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NeX [460]
3 years ago
15

An object is moving north with an initial velocity of 14 m/s accelerates 5m/s for 20 seconds. What is the final velocity of the

object
Physics
1 answer:
olga2289 [7]3 years ago
6 0

Use the kinematic equation: Vf=Vi+at

Then plug;

Vi=14 m/s

a=5 m/s²

t=20 s. Therefore;

Vf=14+(5*20)

Vf=114 m/s.

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How do you write a hypothesis
Mashcka [7]
It's and if, then statement!
5 0
3 years ago
David is investigating the properties of soil using the sample shown.
xenn [34]

Although the sample is not shown in this question, we can conclude that it would be reasonably easy for David to provide evidence of the color, consistency, temperature, and texture of the soil.

<h3 />

<h3>What are these properties an example of?</h3>

These are all examples of the physical properties of a sample. Since we cannot see the sample that David is using, it would be safest to assume that he would have no trouble providing evidence as to the physical properties of the soil, the:

  • Color
  • Consistency
  • Temperature
  • Texture

are all examples of this.

Therefore, we can confirm that David can provide evidence of the color, consistency, temperature, and texture of the soil.

To learn more about physical properties visit:

brainly.com/question/24632287?referrer=searchResults

7 0
3 years ago
1 point
Tju [1.3M]

Answer:

Option A nuclear

Explanation:

The rate of electricity production in nuclear power plant is much higher as compared to the rate of electricity generation in gas, wind and solar power plants.

Thus, in case where large amount of electricity is to be produced in a short period then one must rely on nuclear power plants.

Therefore, option A is correct

7 0
3 years ago
At what rate does gravity cause objects to accelerate at sea level on earth
VLD [36.1K]

The acceleration due to gravity on Earth is 9.8 m/s per second.

5 0
3 years ago
A small logo is embedded in a thick block of crown glass (n = 1.52), 4.70 cm beneath the top surface of the glass. The block is
harkovskaia [24]

The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})

Where,

d_g = Depth of glass

n_w = Refraction index of water

n_g = Refraction index of glass

n_{air} = Refraction index of air

d_w = Depth of water

I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to

d'w = d_w (\frac{n_{air}}{n_w})+d_g (\frac{n_{air}}{n_g})

d'w = (1.7cm) (\frac{1}{1.33})+(4.2cm)(\frac{1}{1.52})

d'w = 4.041cm

Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm

3 0
3 years ago
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