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melomori [17]
3 years ago
14

A car left skid marks 78 m long on a road as it slid to a stop. If the car's acceleration was -3.9 m/s2, what was its velocity b

efore it began to skid, to the nearest m/s?
Physics
1 answer:
spayn [35]3 years ago
7 0
It is 25 I’m sure of it
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I drop an egg from a certain distance and it takes 3.74 seconds to reach the ground. How high up was the egg?
OverLord2011 [107]

Answer:

68.5 meters

Explanation:

Given:

v₀ = 0 m/s

a = 9.8 m/s²

t = 3.74 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0) (3.74) + ½ (9.8) (3.74)²

Δy = 68.5

The egg fell 68.5 meters.

7 0
3 years ago
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prisoha [69]

Answer:

b

Explanation:

the gravitational pull also helps with that but

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The ability to clearly see objects at a distance but not close up is properly called ________. The ability to clearly see object
melamori03 [73]

Answer:

Hyperopia

Explanation:

     In hyperopia ,people face difficulties  to see close up object , but can see object easily which are at a distance.

The main reason of hyperopia is our eyeball.When our eyeball become too  short , then light focus behind the retina. Sowe will face problem to see near object but we can see distance object easily. Hyperopia is the opposite of nearsightedness. Hyperopia can be corrected by using contact lenses.

5 0
3 years ago
The pressure of a gas produces a net force (the sum of all the forces) at ____ angles to the wall of a container.
ivanzaharov [21]

Answer: RIGHT angles

Explanation:

7 0
3 years ago
31. If you threw a baseball straight out at 45 m/s from a height of 1.5 meters (A) how long would it be in the air? B) How far o
coldgirl [10]

Answer:

A) t = 0.55 s

B) x = 24.8 m

Explanation:

A) We can find the time at which the ball will be in the air using the following equation:

y_{f} = y_{0} + v_{0y}t - \frac{1}{2}gt^{2}    

Where:

y_{f} is the final height= 0  

y_{0} is the initial height= 1.5 m

v_{0y} is the component of the initial speed in the vertical direction = 0 m/s        

t: is the time =?      

g: is the gravity = 9.81 m/s²

0 = 1.5 m - \frac{1}{2}9.81 m/s^{2}t^{2}

By solving the above equation for t we have:

t = \sqrt{\frac{2*1.5 m}{9.81 m/s^{2}}} = 0.55 s  

Hence, the ball will stay 0.55 seconds in the air.

                             

B) We can find the distance traveled by the ball as follows:

x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}

Where:  

a: is the acceleration in the horizontal direction = 0  

x_{f} is the final position =?  

x_{0} is the initial position = 0      

v_{0x} is the component of the initial speed in the horizontal direction = 45 m/s                                                                                            

x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}

x_{f} = 0 + 45 m/s*0.55 s + 0 = 24.8 m

Therefore, the ball will travel 24.8 meters.

I hope it helps you!

3 0
3 years ago
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