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3 years ago

Discuss how the following factors can affect the stopping

Physics

1 answer:

lys-0071 [83]3 years ago

8 0

Answer:

stopping distance = thinking distance + braking distance

Condition of the car : if the condition is bad it will increase the stopping distance as if the car brakes or tyres are in poor condition <u>the braking distance will increase</u>

Condition of the driver : if he is on drug or alcohol or he is tires or he has distractions then ( basically in bad condition ) . The stopping distance will increase as <u>thinking distance will increase .</u>

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Weather : if it is snowing or there is bad weather conditions then stopping distance will increase as braking distance will increase

Road surface : bad conditions( broken or dust) will increase braking distance so therefore the stopping distance will increase

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A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 73° above the hor

Brrunno [24]

Answer:

7 s

Explanation:

u = Initial velocity of rock = 12.2 m/s

$\theta$ = Angle of throw = $73^{\circ}$

x = Displacement in x direction = 25 m

Displacement in x direction is given by

$x=u\cos\theta t\\\Rightarrow t=\dfrac{x}{u\cos\theta}\\\Rightarrow t=\dfrac{25}{12.2\times \cos73^{\circ}}\\\Rightarrow t=7\ \text{s}$

Time taken to reach the ground is 7 s.

4 0

3 years ago

A jumbo jet must reach a speed of 360 km/h on the runway for takeoff. What is the lowest constant acceleration needed for takeof

weeeeeb [17]

The lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².

To find the answer, we need to know about the Newton's equation of motion.

<h3>What's the Newton's equation of motion to find the acceleration in term of initial velocity, final velocity and distance?</h3>

The Newton's equation of motion that connects velocity, distance and acceleration is V² - U²= 2aS
V= final velocity, U= initial velocity, S= distance and a= acceleration

<h3>What's the acceleration, if the initial velocity, final velocity and distance are 0 m/s, 360km/h and 1.8 km respectively?</h3>

Here, S= 1.8 km or 1800 m, V= 360km/h or 100m/s , U= 0 m/s
So, 100²-0= 2×a×1800

=> 10000= 3600a

=> a= 10000/3600 = 2.8 m/s²

Thus, we can conclude that the lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².

Learn more about the Newton's equation of motion here:

brainly.com/question/8898885

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6 0

2 years ago

A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =

Mila [183]

Answer:

$v=6.65m/sec$

Explanation:

From the Question we are told that:

Mass $m=97.6$

Coefficient of kinetic friction $\mu k=0.555$

Generally the equation for Frictional force is mathematically given by

$F=\mu mg$

$F=0.555*97.6*9.8$

$F=531.388N$

Generally the Newton's equation for Acceleration due to Friction force is mathematically given by

$a_f=-\mu g$

$a_f=-0.555 *9.81$

$a_f=-54455m/sec^2$

Therefore

$v=u-at$

$v=0+5.45*1.22$

$v=6.65m/sec$

4 0

3 years ago

Determine the centripetal force on a vehicle rounding a circular curve with a radius of 80 m at a constant speed of 90 km/h if t

quester [9]

Mass of the vehicle = 2000 kg
Velocity of the vehicle = 90 km/hr
= 25 m/s
Radius of the curve = 80 m
Then
Centripetal force = [2000 * (25)^2]/80
= (2000 * 625)/80
= 15625 kg m/s
= 15625 Newton second
I hope that this is the answer that you were looking for and the answer has come to your desired help.

3 0

3 years ago

You are standing 1 meter from a squawking parrot. If you move to a distance three meters away, the sound

spayn [35]

Use the inverse square law, thus if you move a distance of 3m away, the sound intensity decrease by 1/3^2= 1/9

5 0

4 years ago

Read 2 more answers