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Zolol [24]
3 years ago
10

What's the number of protons

Physics
1 answer:
ad-work [718]3 years ago
5 0

It would be equal to atomic number

OR mass number - No. of neutrons

You might be interested in
Consider three force vectors F~ 1 with magnitude 43 N and direction 38◦ , F~ 2 with magnitude 26 N and direction −140◦ , and F~
Rama09 [41]

Answer:

34.70 N

Explanation:

Given :

F~ 1 = 43 N in direction 38◦

F~ 2 = 26 N in direction −140◦

F~ 3 = 27 N in direction 110◦

Therefore,

F~x = 43 cos (38) + 26 cos (-140) + 27 cos (110)

      = 43  (0.7) + 26  (-0.7) + 27  (-0.3)

      =  3.8

F~y = 43 sin (38) + 26 sin (-140) + 27 sin (110)

      = 43  (0.6) + 26  (-0.6) + 27  (0.9)

      = 34.5

so, F~ = $ \sqrt{3.8^2 + 34.5^2}$

          = 34.70 N

6 0
3 years ago
This is the change in kinetic energy of a system in which a 16 kg object moving at 25 m/s slows to a velocity of 20 m/s
Dennis_Churaev [7]

The kinetic energy of an object is given by

KE = 0.5mv²

where m is the mass and v is the velocity.

To calculate the change in kinetic energy...

Initial KE:

KEi = 0.5mVi²

where Vi is the initial velocity.

Final KE:

KEf = 0.5mVf²

where Vf is the final velocity.

ΔKE = KEf - KEi

ΔKE = 0.5mVi² - 0.5mVf²

ΔKE = 0.5m(Vf²-Vi²)

Given values:

m = 16kg

Vi = 25m/s

Vf = 20m/s

Plug in the given values and solve for ΔKE:

ΔKE = 0.5×16×(20²-25²)

ΔKE = -1800J

5 0
3 years ago
1. Two astronauts are 2.00 m apart in their spaceship. One speaks to the other. The conversation is transmitted to earth via ele
Veronika [31]

Answer:

1.6949*10^6 m

Explanation:

Our values are

d=2m

v=354m/s

We can find the time through

t=\frac{d}{v}

t=\frac{2}{354}

t=5.64*10^{-3}s

The expression for the distance between the Earth and the spaceship is as follow:

d=ct

Where c is Light speed, and t our previous time.

d= (3*10^8)(5.64*10^{-3})

d= 1.6949*10^6m

Therefore the distance between the Eath and the Spaceship is 1.6949*10^6 m

4 0
4 years ago
A pupil adds 37g of ice at 0°C to 100g of water at 30°C. The final temperature of the water and melted ice is 0°C. no heat is lo
mojhsa [17]

Answer:

The specific latent heat of ice is approximately 341 J/g

The correct option is;

b) 341 J/g

Explanation:

The given parameters are;

The mass of ice the pupil adds to the water, m₁ = 37 g

The initial temperature of the ice, T₁ = 0°C

The mass of the water to which the ice is added, m₂ = 100 g

The initial temperature of the water, T₂₁ = 30°C

The final temperature of the water and the melted ice, T₂₂ = 0°C

The specific heat capacity of the water, c₂ = 4.2 J/(g·°C)

By the principle of conservation of energy, we have;

The heat gained by the ice = The heat lost by the water = ΔQ₂

Given that the ice is only melted with no change in temperature, we have;

The heat gained by the ice = The latent heat needed for melting the ice

ΔQ₂ = m₂ × c₂ × (T₂₂ - T₂₁) = 100 × 4.2 × (0 - 30) = -12,600 J

The heat gained by the ice = m₁ × L_f

Where;

L_f represents the specific latent heat of fusion of ice;

We have;

12,600 = 37 × L_f

L_f = 12,600/37 = 340.54 J/g ≈ 341 J/g

The specific latent heat of fusion of ice = L_f ≈ 341 J/g.

5 0
3 years ago
An archer aims his arrow directly at an orange hanging on an orange tree. At the moment he releases the arrow, the orange drops
Elena L [17]

Answer:

Yes it would

Explanation:

As the arrow and the orange are in the same position initially, vertically speaking. They are also subjected to the same gravitational acceleration g only in the vertical direction. They also start their motion at the same time. So their equation of motion can be written as

y = h + gt^2/2

Where h is the initial height of both of them. Since their g, h, and t are the same, their vertical position must be the same at the same time. As the arrows progress horizontally, it would hit the orange.

6 0
3 years ago
Read 2 more answers
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