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LUCKY_DIMON [66]
2 years ago
14

Help! Help!

Physics
1 answer:
andreyandreev [35.5K]2 years ago
4 0

Answer

I feel the answer is C because it could cause mental and physical trauma

Explanation:

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A car starts from rest and accelerates to 14 m/s in 2 seconds. What was its acceleration
Romashka [77]

Answer:

7m/s^2

Explanation:

using v=u+at

since the car started from rest, u=0 , v=14m/s t=2s

a =acceleration.

14=0+a×2

14=0+2a

14=2a

a= 14/2 =7

a=7m/s^2

7 0
2 years ago
In what way are liquids different from solids?.
puteri [66]

Answer:

solid molecules are packed closely and liquid molecules are losely packed

7 0
2 years ago
A 4.5 g coin sliding to the right at 23.8 cm/s makes an elastic head-on collision with a 13.5 g coin that is initially at rest.
Airida [17]

Answer:

a) v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} ), b) \Delta K = 9.559\times 10^{-5}\,J

Explanation:

a) The final velocity of the 13.5 g coin is found by the Principle of Momentum Conservation:

(4.5\times 10^{-3}\,kg)\cdot (23.8\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot (0\,\frac{m}{s} ) = (4.5\times 10^{-3}\,kg)\cdot (-11.9\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot v

The final velocity is:

v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )

b) The change in the kinetic energy of the 13.5 g coin is:

\Delta K = \frac{1}{2}\cdot (13.5\times 10^{-3}\,kg)\cdot \left[(11.9\times 10^{-2}\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}\right]

\Delta K = 9.559\times 10^{-5}\,J

4 0
3 years ago
Read 2 more answers
Which has more momentum, a 2.0 kg dog at 41 m/s or a 75 kg pony at 1.0 m/s?
Nesterboy [21]

Answer:

The dog has more momentum than the pony.

Explanation:

To solve this problem, we must remember the formula for calculating momentum, which is given below:

momentum = p = m*v

where m represents the mass of the object and v represents the velocity of the object

Using this knowledge, let's calculate the momentum for the dog and the pony.

Dog: p = m*v = (2kg)*(41 m/s) = 82 kg*m/s

Pony: p = m*v = (75kg)*(1 m/s) = 75 kg*m/s

Since 82 > 75, we can conclude that the dog has more momentum.

Hope this helps!

3 0
3 years ago
The intensity of light from a star (its brightness) is the power it outputs divided by the surface area over which it’s spread:
kow [346]

Answer:

\frac{d_{1}}{d_{2}}=0.36

Explanation:

1. We can find the temperature of each star using the Wien's Law. This law is given by:

\lambda_{max}=\frac{b}{T}=\frac{2.9x10^{-3}[mK]}{T[K]} (1)

So, the temperature of the first and the second star will be:

T_{1}=3866.7 K

T_{2}=6444.4 K

Now the relation between the absolute luminosity and apparent brightness  is given:

L=l\cdot 4\pi r^{2} (2)

Where:

  • L is the absolute luminosity
  • l is the apparent brightness
  • r is the distance from us in light years

Now, we know that two stars have the same apparent brightness, in other words l₁ = l₂

If we use the equation (2) we have:

\frac{L_{1}}{4\pi r_{1}^2}=\frac{L_{2}}{4\pi r_{2}^2}

So the relative distance between both stars will be:

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{L_{1}}{L_{2}} (3)

The Boltzmann Law says, L=A\sigma T^{4} (4)

  • σ is the Boltzmann constant
  • A is the area
  • T is the temperature
  • L is the absolute luminosity

Let's put (4) in (3) for each star.

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{A_{1}\sigma T_{1}^{4}}{A_{2}\sigma T_{2}^{4}}

As we know both stars have the same size we can canceled out the areas.

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{T_{1}^{4}}{T_{2}^{4}}

\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}

\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}

\frac{d_{1}}{d_{2}}=0.36

I hope it helps!

5 0
3 years ago
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