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Alla [95]
3 years ago
8

The coefficients of friction between a race cars tyres and the track surface are

Physics
1 answer:
Luba_88 [7]3 years ago
5 0

the question is about tyres of a race car, which are made of rubber and will be in contact with a race track, which is generally made from asphalt, the static coefficient of friction is in the range of (0.5–0.8), in dry conditions (Source: Friction and Friction Coefficients ).

Explanation:

please mark me as a brainlieast

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The boom of a fire truck raises a fireman (and his equipment – total weight 280 lbf) 60 ft into the air to fight a building fire
Katen [24]

Answer:

a)Work done by fireman=   2.15  Btu

b) Time t= 0.86 sec

Explanation:

Given that

Weight = 280 lbf

We know that 1 lbf = 4.44 N

so 280 lbf = 1245.5 N

Weight =1245.5 N

Height h = 60 ft

We know that

1 ft = 0.3048 m

So 60 ft = 18.28 m

 h =18.28 m

Power = 3.5 hp

We know that

1 hp =0.74 KW

So 3.5 hp = 2.61 KW

Power = 2.61 KJ/s

So the work done by fireman = Weight x h

Now by putting the values

Work done by fireman= 1245.5 x 18.28 J

Work done by fireman=   2267.74 J

Work done by fireman=   2.26774  KJ

We know that 1 Btu= 1.05 KJ

So   2.266 KJ = 2.15 Btu

Work done by fireman=   2.15  Btu

We know that ,rate of work is called power.

Power x time =  work

2.61 x t = 2.26

So t= 0.86 sec

6 0
3 years ago
The starship Enterprise approaches the planet Risa at a speed of 0.8c relative to the planet. On the way, it overtakes the inter
Alchen [17]

Answer:

<em>0.3c</em>

<em></em>

Explanation:

The speed of Enterprise relative to Risa is 0.8c

Relative speed of both ships as measured from Enterprise is 0.5c

therefore, relative speed of Astra to Enterprise is 0.8c - 0.5c = <em>0.3c</em>

<em>this is also the relative speed with which Astra approaches the planet Risa since Enterprise's speed was calculated relative to Risa.</em>

8 0
3 years ago
A train reaches a speed of 35.0 m/s after accelerating at a rate of 5.00 m/s2 over a distance of 40.0 m. What was the train’s in
MatroZZZ [7]

Answer:

Initial velocity, U = 28.73m/s

Explanation:

Given the following data;

Final velocity, V = 35m/s

Acceleration, a = 5m/s²

Distance, S = 40m

To find the initial velocity (U), we would use the third equation of motion.

V² = U² + 2aS

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement measured in meters.

Substituting into the equation, we have;

35² = U + 2*5*40

1225 = U² + 400

U² = 1225 - 400

U² = 825

Taking the square root of both sides, we have;

Initial velocity, U = 28.73m/s

5 0
3 years ago
Maintenance __________ are needed in great numbers to service all sorts of technical equipment.
damaskus [11]
Maintenance spanner are needed in great numbers to service all sorts of technical equipment
5 0
3 years ago
A projectile is fired horizontally from a height of 78.4 m at a speed of 300 m/sec. How far did it travel horizontally before hi
Mice21 [21]

Answer:

Explanation:

Using the formula for calculating range expressed as;

R = U√2H/g

U is the speed = 300m/s

H is the maximum height = 78.4m

g is the acceleration due to gravity = 9.8m/s²

Substitute into the fromula;

R = 300√2(78.4)/9.8

R = 300 √(16)

R = 300*4

R = 1200m

Hence the projectile travelled 1200m before hitting the ground

3 0
3 years ago
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