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3 years ago

Explain what relative weight is and why it is important to the calculation of atomic mass.

Physics

1 answer:

Colt1911 [192]3 years ago

7 0

Relative weight- weight that is relevant, and it is important to calculate atomic mass because that's the answer

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The 20 or so minerals that make up most of the rocks of earth's crust are know as what

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3 years ago

What is the x component of (+3m)ι^ And (+3m/s) i^?

expeople1 [14]

The "i" component of a vector is in the x-direction. Therefore, the x-component is 3 m/s.

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3 years ago

Which of these is an example of qualitative observation?

Zanzabum

Iron nails are attracted to the magnet

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3 years ago

Read 2 more answers

In an elastic collision, a 400-kg bumper car collides directly from behind with a second, identical bumper car that is traveling

kirza4 [7]

Answer:

v₁ = 3.5 m/s

v₂ = 6.4 m/s

Explanation:

We have the following data:

m₁ = mass of trailing car = 400 kg

m₂ = mass of leading car = 400 kg

u₁ = initial speed of trailing car = 6.4 m/s

u₂ = initial speed of leading car = 3.5 m/s

v₁ = final speed of trailing car = ?

v₂ = final speed of leading car = ?

The final speed of the leading car is given by the following formula:

$v_2=\frac{2m_1}{m_1+m_2}u_1-\frac{m_1-m_2}{m_1+m_2}u_2\\\\v_2=\frac{(2)(400\ kg)}{400\ kg+400\ kg}(6.4\ m/s)-\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(3.5\ m/s)$

The final speed of the leading car is given by the following formula:

$v_1=\frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2\\\\v_1=\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(6.4\ m/s)+\frac{(2)(400\ kg)}{400\ kg+400\ kg}(3.5\ m/s)$

4 0

3 years ago

. A mass m is traveling at an initial speed of 25.0 m/s. It is brought to rest in a distance of 62.5 m by a net force of 15.0 N.

harkovskaia [24]

Answer:

m = 3 kg

The mass m is 3 kg

Explanation:

From the equations of motion;

s = 0.5(u+v)t

Making t thr subject of formula;

t = 2s/(u+v)

t = time taken

s = distance travelled during deceleration = 62.5 m

u = initial speed = 25 m/s

v = final velocity = 0

Substituting the given values;

t = (2×62.5)/(25+0)

t = 5

Since, t = 5 the acceleration during this period is;

acceleration a = ∆v/t = (v-u)/t

a = (25)/5

a = 5 m/s^2

Force F = mass × acceleration

F = ma

Making m the subject of formula;

m = F/a

net force F = 15.0N

Substituting the values

m = 15/5

m = 3 kg

The mass m is 3 kg

7 0

3 years ago