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mars1129 [50]
4 years ago
5

On a road trip, a driver achieved an average speed of (48.0+A) km/h for the first 86.0 km and an average speed of (43.0-B) km/h

for the remaining 54.0 km. What was her average speed (in km/h) for the entire trip? Round your final answer to three significant figures.
Physics
1 answer:
tensa zangetsu [6.8K]4 years ago
6 0

Answer:

45.94 km/h

Explanation:

Let the average speed be 48km/h for the first 86 km and 43 km/h for the remaining 54 km. We can calculate the time it takes to travel each of the segment:

For the first 86 km segment:

t_1 = s_1/v_1 = 86 / 48 = 1.79 hours

For the remaining 54 km segment:

t_2 = s_2/v_2 = 54 / 43 = 1.256 hours

So the total times it takes is

t = t_1 + t_2 = 1.79 + 1.256 = 3.047 hours

And the total distance she travels:

s = s_1 + s_2 = 86 + 54 = 140 km

The average speed for the entire trip would be total distance divided by total time:

v = s / t = 140 / 3.047 = 45.94 km/s

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If the value of static friction for a couch is 400N, what could be a possible value of its kinetic friction? *
Dafna11 [192]

Answer:

kinetic friction may be greater than 400 N or smaller than 400 N

Explanation:

As we know that maximum value of static friction on the rough surface is known as limiting friction and the formula of this limiting friction is known as

F_s = \mu_s N

now when object is sliding on the rough surface then the friction force on that surface is known as kinetic friction and the formula of kinetic friction is known as

F_k = \mu_k N

now we know that

\mu_k < \mu_s

so here value of limiting static friction force is always more than kinetic friction

also we know that

initially when body is at rest then static friction value will lie from 0 N to maximum limiting friction

and hence kinetic friction may be greater than static friction or if the static friction is maximum limiting friction then kinetic friction is smaller than static friction

so kinetic friction may be greater than 400 N or smaller than 400 N

5 0
4 years ago
A baseball of mass m1 = 0.45 kg is thrown at a concrete block m2 = 7.25 kg. The block has a coefficient of static friction of μs
miskamm [114]

Answer:

a. v = \frac{mu_sm_2g\Delta t}{m_1}

b. 21.64 m/s

Explanation:

Let g = 9.81m/s2

a. The weight of the block is product of its mass and gravitational acceleration

W = m_2g = 7.25*9.81 = 71.1225N

which is also the normal force acting on the block from the floor so it stays balanced.

N = 71.225N

The static friction of the block is product of its normal force from the floor and the friction coefficient

F_s = \mu_sN = \mu_sW = mu_sm_2g

For the block to move, the force generated by the impact must be at least equal to the static friction.

F = F_s = mu_sm_2g

The impulse is product of this force and time duration of impact.

I = F\Delta t = mu_sm_2g\Delta t

As impulse is generated by change in momentum of the ball, which is product of its mass and velocity v

I = \Delta p = m_1\Delta v

mu_sm_2g\Delta t = m_1 v

v = \frac{mu_sm_2g\Delta t}{m_1}

b. v = \frac{mu_sm_2g\Delta t}{m_1} = \frac{0.74*7.25*9.81*0.185}{0.45} = 21.64 m/s

8 0
3 years ago
Which trait would most likely lead a scientist to repeat an experiment and verify the data?
-BARSIC- [3]
If the results of the experiment differ; so u repeat the experiment to notice a pattern.
6 0
3 years ago
Read 2 more answers
A 63.0kg sprinter starts a race with an acceleration of 24.0m/s/s . what is the net force of him
shusha [124]

Answer:

1512N

Explanation:

f = ma

here, m=63kg

a=24ms^-²

so, f=ma=63×24=1512N (Newton)

8 0
3 years ago
1134567 A jet with mass m = 1.1 Ă— 105 kg jet accelerates down the runway for takeoff at 2 m/s2. 1) What is the net horizontal f
Kipish [7]

Initially jet is moving horizontally

so we know that

m = 1.1 \times 10^5 kg

a = 2 m/s^2

now we have

PART a)

Net horizontal force is given as

F_x = ma_x

F_x = (1.1\times 10^5)(2)

F_x = 2.2 \times 10^5 N

Part b)

since the acceleration in vertical direction is zero

so as per Newton's law we know

F = ma

F = 0 N

PART c)

speed of the jet increases in vertical direction from ZERO to 21 m/s in 20 seconds and speed increases in horizontal direction from 95 m/s from 80 m/s in same time

vertical acceleration

a_y = \frac{21 - 0}{20} = 1.05 m/s^2

horizontal acceleration

a_x = \frac{95 - 80}{20} = 0.75 m/s^2

now net horizontal force will be

F_x = ma_x

F_x = (1.1 \times 10^5)(0.75) = 8.25 \times 10^4 N

PART d)

Now for vertical force we have

F_y = ma_y

F_y = (1.1 \times 10^5)(1.05) = 1.155 \times 10^5 N

PART e)

Now after reaching the cruising level the horizontal speed becomes constant and in vertical direction speed decrease to ZERO from 21 m/s in 13 s

So we have

a_x = 0

a_y = \frac{0 - 21}{13} = -1.62 m/s^2

now we have horizontal force as

F_x = 0

Part f)

F_y = ma_y

F_y = (1.1 \times 10^5)(-1.62) = - 1.78 \times 10^5 N

7 0
4 years ago
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