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MrRa [10]
2 years ago
13

the shock absorbers in a car act as a big spring with k= 21900 N/m. when a 92.5 kg person gets in, how far does the spring stret

ch?
Physics
1 answer:
r-ruslan [8.4K]2 years ago
4 0

Answer: 0.04139m

Explanation:

First, we need to calculate the weight of the man which will be:

Weight = mass × acceleration due to gravity

Weight = mg

Weight = 92.5 × 9.8

Weight = 906.5N

Then, we calculate the force which will be:

F = kx

mg = kx

x = mg/k

x = 906.5/21900

x = 0.04139m.

The spring stretched for 0.04139m.

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2 years ago
The length of a wire 2.00 m is measured as 2.02m. What is the percentage error in the measurement?
n200080 [17]

Answer:

1%

Explanation:

Percent error can be found by dividing the absolute error (difference between measure and actual value) by the actual value, then multiplying by 100.

Percent Error=\frac{V_{measured}- V_{true} } {V_{true}} *100

The measured value is 2.02 meters and the actual value is 2.00 meters.

V_{measured}=2.02\\\\V_{true}=2.00

Percent Error=\frac{2.02-2.00}{2.00} *100

First, evaluate the fraction. Subtract 2.00 from 2.02

Percent Error=\frac{0.02}{2.00}*100

Next, divide 0.02 by 2.00

PercentError=0.01 *100

Finally, multiply 0.01 and 100.

Percent  Error=1\\Percent  Error= 1 \%

The percent error is 1%.

6 0
3 years ago
In a historical movie, two knights on horseback start from rest 86 m apart and ride directly toward each other to do battle. Sir
Harlamova29_29 [7]

Answer:

Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

Explanation:

Let the distance covered by Sir George be S_{1}

and the distance covered by Sir Alfred be S_{2}

Since the knights collide, hence they must have traveled for the same amount of time just before collision

From one of the equations of motion for linear motion

S = ut + \frac{1}{2}at^{2}

Where S is the distance traveled

u is the initial velocity

a is the acceleration

and t is the time

For Sir George,

S = S_{1}

u = 0 m/s (Since they start from rest)

a =0.21 m/s²

Hence,

S = ut + \frac{1}{2}at^{2} becomes

S_{1}  = (0)t + \frac{1}{2}(0.21)t^{2}\\S_{1}  = 0.105 t^{2}\\

t^{2} = \frac{S_{1}}{0.105}

Now, for Sir Alfred

S = S_{2}

u = 0 m/s (Since they start from rest)

a =0.26 m/s²

Hence,

S = ut + \frac{1}{2}at^{2} becomes

S_{2}  = (0)t + \frac{1}{2}(0.26)t^{2}\\S_{2}  = 0.13 t^{2}\\

t^{2} = \frac{S_{2}}{0.13}

Since, they traveled for the same time, t just before collision, we can write

\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that

S_{1} + S_{2} = 86 m

Then, S_{2} = 86 - S_{1}

Then,

\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13} becomes

\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}

0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}

S_{1} = 38.43 m

∴ Sir George covered a distance of 38.43 m just before collision.

Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

6 0
2 years ago
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