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3 years ago

Four balls with different masses are dropped simultaneously from the heights

Physics

1 answer:

Scorpion4ik [409]3 years ago

7 0

D will hit last because if air resistance is null then the only force enacting on the balls is the force of gravity. Force of Gravity has an acceleration of 9.81 ms^2 and so every ball had the same acceleration and ball D is the furthest away from the floor.

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While watching a small fireworks display from a building 204 feet tall, you observe a rocket that travels from ground to eye lev

faltersainse [42]

204 ft = 62.2 m
We know that the initial velocity of the rocket was 0, we can find its acceleration using:
s = ut + 1/2 at²
62.2 = 0.5a(6)²
a = 3.46 m/s²
Now, the final velocity of the rocket:
v = u + at
v = 3.46 x 6
v = 20.76 m/s = 68.1 ft/s

4 0

3 years ago

Suppose the tip of the minute hand of a clock is two inches from the center of the clock. Determine the distance traveled by the

sashaice [31]

1) The minute travels the circumference of a 4 inches circle is 60 minutes:

Circumference = 2πr = 2π(2in) = 4π in
Time = 60 minutes

2) Constant angular velocity => (4π / 60) = (x / 20) => x = 4π / 3 inches

x = 4.19 inches

Answer: 4.19 inches

4 0

3 years ago

A hiker walks 20.51 m at 33.16 degrees. What is the Y component of his displacement?

Serhud [2]

Answer:

<em>The y component of his displacement is 11.22 meters</em>

Explanation:

<u>Components of the displacement</u>

The displacement is a vector because it has a magnitude and a direction. Let's suppose a displacement has a magnitude r and a direction θ, measured with respect to the positive x-direction. The horizontal component of the displacement is calculated by:

$x=r\cos\theta$

The vertical component is calculated by:

$y=r\sin\theta$

The hiker has a displacement with magnitude r = 20.51 m at an angle of 33.16 degrees. Substituting in the above equation:

$y=20.51\sin(33.16^\circ)$

$y=11.22\ m$

The y component of his displacement is 11.22 meters

7 0

3 years ago

a car travelling at 50m/h on a horizontal highway (a) if the coefficient of static friction between road and tyres on a rainy da

aleksandrvk [35]

Hope this helps!

-Lilly

3 0

3 years ago

How far apart would you have to place the poles of a 1. 5 v battery to achieve the same electric field?

Zarrin [17]

To place the poles of a 1. 5 v battery to achieve the same electric field is 1.5×10−2 m

The potential difference is related to the electric field by:

∆V=Ed

where,

∆V is the potential difference

E is the electric field

d is the distance

what is potential difference?

The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.

We want to know the distance the detectors have to be placed in order to achieve an electric field of

E=1v/cm=100v/cm

when connected to a battery with potential difference

∆v=1.5v

Solving the equation,we find

$d = \frac{ \:Δv}{e}$