A 25.0 g marble sliding to the right at 20.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same dir
1 answer:
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Answer:
a)
b)
c)
Explanation:
From the exercise we know the initial velocity of the projectile and its initial height
To find what time does it take to reach maximum height we need to find how high will it go
b) We can calculate its initial height using the following formula
Knowing that its velocity is zero at its maximum height
So, the projectile goes 1024 ft high
a) From the equation of height we calculate how long does it take to reach maximum point
Solving the quadratic equation
So, the projectile reach maximum point at t=2s
c) We can calculate the final velocity by using the following formula:
Since the projectile is going down the velocity at the instant it reaches the ground is:
Answer:
As Per Provided Information
Moving body has 2m/s² acceleration
Time taken by body is 4 second
We are asked to find the 'change in velocity' ( ∆V) by the body.
<u>Formula Used here</u>
<u>Substituting </u><u>the </u><u>given </u><u>value</u>
<u></u>
<u>Therefore</u><u>,</u>
- <u>Change </u><u>in </u><u>velocity </u><u>is </u><u>8</u><u> </u><u>m/</u><u>s</u>
Answer:
electrons
Explanation:
The magnitude of the electric field outside an electrically charged sphere is given by the equation
where
k is the Coulomb's constant
Q is the charge stored on the sphere
r is the distance (from the centre of the sphere) at which the field is calculated
In this problem, the cloud is assumed to be a charged sphere, so we have:
is the maximum electric field strength tolerated by the air before breakdown occurs
is the radius of the sphere
Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:
Assuming that the cloud is negatively charged, then
And since the charge of one electron is
The number of excess electrons on the cloud is