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JulsSmile [24]
3 years ago
9

Using clay electric rates, how much does it cost to watch Netflix for 4 hours on a smart tv?

Physics
1 answer:
OverLord2011 [107]3 years ago
7 0

Answer: TVs cost between $0.0015 and $0.0176

Explanation:

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Una partícula se mueve en el plano XY efectúa un desplazamiento mientras actúa sobre ella una fuerza constante. X= (4i + 3j) m,
dsp73

Answer:

a) La magnitud del desplazamiento es de 5 m

La magnitud de la fuerza es 20 N

b) El trabajo realizado por la fuerza es de 100 J

c) El ángulo entre la fuerza y el plano es 0 °

Explanation:

a) La magnitud del desplazamiento se encuentra por la relación;

\left | X \right | = \sqrt{X_{x}^{2}+X_{y}^{2}}

Lo que da;

\left | X \right | = \sqrt{4^{2}+3^{2}} = 5 \ m

De manera similar, la magnitud de la fuerza, F, se encuentra como sigue;

\left | F \right | = \sqrt{F_{x}^{2}+F_{y}^{2}}

Lo que da;

\left | F \right | = \sqrt{16^{2}+12^{2}} = 20 \ N

b) El trabajo, W, realizado por la fuerza = Fuerza, F × Distancia, X

∴ Ancho = 20 N × 5 m = 100 N · m = 100 J

c) La dirección de la fuerza viene dada por la siguiente fórmula;

tan^{-1} \left (\dfrac{F_y}{F_x} \right ) = tan^{-1} \left (\dfrac{12}{16} \right )  = 38.9^{\circ}

La dirección del plano viene dada por la siguiente fórmula;

tan^{-1} \left (\dfrac{X_y}{X_x} \right ) = tan^{-1} \left (\dfrac{3}{4} \right )  = 38.9^{\circ}

Por tanto, el ángulo entre la fuerza y el plano = 0 °

La fuerza actúa a lo largo del plano.

6 0
3 years ago
A cylindrical shell of radius 7.00 cm and length 2.21 m has its charge uniformly distributed on its curved surface. The magnitud
polet [3.4K]

Answer:

The net charge on the shell is 30x10^-9C

Explanation:

Pls see attached file

8 0
3 years ago
Calculate the time (in seconds) needed for a car to accelerating from 0 m/s to 10 m/s at 5 m/s^2?
bazaltina [42]

Answer: 2 s

Explanation: In order to solve this problem we have to use the formule of the final speed  getting with a constant acceleration, it is given by;

Vfinal=Vo+a*t    where Vo is zero.

so then  t=Vfinal/a = (10 m/s)/5 m/s^2= 2 s

4 0
3 years ago
A 1.89 kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 4.77 N
Maksim231197 [3]

Answer:

Magnitude F(t)=26.6 N

Direction: -x

Explanation:

Given data

Spring constant K=4.77 N/m

Mass m=1.89 kg

Displace A=5.56m

Time t=3.96s

To find

Magnitude of force F

Solution

The angular frequency is given as

w=\sqrt{\frac{K}{m} } \\w=\sqrt{\frac{4.77N/m}{1.89kg} }\\w=1.59rad/s

Force on object is

F(t)=-mAw^{2}Cos(wt)

Substitute given values

So

F(t)=-(1.89kg)(5.56m)(1.59rad/s)^{2}Cos(1.59*3.96)\\F(t)=-26.6N

So

Magnitude F(t)=26.6 N

Direction: -x

4 0
3 years ago
9. A 0.500 kg ball with a speed of 4.0 m/s strikes a stationary 1.0 kg target. The ball and target
Virty [35]

Answer:

D. 2.0 kg m/s

Explanation:

Momentum after = momentum before

p = (0.500 kg) (4.0 m/s) + (1.0 kg) (0 m/s)

p = 2.0 kg m/s

4 0
3 years ago
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