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yawa3891 [41]
2 years ago
14

Yall is this right

Physics
1 answer:
Aleksandr [31]2 years ago
3 0

Answer:

(A)

Explanation:

ITS A ok ok bye

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Two parallel plates that are initially uncharged are separated by 1.2 mm. What charge must be transferred from one plate to the
Gnoma [55]

Answer: 55.52 *10^-6 C= 55.52 μC

Explanation: In order to solve this question we have to take into account the following expressions:

potential energy stired in a capacitor is given by:

U=Q^2/(2*C)  where Q and C are the charge and capacitance of the capacitor.

then we have:

Q^2= 2*C*U=

C=εo*A/d where A and d are the area and separation of the parallel plates capacitor

Q^2=2*εo*A*U/d=2*8.85*10^-12*1.9*10^-5*11*10^3/(1.2*10^-3)=

=55.52 *10^-6C

4 0
3 years ago
An 85.0-kg mountain climber plans to swing down, starting from rest, from a ledge using a light rope 6.50 m long. he holds one e
Rainbow [258]

Understanding the given:
85 kg mountain climber
6.50 m long rope
gravity = 10m/s2

If we want to identify the work done on this scenario 
we get f = 85kg x 10m/s2 = 850 N
w = 850N x 6.5 m = 5525 J

Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
4 0
3 years ago
A boat loaded with a barrel of water floats in a swimming pool. When the water in the barrel is poured overboard, the swimming p
Lady bird [3.3K]

Answer: Remain unchanged

Explanation:

The boat with water barrel overboard floats in swimming pool when weight of the water displaced by the boat is equal to the buoyant force acting on the boat.

When the water in the barrel is poured overboard, the level of the swimming pool level would remain unchanged as the weight of the boat  with the water and barrel would remain unchanged ( as the density and volume of the whole system remains same) and hence, the weight of the water (of the swimming pool) displaced by the boat would remain same.

A boat loaded with a barrel of water floats in a swimming pool. When the water in the barrel is poured overboard, the swimming pool level will <u>remain unchanged. </u>

5 0
3 years ago
Un proyectil es lanzado horizontalmente desde una altura de 12 metro con una velocidad de 80 m/sg. a.Calcular el tiempo de vuelo
Naily [24]

Answer:

t= 1,56 s ,   x= 124,8 m , v = (80 i^ - 15,288 j ) m/s

Explanation:

Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso

        y = y₀ + v_{oy} t – ½ g t²

en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)

       0 = y₀ – ½ g t²

       t= √ (2 y₀/g)

calculemos

       t= √ ( 2 12 / 9,8)

       t= 1,56 s

El alcance del proyectil es la distancia horizontal recorrida  

        x = v₀ₓ t

        x = 80 1,56

        x= 124,8 m

La velocidad de impacto cuando toca el suelo

        vx = v₀ₓ = 80 ms

        v_{y} = v_{oy} – gt

        v_{y} = - 9,8 1,56

        v_{y} = - 15,288 m/s

la velocidad es

       v = (80 i^ - 15,288 j ) m/s

Traducttion  

This is a projectile launching exercise, let's start by finding the time it takes to reach the ground

        y = y₀ + v_{oy} t - ½ g t²

in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero (v_{oy} = 0)

       0 =y₀I - ½ g t²

       t = √ (2y₀ / g)

let's calculate

       t = √ (2 12 / 9.8)

       t = 1.56 s

Projectile range is the horizontal distance traveled

        x = v₀ₓ t

        x = 80 1.56

        x = 124.8 m

Impact speed when it hits the ground

        vₓ = v₀ₓ = 80 ms

        v_{y} = v_{oy} - gt

        v_{y} = - 9.8 1.56

        v_{y} = - 15,288 m / s

the speed is

       v = (80 i ^ - 15,288 j) m / s

7 0
3 years ago
The escape velocity is defined to be the minimum speed with which an object of mass m must move to escape from the gravitational
s344n2d4d5 [400]

Answer:

v = √2G M_{earth} / R

Explanation:

For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)

        Eo = K + U = ½ m1 v² - G m1 m2 / r1

        Ef = - G m1 m2 / r2

When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf

       Eo = Ef

       ½ m1v² - G m1 M_{earth} / R = - G m1 M_{earth} / R

      v² = 2G M_{earth} (1 / R - 1 / Rinf)

If we do Rinf = infinity     1 / Rinf = 0

       v = √2G M_{earth} / R

      Ef = = - G m1 m2 / R

The mechanical energy is conserved  

 

      Em = -G m1  M_{earth} / R

      Em = - G m1  M_{earth} / R

     R = int        ⇒  Em = 0

6 0
3 years ago
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