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aliya0001 [1]
3 years ago
13

E mirror.wWWW+9の10 V​

Physics
2 answers:
gladu [14]3 years ago
5 0

Answer:

Explanation:

Your question did not get pasted correctly. Please take a photo of the question and add to the original post. I will correct my answer once that is done. Thank you.

almond37 [142]3 years ago
4 0

Answer:

Explanation: dont put fake answers on people's pages.

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Which form of energy does a plant store when light is transformed during photosynthesis
Stolb23 [73]
The form of energy a plant stores when light is transformed during photosynthesis is chemical energy. Hope this helps!
8 0
3 years ago
In a heat engine if 1000 j of heat enters the system the piston does 500 j of work, what is the final internal energy of the sys
nydimaria [60]

Answer : The final energy of the system if the initial energy was 2000 J is, 3500 J

Solution :

(1) The equation used is,

\Delta U=q+w\\\\U_{final}-U_{initial}=q+w

where,

U_{final} = final internal energy

U_{initial} = initial internal energy

q = heat energy

w = work done

(2) The known variables are, q, w and U_{initial}

initial internal energy = U_{initial} = 2000 J

heat energy = q = 1000 J

work done = w = 500 J

(3) Now plug the numbers into the equation, we get

U_{final}-(2000J)=(1000J)+(500J)

(4) By solving the terms, we get

U_{final}-(2000J)=(1000J)+(500J)

U_{final}-(2000J)=1500J

U_{final}=2000J+1500J

U_{final}=3500J

(5) Therefore, the final energy of the system if the initial energy was 2000 J is, 3500 J

5 0
3 years ago
A solid disk of mass 2 kg and radius 2 m is given a horizontal push of 20N at a point .3 m above its center. a. What is the mini
Margaret [11]

Answer:

\mu_s=1.0205

Explanation:

Given:

  • mass of solid disk, m=2\ kg
  • radius of disk, r=2\ m
  • force of push applied to disk, F=20\ N
  • distance of application of force from the center, s=0.3\ m

<em>For the condition of no slip the force of  static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.</em>

\therefore F

where:

f_s = static frictional force

\Rightarrow 20

\Rightarrow 20

\Rightarrow 20

\mu_s>1.0204

7 0
3 years ago
Un cable está tendido sobre dos postes colocados con una separación de 10 m. A la mitad del cable se cuelga un letrero que provo
lisabon 2012 [21]

Answer:

El peso del cartel es 397,97 N

Explanation:

La tensión dada en cada segmento del cable = 2000 N

El desplazamiento vertical del cable = 50 cm = 0,5 m

La distancia entre los polos = 10 m

La posición del letrero en el cable = En el medio = 5

El ángulo de inclinación del cable a la vertical = tan⁻¹ (0.5 / 5) = 5.71 °

El peso del letrero = La suma del componente vertical de la tensión en cada lado del letrero

El peso del signo = 2000 × sin (5.71 grados) + 2000 × sin (5.71 grados) = 397.97 N

El peso del signo = 397,97 N.

8 0
4 years ago
An inductor with an inductance of .5 henrys (H) is to be connected to a 60 Hz circuit. What will the inductive reactance (X L) b
Anika [276]

Answer:

1885.2 ohms

Explanation:

Step one:

given data

L=5H

f=60Hz

Required

The inductive reactance of the inductor

Step two:

Applying the expression

XL= 2πfL

substitute

XL=2*3.142*60*5

XL=1885.2 ohms

4 0
3 years ago
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