GMm}{(R+h)^2}

we need to find "h"

taking the ratio of two:

\frac{mg}{(mg)'}=\frac{(R+h)^2}{R^2}\\ \Rightarrow \frac{90kg \times 9.8 m/s^2}{134.4 N}=\frac{(R+h)^2}{R^2}\\ \Rightarrow 6.56 R^2= (R+h)^2 \Rightarrow h= (2.56-1)R\\ \Rightarrow h = 1.56 R = 1.56 \times 6371 km = 9938. 8 km

Hence, Pete would weigh 30 pounds at 9938.8 km above the surface of the Earth.

5 0
2 years ago
A student lifts a set of books off a table and places them in the upper shelf of a book case which is 2 meters above the table.
AfilCa [17]
The work done is the product between the intensity of the force applied F, the amount of the displacement d of the book and the cosine of the angle \theta between the direction of the force and the direction of the displacement:
W=Fd \cos \theta
In our problem, the student is lifting the book, so he is applying a force directed upward, and the book is moving upward, so F and d are parallel and therefore the angle is zero, so \cos \theta = \cos 0=1
Therefore, the work done is
W=Fd=(5 N)(2 m)=10 J
6 0
1 year ago
Read 2 more answers
The Lamborghini Huracan has an initial acceleration of 0.85g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode al
SashulF [63]

Answer:

7.9 \frac{m}{s^{2} }

Explanation:

Take the fact that mass is inversely proportional to accelertation:

m ∝ a

Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:

\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\

Using algebra, we can rearrange our equation to find the final acceleration, a_{2}:

a_{2}  = \frac{a_{1}*m_{1}  }{m_{2} } \\

Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:

0.85g * 9.8 \frac{m }{s^{2} } = 8.33 \frac{m }{s^{2} }

Plug everything in:

7.9 \frac{m }{s^{2} } = \frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}

(1590kg the initial weight plus the weight of the added passenger)

8 0
1 year ago
Alex goes cruising on his dirt bike. He rides 700m north, 300m east, 400 m north, 600m west, 1200m south, 300m east and finally
Bess [88]

Find Displacement and Distance

displacement ...

north is 700+400+100 =1200m n

south=1200m

1200-1200=0


east is 300+300=600m

west is 600m

600-600=0

back at dtart. displ zero


distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m  = 3600m


3 0
2 years ago
You create a plot of voltage (in V) vs. time (in s) for an RC circuit as the capacitor is charging, where V=V_{0} \cdot \left(1-
creativ13 [48]

Answer:

The time constant and its uncertainty is t ± Δt = 0.526 ± 0.057 s

Explanation:

If we make a comparison we have to:

y = A*(1-e^-(C*x)) + B

If the time remains constant we have to:

t = R*C = 1/C

In this way we calculate the time constant and its uncertainty. this will be equal to:

t ± Δt = (1/1.901) ± (0.2051/1.901)*(1/1.901) = 0.526 ± 0.057 s

3 0
1 year ago
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