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makvit [3.9K]
3 years ago
12

R: Law of Conservation of Energy Pre-Write

Physics
1 answer:
Bad White [126]3 years ago
3 0

Answer:

V at C is 3.6 m/s

Explanation:

At A kinetic energy is zero and potential energy=mgh=0.5*9.81*0.6=2.943 J

By conservation of energy.

KE+PE=Constant

At C PE=0.6 J

the KE=2.943-0.6=2.343 J

KE=0.5*m*v^2

v=√[KE/(0.5*m)]=3.06 m/s

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Anna [14]

Answer:

D

Explanation:

6 0
3 years ago
Which list of properties of alternation current is the most likely reason it was chosen over direct current to provide electrici
Paraphin [41]

Answer:

- Direct current is a current in which electrons flow in one direction only

- Alternating current is a current in which the direction of the electron flow reverses periodically - so, half a cycle forward, half a cycle backward

There are several advantages of using alternating currents for the transmission of electricity across a country, over large distances. The main advantages are:

- The voltage of alternating currents can be easily increased/decreased by using transformers. For instance, a transformer is used at the beginning of the transmission line to increase the voltage (electricity is transmitted at high voltage in order to reduce dissipated power), and then another transformer is used before the electricity enters the houses, in order to decrease the voltage. Transformers only work with alternating currents.

- It is easy to interrupt the flow of an alternating current, because its value naturally becomes zero every half a cycle, so this is useful in case the current must be interrupted.

8 0
3 years ago
A truck driving east on a highway goes from the 42 mi marker to the 48 mi
Leona [35]

Average velocity is 1..2 mi/min east

Explanation:

  • Velocity = Displacement/Time

Here, displacement = 48 mi - 42 mi = 6 miles

Time = 5 minutes

⇒ Average Velocity = 6/5 = 1.2 mi/min east

3 0
3 years ago
O
Sergio [31]

see

below

Explanation:

refractive index = speed of light in vacuum / speed of light in medium

light travels at a speed of 3.0 x 10^8 m/s in vacuum

refractive index = 3.0 x 10^8 / 2.0 x 10^8

refractive index = 1.5

hope this helps, please mark it

4 0
3 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
3 years ago
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