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3 years ago

Một vật dao động điều hoà trên quỹ đạo dài 20cm. Ở ví trí cân bằng, vật có vận tốc 20(cm/s). Chu kỳ dao động của vật là

Physics

1 answer:

satela [25.4K]3 years ago

8 0

Answer:

20cm

Explanation:

Im not sure so double check it

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What is the change in momentum of a 0.18 kg arrow that is traveling at 100 m/s and is stopped by a hay

Kryger [21]

The 2 seconds part isn’t relèvent to the question.
So momentum = mass x velocity
So 0.18 x 100 = 18

3 0

3 years ago

The steering wheel of a car has a radius of 0.19 m, and the steering wheel of a truck has a radius of 0.25 m. The same force is

kodGreya [7K]

Answer:

$\frac{T_t}{T_c} = 1.32$

Explanation:

The torque applied on an object can be calculated by the following formula:

$T = Fr$

where,

T = Torque

F = Applied Force

r = radius of the wheel

For car wheel:

$T_c = Fr_c\\$

For truck wheel:

$T_t = Fr_t$

Dividing both:

$\frac{T_t}{T_c} = \frac{Fr_t}{Fr_c}$

for the same force applied on both wheels:

$\frac{T_t}{T_c} = \frac{r_t}{r_c} \\$

where,

rt = radius of the truck steering wheel = 0.25 m

rc = radius of the car steering wheel = 0.19 m

Therefore,

$\frac{T_t}{T_c} = \frac{0.25\ m}{0.19\ m} \\$

$\frac{T_t}{T_c} = 1.32$

8 0

3 years ago

HELP PLS LIKE RIGHT NOW PLSSSSSS

Anit [1.1K]

The answer is A for this one

5 0

3 years ago

A coil is connected in series with a 12.6 kΩ resistor. An ideal 65.2 V battery is applied across the two devices, and the curren

ivolga24 [154]

To solve the problem, start by applying the concepts related to current in an RL circuit. The current is defined exponentially and using Ohm's law we can put the initial current in terms of the voltage and resistance. Consecutively with the calculated time constant we can find the respective inductance. For the second part we will apply the electrical potential energy connectors to find the amount of stored energy.

PART A)

$i = i_0 (1-e^{-t/T})$

$i = (V/R)(1-e^{-t/T})$

$0.00402 = (65.2/12600)(1-e^{-0.00402/T})$

$T = 0.002679s$

$T = L/R$

Inductance can be defined then,

$L = RT$

$L = (12600)(0.002679)$

$L = 33.75H$

PART B) Now the energy is given under the terms:

$E = \frac{1}{2}Li^2$

$E = \frac{1}{2} (33.75)(0.00402)^2$

$E = 0.0002727J$

Therefore the energy stored in the coil at this same moment is 0.0002727J

3 0

3 years ago

A positively charged particle is in the center of a parallel plate capacitor that has charge +/- Q on it's plates. Suppose the d

ivolga24 [154]

Answer:

the force remains constant if the charge does not change

Explanation:

In a capacitor the capacitance is given by

C = ε₀ A / d

Where ε₀ is the permissiveness of emptiness, A is about the plates and d the distance between them.

The charge on the capacitor is given by the ratio

Q = C ΔV

Let's apply these expressions to our problem, if the load remains constant

C = Q / ΔV = ε₀ A / d

ΔV / d = Q / ε₀ A

If the distance increases the capacitance should decrease, therefore if the charge is a constant the voltaje difference must increase

Now we can analyze the force on the test charge in the center of the capacitor

ΔV = E d

E= ΔV/d

F = q E

F = q ΔV / d

Let's replace

F = q Q /ε₀ A

From this expression we see that the force is constant since the voltage increase is compensated by increasing the distance, therefore the correct answer is that the force remains constant if the charge does not change

8 0

3 years ago