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Paul [167]
3 years ago
6

A force of 8,480 N is applied to a cart to decelerate it at a rate of 32.0 m/s 2 . What is the mass of the cart?

Physics
1 answer:
Vinvika [58]3 years ago
5 0
The force is given by the product of mass and the acceleration thus the force is given
8480/32 is equal to 265 kg
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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
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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)=

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________ is the average kinetic energy of each atom.<br><br> radiation<br> temperature<br> potential
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When light is incident normally on the interface between two transparent optical media, the intensity of the reflected light is
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3 years ago
Uma carga puntiforme de + 3,0uC é colocada em um ponto P de um campo elétrico gerado por uma partícula eletrizada com carga desc
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Responda:

1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5

Explicação:

Dado o seguinte:

Carga (q) = 3uC = 3 × 10 ^ -6C

Força elétrica (Fe) = 18N

Intensidade do campo elétrico (E) =?

1)

Lembre-se:

Força elétrica (Fe) = carga (q) * Intensidade do campo elétrico (E)

Fe = qE; E = Fe / q

E = 18N / (3 × 10 ^ -6C)

E = 6N / 10 ^ -6C

E = 6 × 10 ^ 6NC ^ -1

2)

Lembre-se:

E = kQ / r ^ 2

E = intensidade do campo elétrico

Q = carga de origem

r = distância de espera = 30cm = 30/100 = 0,3m

K = 9,0 × 10 ^ 9

6 × 10 ^ 6 = (9,0 × 10 ^ 9 * Q) / 0,3 ^ 2

9,0 × 10 ^ 9 * Q = 6 × 10 ^ 6 * 0,09

Q = 0,54 × 10 ^ 6 / 9,0 × 10 ^ 9

Q = 0,06 × 10 ^ (6-9)

Q = 0,06 × 10 ^ -3

Q = 6 × 10 ^ -5 = 60 × 10 ^ -6 = 60μC

7 0
4 years ago
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