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rodikova [14]
3 years ago
8

A high-performance sports car can go from 0 to 100 mph in 5.8 s. (Assume the car travels in the positive direction. Indicate the

direction with the sign of your answer.) (a) What is the car's average acceleration (in SI units; meter-kilogram-second)The same car can come to a complete stop from 34 m/s in 5.1 s. What is its average acceleration?
Physics
1 answer:
sasho [114]3 years ago
7 0

Answer: a) 7.71 m/s², b) - 6.67 m/s²

Explanation: first thing to note is that

1 mile = 1609.34

1 hour = 3600s

Hence, 100mph to m/s = (100 × 1609.34)/3600 = 44.71 m/s

Initial velocity (u) = 0, final velocity (v) = 44.71 m/s, t = 5.8s, a = acceleration =?

By using newton's laws of motion

v = u + at

44.71 = 0 + a(5.8)

44.71 = 5.8a

a = 44.71/5.8

a = 7.71 m/s²

Question b)

The car is completing a stop which implies that the car is coming to rest, and when a car is coming to rest, the final velocity (v) is zero.

Hence u = 34 m/s, v = 0, a =?, t = 5.1 s

v = u + at

0 = 34 + a(5.1)

a(5.1) = - 34

a = - 34/5.1

a = - 6.67 m/s².

The negative sign beside the acceleration shows that the body is decelerating

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. If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect tim
inn [45]

Answer:

The appropriate response will be "Length must be increased by 0.012%".

Explanation:

The given values is:

ΔT = 5 s/day

Now,

⇒ \frac{\Delta T}{T} =\frac{5}{24\times 60\times 60}

On multiplying both sides by  "100", we get

⇒ \frac{\Delta T}{T}\times 100 =\frac{500}{24\times 60\times 60}

⇒ \frac{\Delta T}{T}\times 100=0.005787 (%)

∵  T=2\pi\sqrt{\frac{l}{g} }

On substituting the values, we get

⇒ \frac{\Delta T}{T}% = \frac{1}{2}\times \frac{\Delta l}{l}%

On applying cross multiplication, we get

⇒ \frac{\Delta l}{l}% = 2\times \frac{\Delta T}{T}%

⇒        = 2\times 0.05787

⇒        = 0.011574

⇒        = 0.012%

6 0
3 years ago
At the outer edge of a rotating space habitat, 130 m from the center, the rotational acceleration is g. What is the rotational a
enyata [817]

Answer:

Explanation:

Given:

R1 = 130 m

R2 = 65 m

w^2R = g

Assume, g = 9.81 m/s^2

w^2 = 9.81/130

w = 0.275 rad/s

At R2 = 65 m

g = w^2R

= (0.275^2) × 65

= 4.905 m/s^2

In conclusion,

g × R = k

g1/R1 = g2/R2

g2 = (g1 × 65)/130

= g1 ×1/2

= g1/2

6 0
3 years ago
A 16 N force is applied to an object and 96 J of work is done. How far was the object moved?
ipn [44]
Your answer would be A. You divide 96 by 16 to find the answer
3 0
3 years ago
Read 2 more answers
The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz. What is the period of the crystal's motion?
Elan Coil [88]

Answer:

Time period, T=3.05\times 10^{-5}\ s

Explanation:

Given that,

The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz, f = 32768 Hz

We need to find the period of the crystal's motion. The relationship between the frequency and the time period is given by :

T=\dfrac{1}{f}

T is the time period of the crystal's motion.

Time period is given by :

T=\dfrac{1}{32768}

T=3.05\times 10^{-5}\ s

So, the time period of the crystal's motion is 3.05\times 10^{-5}\ s. Hence, this is the required solution.

8 0
3 years ago
A 0.700-kg ball is on the end of a rope that is 2.30 m in length. The ball and rope are attached to a pole and the entire appara
otez555 [7]

Answer:

The tangential speed of the ball is 11.213 m/s

Explanation:

The radius is equal:

r=2.3*sin70=2.161m (ball rotates in a circle)

If the system is in equilibrium, the tension is:

Tcos70=mg\\Tsin70=\frac{mv^{2} }{r}

Replacing:

\frac{mg}{cos70} sin70=\frac{mv^{2} }{r} \\Clearing-v:\\v=\sqrt{rgtan70}

Replacing:

v=\sqrt{2.161x^{2}*9.8*tan70 } =11.213m/s

7 0
3 years ago
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