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Valentin [98]
2 years ago
15

If runner A is running at 7.50 m/s and runner B is running at 7.90 m/s, how long will it take runner B to catch runner A if runn

er A has a
55.0-m head start?
Physics
1 answer:
yaroslaw [1]2 years ago
3 0

Answer:

t= 137.5 s

Explanation:

So if we are wanting to figure out how long it takes runner B to catch runner A. we must first set the slope of each runner equal to one another

<u>Slopes:</u>

Runner A:    y = 7.50x + 55

Runner B:    y = 7.90 x

sooooo

  7.50 x + 55 = 7.90 x

- 7.50 x          - 7.50 x

 55 = .40 x

55/.40 = .40 x / .40

x = 137.5 s

t= 137.5 s

7.50 * 137.5 + 55 =1086.25 m

7.90 * 137.5 =  1086.25 m

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Electric Field of Ring of Charge A ring shaped charge has a radius of 0.250 meters and bears a total charge of +5.00 C What is t
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Answer:

6.44 × 10^10 N/C

Explanation:

Electric field due to the ring on its axis is given by

E = K q r / (r^2 + x^2)^3/2

Where r be the radius of ring and x be the distance of point from the centre of ring and q be the charge on ring.

r = 0.25 m, x = 0.5 m, q = 5 C

K = 9 × 10^9 Nm^2/C^2

E = 9 × 10^9 × 5 × 0.25 / (0.0625 + 0.25)^3/2

E = 6.44 × 10^10 N/C

8 0
3 years ago
You are in a car moving forward at 12 m/s. You throw a ball in the direction the car is moving. From your point of view, what do
4vir4ik [10]

Answer: it goes the same speed as the car

Explanation:

3 0
3 years ago
An inventor claims to have invented a heat engine that receives 750kJ of heat from a source at 400K and produces 250kJ of net wo
IRISSAK [1]

Answer:

the claim is not valid or reasonable.

Explanation:

In order to test the claim we will find the maximum and actual efficiencies. maximum efficiency of a heat engine can be found as:

η(max) = 1 - T₁/T₂

where,

η(max) = maximum efficiency = ?

T₁ = Sink Temperature = 300 K

T₂ = Source Temperature = 400 K

Therefore,

η(max) = 1 - 300 K/400 K

η(max) = 0.25 = 25%

Now, we calculate the actual frequency of the engine:

η = W/Q

where,

W = Net Work = 250 KJ

Q = Heat Received = 750 KJ

Therefore,

η = 250 KJ/750 KJ

η = 0.333 = 33.3 %

η > η(max)

The actual efficiency of a heat engine can never be greater than its Carnot efficiency or the maximum efficiency.

<u>Therefore, the claim is not valid or reasonable.</u>

3 0
3 years ago
Bro what does this mean ? I don’t get this
zhannawk [14.2K]

CO2 is carbon dioxide a colorless gas

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3 years ago
Read 2 more answers
A(n) ________ microscope keeps an object in focus when the objective lens is changed.
Anvisha [2.4K]

Parfocal is the term used to describe a microscope that maintains focus when the objective lenses are replaced.

<h3>What is the name of the objective lens ?</h3>

For observing minute features within a specimen sample, a high-powered objective lens, often known as a "high dry" lens, is perfect. You can see a very detailed image of the specimen on your slide thanks to the 400x total magnification that a high-power objective lens and a 10x eyepiece provide.

The four objective lenses on your microscope are for scanning (4x), low (10x), high (40x), and oil immersion (100x).

The first-stage lens used to create a picture from electrons leaving the specimen is referred to as the "objective lens." The objective lens is the most crucial component of the imaging system since the quality of the images is determined by how well it performs (resolution, contrast, etc.,).

To learn more than objective lens , visit

brainly.com/question/17307577

#SPJ4

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2 years ago
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