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Angelina_Jolie [31]

3 years ago

13

A solar panel is used to collect energy from the sun and change it into other forms of energy. The picture below shows some sola

r panels on the roof of a building. Which form of energy to collected by the solar panels?
A. Wind

B. sound

C. Magnetic

D. Light

1 answer:

irakobra [83]3 years ago

4 0

C I’m pretty sure!!!!!

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blagie [28]

Assuming Adam is on earth g= 9.8 m/s and m= weight/ gravity = 667/9.8 = 68 kg

8 0

2 years ago

Un móvil se desplaza con movimiento uniforme con una rapidez de 36 Km/h ¿Cuál es la distancia recorrida al cabo de 0,5 horas?

Diano4ka-milaya [45]

Answer:

Distancia = 17,5 kilómetros.

Explanation:

Dados los siguientes datos;

Velocidad = 36 km/h

Tiempo = 0.5 horas

Para encontrar la distancia recorrida;

Distancia = velocidad * tiempo

Distancia = 35 * 0.5

Distancia = 17,5 kilómetros.

Por tanto, la distancia recorrida por el automóvil es de 17,5 kilómetros.

4 0

3 years ago

If a 0.08 kg cell phone falls off a table at 15 m/s, then what is its kinetic energy right before it hits the ground?

Mariana [72]

The kinetic energy of the phone right before it hits the ground is 9J.

<h3>Kinetic energy of the phone</h3>

The kinetic energy of the phone right before it hits the ground is calculated as follows;

K.E = ¹/₂mv²

where;

m is mass of the phone
v is velocity of the phone

K.E = ¹/₂(0.08)(15)²

K.E = 9 J

Thus, the kinetic energy of the phone right before it hits the ground is 9J.

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

7 0

1 year ago

How many charges are required to have an electric field

Liono4ka [1.6K]

One charge is enough in order to have an electric field.

In fact, every charge generates an electric field: for example, the electric field generated by a single point positive charge is radial, as shown in the attached figure. More complicate electric field configurations can be obtained adding charges or using more complicate charge distributions, but one charge is enough to have an electric field.

7 0

3 years ago

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates

melisa1 [442]

Answer:

$v_{2.6b}=11.18\ m.s^{-1}$

$v_{7.2b}=14.19\ m.s^{-1}$

$s_{bb}=226.3305\ m$

$a_{db}=-3.7386\ m.s^{-2}$ negative sign denotes deceleration.

$t_b=21.3956\ s$

$a_y=1.1065\ m.s^{-2}$

Explanation:

Given:

initial speed of blue car, $u_b=0\ m.s^{-1}$

initial speed of yellow car, $u_y=0\ m.s^{-1}$

acceleration rate of blue car, $a_b=4.3\ m.s^{-2}$

time for which the blue car accelerates, $t_{ab}=3.3\ s$

time for which the blue car moves with uniform speed before decelerating, $t_{ub}=14.3\ s$

total distance covered by the blue car before coming to rest, $s_b=253.26 \ m$

distance at which the the yellow car intercepts the blue car just as the blue car come to rest, $s_y=253.26 \ m$

1)

<u>Speed of blue car after 2.6 seconds of starting the motion:</u>

Applying the equation of motion:

$v_{2.6b}=u_b+a_b.t$

$v_{2.6b}=0+4.3\times 2.6$

$v_{2.6b}=11.18\ m.s^{-1}$

<u>Speed of blue car after 7.2 seconds of starting the motion:</u>

∵The car accelerates uniformly for 3.3 seconds after which its speed becomes uniform for the next 14.3 second before it applies the brake.

so,

$v_{7.2b}=u+a_b\times t_{ab}$

$v_{7.2b}=0+4.3\times 3.3$

$v_{7.2b}=14.19\ m.s^{-1}$

<u>Distance travelled by the blue car before application of brakes:</u>

This distance will be $s_{bb}=$ (distance travelled during the accelerated motion) + (distance travelled at uniform motion)

<em>Now the distance travelled during the accelerated motion:</em>

$s_{ab}=u_b.t_{ab}+\frac{1}{2} a_{b}.t_{ab}^2$

$s_{ab}=0\times 3.3+0.5\times 4.3\times 3.3^2$

$s_{ab}=23.4135\ m$

<em>Now the distance travelled at uniform motion:</em>

$s_{ub}=14.19\times 14.3$

$s_{ub}=202.917\ m$

Finally:

$s_{bb}=s_{ab}+s_{ub}$

$s_{bb}=23.4135+202.917$

$s_{bb}=226.3305\ m$

<u>Acceleration of the blue car once the brakes are applied</u>

Here we have:

initial velocity, $u=14.19\ m.s^{-1}$

final velocity, $v=0\ m.s^{-1}$

distance covered while deceleration, $s_{db}=s_b-s_{bb}$

$\Rightarrow s_{db}=253.26 -226.3305=26.9295\ m$

Using the equation of motion:

$v^2=u^2+2a_{db}.s_{db}$

$0^2=14.19^2+2\times a_{db}\times 26.9295$

$a_{db}=-3.7386\ m.s^{-2}$ negative sign denotes deceleration.

<u>The total time for which the blue car moves:</u>

$t_b=t_a+t_{ub}+t_{db}$ ........................(1)

<em>Now the time taken to stop the blue car after application of brakes:</em>

Using the eq. of motion:

$v=u+a_{db}.t_{db}$

$0=14.19-3.7386\times t_{db}$

$t_{db}=3.7956\ s$

Putting respective values in eq. (1)

$t_b=3.3+14.3+3.7956$

$t_b=21.3956\ s$

<u>For the acceleration of the yellow car:</u>

We apply the law of motion:

$s_y=u_y.t_y+\frac{1}{2} a_y.t_y^2$

<em>Here the time taken by the yellow car is same for the same distance as it intercepts just before the stopping of blue car.</em>

Now,

$253.26=0\times 21.3956+0.5\times a_y\times 21.3956^2$

$a_y=1.1065\ m.s^{-2}$

7 0

3 years ago