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3 years ago

10

Arrange the following in order of their appearance in the life cycle of a star: white dwarf, red giant, main-sequence star. Expl

ain your answer.

1 answer:

Goryan [66]3 years ago

6 0

Main-sequence star, red giant, and white dwarf

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A force of 10.0 newtons acts at an angle 20.0 degrees from vertical. What are the horizontal and vertical components of the forc

erik [133]

Horizontal component = (10N) · sin (20°) = 3.42... N (rounded)

Vertical component = (10N) · cos (20°) = 9.39... N (rounded)

8 0

3 years ago

What is the speed shown in the graph? which number do i start on?

den301095 [7]

Answer:

4 m/s

Explanation:

speed = distance/time

speed= 20/5 = 4

similarly for all no. the answer is constant,i.e. 4

3 0

3 years ago

si un disco fue lanzado con una fuerza de 1000 N, y el disc recorrió una distancia e o.6 m ¿Qué trabajo efectuó el lanzador ?

Alexeev081 [22]

Answer:

Trabajo realizado = 600 Nm

Explanation:

Dados los siguientes datos;

Fuerza = 1000 Newton

Distancia = 0.6 metros

Para encontrar el trabajo realizado;

$Trabajo \; realizado = fuerza * distancia$

Sustituyendo en la ecuación, tenemos;

$Trabajo \; realizado = 1000 * 0.6$

Trabajo realizado = 600 Nm

4 0

3 years ago

Two small charged spheres are located on the y-axis. One is at y = 1.00 m, the other is at y = −1.00 m, and they both have a cha

yuradex [85]

Answer:

(a) 23.946 kV

(b) -0.077 J

Explanation:

(a) The electric potential is given by the following formula:

$V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}$ (1)

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1 = q2 = 1.60*10^{-6}C

r1 and r2 are the distance from the charges to the point in which electric potential is evaluated.

Firs, you calculate the distance r1 and r2 by taking into account the position of the charges

$r_1=\sqrt{(1.00)^2+(0.670)^2}m=1.20m\\\\r_2=\sqrt{(-1.00)^2+(0.670)^2}m=1.20m$

Next, you replace the values of the parameters to calculate V:

$V=(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}+(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}\\\\V=23946.66\ V=23.946\ kV$

(b) The potential electric energy is given by:

$U_T=U_{1,2}+U_{1,3}+U_{2,3}\\\\U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}\\\\r_{1,2}=2.00m\\\\r_{1,3}=1.20m\\\\r_{2,3}=1.20m\\\\U_T=(8.98*10^9)[\frac{(1.6*10^{-6})^2}{2.00m}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}]J\\\\U_T=-0.077J$

5 0

3 years ago

A truck skids for a distance of 25m with the road pushing on its tires with force of 1500N as its brakes are applied. What is th

Daniel [21]

Answer:

37500 J

Explanation:

Law of conservation of energy: Its states that energy can neither be created nor destroyed but it can be transformed from one form to another.

Applying the law of conservation of energy.

The change in kinetic energy for the truck = Net work done by the truck skids

ΔK.E = F×d.................... Equation 1

Where ΔK.E = Change in kinetic energy, F = Force, d = distant.

Given: F = 1500 N, d = 25 m

Substitute these values into equation 1

ΔK.E = 1500×25

ΔK.E = 37500 J

3 0

3 years ago