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laila [671]
3 years ago
9

We have a 10kg ball and a 15kg ball. We drop them from the top of the science building while objects are at rest and there is no

wind resistance. Which one hits the ground first?​
Physics
1 answer:
Mashcka [7]3 years ago
3 0
They will hit the ground at the same time, as mass is negligible when calculating the acceleration of gravity when there is no air resistance
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Why is the answer B?
djyliett [7]

Answer:

Explanation:

The center of mass lies on a line that joins position 4 of one start with position 4 of the other star.  The shortest distance between these two points will produce the largest velocity. You are using F = m v^2/R

Small R = large force.

Large Force = increased speed.

The masses don't have any effect on the outcome: they remain constant.

7 0
4 years ago
A light bulb dissipates 100 Watts of power when it is supplied a voltage of 220 volts.
ycow [4]

Given Information:

Power = P = 100 Watts

Voltage = V = 220 Volts

Required Information:

a) Current = I = ?

b) Resistance = R = ?

Answer:

a) Current = I = 0.4545 A

b) Resistance = R = 484 Ω

Explanation:

According to the Ohm’s law, the power dissipated in the light bulb is given by

P = VI

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.

Re-arranging the above equation for current I yields,

I = \frac{P}{V}  \\\\I = \frac{100}{220} \\\\I = 0.4545 \: A \\\\

Therefore, 0.4545 A current is flowing through the light bulb.

According to the Ohm’s law, the voltage across the light bulb is given by

V = IR

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.

Re-arranging the above equation for resistance R yields,

R = \frac{V}{I} \\\\R = \frac{220}{0.4545} \\\\R = 484 \: \Omega

Therefore, the resistance of the bulb is 484 Ω

3 0
3 years ago
Read 2 more answers
Which of the following describes an accelerated motion
Naily [24]
Accelerated motion such as a vehicle (car) or a moving item such as a (football thrown in the air)
3 0
3 years ago
Water flows through a 2.5cm diameter pipe at a rate of
My name is Ann [436]

Answer:

From the Bernoulli energy principle,

ΔP + 1/2ρΔv² = 0 -------------------------- eqn 1

where

ΔP = pressure drop = P2 - P1 = (1 - 0.25)x10⁵ N/m =7.5 x 10⁴N/m

Δv²= velocity change = v₂² - v₁²

ρ = water density = 1kg/m3

Recall volumetric flow rate, Q=A v = constant

A = cross sectional area = πr²=πd²/4

d=pipe diameter at point 2 = 2.5cm = 0.025m and Q =0.20m³/min = 0.00333m³/s

So A= 0.000491m²

we can get v2 = Q/A = 6.79m/s

From eqn 1, v₁² = 2(P2 - P1)/ρ + v₂²

v₁² =  (2 x 7.5 x 10⁴)/1000 + 6.79²

v₁² = 196

v₁ = 14m/s

we can now get the area of the constriction point 2, A₁ = Q/v₁

A₁ = 0.000238m² and the diameter now will be d₁

d₁² = 4 x A₁ / π

     = 4 x 0.000238/3.14 = 0.000303m²

d₁ = √0.000303 = 0.0174m

Therefore, the diameter of  a constriction in the pipe  at the new pressure = 0.0174m = 1.74cm

6 0
3 years ago
Please help help me please
vodomira [7]
The answer is either c or d but c is the best answer
4 0
3 years ago
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