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leva [86]
3 years ago
12

You drop a ball from the top of a building, how long does it take to reach the ground

Physics
1 answer:
irakobra [83]3 years ago
5 0
Need more details to the question
You might be interested in
The acceleration due to gravity on Jupiter is 2.5 times what it is here on earth. An object weighing 347.9 N here on earth will
inessss [21]

Answer:  weight on Jupiter = 869.75 N

              mass  on Earth = mass on Jupiter = 35.5 Kg

Explanation:

W = mg

W = weight

m = mass

g = gravitational acceleration [ on the Earth, g₁ = 9,8 N/kg ]

On the Earth,

G₁ = m x g₁  = 347,9 N

On the Jupiter,

G₂ = mg₂  

mass on the Earth = mass on the Jupiter !  

m = G₁ : g = 347.9 N : 9,8 N/kg = 35.5 kg

G2 : G1 = 2.5

G₂ = 2,5 G₁ = 2,5 x 347.9 N =  869,75 N

5 0
3 years ago
If points a and b are connected by a wire with negligible resistance, find the magnitude of the current in the 12.0 v battery.
Marizza181 [45]
V = I * R
Where V is the voltage, I is the current and R is the resistance. Using Ohm's law, you require resistance to find the current through the wire. Technically, if the wire has a resistance of 0, you will get infinite current. But this isn't possible. Maybe the negligible resistance refers to the battery's internal resistance - not the wire's resistance. 
7 0
3 years ago
When you turn brownie mix into brownies is that a physical change?
kotegsom [21]
No because for example you get a brownie mix and you want to make them well you put all of the mixture in the blender and you finish making them. Well now the brownies are done. well are they still the brownie mix?
5 0
3 years ago
While bats emit a wide variety of sounds, one type emits pulses of sound at a frequency between 39 kHz and 78 kHz. What is the r
sdas [7]

Answer:

The range of wavelengths of the sound is 7692.30 m and 3846.15 m

Explanation:

A bat emits pulses of sound at a frequency between 39 kHz and 78 kHz. It is required to find the range of wavelengths of this sound.

Bat uses ultrasonic waves. It moves with the speed of light.

If f = 39 kHz,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{39\times 10^3}\\\\\lambda=7692.30\ m

If f = 78 kHz,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{78\times 10^3}\\\\\lambda=3846.15\ m

So, the range of wavelengths of the sound is 7692.30 m and 3846.15 m.

6 0
3 years ago
Which best explains how the diffraction pattern observed in young’s experiment supports the wave theory of light?.
KIM [24]

Answer: the airy pattern can only arise from wave propagation

Explanation:if particles went in straight lines through a slit, they would progate linearly and not interfere. The airy pattern arises from diffraction as waves interfere, producing peaks (constructive interference where peaks of waves from each slit coincide) and troughs (destructive interference where peaks and troughs of waves from each slit cancel out). If intensity rather than field is measured nodes occur where 0 values line up instead of troughs

6 0
2 years ago
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