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2 years ago

8

A propagating wave in space with electric and magnetic components. These components oscillate at right angles to each other. It

may travel in a vacuum.

2 answers:

zimovet [89]2 years ago

8 0

Answer:

electromagnetic

Explanation:

Lilit [14]2 years ago

4 0

Answer: electromagnetic

Explanation:

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A boy is pulling a load of 150N with a string inclined at an angle 30 to the horizontal if the tension of string is 105N the for

Lorico [155]

The force tending to lift the load (vertical force) is equal to <u>22.5N.</u>

Why?

Since the boy is pulling a load (150N) with a string inclined at an angle of 30° to the horizontal, the total force will have two components (horizontal and vertical component), but we need to consider the given information about the tension of the string which is equal to 105N.

We can calculate the vertical force using the following formula:

$VerticalForce=Force*Sin(30\° )=(BoysForce-StringForce)*\frac{1}{2}\\\\VerticalForce=(150N-105N)*\frac{1}{2}=VerticalForce=45N*\frac{1}{2}=22.5N$

Hence, we can see that <u>the force tending to lift the load</u> off the ground (vertical force) is equal to <u>22.5N.</u>

Have a nice day!

8 0

2 years ago

A ball on the end of a rope is moving in a vertical circle near the surface of the earth. Point A is at the top of the circle; C

mrs_skeptik [129]

Answer:

Tension in the string will increase

Explanation:

As we know that tension in the string at any angle with the vertical is given as

$T - mgcos\theta = m\omega^2 R$

now we have

$T = mgcos\theta + m\omega^2 R$

also we know that

angular speed of the stone is directly depending on the time period of the motion

so it is given as

$\omega = \frac{2\pi}{T}$

since the frequency of the revolution is increased from n = 1 rev/s to 2 rev/s

so the angular speed would be doubled

So here we can say that

tension in the string will increase when we will increase the frequency of revolution.

3 0

1 year ago

Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit

snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity= (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

is

$Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s$

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

$Inlet volume = inlet velocity*area of circle=\pi r^{2}*inlet velocity$

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

$Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}$

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

4 0

1 year ago

Water is boiled at 300 kPa pressure in a pressure cooker. The cooker initially contains 3 kg of water. Once boiling started, it

Inessa [10]

Answer:

The average rate of energy transfer to the cooker is 1.80 kW.

Explanation:

Given that,

Pressure of boiled water = 300 kPa

Mass of water = 3 kg

Time = 30 min

Dryness friction of water = 0.5

Suppose, what is the average rate of energy transfer to the cooker?

We know that,

The specific enthalpy of evaporate at 300 kPa pressure

$h_{f}=561.47\ kJ/kg$

$h_{fg}=2163.8\ kJ/kg$

We need to calculate the enthalpy of water at initial state

$h_{1}=h_{f}$

$h_{1}=561.47\ kJ/kg$

We need to calculate the enthalpy of water at final state

Using formula of enthalpy

$h_{2}=h_{f}+xh_{fg}$

Put the value into the formula

$h_{2}=561.47+0.5\times2163.8$

$h_{2}=1643.37\ kJ/kg$

We need to calculate the rate of energy transfer to the cooker

Using formula of rate of energy

$Q=\dfrac{m(h_{2}-h_{1})}{t}$

Put the value into the formula

$Q=\dfrac{3\times(1643.37-561.47)}{30\times60}$

$Q=1.80\ kW$

Hence, The average rate of energy transfer to the cooker is 1.80 kW.

3 0

2 years ago

An organ pipe is tuned to exactly 384 Hz when the temperature in the room is 20°C. Later, when the air has warmed up to 25°C, th

maksim [4K]

Answer: A. Greater than 384 Hz

Explanation:

The velocity of sound is directly related to the temperature rather it is directly proportional meaning if the temperature decreases the velocity decreases and if temperature increases the velocity increases.

Now, we are given that temperature has risen from 20°C to 25°C meaning it has increases. So it implies that velocity must also increase.

Also, the velocity for organ pipe is directly proportional to its frequency. Now if velocity increases frequency must also increase. In this case, the original frequency is 384 Hz. Now increasing the temperature resulted in increase in velocity and thus increase in frequency.

So option a is correct. i.e. now frequency will be greater than 384 Hz.

3 0

2 years ago