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3 years ago

11

Decribe the word deceleration . Does it describe a vector quantity?

1 answer:

Komok [63]3 years ago

7 0

Answer: deceleration is considered to describe a decrease or negative change of speed or velocity. ... Three of these namely time, distance and speed are scalar quantities, whereas the remaining three attributes namely displacement, velocity and acceleration are vectors.

Explanation:

Read it online

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What is the equivalent resistance between points A and C if R1=1430, R2=1350, R3=1100, R4=1350, and R5=1150.

Marianna [84]

R1 + R4 = 1430 + 1350 = 2780 = R14 series combination of R1 & R4

R2 + R5 = 1350 + 1150 = 2500 = R25

The circuit has been reduced to 3 resistors in parallel

R314 = 2780 * 1100 / (2780 + 1100) = 788 this is the resistance of the parallel combination of R14 and R3

R31425 = 2500 * 788 / (2500 + 788) = 599 which is the equivalent of the circuit - you can also use the formula for 3 resistors in parallel but this seems simpler

7 0

3 years ago

a bicycle uniformly from rest at time t the velocity of the bicycle is v at what time will the bicycle have a velocity of 4v

sesenic [268]

Here

Acceleration and initial velocities are constant.

According to first equation of kinematics.

$\\ \sf\longmapsto v=u+at$

$\\ \sf\longmapsto v=0+at$

$\\ \sf\longmapsto v=at$

$\\ \sf\longmapsto v\propto t$

Time was t at velocity v
Time will be 4t at velocity 4v

7 0

3 years ago

A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap

viktelen [127]

Answer:

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

$q = 3 \times 10^{-6} C$

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

$V_r + V_L + V_c = V_{net}$

now we will have

$iR + L\frac{di}{dt} + \frac{q}{C} = 12 V$

now we have

$1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12$

So we will have

$q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}$

at t = 0 we have

q = 0

$0 = 3\times 10^{-6} + c_1 + c_2$

also we know that

at t = 0 i = 0

$0 = -4000 c_1 - 1000c_2$

$c_2 = -4c_1$

$c_1 = 1 \times 10^{-6}$

$c_2 = -4 \times 10^{-6}$

so we have

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

$q = 3 \times 10^{-6} C$

5 0

3 years ago

What type of wave is infrared light

GrogVix [38]

Answer: Heat waves? I’m not %100 sure

4 0

3 years ago

Read 2 more answers

Please help with physics 2 part question giving 50 points

AveGali [126]

Answer:

2.88×10⁻⁹ s

2.40×10¹⁵ m/s²

Explanation:

Given:

v₀ = 12300 m/s

v = 6.92×10⁶ m/s

Δx = 0.997 cm = 0.00997 m

Part 1) Find: t

Δx = ½ (v + v₀)t

0.00997 m = ½ (6.92×10⁶ m/s + 12300 m/s) t

t = 2.88×10⁻⁹ s

Part 2) Find: a

(6.92×10⁶ m/s)² = (12300 m/s)² + 2a (0.00997 m)

a = 2.40×10¹⁵ m/s²

5 0

4 years ago