Ask question

Ask question

All categories

3 years ago

8

The temperature light will come on in your vehicle when

1 answer:

nikitadnepr [17]3 years ago

5 0

If your engine is not at the right temprature, usually when it is too hot.

You might be interested in

Which statement describes the movement of carbon as it cycles through the environment?

frosja888 [35]

Answer:

C

Explanation:

7 0

2 years ago

How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 210 turns

Tasya [4]

Complete Question

How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 210 turns.

The output from the secondary coil is 12 V

Answer:

The value is $N_s = 21 \ turns$

Explanation:

From the equation we are told that

The input voltage is $V_{in} = 120 \ V$

The number of turns of the primary coil is $N_p = 210 \ turn$

The output from the secondary is $V_o = 12V$

From the transformer equation

$\frac{N_p}{V_{in}} =\frac{N_s}{V_o}$

Here $N_s$ is the number of turns in the secondary coil

=> $N_s = \frac{N_p}{V_{in}} * V_s$

=> $N_s = \frac{210}{120} * 12$

=> $N_s = 21 \ turns$

4 0

3 years ago

A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav

Vinvika [58]

<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

$y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}$ (1)

$V=V_{o}-g.t$ (2)

Where:

$y$ is the rock's final position

$y_{o}=0$ is the rock's initial position

$V_{o}=30\frac{m}{s}$ is the rock's initial velocity

$V$ is the final velocity

$t$ is the time the parabolic movement lasts

$g=1.62\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of the moon

As we know $y_{o}=0$ , equation (2) is rewritten as:

$y=V_{o}.t+\frac{1}{2}g.t^{2}$ (3)

On the other hand, the maximum height is accomplished when $V=0$ :

$V=V_{o}-g.t=0$ (4)

$V_{o}-g.t=0$

$V_{o}=g.t$ (5)

Finding $t$ :

$t=\frac{V_{o}}{g}$ (6)

Substituting (6) in (3):

$y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}$ (7)

$y_{max}=\frac{{V_{o}}^{2}}{2g}$ (8) Now we can calculate the maximum height of the rock

$y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}$ (9)

Finally:

$y_{max}=277.777m$

4 0

3 years ago

Which statement best describes the energy changes that occur while a child is riding on a sled down a steep, snow-covered hill?

Degger [83]

B. kinetic energy increases and potential energy decreases

5 0

3 years ago

Limestone formations in caves are considered what kind of weathering?

Zanzabum

Answer:

A.

Explanation:

because the carbonic acid reacts to the limestone.

8 0

2 years ago