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2 years ago

Which of the following definitions BEST defines a satellite?

Physics

1 answer:

Temka [501]2 years ago

5 0

Answer:

D, any object that revolves around another object.

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Write two characteristics of an educated society

almond37 [142]

Answer:

Different and better?

Explanation:

i dont think that helps lol

7 0

2 years ago

what is the magnitude of the vector described below? 13 m/s to the east a. east b. meters per second c. 13 m/s d. meters

Katen [24]

Answer:

Explanation:

Magnitude of any quantity is the measurable value of the quantity. While the direction of the given quantity is the specific pointing direction of position or the angle at which it move.

The magnitude of the vector described below? 13 m/s to the east will be 13 m/s

While the direction will be eastward.

Therefore, the magnitude is 13 m/s

The correct answer is option C

3 0

2 years ago

Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 10

11111nata11111 [884]

Answer:

$1.62\times 10^{-8}\ \text{s}$

Explanation:

$\epsilon_0$ = Vacuum permittivity = $8.854\times 10^{-12}\ \text{F/m}$

$A$ = Area = $10\times 2\times 10^{-4}\ \text{m}^2$

$d$ = Distance between plates = 1 mm

$V_c$ = Changed voltage = 60 V

$V$ = Initial voltage = 100 V

$R$ = Resistance = $1000\ \Omega$

Capacitance is given by

$C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.854\times 10^{-12}\times 10\times 2\times 10^{-4}}{1\times 10^{-3}}\\\Rightarrow C=1.7708\times 10^{-11}\ \text{F}$

We have the relation

$V_c=V(1-e^{-\dfrac{t}{CR}})\\\Rightarrow e^{-\dfrac{t}{CR}}=1-\dfrac{V_c}{V}\\\Rightarrow -\dfrac{t}{CR}=\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-CR\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-1.7708\times 10^{-11}\times 1000\ln(1-\dfrac{60}{100})\\\Rightarrow t=1.62\times 10^{-8}\ \text{s}$

The time taken for the potential difference to reach the required level is $1.62\times 10^{-8}\ \text{s}$ .

5 0

2 years ago

A person with mass 65 kg, standing still, throws an object at 4 m/s. If the

fenix001 [56]

Explanation:

Momentum before = momentum after

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(65 kg) (0 m/s) + m (0 m/s) = (65 kg) (-3.5 m/s) + m (4 m/s)

m ≈ 57 kg

7 0

3 years ago

The position of a 55 g oscillating mass is given by x(t)=(2.0cm)cos(10t), where t is in seconds. determine the velocity at t=0.4

NNADVOKAT [17]

The position of the mass is given by (in cm):
$x(t)=2 \cos (10 t)$
The velocity is the derivative of the position:
$v(t) = \frac{dx(t)}{dt} =-10\cdot 2 \sin (10t)=-20 \sin (10t)$
Substituting t=0.40 s, we can find the velocity at this time:
$v(0.40 s)= -20 \sin (10 \cdot 0.4)=15 cm/s=15 \cdot 10^{-2}m/s$

5 0

2 years ago

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