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3 years ago

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The position of a 60 g oscillating mass is given by x(t)=(2.0cm)cos(10t), where t is in seconds. determine the velocity at t=0.4

0s.

1 answer:

mrs_skeptik [129]3 years ago

3 0

The position of an oscillating mass is given by:

$x(t)=A cos (\omega t)$

where A is the amplitude of the oscillation, $\omega$ the angular frequency and t the time.

The velocity of the oscillating mass can be found by calculating the derivative of the position:

$v(t)=x'(t)=-\omega A sin (\omega t)$

In this problem, A=2.0 cm and $\omega=10 rad/s$ , so if we substitute these data and t=0.4 s we can find the velocity at t=0.4 s:

$v(t)=-(10 rad/s)(2.0 cm) sin ((10 rad/s)(0.4 s))=-13.07 cm/s=-0.13 m/s$

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Determine the magnitude of the effective value of g⃗ at a latitude of 60 ∘ on the earth. assume the earth is a rotating sphere.

dezoksy [38]

In addition to acceleration of gravity we experience centrifugal acceleration away from the axis of rotation of the earth. this additional acceleration has value ac = r w^2 where w = angular velocity and r is distance from your spot on earth to the earth's axis of rotation so r = R cos(l) where l = 60 deg is the lattitude and R the earth's radius and w = 1 / (24hr x 3600sec/hr)
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5 0

3 years ago

Aloop of wire of area 71 cm^2 is placed with its plane parallel to a 16 mt magnetic field. the loop is then rotated so that its

kkurt [141]

Answer:

Approximately 1.62 × 10⁻⁴ V.

Explanation:

The average EMF in the coil is equal to

$\displaystyle \frac{\text{Final Magnetic Flux} - \text{Initial Magnetic Flux}}{2}$ ,

Why does this formula work?

By Faraday's Law of Induction, the EMF $\epsilon$ induced in a coil (one loop) is equal to the rate of change in the magnetic flux $\Phi$ through the coil.

$\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t))$ .

Finding the average EMF in the coil is similar to finding the average velocity.

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt$ .

However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

$\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0)$ .

Hence the equation

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}$ .

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in $\Phi(t)$ won't matter.

Apply this formula to this question. Note that $\Phi$ , the magnetic flux through the coil, can be calculated with the equation

$\Phi = B \cdot A \cdot N \; \sin{\theta}$ .

For this question,

$B = \rm 16\; mT = 16\times 10^{-3}\; T$ is the strength of the magnetic field.
$A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2}$ is the area of the coil.
$N = 1$ is the number of loops in the coil.
$\theta$ is the angle between the field lines and the coil.
At $\rm 0\;s$ , the field lines are parallel to the coil, $\theta = 0^{\circ}$ .
At $\rm 0.7\; s$ , the field lines are perpendicular to the coil, $\displaystyle \theta = 90^{\circ}$ .

Initial flux: $\Phi(0)= 0$ .

Final flux: $\Phi(0.7) = \rm 1.1136\times 10^{-4}\; Wb$ .

Average EMF, which is the same as the average rate of change in flux:

$\displaystyle \frac{\Phi(0.7) - \Phi(0)}{0.7} \approx\rm 1.62\times 10^{-4}\; V$ .

8 0

3 years ago

Explain the difference between vehicular kinematics and occupant kinematics

belka [17]

Vehicle Kinematics: a vehicle sliding sideways into a gravel pit, a vehicle driving down a bank, a vehicle driving up a ramp and a vehicle sliding laterally against a curb. "sliding laterally against a curb" and "sliding sideways into a gravel pit".

8 0

3 years ago

Read 2 more answers

A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130

konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(c) W = 0J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

$W=Fd$ (1)

F: applied force = 130N

d: distance = 5.0m

$W=(130N)(5.0m)=650J$

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

$W_f=F_fd=\mu Mgd$ (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

$W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J$

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

$W = 0J$

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

$\Delta K=W_T$ (3)

You calculate the total work:

$W_T=Fd-F_fd=(F-F_f)d$ (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

$F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N$

Next, you replace the values of all parameters in the equation (4):

$W_T=(130N-105.84N)(5.00m)=120.80J$

$\Delta K=120.80J$

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

$W_T=\frac{1}{2}M(v_2^2-v_1^2)$ (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

$v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}$

The final speed of the box is 2.58m/s

5 0

3 years ago

During typical urination, a man releases about 400 mL of urine in about 30 seconds through the urethra, which we can model as a

gregori [183]

Answer:

Explanation:

Given:

volume of urine discharged, $V=400~mL=0.4~L=4\times 10^{-4}~m^3$

time taken for the discharge, $t=30~s$

diameter of cylindrical urethra, $d=4\times10^{-3}~m$

length of cylindrical urethra, $l=0.2~m$

density of urine, $\rho=1000~kg/m^3$

a)

we have volume flow rate Q:

$Q=A.v$ & $Q=\frac{V}{t}$

where:

$A=$ cross-sectional area of urethra

$v=$ velocity of flow

$A.v=\frac{V}{t}$

$\frac{\pi d^2}{4}\times v=\frac{4\times 10^{-4}}{30}$

$v=\frac{4\times4\times 10^{-4}}{30\times \pi (4\times 10^{-3})^2}$

$v=1.06~m/s$

b)

The pressure required when the fluid is released at the same height as the bladder and that the fluid is at rest in the bladder:

$P=\rho.g.l$

$P=1000\times 9.8\times 0.2$

$P=1960~Pa$

5 0

3 years ago