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iragen [17]
3 years ago
15

Мета: :

Physics
1 answer:
Ierofanga [76]3 years ago
3 0
I can’t answer this
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The force of gravity acting<br> on a child's mass on earth is 490 newtons. what's the child's mass?
UNO [17]
F = m • a

What we know:
- Gravity: 9.8 m/s
- Force: 490 N

Equation derived:
m = F/a
m = 490/9.8
= 50 kg
6 0
3 years ago
Read 2 more answers
A 5.00 kg mass is placed on top of a vertical spring, which compresses a distance of 3.13 cm. Calculate the force constant (in N
77julia77 [94]

Answer:

<h2>1567.09 N/m</h2>

Explanation:

Step one:

given data

mass m=5kg

compression x= 3.13cm to m= 0.0313m

<em>According to Hooke's law, provided the elastic limit of an elastic material is not exceeded the extension e is directly proportional to the applied force</em>

F=ke

where

k= spring constant in N/m

e= extension/compression in

Step two:

assume g= 9.81m/s^2

F=mg

F=5*9.81

F=49.05N

substitute in the expression F=ke

49.05=k*0.0313

k=49.05/0.0313

k=1567.09 N/m

<u>The force constant (in N/m) of the spring is 1567.09 N/m</u>

8 0
3 years ago
2. What is the standard value of acceleration due to gravity or "g"?
svp [43]

Answer:

The standard acceleration due to gravity (or standard acceleration of free fall), sometimes abbreviated as standard gravity, usually denoted by ɡ0 or ɡn, is the nominal gravitational acceleration of an object in a vacuum near the surface of the Earth. It is defined by standard as 9.80665 m/s2 (about 32.17405 ft/s2).

Explanation:

6 0
4 years ago
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What does Newton's law all about?<br>​
KengaRu [80]
Newton's law is all about motion
5 0
3 years ago
New 5G networks utilize millimeter-wave radiation. Millimeter-wave radiation refers to electromagnetic waves with frequencies in
seraphim [82]

Answer:

It corresponds to 1mm-10 mm range.

Explanation:

  • Electromagnetic waves (such as the millimeter-wave radiation) travel at the speed of light, which is 3*10⁸ m/s in free space.
  • As in any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:

        v = \lambda * f  (1)

  • Replacing v= c=3*10⁸ m/s, and the extreme values of f (which are givens), in (1) and solving for λ, we can get the free-space wavelengths that correspond to the 30-300 GHz range, as follows:

       \lambda_{low} = \frac{c}{f_{high}}  = \frac{3e8m/s}{300e9Hz} = 1 mm (2)

      \lambda_{high} = \frac{c}{f_{low}}  = \frac{3e8m/s}{30e9Hz} = 10 mm (3)

4 0
3 years ago
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