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ozzi
3 years ago
10

8. The copper rod expands when its temperature

Physics
1 answer:
olchik [2.2K]3 years ago
4 0

Answer:

c

Explanation:

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Although all sports require a person to be in shape to some degree, some sports demand more from the body. Which sport does not
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I  think golf does not demand high level of physical fitness. A golfer just need be skillful and able to play under pressure. In terms of physical exercise i think he/she just need to do some muscle stretches.

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What causes reactants to gain potential energy?
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Give two examples of Newton's second law of motion
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Falling objects drop with an average acceleration of 9.8 m/s2. An arrow is shot with a velocity of 11.76 m/s straight down from
In-s [12.5K]

Answer:

3.8 secs

Explanation:

Parameters given:

Acceleration due to gravity, g = 9.8 m/s^2

Initial velocity, u = 11.76 m/s

Final velocity, v = 49 m/s

Using one of Newton's equations of linear motion, we have that:

v = u + gt

where t = time of flight of arrow

The sign is positive because the arrow is moving downward, in the same direction as gravitational force.

Therefore:

49 = 11.76 + 9.8*t\\\\\\\49 - 11.76 = 9.8t\\\\=> 9.8t = 37.24\\\\\\t = \frac{37.24}{9.8} \\\\\\t = 3.8 secs

The arrow was in flight for 3.8 secs

6 0
4 years ago
Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s
Nezavi [6.7K]

Answer: 1.124 m

Explanation:

This situation is a good example of the projectile motion or parabolic motion, and the main equations that will be helpful in this situations are:  

x-component:  

x=V_{o}cos\theta t   (1)  

Where:  

V_{o}=2.80 m/s is the initial speed  

\theta=39\° is the angle at which the venom was shot

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

y-component:  

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

Where:  

y_{o}=0.4 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

Knowing this, let's begin:

First we have to find t from (2):

0=0.4 m+2.8 m/s sin(39\°) t+\frac{-9.8m/s^{2}t^{2}}{2}   (3)

Rearranging (3):  

-4.9 m/s^{2} t^{2} + 1.762 m/s t + 0.4 m=0   (4)  

This is a <u>quadratic equation</u> (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  

b=1.762  

c=0.4  

Substituting the known values:  

t=\frac{-1.762 \pm \sqrt{(1.762)^{2} - 4(-4.9)(0.4)}}{2(-4.9)} (6)  

Solving (6) we find the positive result is:  

t=0.517 s (7)

Substituting (7) in (1):

x=2.8 m/s cos(39\°) (0.517 s)   (8)

Finally:

x=1.124 m   (9)

4 0
4 years ago
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