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AnnZ [28]
3 years ago
10

What is the difference in force of a car and a tractor

Physics
1 answer:
vodka [1.7K]3 years ago
6 0

Answer:

See a tractor is more slow but has a greater force and a car is fast but has a slower force

Explanation:

so your answer is Tractors have more force then cars

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A cube with 30-cm-long sides is sitting on the bottom of an aquarium in which the water is one meter deep. (Round your answers t
OleMash [197]

Answer:

A) hydrostatic force on top of cube = 882.9N

B) hydrostatic force on sides of cube = 0N

Explanation:

Detailed explanation and calculation is shown in the image below

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3 years ago
The surface tension of isopropanol in air has a value of 23.00 units and the
Y_Kistochka [10]

Answer:

It's A & C

Explanation:

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7 0
3 years ago
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27b. You're in an airplane that flies horizontally with speed 1000 km/h (280 m/s) when an engine falls off. Ignore air resistanc
jonny [76]

Answer:

8400m

Explanation:

The engine that falls off would have the same constant horizontal velocity as the airplane's when if falls off if we ignore air resistance. So it would have a horizontal velocity of 280m/s for 30seconds before it hits the ground.

Therefor the horizontal distance the engine travels during its fall is

280 * 30 = 8400m

6 0
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katen-ka-za [31]
Power is the rate at which work is done (2nd option)
8 0
3 years ago
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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
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