a boat accelerates at a rate of 6.0m/s down a river. how much time will it take the boat to speed up to 7.0m/s?

Adjacent antinodes of a standing wave on a string are 15.0 cm apart. A particle at an antinode oscillates in simple harmonic mot

timama [110]

Answer:

Explanation:

A ) Distance between two adjacent anti-node will be equal to distance between two adjacent nodes . So the required distance is 15 cm .

B ) wave-length, amplitude, and speed of the two traveling waves that form this pattern are as follows

wave length = same as wave length of wave pattern formed. so it is 30 cm

amplitude = 1/2 the amplitude of wave pattern formed so it is .850 / 2 = .425 cm

Speed = frequency x wavelength ( frequency = 1 / time period )

= 1 / .075) x 30 cm

400 cm / m

C ) maximum speed

= ω A

= (2π / T) x A

= 2 X 3.14 x .85 / .075 cm / s

= 71.17 cm / s

minimum speed is zero.

D ) The shortest distance along the string between a node and an antinode

= Wavelength / 4

= 30 / 4

= 7.5 cm

3 0

3 years ago

Which of the following represents thermal energy transfer through radiation

Bogdan [553]

C. since the the heat from the heater is going to the child in <u>waves</u>, it’s<u> radiating </u>

3 0

3 years ago

To practice Problem-Solving Strategy 27.1: Magnetic Forces. A particle with mass 1.81×10−3 kgkg and a charge of 1.22×10−8 CC has

Roman55 [17]

Answer:

The magnitude of the acceleration is $a = 0.33 m/s^2$

The direction is $- \r k$ i.e the negative direction of the z-axis

Explanation:

From the question we are that

The mass of the particle $m = 1.8*10^{-3} kg$

The charge on the particle is $q = 1.22*10^{-8}C$

The velocity is $\= v = (3.0*10^4 m/s ) j$

The the magnetic field is $\= B = (1.63T)\r i + (0.980T) \r j$

The charge experienced a force which is mathematically represented as

$F = q (\= v * \= B)$

Substituting value

$F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \ ( 1.63 \r i + 0.980 \r j )T)$

$= 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \ X \ \ \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \ \r j))$

$= (-5.966*10^4 N) \r k$

Note :

$i \ \ X \ \ j = k \\\\j \ \ X \ \ k = i\\\\k \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \ X \ \ j = -i\\\\i \ \ X \ \ k = - j\\$

Now force is also mathematically represented as

$F = ma$

Making a the subject

$a = \frac{F}{m}$

Substituting values

$a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}$

$= (-0.33m/s^2)\r k$

$= 0.33m/s^2 * (- \r k)$

6 0

3 years ago

A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge

Alexxx [7]

Explanation:

It is given that, a long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire.

The charge per unit length of the wire is $\lambda$ and the net charge per unit length is $2 \lambda$ .

We know that there exist zero electric field inside the metal cylinder.

(a) Using Gauss's law to find the charge per unit length on the inner and outer surfaces of the cylinder. Let $\lambda_i\ and\ \lambda_o$ are the charge per unit length on the inner and outer surfaces of the cylinder.

For inner surface,

$\phi=\dfrac{q_{enclosed}}{\epsilon_o}$

$E.A=\dfrac{q_{enclosed}}{\epsilon_o}$

$0=\dfrac{\lambda_i+\lambda}{\epsilon_o}$

$\lambda_i=-\lambda$

For outer surface,

$\lambda_i+\lambda_o=2\lambda$

$-\lambda+\lambda_o=2\lambda$

$\lambda_o=3\lambda$

(b) Let E is the electric field outside the cylinder, a distance r from the axis. It is given by :

$E_o=\dfrac{\lambda_o}{2\pi \epsilon_o r}$

$E_o=\dfrac{3\lambda}{2\pi \epsilon_o r}$

Hence, this is the required solution.

6 0

3 years ago

What do you think causes the rising and sinking of molten rock in the mantle?

Svet_ta [14]

Answer:

Heyyy hope this helps

Convection currents describe the rising, spread, and sinking of gas, liquid, or molten material caused by the application of heat.

3 0

2 years ago