Which set of arrows represents the induced magnetic force? A B C

Norma-Jean [14]

Answer:

physical quantities that cannot be defined in terms of other quantities.

Examples: Length - meter (m)Time - second (s)Amount of substance - mole (mole)Electric current - ampere (A)Temperature - kelvin (K)Luminous intensity - candela (cd)Mass - kilogram (kg)

6 0

3 years ago

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Eva is in a closed, dark room. She uses her arm muscles to turn on a lamp. When she moves her hand closer to the lamp, the light

12345 [234]

Well, we know that the total energy in a closed system remains constant.

The problem with the story of Eva is that she is not in a closed system.
If the dark room were really a closed system, then she could press the
button or turn the switch all day, and the lamp could not light. It needs
electrical energy coming in from somewhere in order to turn on.

Let's say that Eva used her arm muscles to strike a match and light the
candle on the table. Then we would have have food energy, muscle
energy, chemical energy in the match, chemical energy in the candle,
heat and light energy coming out of the candle, heat energy soaking into
her hand, light energy bouncing off of the book and into her eyes ... all
going on during the story, and the sum total of all of them would remain
constant.

4 0

4 years ago

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On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-f

Sati [7]

On the moon, the gravitational acceleration is 1/6 of 9.8 m/s², so
g = 9.8/6 = 1.633 m/s²

Launch speed = 35 m/s
Launch angle = 27° above the horizontal.
Therefore,
The horizontal velocity is
u = 35*cos(27) = 31.1852 m/s
The vertical launch velocity is
v = 35*sin(27) = 15.8897 m/s

Part A
When the ball reaches maximum height, the time requires is given by
0 = v - gt
t = v/g = 15.8897/1.6333 = 9.7286 s
This is one half of the time of flight, which is
2*9.7286 = 19.457 s

Answer: 19.46 s (2 sig. figs)

8 0

4 years ago

a projectile leaves the ground with a velocity of 35 meters per second at an angle 32° what is the maximum height

Nina [5.8K]

To determine the maximum height for the projectile, we can use the timeless kinematics equation $V _{f} ^{2} -V_{i}^{2}=2a \Delta y$ and plug in what we know. For $V _{f} ^{2}$ , we know it is zero because at the highest location of flight, there is no vertical velocity. For $V _{i} ^{2}$ , we know it is $(35sin(32))^{2}$ since the vertical component is the sine of the velocity. We need to solve for $y$ , so it is left as-is. Since the only acceleration is $g$ , we can substitute -9.81 into it. Solving for the equation yields a solution of 17.55 m for $\Delta y$ .