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ser-zykov [4K]
3 years ago
13

A toy car with an initial velocity of 5 m/s slows to a stop with an acceleration of -1.5 m/s^2.

Physics
1 answer:
enot [183]3 years ago
4 0

Car's Initial velocity (u) = 5 m/s

Car's Final velocity (v) = 0 m/s (Stop)

Acceleration of the car (a) = -1.5 m/s²

Equation used to solve this problem:

\boxed{ \bf{ {v}^{2}  =  {u}^{2}  + 2as}}

By substituting values in the equation, we get:

\rm \longrightarrow  {0}^{2}  =  {5}^{2}  +2  \times ( - 1.5) \times s \\  \\   \rm \longrightarrow 0 = 25 - 3s \\  \\  \rm \longrightarrow 25 - 3s = 0 \\  \\  \rm \longrightarrow 25 - 25 - 3s =0  - 25 \\  \\  \rm \longrightarrow  - 3s =  - 25 \\  \\  \rm \longrightarrow 3s = 25 \\  \\  \rm \longrightarrow  \dfrac{3s}{3}  =  \dfrac{25}{3}  \\  \\  \rm \longrightarrow s = 8.3 \: m

\therefore Displacement of the toy car (s) = 8.3 m

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If you double the distance between you and the center of Earth, what happens to the strength of the gravitational field you expe
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The strength of gravity decreases.

An example of that would be if you were in space; you float around because there's no gravity.

7 0
3 years ago
A 12.0 μF capacitor is charged to a potential of 50.0 V and then discharged through a 265 Ω resistor. A)How long does the capaci
larisa [96]

(a) The time for the capacitor to loose half its charge is 2.2 ms.

(b) The time for the capacitor to loose half its energy is 1.59 ms.

<h3>Time taken to loose half of its charge</h3>

q(t) = q₀e-^(t/RC)

q(t)/q₀ = e-^(t/RC)

0.5q₀/q₀ = e-^(t/RC)

0.5 = e-^(t/RC)

1/2 =  e-^(t/RC)

t/RC = ln(2)

t = RC x ln(2)

t = (12 x 10⁻⁶ x 265) x ln(2)

t = 2.2 x 10⁻³ s

t = 2.2 ms

<h3>Time taken to loose half of its stored energy</h3>

U(t) = Ue-^(t/RC)

U = ¹/₂Q²/C

(Ue-^(t/RC))²/2C = Q₀²/2Ce

e^(2t/RC) = e

2t/RC = 1

t = RC/2

t = (265 x 12 x 10⁻⁶)/2

t = 1.59 x 10⁻³ s

t = 1.59 ms

Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.

Learn more about energy stored in capacitor here: brainly.com/question/14811408

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6 0
1 year ago
Observe the given figure and find the the gravitational force between m1 and m2.​
Leno4ka [110]

Answer:

The gravitational force between m₁ and m₂, is approximately 1.06789 × 10⁻⁶ N

Explanation:

The details of the given masses having gravitational attractive force between them are;

m₁ = 20 kg, r₁ = 10 cm = 0.1 m, m₂ = 50 kg, and r₂ = 15 cm = 0.15 m

The gravitational force between m₁ and m₂ is given by Newton's Law of gravitation as follows;

F =G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}

Where;

F = The gravitational force between m₁ and m₂

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

r₂ = 0.1 m + 0.15 m = 0.25 m

Therefore, we have;

F = 6.67430 \times 10^{-11} \ N \cdot m^2/kg \times \dfrac{20 \ kg\times 50 \ kg}{(0.1 \ m+ 0.15 \ m)^{2}} \approx 1.06789 \times 10^{-6} \ N

The gravitational force between m₁ and m₂, F ≈ 1.06789 × 10⁻⁶ N

8 0
2 years ago
Draw a schematic diagram of a circuit consisting of a battery of five 2 V cells, a 5 ohm resistor, a 10 ohm resistor, a 15 ohm r
MakcuM [25]

Answer:

  • Current = 0.33 A

Explanation:

  • For diagram refer the attachment.

It is given that five cells of 2V are connected in series, so total voltage of the battery:

\dashrightarrow \: \:  \sf V = 2 \times 5 = 10 V

Three resistor of 5\Omega, 10\Omega, 15\Omega are connected in Series, so the net resistance:

\dashrightarrow \: \: \sf R_{n} = R_{1} + R_{2} + R_{3}

\dashrightarrow \: \:  \sf R = 5 + 10 + 15

{ \pink{\dashrightarrow \sf \: \: { \underbrace{R = 30 \:  \Omega}}}}

According to ohm's law:

\dashrightarrow  \sf\: \: V = IR

\dashrightarrow  \sf \: \: I = \dfrac{V}{R}

On substituting resultant voltage (V) as 10 V and resultant resistant, as 30 {\pmb{\sf{\Omega}}} we get:

\dashrightarrow \sf \: \: I = \dfrac{10V}{30\Omega}

{ \pink{\dashrightarrow \sf \: \: { \underbrace{I = 0.33 A}}}}

\thereforeThe electric current passing through the above circuit when the key is closed will be <u>0.33 A</u>

4 0
2 years ago
Read 2 more answers
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