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Maurinko [17]
2 years ago
10

g A ping pong ball (thin shell sphere) rolls down an incline at 30° from rest. What is its acceleration

Physics
1 answer:
Aleonysh [2.5K]2 years ago
3 0

Answer:

Explanation:

According to newtons second law of motion;

F = ma .... 1

Also the force acting aong the inclines is expressed as;

F = mgsintheta

m is the mass of the object

a is the acceleration

theta is angle of inclination

Equate 1 and 2

ma = mg sin theta

a = gsin(theta)

a = 9.8sin30

a = 9.8(0.5)

a = 4.9m/s²

Hence the acceleration of the ping pong is 4.9m/s²

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The planet Krypton has a mass of 7.6 × 1023 kg and radius of 1.7 × 106 m. What is the acceleration of an object in free fall nea
olga55 [171]

Answer:

17.55 m/s²

Explanation:

Parameters given:

Mass of Krypton, M = 7.6 * 10^23 kg

Radius, R = 1.7 * 10^6 m

Gravitational constant, G = 6.6726 * 10^(-11) Nm²/kg²

Acceleration due to gravity of planet of mass M is given as:

g = GM/R²

Since the object is close to the surface of Krypton, we can say that the distance from the Centre of Krypton is the radius of the planet Krypton.

Therefore,

g = (6.6726 * 10^(-11) * 7.6 * 10^23)/(1.7 * 10^6)²

g = 17.55 m/s²

5 0
3 years ago
During a 0.001 s interval while it is between the plates, the change of the momentum of the electron Δp with arrow is < 0, -8
Delvig [45]

Answer:

5.5 x 10^5 N/C

Explanation:

t = 0.001 s

Δp = - 8.8 x 10^-17 kg m /s

Force is equal to the rate of change of momentum.

F = Δp / Δt

F = (8.8 x 10^-17) / 0.001 = 8.8 x 10^-14 N

q = 1.6 x 10^-19 C

Electric field, E = F / q = (8.8 x 10^-14) / (1.6 x 10^-19)

E = 5.5 x 10^5 N/C

8 0
3 years ago
Which of the following sensory receptors would lead you to squint in bright light?. A. thermoreceptors B. mechanoreceptors C. ph
kipiarov [429]
<span>If my memory serves me well, sensory receptors which would lead you to squint in bright light are called </span><span>C. photoreceptors</span>
3 0
3 years ago
Which type of physical activity is being performed in the picture?
mafiozo [28]
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3 0
3 years ago
Read 2 more answers
Uma carga puntiforme de + 3,0uC é colocada em um ponto P de um campo elétrico gerado por uma partícula eletrizada com carga desc
expeople1 [14]

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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