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Pachacha [2.7K]

3 years ago

13

They are 4 springs stretched by the same mass. Spring a stretches 25cm, spring B stretches 10cm, spring C stretches 100cm, and s

pring B stretches 1 cm, which spring has the largest spring constant

1 answer:

Wewaii [24]3 years ago

6 0

Answer:

Spring C

Explanation:

Multiply 25 and 4

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Draw a free-body diagram of the rod ab. Assume the contact surface at b is smooth.

Romashka [77]

Answer:

See attachment

Explanation:

5 0

3 years ago

Suppose you receive $15.00 at the beginning of a week and spend $2.50 each day for lunch. You prepare a graph of the amount you

cestrela7 [59]

Answer:

positive or zero

Explanation:

if you say the amount can be accumulated, then the slope should be positive

if the amount can't be accumulated , the slope should be zero

8 0

3 years ago

In the electric current flow, it is found that the resistance (measured in the units called ohms) offered by a fixed length of w

Likurg_2 [28]

<span>Here's your explanation:
R = k / d^2 (k is a constant, unknown)
</span><span>d1 = 0.01 and R1 = 0.306
d2 = 0.0299 R2 = ?
</span>
<span>Use the equation in line 1 for both case and divide and k will go away and you can find R2</span>

8 0

3 years ago

As a result of friction, the angular speed of a wheel changes with time according to dθ/dt = ω0e^−σt where ω0 and σ are constant

NNADVOKAT [17]

Hi there!

a.

We can use the initial conditions to solve for w₀.

It is given that:

$\frac{d\theta}{dt} = w_0e^{-\sigma t}$

We are given that at t = 0, ω = 3.7 rad/sec. We can plug this into the equation:

$\omega(0)= \omega_0e^{-\sigma (0)}\\\\3.7 = \omega_0 (1)\\\\\omega_0 = \boxed{3.7 rad/sec}$

Now, we can solve for sigma using the other given condition:

$2 = 3.7e^{-\sigma (8.6)}\\\\.541 = e^{-\sigma (8.6)}\\\\ln(.541) = -\sigma (8.6)\\\\\sigma = \frac{ln(.541)}{-8.6} = \boxed{0.0714s^{-1}}$

b.

The angular acceleration is the DERIVATIVE of the angular velocity function, so:

$\alpha(t) = \frac{d\omega}{dt} = -\sigma\omega_0e^{-\sigma t}\\\\\alpha(t) = -(0.0714)(3.7)e^{-(0.0714) (3)}\\\\\alpha(t) = \boxed{-0.213 rad\sec^2}$

c.

The angular displacement is the INTEGRAL of the angular velocity function.

$\theta (t) = \int\limits^{t_2}_{t_1} {\omega(t)} \, dt\\\\\theta(t) = \int\limits^{2.5}_{0} {\omega_0e^{-\sigma t}dt\\\\$

$\theta(t) = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=2.5} \atop {t_1=0}} \right.$

$\theta = -\frac{3.7}{0.0714}e^{-0.0714 t}\left \| {{t_2=2.5} \atop {t_1=0}} \right. \\\\\theta= -\frac{3.7}{0.0714}e^{-0.0714 (2.5)} + \frac{3.7}{0.0714}e^{-0.0714 (0)}$

$\theta = 8.471 rad$

Convert this to rev:

$8.471 rad * \frac{1 rev}{2\pi rad} = \boxed{1.348 rev}$

d.

We can begin by solving for the time necessary for the angular speed to reach 0 rad/sec.

$0 = 3.7e^{-0.0714t}\\\\t = \infty$

Evaluate the improper integral:

$\theta = \int\limits^{\infty}_{0} {\omega_0e^{-\sigma t}dt\\\\$

$\lim_{a \to \infty} \theta = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=a} \atop {t_1=0}} \right.$

$\lim_{a \to \infty} \theta = -\frac{3.7}{0.0714}e^{-0.0714a} + \frac{3.7}{0.0714}e^{-0.0714(0)}\\\\ \lim_{a \to \infty} \theta = \frac{3.7}{0.0714}(1) = 51.82 rad$

Convert to rev:

$51.82 rad * \frac{1rev}{2\pi rad} = \boxed{8.25 rev}$

8 0

3 years ago

Given a 3.00 μF capacitor, a 7.75 μF capacitor, and a 5.00 V battery, find the charge on each capacitor if you connect them in t

USPshnik [31]

Answer:

a) Q1= Q2= 11.75×10^-6Coulombs

b) Q1 =15×10^-6coulombs

Q2 = 38.75×10^-6coulombs

Explanation:

a) For a series connected capacitors C1 and C2, their equivalent capacitance C is expressed as

1/Ct = 1/C1 + 1/C2

Given C1 = 3.00 μF C2 = 7.75μF

1/Ct = 1/3+1/7.73

1/Ct = 0.333+ 0.129

1/Ct = 0.462

Ct = 1/0.462

Ct = 2.35μF

V = 5.00Volts

To calculate the charge on each each capacitors, we use the formula Q = CtV where Cf is the total equivalent capacitance

Q = 2.35×10^-6× 5

Q = 11.75×10^-6Coulombs

Since same charge flows through a series connected capacitors, therefore Q1= Q2=

11.75×10^-6Coulombs

b) If the capacitors are connected in parallel, their equivalent capacitance will be C = C1+C2

C = 3.00 μF + 7.75 μF

C = 10.75 μF

For 3.00 μF capacitance, the charge on it will be Q1 = C1V

Q1 = 3×10^-6 × 5

Q1 =15×10^-6coulombs

For 7.75 μF capacitance, the charge on it will be Q2 = 7.75×10^-6×5

Q2 = 38.75×10^-6coulombs

Note that for a parallel connected capacitors, same voltage flows through them but different charge, hence the need to use the same value of the voltage for both capacitors.

7 0

3 years ago