What type of interaction occurs when a wave hits an object that it cannot pass?

A pickup truck is being driven down the highway carrying eight 100 gallon fish tank full of water. A hole is punctured in each t

irakobra [83]

Answer:

Explanation:

The velocity of the vehicle would increase because the the tanks (when filled with water) must have exerted a force which would reduce the velocity of the vehicle at a certain pressure on the gas pedal. Note that force equals mass multiplied by acceleration; as the mass decreases, so the force decreases. Thus, when the mass exerted by this tanks (on the vehicle) decrease as a result of the hole punctured in them, the force exerted by the tanks would also decrease causing an increase in velocity of the pick up truck when the same pressure is applied on the gas pedal throughout (before and after the puncture).

The conservation law that applied here is the law of conservation of energy which states that energy can neither be created nor destroyed but can be transformed from one form to another. This is because the energy the vehicle used in carrying the load (the tanks) was transformed to the energy that resulted in increasing it's velocity (no new energy was formed as the pressure on the gas pedal remained the same).

6 0

2 years ago

Ch 27 HW Exercise 27.12 10 of 20 Constants A horizontal rectangular surface has dimensions 2.80 cm by 3.15 cm and is in a unifor

deff fn [24]

Answer:

Magnetic field, B = 0.88 T

Explanation:

It is given that,

The dimension of rectangular surface is 2.80 cm by 3.15 cm. The area of rectangular surface is, $A=8.82\ cm^2=0.000882\ m^2$

Angle between the uniform magnetic field and the horizontal, $\theta=31$

Magnetic flux, $\phi=4\times 10^{-4}\ Wb$

Let B is the magnitude of magnetic field in which the rectangular surface is placed. It is given by :

$\phi=BA\ cos\theta$

$\theta$ is the angle between magnetic field and the area

Here, $\theta=90-31=59^{\circ}$

$B=\dfrac{\phi}{A\ cos\theta}$

$B=\dfrac{4\times 10^{-4}}{0.000882\times cos(59)}$

B = 0.88 T

So, the magnitude of magnetic field is 0.88 T. Hence, this is the required solution.

7 0

2 years ago

tyhe mass of an object is 117 g adding 1200j of heat will raise the temperture of the object by 12 celsius what is the specifc h

Sati [7]

Answer:

C 0.85 j/g*k

Explanation:

The specific heat capacity of a material is given by:

$C_s = \frac{Q}{m \Delta T}$

where

Q is the amount of heat supplied to the object

m is the mass of the object

$\Delta T$ is the increase in temperature of the object

For the object in this problem, we have

m = 117 g is the mass

Q = 1200 J is the heat supplied

$\Delta T=12^{\circ}$ is the increase in temperature

Substituting into the formula, we find the specific heat:

$C_s = \frac{1200 J}{(117 g)(12^{\circ})}=0.85 J/gC$

8 0

3 years ago

Tarzan (75 kg) swings from 4 metres high on a vine down to ground level and catches Jane (50 kg). How fast are they moving after

Kruka [31]

Try this solution (if it is possible, check it in other sources):

1. for m_Tarzan=75kg., initial_height=4m., end_height=0 m. and g=10 N/kg. Energy is:

$E=m_{Tarzan}*(H_{initial}-H_{end})*g=75*4*10=3000(J).$

2. The same value of Energy is applied for m_Tarzan+Jane=75+50=125 kg.:

$E=\frac{m_{Tarzan+Jane}* Speed^2}{2}; \ => \ Speed=\sqrt{ \frac{2*E}{m_{Tarzan+Jane}}};$

3. According to the formula of the Speed:

Speed=sqrt(6000/125)=sqrt(48)=4sqrt(3)≈4*1.71=6.84 (m/s)

Answer: 6.84 (m. per sec.)

5 0

2 years ago

Two rings of radius 5 cm are 20 cm apart and concentric with a common horizontal x-axis. The ring on the left carries a uniforml

Yanka [14]

Answer:

The electric field due to the right ring at a location midway between the two rings is $2.41\times10^{3}\ V/m$

Explanation:

Given that,

Radius of first ring = 5 cm

Radius of second ring = 20 cm

Charge on the left of the ring = +30 nC

Charge on the right of the ring = -30 nC

We need to calculate the electric field due to the right ring at a location midway between the two rings

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{qx}{(x^2+R^2)^{\frac{3}{2}}}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times30\times10^{-9}\times0.1}{((0.1)^2+(0.2)^2)^{\frac{3}{2}}}$

$E=2.41\times10^{3}\ V/m$

Hence, The electric field due to the right ring at a location midway between the two rings is $2.41\times10^{3}\ V/m$

3 0

2 years ago