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garri49 [273]
3 years ago
8

You are driving in rainy weather, on a highway with bushes and trees along the sides of the road. If you are driving too fast an

d your vehicle goes off the road, you should
Physics
2 answers:
maxonik [38]3 years ago
7 0

Answer:

Personally I would pull over and wait the rain out in my car, preferably parked under a tree.

GuDViN [60]3 years ago
7 0

Answer:

when you are driving on the road in rainy weather on a highway with bushes and trees on both the sides and you are moving on the road with high speed and your vehicle goes off the road what you should is, first of all, apply the brake and park on the side of the road and wait for the weather to clear or drive on the road slowly so that vehicle can make better contact with the road  and vehicle don't slip again.

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Suppose electric power is supplied from two independent sources which work with probabilities 0.4, 0.5, respectively. if both so
sattari [20]

Answer:

The answers to the questions are as follows

a)  k = 0, P = 0.3

k = 1, P = 0.5

k = 0, P = 0.2

b) The probability that enough power will be available is 0.5.

Explanation:

To solve the question we write the parameters as follows

Probability that the first power source works = P(A) = 0.4

Probability that the second power source works = P(B) = 0.5

When both sources are supplying power  we have the probability = 1

If non of them is producing the probability = 0

a) The probability that exactly k sources work for k=0,1,2 is given by

For k = 0, probability = (1- P(A))× (1- P(B)) = 0.6 × 0.5 =0.3

Therefore the probabilities that exactly 0 source work  = 0.3

for k = 1 we have the probability = P(A)(1-P(B)) + P(B)(1-P(A)

= 0.4(1-0.5)+0.5(1-0.4) = 0.2 + 0.3 = 0.5

The probabilities that exactly 1 source work  = 0.5

for k = 2 we have the probability given by = P(A) × P(B) = 0.4 × 0.5 = 0.2

Therefore the  probability that exactly 2 sources work  = 0.2

b)  The probability that enough power will be available is

0 × P(k = 0) + 0.6 × P(k = 1) + 1 × P(k = 2)

0 × 0.2 + 0.6 × 0.5 + 1 × 0.2 = 0.5

The probability that enough power will be available is 0.5.

3 0
3 years ago
What is the difference between the equilibrium vector and the resultant vector
Tresset [83]

the difference between a resultant and equilibrant vector is that resultant vector is a direct quantity, one with both magnitude and direction, while the equilibrant vector is a force equal to, but opposite of, the resultant sum of vector forces, that force which balances other forces.

8 0
4 years ago
A block of mass 10 kg slides down an inclined plane that has an angle of 30. If the inclined plane has no friction and the block
Helga [31]
No friction present means: Ek = Ep

So Ek = mgh = 10 * 9.8 * 2 = 196 J
4 0
3 years ago
Read 2 more answers
"My distance to the center of the earth is about 4000 miles when I am on the surface. If I go to a height of 8000 miles above th
lord [1]

Given,

Distance from the surface to the center of the earth, d=4000 miles

Distance from the center to you at a height of 8000 miles, a= 8000+4000=12000 miles

The gravitational force acting on a person at the surface is equal to his weight.

From Newton's Universal Law of Gravitation, the gravitational force is

F=\frac{G\times M\times m}{r^2}

Where G is the gravitational constant, M is the mass of the earth, m is the mass of the object/person, r is the distance between the center of the earth and the object/person

At the surface, this force is equal to the weight of the person, W=mg

i.e.

F_s=\frac{G\times M\times m}{d^2}=W

On substituting the of d,

W=\frac{\text{GMm}}{4000^2}

At a height of 8000 miles from the surface, the gravitational force is equal to,

F_a=\frac{GMm}{12000^2}

On dividing the above two equations,

\frac{F_a}{W}=\frac{4000^2^{}}{12000^2}=\frac{1}{9}

Therefore,

F_a=\frac{1}{9}W

Therefore at a height of 8000 miles above the surface of the earth, the force of gravity becomes 1/9 time your weight.

5 0
1 year ago
The three forces acting on a hot-air balloon that is moving vertically are its weight, the force due to air resistance and the u
irinina [24]

Explanation :

The forces acting on hot- air balloon are:

Weight, (W)

Force due to air resistance, (F)

Upthrust force, (U)

Its weight W is acting in downward direction. The upthrust force U acts in upward direction. When the balloon is moving upward, the air resistance is in downward and vice versa.

In this case, the hot-air balloon descends vertically at constant speed.

so, a=0

and F=ma=0

so, W = F + U ....................(1)

when it is ascending let the weight that it is releasing is R, so

(W-R) + F = U..........(2)

solving equation (1) and (2)

(W-R)+F=W-F

R=2F            

2F is the weight of material that must be released from the balloon so that it ascends vertically at the same constant speed.

7 0
3 years ago
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