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3 years ago

Which type of surface would most likely be the best reflector of electromagnetic energy?

1 answer:

5 0

<span>light colored and smooth surface would most likely be the best reflector of electromagnetic energy.Light, shiny surfaces are the best reflectors of radiation and they will allow the waves to reflect and bounce off rather than absorb. we can consider mirror as the example ,it will only reflect the light energy falling on them and it will not absorb. The darker coloured and rough surfaced substances will definitely absorb some amount of light falling on it. so light coloured smooth or shiny surfaced material would be the best reflector for electromagnetic energy.</span>

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John is carrying a shovelful of snow. The center of mass of the 3.00 kg of snow he is holding is 15.0 cm from the end of the sho

Andru [333]

Answer:

James is correct here as the force of hand pushing upwards is always more than the force of hand pushing down

Explanation:

Here we know that one hand is pushing up at some distance midway while other hand is balancing the weight by applying a force downwards

so here we can say

Upwards force = downwards Force + weight of snow

while if we find the other force which is acting downwards

then for that force we can say that net torque must be balanced

so here we have

$F_{down} L_1 = W_{snow} L_2$

so here we have

$F_{down} = \frac{L_2}{L_1} (W_{snow})$

so here we can say that upward force by which we push up is always more than the downwards force

8 0

3 years ago

Consider the minute hand on a clock. (a) Compute the frequency of its motion in cycles per second. State your answer to three si

kicyunya [14]

Answer:

(a)0.0002778

(b) $1.16\times10^{-5}$

Explanation:

(a) The minute hand has a period of 60 minutes ( or 60 * 60 = 3600 seconds) for 1 circle. Its frequency per second would be

1 / 3600 = 0.0002778

(b) The hour hand has a period of 24 hours ( or 24*60 * 60 = 86400

seconds) for 1 circle. Its frequency per second would be

$1 / 86400 = 1.16\times10^{-5}$

5 0

3 years ago

When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0

lora16 [44]

Answer:

$d=0.165m$

Explanation:

Given

$m=0.25kg$ , $x_{1}=5cm*\frac{1m}{100cm}=0.05m$ , $x_{2}=4cm*\frac{1m}{100cm}=0.04m$ , $v=2\frac{rev}{s}$

The tension of the spring is

$F_{k}=K*x_{1}=m*g$

$K=\frac{m*g}{x_{1}}$

$K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m$

The force in the spring is equal to centripetal force so

$F_{c}=\frac{m*v^2}{r}$

$v=w*r=2\pi*r$

But Fc is also

Fc=KxΔr

$F_{c}=K*(r-x_{2})$

Replacing

$m*4\pi^2*r=K*(r-x_{2})$

$0.25kg*4\pi^2*r=49*(r-0.04m)$

$r=0.205m$

total distance is

$d=0.205-0.04=0.165m$

3 0

3 years ago

During metamorphism, what is the major effect of chemically active fluids?

xz_007 [3.2K]

Answer:

Option b

Explanation:

Metamorphism is the process where the variation of the geological texture resulting from the different arrangement of the minerals or the variation of minerals in protoliths, i.e., pre- existing rocks take place such that there occurs no change in state of the protolith, i.e., it does not melt into magma.

The change takes place as a result of the presence of chemically active fluids, heat and pressure.

There is a reaction between the chemically active fluid and the rock through which it passes and promotes the movement of the dissolved ions of silicate and promotes the growth of the mineral grains.

7 0

3 years ago

Air is cooled in a process with constant pressure of 150 kPa. Before the process begins, air has a specific volume of 0.062 m^3/

Mama L [17]

Answer:

The pressure is constant, and it is P = 150kpa.

the specific volumes are:

initial = 0.062 m^3/kg

final = 0.027 m^3/kg.

Then, the specific work can be written as:

$W = \int\limits^{vf}_{vi} {Pdv} \, = P(vf - vi) = 150kPa*(0.0027 - 0.062)m^3/kg = -5.25 kPa*m^3/kg.$

The fact that the work is negative, means that we need to apply work to the air in order to compress it.

Now, to write it in more common units we have that:

1 kPa*m^3 = 1000J.

-5.25 kPa*m^3/kg = -5250 J/kg.

7 0

3 years ago