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Nookie1986 [14]
2 years ago
6

Most of the countries in Europe use SI measurments. How could this be a problem if you went on a trip to Europe?

Physics
2 answers:
Illusion [34]2 years ago
8 0
The answer is because we use different measurements than any other country (assuming you live int he U.S) so you would become easlily confused

faust18 [17]2 years ago
5 0
Not knowing how to convert the type of measurement according to your way of learning from where ever you come from.
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A cannon is fired with an initial horizontal velocity of 20 m/s and initial vertical velocity of 25 m/s. After 3s in the air, th
Vladimir [108]

Answer:

60 m

Explanation:  

After 3 seconds of travel at 20 m/s, the projectile is 3·20 = 60 meters horizontally from the cannon.

__

The vertical height after 3 seconds is 0.9 m, so the straight-line distance from cannon to target is √(60^2 +0.9^2) ≈ 60.007 meters.

7 0
2 years ago
Explain relative velocity.
xxTIMURxx [149]

Answer:

The relative velocity of an object A with respect to another object B.

Explanation:

The relative velocity of an object A with respect to another object B is the velocity that object A would appear to have to an observer situated on object B moving along with it.

4 0
3 years ago
Evaluate the scenarios and select the one that demonstrates the training principle overload.
matrenka [14]

Answer:

B :)

Explanation:

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5 0
2 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
3 years ago
Alan rents a private jet for the weekend. He flies 400 km south to New York, then flies 700 km west to Chicago, then 1200 km to
Ratling [72]

Answer:

Total distance = 400+700+1200= 2300km

Explanation:

the resultant of d 1st right angle triangle + 1200

= 806.2 + 1200 = 2006.2km

7 0
3 years ago
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