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Artyom0805 [142]
4 years ago
13

A snowball is thrown with an initial x velocity of 7.5 m/s and an initial y velocity of 8.4 m/s . Part A How much time is requir

ed for the snowball to reach its highest point
Physics
1 answer:
elena55 [62]4 years ago
8 0

In order to calculate the time taken by the snowball to reach the highest point in its journey, we need to consider the variables along the y-direction.

Let us list out what we know from the question so that we can decide on the equation to be used.

We know that Initial Y VelocityV_{iy} = 8.4 m/s

Acceleration in the Y direction a_{y} = -9.8 m/s^{2}, since the acceleration due to gravity points in the downward direction.

Final Y Velocity V_{fy} = 0 because at the highest point in its path, an object comes to rest momentarily before falling down.

Time taken t = ?

From the list above, it is easy to see that the equation that best suits our purpose here is V_{fy} = V_{iy} + a_{y}t

Plugging in the numbers, we get 0 = 8.4 - (9.8)t

Solving for t, we get t = 0.857 s

Therefore, the snowball takes 0.86 seconds to reach its highest point.

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Answer:

El diálogo intercultural es un intercambio de opiniones abierto y respetuoso entre individuos y grupos pertenecientes a diferentes culturas que conduce a una comprensión más profunda de la percepción global del otro

Explanation:

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5 0
3 years ago
Two equally charged, 1.00 g spheres are placed with 2.00 cm between their centers. when released, each begins to accelerate at 2
Leya [2.2K]
1) Force = m*a = 1.00 g * (1kg / 1000 g) * 225 m/s^2 = 0.225 N

2) Charge

Force = K (charge)^2 /(distance)^2 => charge = √ [Force * distance^2 / k]

k = 9.00 * 10^9 N*m^2 / C^2

charge = √ [0.225 N * (0.02 m)^2 / 9.00* 10^9 N*m^2 / C^2 ]

charge = 0.0000001 C = 0.0001 mili C
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3 years ago
Please help fast science homework
Leokris [45]

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The solar cells transfer light energy to thermal energy.

When the battery is being charged up, chemical energy is transferred to electrical energy.

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3 years ago
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a luminous ball of plasma

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3 years ago
A coin is dropped in a 15.0 m deep well.
labwork [276]

Answer:

t = 1.75

t = 0.04

Explanation:

a)

For part 1 we want to use a kenamatic equation with constant acceleration:

X = 1/2*a*t^2

isolate time

t = sqrt(2X / a)

Plugin known variables. Acceleration is the force of gravity which is 9.8 m/s^2

t = sqrt(2*15m / 9.8m/s^2)

t = 1.75 s

b)

The speed of sound travels at a constant speed therefore we don't need acceleration and can use the equation:

v = d / t

isolate time

t = d / v

plug in known variables

t = 15m / 340m/s

t = 0.04 s

7 0
3 years ago
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