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hjlf
4 years ago
6

A car starts from rest and after 7 seconds it is moving at 42 m/s. What is the car’s average acceleration? A. 0.17 m/s2 B. 1.67

m/s2 C. 6 m/s2 D. 7 m/s2 E. none of the above
Physics
2 answers:
aliya0001 [1]4 years ago
5 0
Acceleration =velocity /time
=42/7
=6
tamaranim1 [39]4 years ago
4 0

C. 6 m/s2 is correct

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When using science to investigate physical phenomena, which characteristic of the event must exist?
Vesna [10]

The correct answer is B. Measurable

Explanation:

The use of science to investigate a phenomenon implies using measurements and observations to better understand a phenomenon or test a hypothesis. Moreover, science focuses on natural phenomena that can be objectively studied through measurement instruments such as a thermometer, balance, hydrometer, etc.

In this context, for a phenomenon to be studied by science this needs to be measurable because the use of precise instruments as wells as numbers allow scientist to analyze and understand a phenomenon. Moreover, phenomena that depend on personal perspectives and cannot be measure is considered as non-scientific.

6 0
3 years ago
-What can you say about the snowboarder’s kinetic energy as he moves?
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Answer:

  His kinetic energy increases, potential energy decreases

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  Snowboarder is starting from a height and moving to the down direction. As he moves down his velocity increases, we know that kinetic energy is given by the expression \frac{1}{2} mv^2, so as he moves his kinetic energy increases.

  When the snowboarder is starting his potential energy is maximum(Potential energy = mgh), as he comes down his potential energy decreases.

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3 years ago
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3 years ago
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ozzi

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The energy stored in the solenoid is 7.078 x 10⁻⁵ J

Explanation:

Given;

diameter of the solenoid, d = 2.80 cm

radius of the solenoid, r = d/2 = 1.4 cm

length of the solenoid, L = 14 cm = 0.14 m

number of turns, N = 200 turns

current in the solenoid, I = 0.8 A

The cross sectional area of the solenoid is given as;

A = \pi r^2\\\\A = \pi (0.014)^2\\\\A = 6.16*10^{-4} \ m^2

The inductance of the solenoid is given by;

L = \frac{\mu_0 N^2A}{l} \\\\L =  \frac{(4\pi*10^{-7})(200^2)(6.16*10^{-4})}{0.14}\\\\L = 2.212*10^{-4} \ H

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E = ¹/₂LI²

E = ¹/₂(2.212 x 10⁻⁴)(0.8)²

E = 7.078 x 10⁻⁵ J

Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J

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