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3 years ago

The speedometer in your car reads 55 mi/h. This represents the what of the car

Physics

2 answers:

lozanna [386]3 years ago

5 0

Speed it's going an hour, how many miles it's going an hour?

frez [133]3 years ago

5 0

That's the car's instantaneous speed.

You might be interested in

The radiation per unit area from the Sun reaching the earth is 1400 W/m2 , approximately the amount of radiative power per unit

natka813 [3]

Answer:

83.2 W/m^2

Explanation:

The radiation per unit area of a star is directly proportional to the power emitted, which is given by Stefan-Boltzmann law:

$P=\sigma A T^4$

where

$\sigma$ is the Stefan-Boltzmann constant

A is the surface area

T is the surface temperature

So, we see that the radiation per unit area is proportional to the fourth power of the temperature:

$I \propto T^4$

So in our problem we can write:

$I_1 : T_1^4 = I_2 : T_2^4$

where

$I_1 = 1400 W/m^2$ is the power per unit area of the present sun

$T_1 = 5800 K$ is the temperature of the sun

$I_2$ is the power per unit area of sun X

$T_2 = 2864 K$ is the temperature of sun X

Solving for I2, we find

$I_2 = \frac{I_1 T_2^4}{T_1^4}=\frac{(1400 W/m^2)(2864 K)^4}{(5800 K)^4}=83.2 W/m^2$

6 0

3 years ago

three point charges are arranged in a line. charge q3=+5.00 nC and is located at the origin. charge q2=-3.00 nC and is located a

valentinak56 [21]

Answer:

$q_1=+0.375\ {10}^{-9}$

Explanation:

Electrostatic Forces

The force exerted between two point charges $q_1$ and $q_2$ separated a distance d is given by Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{d^2}$

The forces are attractive if the charges have different signs and repulsive if they have equal signs.

The problem described in the question locates three point charges in a straight line. The charges have the values shown below

$\displaystyle q_3=+5\ 10^{-9}\ c$

$\displaystyle q_2=-3\ 10^{-9}\ c$

The distance between $q_3$ and $q_2$ is

$\displaystyle d_2=4cm=0.04\ m$

The distance between $q_3$ and $q_1$ is

$\displaystyle d_1=2cm=0.02\ m$

We must find the value of $q_1$ such that

$\displaystyle |F_3|=0$

Applying Coulomb's formula for $q_1$ is

$\displaystyle F_{13}=\frac{k\ q_1\ q_3}{d_1^2}$

Now for $q_2$

$\displaystyle F_{23}=\frac{k\ q_2\ q_3}{d_2^2}$

If the total force on $q_3$ is zero, both forces must be equal. Note that being q2 negative, the force on q3 is to the right. The force exerted by q1 must go to the left, thus q1 must be positive. Equating the forces we have:

$\displaystyle F_{13}=F_{23}$

$\displaystyle \frac{k\ q_1\ q_3}{d_1^2}=\frac{k\ q_2\ q_3}{d_2^2}$

Simplfying and solving for $q_1$

$\displaystyle q_1=\frac{q_2\ d_1^2}{d_2^2}$

$\displaystyle q_1=\frac{3.10^{-9}\ 0.02^2}{0.04^2}$

$\boxed{\displaystyle q_1=+0.375\ {10}^{-9}}$

5 0

3 years ago

Part C: Quantitative Problems when vf is not 0

Alina [70]

Answer:

(a)

$\triangle v=-8\ m/s\\\triangle mv=-56\ kg.m/s$

(b)

1120 N

Explanation:

Change in velocity, $\triangle v$ is given by subtracting the initial velocity from the final velocity and expressed as $\triangle v= v_f -v_i$

Where v represent the velocity and subscripts f and i represent final and initial respectively. Since the ball finally comes to rest, its final velocity is zero. Substituting 0 for final velocity and the given figure of 8 m/s for initial velocity then the change in velocity is given by

$\triangle v=0-8=-8\ m/s$

To find $m\triangle v$ then we substitute 7 kg for m and -8 m/s for $\triangle v$ therefore $\triangle\ v=7 Kg\times -8 m/s=-56\ Kg.m/s$

(b)

The impact force, F is given as the product of mass and acceleration. Here, acceleration is given by dividing the change in velocity by time ie

$a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}$

Substituting t with 0.05 s then $a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}=\frac {-8}{0.05}=-160 m/s^{2}$

Since F=ma then substituting m with 7 Kg we get that F=7*-160=-1120 N

Therefore, the impact force is equivalent to 1120 N

3 0

3 years ago

Place /our live /music / important / has / in / an

oksian1 [2.3K]

Answer:

Music has an important place in our lives.

3 0

3 years ago

An object ends up at a final position of x=-55.25 meters after a displacement of -189.34 meters after a displacement of -189.34

Travka [436]

The initial position of the object was found to be 134.09 m.

<u>Explanation:</u>

As displacement is the measure of difference between the final and initial points. In other words, we can say that displacement can be termed as the change in the position of the object irrespective of the path followed by the object to change the path. So

Displacement = Final position - Initial position.

As the final position is stated as -55.25 meters and the displacement is also stated as -189.34 meters. So the initial position will be

Initial position of the object = Final position-Displacement

Initial position = -55.25 m - (-189.34 m) = -55.25 m + 189.34 m = 134.09 m.

Thus, the initial position for the object having a displacement of -189.34 m is determined as 134.09 m.

4 0

3 years ago