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3 years ago

If you are asked to determine an object’s speed, what information must you have?

1 answer:

5 0

A) distance and period of time

The speed of an object is equal to the distance it travels divided by the amount of time it takes to travel the distance.

This measures the average speed

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A parallel-plate capacitor connected to a battery becomes fully charged. After the capacitor from the battery is disconnected, t

Valentin [98]

Answer:

C.)The energy stored in the capacitor doubles its original value.

Explanation:

The capacitance of a capacitor is given by:

C₁ = ∈A/d...............................(1)

If the distance between the plates is doubled

C₂ = ∈A/2d.............................(2a)

From equations (1) and (2), the relationship between C₁ and C₂ is:

C₂ = C₁/2....................................(2b)

The original Energy stored in the capacitor is given by :

E₁ = Q²/2C₁...............................(3)

On doubling the separation between the capacitance plates

E₂ = Q²/2C₂...............................(4a)

E₂ = Q²/2 * 1/C₂........................(4b)

Putting equation (2b) into (4b)

E₂ = Q²/2 * 2/C₁

E₂ = Q²/C₁.....................................(5)

Comparing equations (3) and (5)

E₂ = 2E₁

Therefore, the energy stored in the capacitor doubles its original value

4 0

4 years ago

Helppppppppppp im dieing

Otrada [13]

It would be 1. B 2. A 3. A

7 0

3 years ago

A rocket travels 1.3 km in 62 ms. What is its average speed in m⋅s−1? Do not give your answer in scientific notation. The answer

hjlf

Answer:

Average speed = 0.35 m/s

Explanation:

Given the following data;

Distance = 1.3 Km

Time = 62 minutes

To find the average speed in m/s;

First of all, we would convert the quantities to their standard unit (S.I) of measurement;

Conversion:

1.3 kilometres to meters = 1.3 * 1000 = 1300 meters

For time;

1 minute = 60 seconds

62 minutes = X

Cross-multiplying, we have;

X = 62 * 60

X = 3720 seconds

Now, we can calculate the average speed in m/s using the formula;

$Speed = \frac {distance}{time}$

$Speed = \frac {1300}{3720}$

Average speed = 0.35 m/s

7 0

3 years ago

A long solenoid with 8.22 turns/cm and a radius of 7.00 cm carries a current of 19.4 mA. A current of 3.59 A exists in a straigh

daser333 [38]

Answer:

a. 3.039cm

b.magnetic field is $B=2.958\times10^{-5}T$

Explanation:

Direction of the solenoid magnetic field is along the axis of the solenoid. and magnetic field due to the wire perpendicular to that due to the solenoid.. Magnetic field at r is given by:

$\overrightarrow B = \overrightarrow B_s+ \overrightarrow B_w,\ \ \ \ \ \overrightarrow B_s\perp \overrightarrow B_w$

Angle of net magnetic field from axial direction is given by:

$tan\ \theta=\frac{B_w}{B_s}$ ,

Field due to solenoid:

$B_s=\mu_onI_s, \ \ \ \ n=(8.22 t/cm)(100cm/m)=822turn/m$

Field due to wire:

$B_w=\frac{\mu_oI_w}{2\pi r}$

Therefore, r:

$tan\ \theta=\frac{B_w}{B_s}\\\\=\frac{\mu_oI_w}{2\pi r(\mu_o nI_s)}\\\\r=\frac{I_w}{2\pi nI_stan \ \theta}\\\\r=\frac{3.59A}{2\pi\times822\times19.4\times10^{-3}A \ tan 49.7\textdegree}\\\\r=3.039cm$

Hence, the radial distance is 3.039cm

b.The magnetic field strength is given by:

$B=\sqrt{B_w^2+B_s^2}\\\\tan 49.7\textdegree=\frac{B_w}{B_s}\\\\1.179=\frac{B_w}{B_s}\\\\B_w=1.179B_s\\\\B=\sqrt{(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{3}A)+1.179(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{-3}A)}\\\\B=2.958\times10^{-5}T$

Hence, the magnetic field is $B=2.958\times10^{-5}T$

7 0

3 years ago

Car A (1750 kg) is travelling due south and car B (1450 kg) is travelling due east. They reach the same intersection at the same

Contact [7]

Consider the east-west direction along x-axis and north-south direction along y-axis. In unit vector notation, velocities can be given as

$\underset{V_{A}}{\rightarrow}$ = velocity of car A before collision = 0 i - $V_{A}$ j

$\underset{V_{B}}{\rightarrow}$ = velocity of car B before collision = $V_{B}$ i + 0 j

$\underset{V_{AB}}{\rightarrow}$ = velocity of combination after collision = (35.8 Cos31.6) i - (35.8 Sin31.6) j = 30.5 i - 18.8 j

$M_{A}$ = mass of car A = 1750 kg

$M_{B}$ = mass of car B = 1450 kg

Using conservation of momentum

$M_{A}$ $\underset{V_{A}}{\rightarrow}$ + $M_{B}$ $\underset{V_{B}}{\rightarrow}$ = ( $M_{A}$ + $M_{B}$ ) ( $\underset{V_{AB}}{\rightarrow}$ )

(1750) (0 i - $V_{A}$ j) + (1450) ( $V_{B}$ i + 0 j) = (1750 + 1450) (30.5 i - 18.8 j)

(1450) $V_{B}$ i - (1750) $V_{A}$ j = 97600 i - 60160 j

Comparing the coefficient of "i" and "j" both side

(1450) $V_{B}$ = 97600 and - (1750) $V_{A}$ = - 60160

$V_{B}$ = 67.3 km/h and $V_{A}$ = 34.4 km/

6 0

4 years ago