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Arte-miy333 [17]
3 years ago
13

Please answer this question I don't know how to do it.

Physics
1 answer:
nlexa [21]3 years ago
7 0
One year = 365 days = 8760 hours = 525,600 minutes = 31,536,000 seconds

One light year = speed of light * 31,536,000 seconds

1340 light years = 1340 * one light year
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Average weather conditions in a specific location for a long time
masya89 [10]

Answer:

B- the climate

7 0
3 years ago
Read 2 more answers
A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o
iVinArrow [24]
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: v=15 m/s
- radius of the hill: r=100 m

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car W=mg (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, m \frac{v^2}{r}, so we can write:
mg-N=m \frac{v^2}{r} (1)
By rearranging the equation and substituting the numbers, we find N:
N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
mg=m \frac{v^2}{r}
from which we find
v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s
7 0
4 years ago
The curved section of a speedway is a circular arc having a radius of 190 m. this curve is properly banked for racecars moving a
Anestetic [448]

The banking angle of the curved part of the speedway is determined as 32⁰.

<h3>Banking angle of the curved road</h3>

The banking angle of the curved part of the speedway is calculated as follows;

V(max) = √(rg tanθ)

where;

  • r is radius of the path
  • g is acceleration due to gravity

V² = rg tanθ

tanθ = V²/rg

tanθ = (34²)/(190 x 9.8)

tanθ = 0.62

θ = arc tan(0.62)

θ = 31.8

θ ≈ 32⁰

Learn more about banking angle here: brainly.com/question/8169892

#SPJ1

8 0
2 years ago
In the picture of the atom above, what subatomic particle does the letter A represent?
ra1l [238]

Answer:

Electron

Explanation:

In the picture, the letter A is pointing to an electron.

4 0
3 years ago
A worker assigned to the restoration of the Washington Monument is checking the condition of the stone at the very top of the mo
Mnenie [13.5K]

Answer:

Your answer is: K.E = 8.3 J

Explanation:

If the height (h) = 169.2 meters (m) and the mass (m) is 0.005 kilograms (kg) the total energy will be kinetic energy which is equal to the potential energy.

K.E = P.E and also P.E equals to mgh

Then you substitute all the parameters into the formula  ↓

P.E = 0.005 × 9.81 × 169.2

P.E = 8.2908 J

So your answer is 8.2908 but if you round it is K.E = 8.3

7 0
3 years ago
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