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Anna [14]
3 years ago
14

When a piece of metal is heated to high temperatures, it begins to glow red,

Physics
1 answer:
salantis [7]3 years ago
7 0

Answer:

The particle model, because the metal is only absorbing radiation

of specific frequencies

Explanation:

The heating of a metal is an example of blackbody radiation. Blackbody radiation refers to the spectrum of light emitted by any heated object; common examples include the heating element of a toaster and the filament of a light bulb. The spectral intensity of blackbody radiation peaks at a frequency that increases with the temperature of the emitting body(Encyclopedia Britianica).

Max Plank proffered explanation to black body radiation by assuming that the oscillators in the heated body emit or absorb energy in discrete frequencies given by E=hf. Hence the energy was directly proportional to the frequency of the oscillator.

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Hi..Please answer questions???​
Nadya [2.5K]

Answer:

A

Explanation:

Becuse its complete number

3 0
3 years ago
A car starts from rest and accelerates uniformly for a five seconds along a straight road. If speed obtained by the car is 72 km
Step2247 [10]

Answer:

50 meters

Explanation:

Let's start by converting to m/s. There are 3600 seconds in an hour and 1000 meters in a kilometer, meaning that 72km/h is 20m/s.

v_f=v_o+at

Since the car starts at rest, you can write the following equation:

20=0+a(5) \\\\a=20\div 5=4 m/s^2

Now that you have the acceleration, you can do this:

d=v_o+\dfrac{1}{2}at^2

Once again, there is no initial velocity:

d=\dfrac{1}{2}(4)(5)^2=2 \cdot 25=50m

Hope this helps!

8 0
3 years ago
PLEASE HELP ME 45 POINTS
sergij07 [2.7K]

Answer:

a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

c) The tension force in the cord is 28.02 newtons.

d) The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is 3.551 meters per second.

Explanation:

a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

b) Both masses are connected to each other by the same cord, the direction of acceleration will be dominated by the mass of greater mass (mass A) and both masses have the same magnitude of acceleration. By the 2nd Newton's Law, we create the following equation of equilibrium:

Mass A

\Sigma F = T - m_{A}\cdot g = -m_{A}\cdot a (1)

Mass B

\Sigma F = T - m_{B}\cdot g = m_{B}\cdot a (2)

Where:

T - Tension force in the cord, measured in newtons.

m_{A}, m_{B} - Masses of blocks A and B, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

a - Net acceleration of the each block, measured in meters per square second.

By subtracting (2) by (1), we get an expression for the acceleration of each mass:

m_{B}\cdot a +m_{A}\cdot a = T-m_{B}\cdot g -T + m_{A}\cdot g

(m_{B}+m_{A})\cdot a = (m_{A}-m_{B})\cdot g

a = \frac{m_{A}-m_{B}}{m_{B}+m_{A}} \cdot g

If we know that m_{A} = 5\,kg, m_{B} = 2\,kg and g = 9.807\,\frac{m}{s^{2}}, then the acceleration of the masses is:

a = \left(\frac{5\,kg-2\,kg}{5\,kg+2\,kg}\right) \cdot\left(9.807\,\frac{m}{s^{2}} \right)

a = 4.203\,\frac{m}{s^{2}}

The acceleration of the masses is 4.203 meters per square second.

c) From (2) we get the following expression for the tension force in the cord:

T = m_{B}\cdot (a+g)

If we know that m_{B} = 2\,kg, g = 9.807\,\frac{m}{s^{2}} and a = 4.203\,\frac{m}{s^{2}}, then the tension force in the cord:

T = (2\,kg)\cdot \left(4.203\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}}  \right)

T = 28.02\,N

The tension force in the cord is 28.02 newtons.

d) Given that system starts from rest and net acceleration is constant, we determine the time taken by the block to cover a distance of 1.5 meters through the following kinematic formula:

\Delta y  = \frac{1}{2}\cdot a\cdot t^{2} (3)

Where:

a - Net acceleration, measured in meters per square second.

t - Time, measured in seconds.

\Delta y - Covered distance, measured in meters.

If we know that a = 4.203\,\frac{m}{s^{2}} and \Delta y = 1.5\,m, then the time taken by the system is:

t = \sqrt{\frac{2\cdot \Delta y}{a} }

t = \sqrt{\frac{2\cdot (1.5\,m)}{4.203\,\frac{m}{s^{2}} } }

t \approx 0.845\,s

The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is calculated by the following formula:

v = a\cdot t (4)

Where v is the final speed of the system, measured in meters per second.

If we know that a = 4.203\,\frac{m}{s^{2}} and t \approx 0.845\,s, then the final speed of the system is:

v = \left(4.203\,\frac{m}{s^{2}} \right)\cdot (0.845\,s)

v = 3.551\,\frac{m}{s}

The final speed of the system is 3.551 meters per second.

8 0
3 years ago
What do ocean waves and sound waves have in common?
Maksim231197 [3]
D. Both exhibit the same particle-to-particle interaction.Because disturbance is propagated with the help of particles. Other than this,[ <span>light waves are electromagnetic waves. ocean waves and sound waves are mechanical waves. they are able to transfer energy. electromagnetic wave and ocean waves are transverse waves while sound waves are the longitudinal wave. they show wave properties: reflection, refraction, diffraction, interference, and plane-polarization. longitudinal waves such as sound waves cannot be plane-polarized]. The one in the box shows different examples of waves with their examples. Hope it helps.</span>
7 0
3 years ago
Read 2 more answers
A shopper pushes a 5.32 kg grocery cart with a 12.7 N force directed at -28.7° below horizontal. A friction force of 8.33 N push
atroni [7]

Answer:

0.8214 m/s^2

Explanation:

Fnet= Fpushed - Ffriction

Fpushed = 12.7N      Ffriction = 8.33N

Fnet = 12.7N - 8.33N = 4.37N

Fnet= mass(acceleration)

Fnet = 4.37N    mass = 5.32 kg

4.37N = 5.32 kg(acceleration)

acceleration= 0.8214 m/s^2

3 0
3 years ago
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