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maria [59]
3 years ago
14

A wave on a string is reflected from a fixed end. The reflected wave 1. Is in phase with the original wave at the end. 2. Has a

larger speed than the original wave. 3. Is 180◦ out of phase with the original wave at the end. 4. Cannot be transverse. 5. Has a larger amplitude than the original wave.
Physics
1 answer:
MrRa [10]3 years ago
5 0

Answer:

3. Is 180◦ out of phase with the original wave at the end.

Explanation:

Here when wave is reflected by the rigid boundary then due to the rigidly bounded particles at the end or boundary they have tendency not to move and remains fixed at their position.

Due to this fixed position we can say when wave reach at that end the particles will not move and they apply equal and opposite force at the particles of string

Due to this the reflected wave is transferred back into the string in opposite phase with respect to the initial wave

so here correct answer will be

3. Is 180◦ out of phase with the original wave at the end.

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svet-max [94.6K]

a) Solids keep shape, liquids take shape of containers but don't spill, gases take container's shape and spill out

b) if you heat gas, speed of its molecules will increase and they'll push the container's walls stronger, so the pressure will increase when the container heated

c) Heat flows from warmer to colder bodies

d) For monatomic gases it's U=1.5nRT only, molecular gas has bonds between atoms so total internal energy increases

e) Of gases

f) v=\sqrt{\frac{3RT}{M}}=321 m/s

g) U=5/2*nRT=37830.85 J

7 0
3 years ago
a race car accelerates uniformly from 18.5 m/s to 46.1m/s in 2.47 seconds detrrmine the acceleration of the car and distance tra
LenaWriter [7]

Because acceleration is constant, the acceleration of the car at any time is the same as its average acceleration over the duration. So

a=\dfrac{\Delta v}{\Delta t}=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.47\,\mathrm s}=11.2\,\dfrac{\mathrm m}{\mathrm s^2}

Now, we have that

{v_f}^2-{v_0}^2=2a\Delta x

so we end up with a distance traveled of

\left(46.1\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(11.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x

\implies\Delta x=79.6\,\mathrm m

6 0
3 years ago
A team of engineering students is testing their newly designed 200 kg raft in the pool where the diving team practices. The raft
elena-s [515]

Answer:

The water level rises more when the cube is located above the raft before submerging.

Explanation:

These kinds of problems are based on the principle of Archimedes, who says that by immersing a body in a volume of water, the initial water level will be increased, raising the water level. That is, the height in the container with water will rise in level. The difference between the new volume and the initial volume of the water will be the volume of the submerged body.

Now we have two moments when the steel cube is held by the raft and when it is at the bottom of the pool.

When the cube is at the bottom of the water we know that the volume will increase, and we can calculate this volume using the volume of the cube.

Vc = 0.45*0.45*0.45 = 0.0911 [m^3]

Now when a body floats it is because a balance is established in the densities, the density of the body and the density of the water.

Ro_{H2O}=R_{c+r}\\where:\\Ro_{H2O}= water density = 1000 [kg/m^3]\\Ro_{c+r}= combined density cube + raft [kg/m^3]

Density is given by:

Ro = m/V

where:

m= mass [kg]

V = volume [m^3]

The buoyancy force can be calculated using the following equation:

F_{B}=W=Ro_{H20}*g*Vs\\W = (200+730)*9.81\\W=9123.3[N]\\\\9123=1000*9.81*Vs\\Vs = 0.93 [m^3]

Vs > Vc, What it means is that the combined volume of the raft and the cube is greater than that of the cube at the bottom of the pool. Therefore the water level rises more when the cube is located above the raft before submerging.

7 0
4 years ago
8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s
allsm [11]

Answer:

a) \omega = 50\,\frac{rad}{s}, b) \omega = 0\,\frac{rad}{s}

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

\tau = F \cdot r

\tau = m\cdot a \cdot r

\tau = m \cdot \alpha \cdot r^{2}

Where \alpha is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}

\alpha = -3\,\frac{rad}{s^{2}}

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)

\omega = 50\,\frac{rad}{s}

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}

t = 66.667\,s

Since t > 66.667\,s, then the angular velocity is equal to zero. Therefore:

\omega = 0\,\frac{rad}{s}

7 0
3 years ago
describe the time of day that an early explorer might have planned to enter a harbor and when he might have planned to leave
inessss [21]
He would try to enter as the tide is rising, and leave as the tide is falling. Those things happen at all different times of day during a month.
4 0
3 years ago
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