One principle of environmental law and policy in the U.S. is to make polluters pay . True or false ?

Turn the ignition switch to start and release the key immediately or you could destroy the______________.

vredina [299]

Starter

Explanation:

Turn the ignition switch to start and release the key immediately or you could destroy the starter.

The car starter is used to cause ignition in the internal combustion engine in order to fire the piston and cause mechanical motion. The starter is used to start the cyclic process of the internal combustion engine.

Once the engine starts by igniting the starter, it is best to release it.
The starter ensures that the spark plug is engaged and the motor is brought into work.
If the ignition is still engaged, the process continues repeatedly and it can damage the starter of the car.

learn more:

Automobile brainly.com/question/2599962

#learnwithBrainly

5 0

3 years ago

How do I find frictional force with force applied?

lions [1.4K]

Answer: Choose the normal force acting between the object and the ground. Let's assume a normal force of 250 N.

Determine the friction coefficient.

Multiply these values by each other: 250 N * 0.13 = 32.5 N .

You just found the force of friction!

Explanation:

8 0

3 years ago

Calculate the speed of a car that has traveled 300 miles in 6 hours.

PilotLPTM [1.2K]

300 miles / 6 hours = 50 miles per hour

5 0

3 years ago

Two bodies fall freely from different heights and reach the ground simultaneously. The time of descent for the first body is 1s

jok3333 [9.3K]

The initial height of the first body is given by:
$h_1 = \frac{1}{2}gt^2$
where
g is the gravitational acceleration
t is the time it takes for the body to reach the ground
Substituting t=1 s, we find
$h_1 = \frac{1}{2}(9.81 m/s^2)(1 s)^2=4.9 m$

The second body takes takes t=2 s to reach the ground, so it was located at an initial height of
$h_2 = \frac{1}{2}(9.81 m/s^2)(2 s)^2=19.6 m$

The second body started its fall 1 second before the first body, therefore when the second body started its fall, the first body was located at its initial height, i.e. at 4.9 m from the ground.

6 0

3 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

3 years ago