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iren2701 [21]
3 years ago
13

In complete sentences describe the motion of the block and the scooter throughout this activity ?

Physics
1 answer:
tangare [24]3 years ago
5 0

Answer:

????

Explanation:

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How much watt is called 1 H. P​
agasfer [191]

Answer:

746 watts

Explanation:

7 0
2 years ago
Why is the curve between 1950 and 1980 relatively flat and centered around zero degrees difference from the baseline? (Hint: how
zimovet [89]

Look at the title of the graph, in small print under it.

Each point is "compared to 1950-1980 baseline". So the set of data for those years is being compared to itself. No wonder it matches up pretty close !

3 0
3 years ago
Instructions:Select all the correct answers.
melomori [17]
The answer would be B: The ground exerts an equal force on the golf ball.
6 0
3 years ago
Read 2 more answers
the focal length of a simple magnifier is 8.50 cmcm . assume the magnifier to be a thin lens placed very close to the eye.
Igoryamba

When the object is at the focal point the angular magnification is 2.94.

Angular magnification:

The ratio of the angle subtended at the eye by the image formed by an optical instrument to that subtended at the eye by the object when not viewed through the instrument.

Here we have to find the angular magnification when the object is at the focal point.

Focal length = 6.00 cm

Formula to calculate angular magnification:

Angular magnification = 25/f

                                            = 25/ 8.5

                                             = 2.94

Therefore the angular magnification of this thin lens is 2.94

To know more about angular magnification refer:: brainly.com/question/28325488

#SPJ4

5 0
1 year ago
A) An automobile light has a 1.0-A current when it is connected to a 12-V battery. Determine the resistance of the light.
kirill [66]

Answer:

The resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V

Explanation:

Given:

(A)

Current I = 1 A

Voltage V = 12 V

For finding the resistance,

  V = IR

  R = \frac{V}{I}

  R = \frac{12}{1}

  R = 12Ω

(B)

For finding power delivered,

  P = I^{2} R

  P = (1) ^{2} \times 12

  P = 12 Watt

(C)

For finding the potential difference,

   V = IR

   V = 5 \times 10^{-3} \times 2

   V = 10 \times 10^{-3}

   V = 0.01 V

Therefore, the resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V

4 0
3 years ago
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