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Ahat [919]
2 years ago
14

You found the distance that you will cover while braking. You can use the expected amount of time for braking to check your answ

er for distance.
Using the information given in the introduction and the value for distance that you found above, determine the time required for you to stop once you press the brakes. Note that the time for braking is just the time interval t3-t2.
This is the information given in the introduction:_______.
You are driving down the highway late one night at 20 m/swhen a deer steps onto the road 35m in front of you. Your reaction time before stepping on the brakes is 0.50s, and the maximum deceleration of your car is 10m/s2. How much distance is between you and the deer when you come to a stop?
Physics
1 answer:
katrin [286]2 years ago
3 0

Answer:

Explanation:

First we need to determine the distance covered during deceleration. According to the equation of motion.

S = ut+1/2at²

Given:

u = 20m/s

t = 0.50s

a = -10m/s (deceleration is negative acceleration)

S = 20²+1/2(-10)(0.5)²

S = 400-5(0.5)²

S = 400-5(0.25)

S = 400-1.25

S = 398.75m

If the deer steps onto the road 35m in front of you, the distance between you and the deer when you come to a stop will be 398.75-35 = 363.75m

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- vocational aim
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6 0
2 years ago
A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh
alekssr [168]

(a) 392 N/m

Hook's law states that:

F=k\Delta x (1)

where

F is the force exerted on the spring

k is the spring constant

\Delta x is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N

- The stretching of the spring is

\Delta x=15 cm-10 cm=5 cm=0.05 m

Solving eq.(1) for k, we find the spring constant:

k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N

And so, the stretch of the spring is

\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm

And since the initial lenght of the spring is

x_0 = 10 cm

The final length will be

x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm

8 0
3 years ago
Read 2 more answers
A horizontal force of 12N is applied to 1.5kg of block which rests on a horizontal surface. If the coefficient friction is 0.3,
Rama09 [41]

Answer:

7 m/s²

Explanation:

From the question given above, the following data were:

Force applied (Fₐ) = 15 N

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Acceleration (a) of block =?

Next, we shall determine the frictional force. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Normal reaction (R) = mg = 1.5 × 10 = 15 N

Frictional force (Fբ) =?

Fբ = μR

Fբ = 0.3 × 15

Fբ = 4.5 N

Next, we shall determine the net force acting on the block. This can be obtained as follow:

Force applied (Fₐ) = 15 N

Frictional force (Fբ) = 4.5 N

Net force (Fₙ) =.?

Fₙ = Fₐ – Fբ

Fₙ = 15 – 4.5

Fₙ = 10.5 N

Finally, we shall determine the acceleration produced in the block. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Net force (Fₙ) = 10.5 N

Acceleration (a) of block =?

Fₙ = ma

10.5 = 1.5 × a

Divide both side by a

a = 10.5 / 1.5

a = 7 m/s²

Therefore, the acceleration produced in the block is 7 m/s²

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3 years ago
True or false? Magnetic reversals are recorded in the newly formed oceanic crust on BOTH sides of a mid-ocean ridge spreading ce
andrey2020 [161]

answer: false is my answer for thi question

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3 years ago
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ANSWER - (1) are constantly moving (2) have volume (3) have intermolecular forces (4) undergo perfectly elastic collisions (5) have an average kinetic energy proportional to the ideal gas’s absolute temperature
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