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3 years ago

If a book has a weight of 23.3 N on earth, what is its mass

Physics

1 answer:

vaieri [72.5K]3 years ago

6 0

Answer:

2.38 kg

Explanation:

by definition, weight in physics is the force of gravity (Fg)

Fg = mass x gravity or mg

gravity is 9.8 m/s^2

N = kg m/s^2

Fg = mg

23.3 kg m/s^2 = 9.8 m/s^2 x m

23.3 kg m/s^2 / 9.8 m/s^2 = m >>> m/s^2 cancel out

23.3/9.8 = 2.37755 kg

m = 2.38 kg

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Answer:

c.the spacing between particles increases

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3 years ago

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A block has two strings attached to it on opposite ends. One string has a force of 5 N,

juin [17]

Unless you have a diagram to include or any other additional info, I'll assume the block is being pulled by two opposing forces along the horizontal surface.

Horizontally, the block is under the influence of

• one rope pulling in one direction with magnitude 15 N,

• the other rope pulling in the opposite direction with mag. 5 N, and

• friction, opposing the direction of the block's motion, with mag. 3 N.

It stands to reason that the block is accelerating in the direction of the larger pulling force.

(A) By Newton's second law, we have

15 N + (-5 N) + (-3 N) = m (1 m/s²)

where m is the mass of the block. Solve for m :

7 N = m (1 m/s²)

m = (7 N) / (1 m/s²)

m = 7 kg

(B) The friction force is proportional to the normal force, so that if f is the mag. of friction and n is the mag. of the normal force, then f = µ n where µ is the coefficient of friction.

The block does not bounce up and down, so its vertical forces are balanced, which means the normal force and the block's weight (mag. w) cancel out:

n + (-w) = 0

n = w

n = m g

where g = 9.8 m/s² is the mag. of the acceleration due to gravity.

n = (7 kg) (9.8 m/s²)

n = 68.6 N

Then

3 N = µ (68.6 N)

µ = (3 N) / (68.6 N)

µ ≈ 0.044

4 0

3 years ago

Which of the following is NOT an example of the 3rd law of motion (for every Action, there is an equal but opposite reaction?

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Answer:

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I believe his is the answer because I don't see any force and not enough reaction

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2 years ago

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Nana76 [90]

The formula used in calculations relating to transformers is:

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Substituting the given values to find for Vs,

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4 years ago

How many photons will be required to raise the temperature of 1.8 g of water by 2.5 k ?'?

tatyana61 [14]

Missing part in the text of the problem:
"Water is exposed to infrared radiation of wavelength 3.0×10^−6 m"

First we can calculate the amount of energy needed to raise the temperature of the water, which is given by
$Q=m C_s \Delta T$
where
m=1.8 g is the mass of the water
$C_s = 4.18 J/(g K)$ is the specific heat capacity of the water
$\Delta T=2.5 K$ is the increase in temperature.

Substituting the data, we find
$Q=(1.8 g)(4.18 J/(gK))(2.5 K)=18.8 J=E$

We know that each photon carries an energy of
$E_1 = hf$
where h is the Planck constant and f the frequency of the photon. Using the wavelength, we can find the photon frequency:
$\lambda = \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{3 \cdot 10^{-6} m}=1 \cdot 10^{14}Hz$

So, the energy of a single photon of this frequency is
$E_1 = hf =(6.6 \cdot 10^{-34} J)(1 \cdot 10^{14} Hz)=6.6 \cdot 10^{-20} J$

and the number of photons needed is the total energy needed divided by the energy of a single photon:
$N= \frac{E}{E_1}= \frac{18.8 J}{6.6 \cdot 10^{-20} J} =2.84 \cdot 10^{20} photons$

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3 years ago