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Aliun [14]
3 years ago
12

A student wants to design an experiment to study the transformation of mechanical energy. Which object can be used to investigat

e how gravitational potential energy transforms into kinetic energy?(1 point)
a spring

a flat race car track

a slide

a rug
Physics
2 answers:
Vsevolod [243]3 years ago
7 0
I guess it’s the 3rd one
raketka [301]3 years ago
5 0

Answer: I think it would be a spring

Explanation: a rug, a slide, and a race track would all have friction and friction is not a conservative force which wouldn’t qualify for mechanical energy.

That’s my guess I’m not very good at this stuff though

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A heating element on an electric range operating at 200. v has a resistance of 32.0 ohms. the current drawn by the element is
LenaWriter [7]

Answer:

6.25 A

Explanation:

We cant find the current drawn by the element by using Ohm's law:

V=RI

where

V is the potential difference across the resistance of the element

R is the resistance

I is the current

In this problem, we know

V = 200 V

R=32.0 \Omega

Therefore we can re-arrange the equation and solve for the current:

I=\frac{V}{R}=\frac{200}{32}=6.25 A

4 0
3 years ago
Why are temperatures of the currents generally colder at the poles than the equator?
patriot [66]

Answer:

the answer is A

Explanation:

6 0
3 years ago
The angular velocity of a flywheel obeys the equa tion w(1) A Br2, where t is in seconds and A and B are con stants having numer
makkiz [27]

Answer:

A \to rad/s

B  \to rad/s^3

Explanation:

\omega_z(t)=A + Bt^2

Required

The units of A and B

From the question, we understand that:

\omega_z(t) \to rad/s

This implies that each of A and Bt^2 will have the same unit as \omega_z(t)

So, we have:

A \to rad/s

Bt^2 \to rad/s

The unit of t is (s); So, the expression becomes

B * s^2 \to rad/s

Divide both sides by s^2

B  \to \frac{rad/s}{s^2}

B  \to rad/s^3

5 0
3 years ago
You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state
posledela

Answer:

a)  y₂ = 49.1 m ,    t = 1.02 s , b)   y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

            v_{y}² = v_{oy}² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

           y-yo = v_{oy}² /2 g
            y- 0 = 10.0²/2 9.8
            y - 0 = 5.10 m
            
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
             y₂ = 5.1 + 44
             y₂ = 49.1 m
Let's use the other equation to find the time
              [tex]v_{y} = v_{oy} - g t

              t = v_{oy} / g

              t = 10 / 9.8

              t = 1.02 s

b) the maximum height

            y- 44.0 = v_{y}² / 2 g

            y - 44.0 = 5.1

            y = 5.1 +44.0

            y = 49.1 m

The time is the same because it does not depend on the initial height

              t = 1.02 s

7 0
3 years ago
What type of visible light spectrum does the sun produce?
VLD [36.1K]
UV Rays. - yung bandz
                       


4 0
3 years ago
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