Help me please Ima give brainiest

Answer:

40 degrees Celsius

Explanation:

Have a great summer :)

Answer:

Refer to the attachment!~

Answer:

Gravitational force -an attractive force that exists between all objects with mass; an object with mass attracts another object with mass; the magnitude of the force is directly proportional to the masses of the two objects and inversely proportional to the square of the distance between the two objects.

Explanation:

hope it helped

Answer:

Kayla. You can calculate it using the formula for momentum: momentum=mass×velocity and find the bigger number between the two momentums

Answer:

Explanation:

[tex]h=-v^2 /2g[/tex]

[tex]with\\g = 9,8 m/s^2 or 10 m/s^2[/tex]

[tex]h= (-13)^2 / 2 * 9,8 = 8,6[/tex]

Answer:

The iron bolt is attracted to the electromagnet. The force of attraction between the electromagnet's core and the iron bolt is greater than the force of the spring, so the iron bolt is pulled out of its housing in the door

Explanation:

Answer:

I am not sure if this is the answer

acceleration: 2.2m/s

Explanation:

here

initial velocity(u): 11m/s

Final velocity(v): 33m/s

time taken(t): 10 s

now

a:v-u/t

or

acceleration:final velocity-initial velocity/time taken

or

a: 33-11/10

or

a:22/10, divide it

: a=2.2m/s#

Explanation:

Gravitational potential energy is energy an object possesses because of its position in a gravitational field. ... The gravitational potential energy is equal to its weight times the height to which it is lifted. PE = kg x 9.8 m/s2 x m = joules. The 9.8 us the gravitational acceleration constant.

so the answer is "the gravitational potential energy is decreasing"

Answer:

a. 0.41 m

b. 5.72 m/s

c. i. For part (a), I chose the hand as the reference level since the penny was thrown from the hand and the height of the penny at the hand is zero and also, it is easier to calculate from a zero reference level.

ii. For part (b), I chose the ground as the reference level since the height of the penny above the ground is positive and the height of the penny when the penny hits the ground is zero and also, it is easier to calculate from a zero reference level.

d. 5.72 m/s

Explanation:

a. Use energy to find how high the penny goes above your hand before stopping.

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy at the hand, E equals the total mechanical energy when the penny stops in the air, E'.

E = E'

U + K = U' + K' where U = initial potential energy at hand level = mgh where h = height at hand level = 0, K = initial kinetic energy at hand level = 1/2mv² where v = speed at hand level = 2.85 m/s, U' = final potential energy at stopping level = mgh' where h' = height at stopping level, K = final kinetic energy at stopping level = 1/2mv'² where v = speed at stopping level = 0 m/s (since the penny momentarily stops)

So, U + K = U' + K'

mgh + 1/2mv² = mgh' + 1/2mv'²

substituting the values of the variables into the equation, we have

mg(0) + 1/2m(2.85 m/s)² = mgh' + 1/2m(0 m/s)²

0 + 1/2m(8.1225 m²/s²) = mgh' + 0

m(4.06125 m²/s²) = mgh'

h' = 4.06125 m²/s² ÷ g

h' = 4.06125 m²/s² ÷ 9.8 m/s²

h' = 0.41 m

(b) The penny then falls to the floor, 1.26 m below your hand. Use energy to find its speed just before it hits the floor.  

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy when the penny stops in the air, E'  equals the total mechanical energy on the ground, E"

E' = E"

U' + K' = U" + K" where U' = initial potential energy at stopping level = mgh" where h' = height at stopping level = height of penny above hand, h' + height of hand above ground = 0.41 m + 1.26 m = 1.67 m, K = initial kinetic energy at stopping level = 1/2mv'² where v = speed at stopping level = 0 m/s (since the penny momentarily stops), U = final potential energy at ground level = mgh₁ where h₁ = height at ground level = 0, K = final kinetic energy at ground level = 1/2mv"² where v" = speed at ground level,

So, U' + K' = U' + K'

mgh" + 1/2mv'² = mgh₁ + 1/2mv"²

substituting the values of the variables into the equation, we have

mg(1.67 m) + 1/2m(0 m/s)² = mg(0) + 1/2mv"²

1.67mg + 0 = 0 + 1/2mv"²

1.67mg = 1/2mv"²

1.67g = 1/2v"²

v"² = 2(1.67g)

v" = √[2(1.67g)]

v" = √[2(1.67 m × 9.8 m/s²)]

v" = √[2(16.366 m²/s²)]

v" = √[32.732 m²/s²)]

v" = 5.72 m/s

(c) Explain your choice of reference level for parts (a) and (b).

i. For part (a), I chose the hand as the reference level since the penny was thrown from the hand and the height of the penny at the hand is zero and also, it is easier to calculate from a zero reference level.

ii. For part (b), I chose the ground as the reference level since the height of the penny above the ground is positive and the height of the penny when the penny hits the ground is zero and also, it is easier to calculate from a zero reference level.

(d) Choose a different reference level and repeat part (b)

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy when the penny stops in the air, E'  equals the total mechanical energy on the ground, E"

E' = E"

U' + K' = U" + K" where U' = initial potential energy at stopping level = mgh' where h' = height at stopping level = 0.41 m, K = initial kinetic energy at stopping level = 1/2mv'² where v' = speed at stopping level = 0 m/s (since the penny momentarily stops), U = final potential energy at ground level = mgh₁ where h₂ = height of hand above the ground level = height of ground below hand = -1.26 m(it is negative since the ground is below the hand), K = final kinetic energy at ground level = 1/2mv"² where v = speed at ground level,

So, U' + K' = U' + K'

mgh' + 1/2mv'² = mgh₂ + 1/2mv"²

substituting the values of the variables into the equation, we have

mg(0.41 m) + 1/2m(0 m/s)² = mg(-1.26 m) + 1/2mv"²

0.41mg + 0 = -1.26 mg + 1/2mv"²

0.41mg + 1.26mg = 1/2mv"²

1.67mg = 1/2mv"²

1.67g = 1/2v"²

v"² = 2(1.67g)

v" = √[2(1.67g)]

v" = √[2(1.67 m × 9.8 m/s²)]

v" = √[2(16.366 m²/s²)]

v" = √[32.732 m²/s²)]

v" = 5.72 m/s

Answer:

Explanation:

This is a classic Law of Momentum Conservation problem. For us the equation will look like this:

[tex][(m_yv_y+m_bv_b)]_b=[(m_y+m_b)v_{both}]_a[/tex] Filling in with our given info:

[tex][(70.0)(0)+(.40)(10.0)]_b=[(70.0+.40)v_{both}]_a[/tex] and

4.0 = 70.4v and

v = .06 m/s

Explanation:

The voltage of the lamp, V = 240 V

Power of the lamp, P = 40 W

It is running for 62 hours.

The cost of running is $2.50k per KWH

Electric power is,

P = 40×62 Wh

= 2480 Wh

P = 2.48 kWh

At the rate of $2.5 per kWh

P = $6.2

So, the cost of running is $6.2 per kWh.

Answer:

In a circular motion, the object just moves in a circle. In rotational motion, the object rotates about an axis. ... For example, Earth rotating on its own axis.

Answer:

Its different

Explanation:

it's different because

Answer:

total distance dibide by total time

Answer:

(D)

Explanation:

When switch C is opened then, Current is not flowing across 3 So bulb 3 will go out.

But current is flowing across 1 and 2 bulb because their switch is closed

therefore bulb 1 and 2 will remain it.

Hence, option (D) will be correct.

Answer:

Figure is not there

Explanation:

Answer:2 protons and 2 neutrons

Explanation:In Nuclei, There are 2 forces. 1 force is electrostatic and acts as repulsion between 2 protons. The other is force of attraction called Nuclear force between 2 neutrons.

By Newton's second law, the net force acting on the crate parallel to the surface is

∑ F = mg sin(10°) - 800 N = ma

where m = 500 kg is the mass of the crate and a is the acceleration.

Solve for a :

a = ((500 kg) (9.80 m/s^2) sin(10°) - 800 N) / (500 kg)

a ≈ 0.102 m/s^2

Explanation:

Uniform velocity is when an object goes an equal amount of space in an equal amount of time whereas non uniform velocity is when the object covers an unequal amount of distance in an equal amount of time.

Answer:

Energy is: the ability to do work

The formula for energy is: power x time = energy

Explanation:

Hope this helps (there isn't really an explanation)

hope this helps

Answer:

If a person is pushing a desk across the room, then there is an applied force acting upon the object. The applied force is the force exerted on the desk by the person.

Answer:

the work done by the woman is 6,000 J

Explanation:

Given;

force applied by the woman, F = 400 N

distance moved by the woman, d = 15 m

The work done by the woman is calculated as follows;

W = F x d

W = 400 N x 15 m

W = 6000 Nm = 6,000 J

Therefore, the work done by the woman is 6,000 J

Answer:

Las limitaciones de un modelo o prototipo son;

1) Los parámetros ambientales (donde se opera el modelo, prototipo o producto) son diferentes y, por lo tanto, pueden producir relaciones y factores ambientales que serán diferentes de los factores ambientales y las relaciones del objeto real.

2) El análisis del problema puede ser inadecuado

3) La posibilidad de falta de satisfacción del cliente con un modelo, preferencia por la demostración real del producto.

4) Reproducción inexacta del entorno del producto durante la prueba del modelo

5) El factor de costo del modelo

6) Mayor complejidad introducida por el modelo / prototipo al análisis de la solución

Explanation:

El modelo o prototipo es la presentación del diseño articulado, construido para demostrar el producto real con el propósito de encontrar la existencia de errores en el diseño que serían corregidos, antes de que se realice la producción real

Explanation:

Glass is the best sound absorber material.

Answer:

The correct option is A. a second action potential is generated until the interval reaches the absolute refractory period.

Explanation:

The inter-stimulus interval (ISI) is the temporal interval between two successive stimuli, measured from the offset of the first stimulus to the commencement of the second.

A cell's refractory period is the time during which it is unable to replicate an action potential. Therefore, the absolute refractory period is the amount of time it takes for a second action potential to be initiated, regardless of how large a stimulus is applied repeatedly.

A second action potential is generated when the gap between the stimuli decreases until the interval reaches the absolute refractory period.

Therefore, the correct option is A. a second action potential is generated until the interval reaches the absolute refractory period.

That is, when the interval between the stimuli decreases, a second action potential is generated until the interval reaches the absolute refractory period.