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3 years ago

Which of the following statements is true?

Physics

1 answer:

Free_Kalibri [48]3 years ago

6 0

Answer:

$Changes \: in \: the \: independent \: variable \: cause \\ changes \: in \: the \: dependent \: variable.$

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Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 23.5 m above water wit

Elodia [21]

As stated in the statement, we will apply energy conservation to solve this problem.

From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as

$\Delta KE = \Delta PE$

$\frac{1}{2}mv_f^2-\frac{1}{2} mv_0^2 = mgh_2-mgh_1$

Where,

m = mass

$v_{f,i}$ = initial and final velocity

g = Gravity

h = height

As the mass is tHe same and the final height is zero we have that the expression is now:

$\frac{1}{2}v_f^2-\frac{1}{2} v_0^2 = gh_2$

$\frac{1}{2} (v_f^2-v_0^2) = gh_2$

$(v_f^2-v_0^2) = 2gh_2$

$v_f = \sqrt{2gh_2+v_0^2}$

$v_f = \sqrt{2(9.8)(23.5)+13.6^2}$

$v_f = 25.4m/s$

7 0

3 years ago

If the mass of an object is 44 kilograms and its velocity is 10 meters per second east, how much Kinetic Energy does it have?

olya-2409 [2.1K]

Given: Mass m = 44 Kg; Velocity v = 10 m/s

Required: Kinetic energy K.E = ?

Formula: K.E = 1/2 mv²

K.E 1/2 (44 Kg)(10 m/s)²

K.E = 2,200 Kg.m²/s²

K.E = 2,200 J Answer is A

8 0

3 years ago

The frequency of a wave is 250hz. What is the period

antoniya [11.8K]

Answer:

Hope it helps for you :)))))

8 0

3 years ago

How does the angle of launch affect the kinetic energy of a rubber band?

Lady_Fox [76]

Answer:

The angle of launch of the rubber band affects the initial velocity. The more the rubber band is stretched the more force it applies to return to equilibrium and the more kinetic energy that results in.

3 0

3 years ago

A very long wire generates a magnetic field of 0.0020x 10^-4 T at a distance of 10 mm. What is the magnitude of the current? A)

zimovet [89]

Answer:

2*10^-5

Explanation:

B=IL

I=B/L

I=0.0020*10^-4/10

I=2*10^5

6 0

4 years ago