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Naily [24]
2 years ago
14

An egg of mass 0.060 kg is dropped from the top of a building. Just before it reaches the ground, it has a total kinetic energy

of 5.9 J. What is the velocity of the egg just before it strikes the ground?
Physics
1 answer:
3241004551 [841]2 years ago
3 0

The velocity is 14 m/s

The parameters given on the question are

mass= 0.060 kg

kinetic energy= 5.9 joules

K.E= 1/2mv²

5.9= 1/2 × 0.060 × v²

5.9= 0.5 × 0.060v²

5.9= 003v²

v²= 5.9/0.03

v²= 196.66

v= √196.66

v= 14 m/s

Hence the velocity of the egg before it strikes the ground is 14 m/s

brainly.com/question/2084569?referrer=searchResults

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2.37eV stopping potential would be required to arrest the current of photoelectrons.

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The minimal negative voltage that must be provided to the anode to halt the photocurrent is known as stopping potential. When expressed in electron volts, the maximal kinetic energy of the electrons is equal to the stopping voltage.

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Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s
Alenkasestr [34]

Answer: 1.289 m

Explanation:

The path the cobra's venom follows since it is spitted until it hits the ground, is described by a parabola. Hence, the equations for parabolic motion (which has two components) can be applied to solve this problem:

<u>x-component: </u>

x=V_{o}cos\theta t  (1)

Where:

x is the horizontal distance traveled by the venom

V_{o}=3.10 m/s is the venom's initial speed

\theta=47\° is the angle

t is the time since the venom is spitted until it hits the ground

<u>y-component: </u>

y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}   (2)

Where:

y_{o}=0.44 m  is the initial height of the venom

y=0  is the final height of the venom (when it finally hits the ground)

g=-9.8m/s^{2}  is the acceleration due gravity

Let's begin with (2) to find the time it takes the complete path:

0=0.44 m+3.10 m/s sin\theta(47\°)+\frac{-9.8m/s^{2} t^{2}}{2}   (3)

Rewritting (3):

-4.9 m/s^{2} t^{2} + 2.267 m/s t + 0.44 m=0   (4)

This is a quadratic equation (also called equation of the second degree) of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a} (5)

Where:

a=-4.9 m/s^{2

b=2.267 m/s

c=0.44 m

Substituting the known values:

t=\frac{-2.267 \pm \sqrt{2.267^{2}-4(-4.9)(0.44)}}{2(-4.9)} (6)

Solving (6) we find the positive result is:

t=0.609 s (7)

Substituting (7) in (1):

x=(3.10 m/s)cos(47\°)(0.609 s)  (8)

We finally find the horizontal distance traveled by the venom:

x=1.289 m  

7 0
3 years ago
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