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KonstantinChe [14]
3 years ago
10

It's important to match your exercise shoes with the type of exercise in which you will be participating.

Physics
1 answer:
Arada [10]3 years ago
6 0

Answer:

True

Explanation:

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A 10 kg block is attached to a light cord that is wrapped around the pulley of an electric motor, as shown above. If the motor r
Anika [276]

Answer:

156.8 Watts

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 10 kg

Height (h) = 8 m

Time (t) = 5 s

Power (P) =?

Next, we shall determine the energy used by the motor to raise the block. This can be obtained as follow:

Mass (m) = 10 kg

Height (h) = 8 m

Acceleration due to gravity (g) = 9.8 m/s²

Energy (E) =?

E = mgh

E = 10 × 9. 8 × 8

E = 784 J

Finally, we shall determine the power output of the motor. This can be obtained as illustrated below:

Time (t) = 5 s

Energy (E) = 784 J

Power (P) =?

P = E/t

P = 784 / 5

P = 156.8 Watts

Therefore, the power output of the motor is 156.8 Watts

7 0
3 years ago
Which of the following gases are the heaviest? <br> O2, CH4, CO2, Cl2
kvasek [131]

the answer is

CI2 because its 70.


5 0
3 years ago
Physics Conversion help!!
stiv31 [10]

[Assuming that you've written 3.40 kg in 'a', and not 3.90 kg]

(a) 3,400 g x <u>0.001</u> = 3.40 kg [converting grams to kilograms]

(b) 220 cm x <u>0.01</u> = <u>2.2</u> m [converting centimeters to meters]

(c) 9.42 kg x <u>1000</u> = <u>9420</u> g [converting kilograms to grams]

(d) 6.53 m x <u>100</u> = <u>653</u> cm [converting meters to centimeters]

5 0
3 years ago
Can someone help me please
Evgen [1.6K]
The last one: meter
5 0
3 years ago
Read 2 more answers
You need to focus a 10 mW, 632.8 nm Gaussian laser beam that is 5.0 mm in diameter into a sample. You have access to a lens with
Anna [14]

Answer:

ee that the lens with the shortest focal length has a smaller object

           

Explanation:

For this exercise we use the constructor equation or Gaussian equation

        \frac{1}{f}  = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image respectively.

Magnification a lens system is

          m = \frac{h'}{h} = - \frac{q}{p}

             h ’= -\frac{h q}{p}

In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞

Let's calculate the distance to the image for each lens

f = 6.0 cm

           \frac{1}{q} = \frac{1}{f }  - \frac{1}{p}

as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf

          q = f = 6.0 cm

for the lens of f = 12.0 cm q = 12.0 cn

to find the size of the image we use

           h ’= h q / p

where p has a high value and is the same for all systems

           h ’= h / p q

Thus

f = 6 cm h ’= fo 6 cm

 

f = 12 cm h ’= fo 12  cm

therefore we see that the lens with the shortest focal length has a smaller object

8 0
3 years ago
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