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yaroslaw [1]
2 years ago
5

What is your initial speed if you accelerate at 5.8 m/s/s for 3.0 seconds and achieve a final speed of 45 m/s?

Physics
1 answer:
Natali5045456 [20]2 years ago
8 0

Answer:

27.6 m/s

Explanation:

hopefully it makes sense and is visible

:)

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What is the efficacy of a 60.0 W incandescent lightbulb that produces 830 lumens?
Alex17521 [72]

η = 13.8 lm/W. The luminous efficacy of a incandescent lightbulb that produces 830 lumens and consumed a power of 60 W is 13.8 lm/W.

The luminous efficacy of a light source is the relationship between the luminous flux (in lumens) emitted by a light source and the power (in watts). The luminous efficacy of a light source or luminous efficiency measures the part of electrical energy that is used to illuminate and is obtained by dividing the luminous flux emitted by the electrical power consumed. Luminous efficiency is expressed in lumens per watt (lm / W). It is given by the relation:

η = F / P.  Where F is the luminous flux, and P is the power consumed by the light source.

The efficacy of a 60.0 W incandescent lightbulb that produces 830 lm is:

η = 830 lm / 60 W

η = 13.8 lm/W

7 0
3 years ago
What are 2applications of radar that you have heard of in your life? What can radar tell an observer about some object, whether
RoseWind [281]

Answer:

Radar is used in space and in military

Explanation:

Radar helps to identify the type of forget that is in military it can be used to identify the enemies. It helps to identify the position of the object i.e helps to determine how far the object or subject is. It can be used to guide objects in space and monitor movements of other space objects.

4 0
2 years ago
The top of a swimming pool is at ground level. If the pool is 3.00 m deep, how far below ground level does the bottom of the poo
Yuri [45]

Answer:

a) 2.25 m

b) 2.625 m

Explanation:

Refraction is the name given to the phenomenon of the speed of light changing the the boundary when it moves from one physical medium to the other.

Refractive index is the ratio of the speed of light in empty vacuum (air is an appropriate substitution) to the speed of light in the medium under consideration.

In terms of real and apparent depth, the refractive index is given by

η = (real depth)/(apparent depth)

a) Real depth = 3.00 m

Apparent depth = ?

Refractive index, η = 1.333

1.333 = 3/(apparent depth)

Apparent depth = 3/1.3333 = 2.25 m.

Hence the bottom of the pool appears to be 2.25 m below the ground level.

b) Real depth = 1.5 m

Apparent depth = ?

Refractive index, η = 1.333

1.3333 = 1.5/(apparent depth)

Apparent depth = 1.5/1.3333 = 1.125 m

But the pool is half filled with water, there is a 1.5 m depth on top of the pool before refraction starts.

So, apparent depth of the pool = 1.5 + 1.125 = 2.625 m below the ground level

4 0
2 years ago
A cylindrical metal bar of diameter 14cm and length 484 cm is to be melted to make a metalic cube.if one side of the box is 220m
asambeis [7]

The number of boxes were produced willl be 7.57.n stands for the no of boxes.

<h3>What is volume?</h3>

The term “volume” refers to the amount of three-dimensional space taken up by an item or a closed surface. It is denoted by V and its SI unit is in cubic cm.

The volume of the cylindrical metal bar;

V₁=πr²h

V₁=3.4 × 7² × 484 cm

V₁=80,634.4 cm³

The one side of the box is,220mm(22 cm)

V₂ = a³

V₂ = (22)³

V₂ = 10,648 cm³

The no of boxes is found as;

n = V₁/V₂

n=80,634.4 cm³/10,648 cm³

n=7.57

Hence, the number of boxes were produced willl be 7.57

To learn more about the volume, refer to brainly.com/question/1578538

#SPJ2

5 0
2 years ago
An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the sp
laila [671]

Answer:

The centripetal acceleration changed by a factor of 0.5

Explanation:

Given;

first radius of the horizontal circle, r₁ = 500 m

speed of the airplane, v = 150 m/s

second radius of the airplane, r₂ = 1000 m

Centripetal acceleration is given as;

a = \frac{v^2}{r}

At constant speed, we will have;

v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1

a₂ = 0.5a₁

Therefore, the centripetal acceleration changed by a factor of 0.5

7 0
3 years ago
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