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andrew-mc [135]
3 years ago
8

In a front-end collision, a 1500 kg car with shock-absorbing bumpers can withstand a maximumforce of 80 000 N before damage occu

rs. If the maximum speed for a non-damaging collision is4.0 km/h, by how much must the bumper be able to move relative to the car
Physics
1 answer:
Harrizon [31]3 years ago
4 0

Answer:

The bumper will be able to move by 0.01155m.

Explanation:

The magnitude of deceleration of the car in the front end collision.

a = \frac{F_m}{m} \\

a = \frac{80000}{1500} \\

a = 53.33

This is the deceleration of the car that is generated to stop due to a front end collision.

4 km/h = 1.11 m/s

Now, the initial speed of the bumper in the relation of car, Vi = 0

Now, the initial speed of the bumper in the relation of car, Vf = 1.11 m/s

Use the below equation:

s = \frac{(Intitial \ speed)^2 – (Final \ speed)^2}{2a} \\

s = \frac{(1.11)^2 – (0)}{2 \times 53.33} \\

s = 0.01155 \\

Thus, the bumper can move relative to the car is 0.01155 m .

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const2013 [10]

Answer:

The answer to your question is    vo = 5.43 m/s

Explanation:

Data

distance = d= 5.8 m

height = 3 m

height 2 = 1.7 m

angle = 60°

vo = ?

g = 9.81 m/s²

Formula

              hmax = vo²sinФ/ 2g

Solve for vo²

              vo² = 2ghmax / sinФ

Substitution

              vo² = 2(9.81)(3 - 1.7) / 0.866

Simplification

              vo² = 19.62(1.3) / 0.866

              vo² = 25.51 / 0.866

              vo² = 29.45

Result

              vo = 5.43 m/s

               

5 0
3 years ago
Acceleration is often measured in what unit? A. newtons B. kilograms C. meters per second D. meters per second squared
spayn [35]

D. meters per second squared

3 0
4 years ago
Read 2 more answers
In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a fur
Margaret [11]

Answer:

D = -4/7 = - 0.57

C = 17/7 = 2.43

Explanation:

We have the following two equations:

3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)

First, we isolate C from equation (2):

2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)

using this value of C from equation (3) in equation (1):

3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}

<u>D = - 0.57</u>

Put this value in equation (3), we get:

C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C =  \frac{17}{7}\\

<u>C = 2.43</u>

5 0
3 years ago
Darcy is going to make raspberry jam for the county fair.
valentinak56 [21]

Answer:

She can make have 30 jars with raspberries in them with 50 left over.

Explanation:

1,700 divided by 55

30 equally

but 50 left over

This means that she can make have 30 jars with raspberries in them with 50 left over.

5 0
3 years ago
The pupil of the human eye can vary in diameter from 2.00 mm in bright light to 8.00 mm in dim light. The eye has a focal length
Zarrin [17]

The indicated data are of clear understanding for the development of Airy's theory. In optics this phenomenon is described as an optical phenomenon in which The Light, due to its undulatory nature, tends to diffract when it passes through a circular opening.

The formula used for the radius of the Airy disk is given by,

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Where,

y_r = Range of the radius

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Our values are given by,

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f= 25mm = 25*10^{-3}m

\lambda = 750nm = 750*10^{-9}m

State 2:

d=8.00mm = 8*10^{-3}m

f= 25mm = 25*10^{-3}m

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Replacing in the first equation we have:

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y_{r1}= 11.4\mu m

And also for,

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y_{r2} = 1.49\mu m

Therefor, the airy disk radius ranges from 1.49\mu m to 11.4\mu m

7 0
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