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rjkz [21]
3 years ago
13

a student wants to keep their coffee hot they have a choice of putting it in a silver drinks container or a blacks drinks contai

ner
Physics
1 answer:
Vaselesa [24]3 years ago
5 0

Answer: a silver cup would be better because it would most Likely be metal and metal. Keeps things hot for a long time

Explanation:

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a 202 kg bumper car moving right at 8.50 m/s collides with a 355 kg car at rest. Find the total momentum of the system.
Hunter-Best [27]

Explanation:It is given that,

Mass of bumper car, m₁ = 202 kg

Initial speed of the bumper car, u₁ = 8.5 m/s

Mass of the other car, m₂ = 355 kg

Initial velocity of the other car is 0 as it at rest, u₂ = 0

Final velocity of the other car after collision, v₂ = 5.8 m/s

Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁

Using the conservation of linear momentum as :

p₁ = m₁v₁ = -342 kg-m/s

So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.

3 0
3 years ago
Read 2 more answers
O
Sergio [31]

see

below

Explanation:

refractive index = speed of light in vacuum / speed of light in medium

light travels at a speed of 3.0 x 10^8 m/s in vacuum

refractive index = 3.0 x 10^8 / 2.0 x 10^8

refractive index = 1.5

hope this helps, please mark it

4 0
3 years ago
An unmarked police car traveling a constant 85 km/h is passed by a speeder. Precisely 2.00 s after the speeder passes, the polic
steposvetlana [31]

Answer:

observe o gráfico abaixo que mostra o número de internação em um hospital no município no período de 1960

6 0
3 years ago
A 1 pF capacitor is connected in parallel with a 2 pF capacitor, the parallel combination then being connected in series with a
asambeis [7]

Hi there!

We can approach this problem in many ways, but to show you how I arrive to the final conclusion, I will begin by solving the circuit with an assigned value for the power source.

Let's use a power source value of 6V (produces nice numbers).

Recall the following rules.

Capacitors in series:

  • Voltage ADDS up.
  • Charge is EQUAL across each.
  • Total capacitance uses the reciprocal rule.

Capacitors in parallel:

  • Voltage is EQUAL across each.
  • Charge ADDS up.
  • Total capacitance is simply the sum.

Solving for total capacitance:
C_P = 1 + 2 = 3 pF\\\\C_T = \frac{1}{\frac{1}{3} + \frac{1}{3}} = 1.5 pF

Using rules for capacitors in series and parallel, the total capacitance of the circuit is 1.5 pF.

Thus, the total charge is:
Q = C_TV\\Q = 1.5 pF * 6 V = 9 pC

9 pC will go through the parallel combination and the individual capacitor in series with the combination.

Since the voltage adds up, we can find the amount of voltage across the 3pF capacitor with the remaining going through the branches of the parallel combination.

V = \frac{Q}{C}\\\\V = \frac{9pC}{3pF} = 3V

Therefore, 3V goes through each branch since 6V - 3V = 3V.

Solving for the charge for each capacitor:
Q  = CV\\Q = (1 pF)(3V) = 3pC\\\\Q = (2pF)(3V) = 6pC\\\\Q = (3pF)(3V) = 9pC

<u>Thus, the capacitor with the greatest charge is the 3 pF capacitor. </u>

To explain without all of the work above, the equivalent capacitance of the parallel combination (1 pF + 2pF = 3pF) is equivalent to the capacitance of the capacitor in series (3pF). Thus, the voltage across the parallel capacitors (since voltage is the same across branches in parallel) and the series capacitor is equal. However, since charge SUMS UP for capacitors in parallel, they would have less charge than the single capacitor in series.

5 0
2 years ago
A mass of 1 slug is suspended from a spring whose spring constant is 9 lb/ft. The mass is initially released from a point 1 foot
FromTheMoon [43]

Answer:

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

Explanation:

To solve the problem/ we first write the differential equation governing the motion. So,

m\frac{d^{2} x}{dt^{2} } = -kx \\ m\frac{d^{2} x}{dt^{2} } + kx = 0\\\frac{d^{2} x}{dt^{2} } + \frac{k}{m} x = 0

with m = 1 slug and k = 9 lb/ft, the equation becomes

\frac{d^{2} x}{dt^{2} } + \frac{9}{1} x = 0\\\frac{d^{2} x}{dt^{2} } + 9 x = 0

The characteristic equation is

D² + 9 = 0

D = ±√-9 = ±3i

The general solution of the above equation is thus

x(t) = c₁cos3t + c₂sin3t

Now, our initial conditions are

x(0) = -1 ft and x'(0) = -√3 ft/s

differentiating x(t), we have

x'(t) = -3c₁sin3t + 3c₂cos3t

So,

x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)

x(0) = c₁cos(0) + c₂sin(0)

x(0) = c₁ × (1) + c₂ × 0

x(0) = c₁ + 0

x(0) = c₁ = -1

Also,

x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)

x'(0) = -3c₁sin(0) + 3c₂cos(0)

x'(0) = -3c₁ × 0 + 3c₂ × 1

x'(0) = 0 + 3c₂

x'(0) = 3c₂ = -√3

c₂ = -√3/3

So,

x(t) = -cos3t - (√3/3)sin3t

Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)

where A = √c₁² + c₂² = √[(-1)² + (-√3/3)²] = √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3 and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.

Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3

x(t) = (2√3/3)sin(3t + 4π/3)

So, the velocity  v(t) = x'(t) = (2√3)cos(3t + 4π/3)

We now find the times when v(t) = 3 ft/s

So (2√3)cos(3t + 4π/3) = 3

cos(3t + 4π/3) = 3/2√3

cos(3t + 4π/3) = √3/2

(3t + 4π/3) = cos⁻¹(√3/2)

3t + 4π/3 = ±π/6 + 2kπ    where k is an integer

3t  = ±π/6 + 2kπ - 4π/3

t  = ±π/18 + 2kπ/3 - 4π/9

t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9

t = π/18 - 4π/9 + 2kπ/3  or -π/18 - 4π/9 + 2kπ/3

t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3

Since t is not less than 0, the values of k ≤ 0 are not included

So when k = 1,

t = 5π/18 and π/6. So,

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

6 0
3 years ago
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