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Nitella [24]
4 years ago
6

T or F. A virtual image can sometimes be seen on a screen; it just depends on the situation.

Physics
1 answer:
irga5000 [103]4 years ago
3 0
It would be completely false to state that a <span>virtual image can sometimes be seen on a screen; it just depends on the situation. A virtual image can never be seen on a screen. The correct option among the two options that are given in the question is the second option. I hope the answer has come to your help.</span>
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Someone plz answer this quickly!!
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B. In the same group

Explanation:

Elements in the same group shared chemical properties.

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What are the factors affecting gravitational force?​
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  • Distance between objects
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  • Force of gravitational field

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The magnetic field of a magnet is strongest
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4 0
3 years ago
Calculate the force of gravity acting on an object that has a mass of 1.3 kg. the object is on earth so use 9.8 m/s^2 as the acc
krok68 [10]

Gravity constantly acts on the apple so it goes faster and faster ... in other words it accelerates.

Ignoring air resistance, its velocity increases by 9.8 meters per second every second. That is two lots of "per second" and is written:

9.8 m/s2

9.8 m/s2 is the acceleration due to gravity near the Earth's surface. Nearly everything in our lives happens near the Earth's surface, so that value gets used a lot, and is written as a little g:

g = 9.8 m/s2

The average value of g is 9.80665 m/s2, but values are different around the world, such as Calcutta at 9.78548, London at 9.81599 and Tokyo at 9.79805.

So most people just use 9.8 m/s2

7 0
4 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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