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Nataly [62]
3 years ago
11

When falling through the air, which of the following objects will hit the ground first?

Physics
2 answers:
Lesechka [4]3 years ago
6 0
They would all fall and hit the ground at the same time
ipn [44]3 years ago
5 0
I think the answer is a penny

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Was work done on a book that fell from a desk to the floor? If so, what force was involved?
IgorLugansk [536]
Sure ! 
The work is (weight of the book) x (height of the desk above the floor).
The force (weight) is the result of gravity.
It was gravity that did the work.
7 0
3 years ago
Read 2 more answers
Can someone please answer this, ill give you brainliest Would be very appreciated.
mina [271]

Water cycle basically involves the change of water into liquid to vapour and again to liquid states through a cyclic manner .

The answers are

  1. Evaporation
  2. Transpiration
  3. Condensation
  4. precipitation
  5. runoff
  6. seapage
  7. ground water
5 0
2 years ago
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A student used a tuning fork of frequency 320 Hz and observed that the speed of sound was 339 m/s. Calculate the wavelength of t
Helen [10]

To solve this problem it is necessary to apply the concepts related to wavelength as a function of speed and frequency. In mathematical terms it can be expressed as

\lambda = \frac{v}{f}

Where,

v = Velocity

f = Frequency

According to our values the frequency (f) is 320Hz and the speed (v) is 339m / s.

Replacing in the given equation we have to,

\lambda = \frac{v}{f}\\\lambda = \frac{339}{320}\\\lambda = 1.059m\approx 1.06m

Therefore the wavelength of this sound wave is 1.06m

5 0
3 years ago
A current of 16.0 mA is maintained in a single circular loop of 1.90 m circumference. A magnetic field of 0.790 T is directed pa
AVprozaik [17]

To solve this problem it is necessary to take into account the concepts related to the magnetic moment and the torque applied over magnetic moments.

For the case of the magnetic moment of a loop we have to,

\mu = IA

Where

I = Current

A = Area of the loop

Moreover the torque exerted by the magnetic field is defined as,

\tau = IAB

Where,

I = Current

A = Area of the loop

B = Magnetic Field

PART A) First we need to find the perimeter, then

P = 2\pi r

r = \frac{P}{2\pi}

r = \frac{1.9}{2\pi}

r = 0.3025m,

The total Area of the loop would be given as,

A = \pi r^2

A = \pi 0.3025^2

A = 0.287m^2

Substituting at the equation of magnetic moment we have

\mu = (16*10^{-3})(0.287)

\mu = 4.58*10^{-3} A.m^2

Therefore the magnetic moment of the loop is 4.58*10^{-3}Am^2

PART B)  Replacing our values at the equation of torque we have that

\tau = IAB

\tau = (16*10^{-3})(0.287)(0.790)

\tau = 3.62*10^{-3}Nm

Therefore the torque exerted by the magnetic field is 3.62*10^{-3}Nm

6 0
3 years ago
Can someone answer these questions for me all 10 I’m confused
satela [25.4K]

Answer:

Is this the whole paper? I'll need to know so I can answer.

Explanation:

7 0
3 years ago
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