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Tasya [4]
3 years ago
8

You run away from a plane mirror at 2.30 m/s. At what speed does your image move away from you?

Physics
1 answer:
ivanzaharov [21]3 years ago
5 0

Answer:

4.60m/s

.............

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A mass weighing 32 pounds stretches a spring 2 feet. Determine the amplitude and period of motion if the mass is initially relea
grandymaker [24]

A mass weighing 32 pounds stretches a spring 2 feet.

(a) Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s.

(b) How many complete cycles will the mass have completed at the end of 4 seconds?

Answer:

A = 1.803 ft

Period = \frac{\pi}{2} seconds

8 cycles

Explanation:

A mass weighing 32 pounds stretches a spring 2 feet;

it implies that the mass (m) = \frac{w}{g}

m= \frac{32}{32}

= 1 slug

Also from Hooke's Law

2 k = 32

k = \frac{32}{2}

k = 16 lb/ft

Using the function:

\frac{d^2x}{dt} = - 16x\\\frac{d^2x}{dt} + 16x =0

x(0) = -1        (because of the initial position being above the equilibrium position)

x(0) = -6          ( as a result of upward velocity)

NOW, we have:

x(t)=c_1cos4t+c_2sin4t\\x^{'}(t) = 4(-c_1sin4t+c_2cos4t)

However;

x(0) = -1 means

-1 =c_1\\c_1 = -1

x(0) =-6 also implies that:

-6 =4(c_2)\\c_2 = - \frac{6}{4}

c_2 = -\frac{3}{2}

Hence, x(t) =-cos4t-\frac{3}{2} sin 4t

A = \sqrt{C_1^2+C_2^2}

A = \sqrt{(-1)^2+(\frac{3}{2})^2 }

A=\sqrt{\frac{13}{4} }

A= \frac{1}{2}\sqrt{13}

A = 1.803 ft

Period can be calculated as follows:

= \frac{2 \pi}{4}

= \frac{\pi}{2} seconds

How many complete cycles will the mass have completed at the end of 4 seconds?

At the end of 4 seconds, we have:

x* \frac{\pi}{2} = 4 \pi

x \pi = 8 \pi

x=8 cycles

5 0
3 years ago
Which body system remove carbon dioxide and waste
antiseptic1488 [7]
The excretory system

7 0
3 years ago
Read 2 more answers
A 15 kg mass is moving at 7.50 meters per second on a horizontal, frictionless surface. What is the total work that must be done
sashaice [31]
Kinetic energy = (1/2) (mass) x (speed)²

At 7.5 m/s, the object's KE is (1/2) (7.5) (7.5)² = 210.9375 joules

At 11.5 m/s, the object's KE is (1/2) (7.5) (11.5)² = 495.9375 joules

The additional energy needed to speed the object up from 7.5 m/s
to 11.5 m/s is (495.9375 - 210.9375) = <em>285 joules</em>.

That energy has to come from somewhere. Without friction, that's exactly
the amount of work that must be done to the object in order to raise its
speed by that much.
8 0
3 years ago
If your speedometer has an uncertainty of 2.5 km/h at a speed of 92 km/h, what is the percent uncertainty
Elis [28]

Answer:

2.7%

Explanation:

Given:

Uncertainty of the speedometer (u)= 2.5km/h

Speed measured at that uncertainty (v) = 92km/h

Percent uncertainty (p) is given as the ratio of the uncertainty to the speed measured then multiplied by 100%. i.e

p = \frac{u}{v} * 100%

p = \frac{2.5}{92} * 100%

p = 2.7%

Therefore, the percent uncertainty is 2.7%

6 0
3 years ago
PLEWSE SOMEONE HELP ILL MARK BRAINLIST
Margaret [11]

Answer:

D)

Explanation:

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Brainliest pls ❤️

3 0
3 years ago
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