Answer:
The standard of efficiency varies by product category
these are called BEE star label and they show how much electricity the appliance consumes in year each appliance gets btw one and five stars with five stars meaning that it's extremely efficienct andis likely to keep your electricity bills in check
hope it's helpful
Select the correct answer. Which characteristic is not a characteristic of plastic? OA easy to mold O'B. good conductor of heat OC. lightweight and durable OD. repels water
Answer:
O'B. Good conductor of heat.
Explanation:
Plastic is not a good conductor of heat because it has no delocalized electrons it's surface to transfer heart.
Answer:
I hope this helps
Explanation:
A constant unbalanced force is applied to an object for a period of time. What will happen to the objects acceleration?
a. It will accelerate in the direction of the net force.
b. It will accelerate opposite the direction net force.
c. There is no acceleration.
d. Not enough information to answer
Answer:
A
Explanation:
The moving object will accelerate as long as a net force is applied. If a net force is not applied, the object will continue to move at a constant speed along the same path in curved spacetime (due to gravity). If the net force continues the velocity of the object will continue to increase until nearly reaching the limit (speed of light).
What process is mostly responsible for the blue appearance of the sky
Answer:reflection by dust particles in air
Differentiate between a derived quantity and a derived unit.
Explanation:
Derived quantities are quantities dependent on fundamental quantities while derived units are the units of these quantities
An 8kg ball travelling at 4m/s collides head on with a 3kg ball travelling at 14m/s. Rhe balls bounce off each other and travel back the way they came. The 8kg ball travels away at 2m/s calculate velocity of 3kg ball after the collision.
Answer:
Explanation:
There are a couple of assumptions I had to make here and also a couple of rules based on what I use in my classroom when I teach the Law of Momentum Conservation. First of all, I am going to call the 8kg ball 1 and say that it is moving to the right (and right is positive), and that means that the 3kg ball is ball 2 and say that it is moving to the left (and left is negative). I had to assume that the 2 balls were moving towards each other; hence, the different signs assigned to their movement. I also added in another significant digit since we have only 1 in most of these values and adding in a .0 is not going to change the value of any number. The Law of Momentum Conservation in this particular instance says
[tex][m_1v_1+m_2v_2]_b=[m_1v_1+m_2v_2]_a[/tex] which is the mathematical way of saying that the momentum after the collision is the same as the momentum before it. Filling in:
[tex][(8.0*4.0)+(3.0*-14)]_b=[(8.0*-2.0)+(3.0*v)]_a[/tex] and doing the math here simplifies to
32 - 42 = -16 + 3.0v and
-10 = -16 + 3.0v and
6.0 = 3.0v so
v = 2.0 (and the positive indicates that ball 2 is now moving to the right)
Un campo magn´etico uniforme B, con magnitud 1.2 mT, apunta verticalmente hacia arriba a lo largo del volumen del sal´on en que usted est´a sentado. Un prot´on de 5.3 MeV se mueve horizontalmente de sur a norte a trav´es de cierto punto en el salon. ¿Qu´e fuerza magn´etica deflectora act´ua sobre el prot´on cuando pasa por este punto? Cu´al es su aceleraci´on, la masa del prot´on es de 1.67x10−27 kg. Resp. 6.1x10−15 N; 3.7x1012 m/s2.
Answer:
F = 6.12 10⁻¹⁵ N, a = 3.66 10¹² m / s²
Explanation:
The magnetic force is
F = q v x B
bold letters indicate vectors, the modulus of this expression is
F = q v B sin θ
as they indicate that the magnetic field is vertical the proton moves horizontally the angle is 90º
suppose that the energy of the proton is totally kinetic
E = 5.3 MeV (10⁶ 1.6 10⁻¹⁹J / 1 eV) = 8.48 10⁻¹³ J
E = K = ½ m v²
v = [tex]\sqrt{2E/m}[/tex][tex]\sqrt{2 \ 8.48 \ 10^{-13} / 1.67 \ 10^{-27}}[/tex]
v = [tex]\sqrt{2 \ 8.48 \ 10^{-13} / 1.67 \ 10^{-27}}[/tex]
v = [tex]\sqrt{10.155 \ 10^{14}}[/tex]
v = 3.19 10⁷ m / s
we calculate
F = 1.6 10⁻¹⁹ 3.19 10⁷ 1.2 10⁻³
F = 6.12 10⁻¹⁵ N
The direction of this out is given by the right hand rule,
thumb in the direction of speed, south to north
fingers extended in the direction of the magnetic field, vertical
the probe points in the direction of the force for a positive charge, East
To calculate the acceleration we use Newton's second law
F = ma
a = F / m
a = 6.12 10⁻¹⁵ / 1.67 10⁻²⁷
a = 3.66 10¹² m / s²
the direction of the acceleration is the same as the direction of the force
As distance increases, force decrease and vise versa, as distance decreases then, force increases. This is a ________________ relationship
A) directly proportional
B) toxic
C) inverse
D) reverse
A box at rest is in a state of equilibrium half way up on a ramp. The ramp has an incline of 42° . What is the force of static friction acting on the box if box has a gravitational force of 112.1 N ?
70 N
80 N
75 N
85 N
Answer:
75 N
Explanation:
Given that,
The angle of incline of the ramp is 42°.
The gravitational force of the box, W = 112.1 N
We need to find the force of static friction on the box.
We have that the component of the gravitational force acting along the inclined plane i.e.
[tex]F=W\sin\theta\\\\F=112.1\times \sin(42)\\\\F=75\ N[/tex]
So, the force of static friction is equal to 75 N.
A solar panel is used to recharge a battery. The solar panel produces 0.80 W of electrical power. The panel is 20% efficient. What is the power input of the sunlight onto the solar panel? * 1 point 0.16 W 4.0 W 8.0 W 16.0 W
Answer:
Input power, Ip = 4 Watts
Explanation:
Given the following data;
Power output = 0.8 Watts
Efficiency = 20%
To find the power input of the sunlight onto the solar panel;
Mathematically, the efficiency of a machine is given by the formula;
[tex] Efficiency = \frac {Out-put \; power}{In-put \; power} * 100 [/tex]
Substituting into the formula, we have;
[tex] 20 = \frac {0.8}{Ip} * 100 [/tex]
[tex] 20 = \frac {80}{Ip} [/tex]
Cross-multiplying, we have;
[tex] 20Ip = 80 [/tex]
[tex] Ip = \frac {80}{20} [/tex]
Input power, Ip = 4 Watts
A light ray in medium 1 striking a boundary at an angle of theta Subscript 1 Baseline. In medium 2 is a second light ray that is at a larger angle from the normal with angle theta Subscript 2 Baseline. Medium 1 has the index of refraction of n Subscript 1 Baseline and in medium 2 has the index of refraction of n Subscript 2 Baseline.
Which medium (1 or 2) is more dense?
Medium
Answer:
The correct answer is 1
Explanation:
A light ray in medium 1 striking a boundary at an angle of theta Subscript 1 Baseline. In medium 2 is a second light ray that is at a larger angle from the normal with angle theta Subscript 2 Baseline. Thus, 1 is the correct option.
What is wavelength?
Wavelength is the distance between two identical or similar positions it means that it is the distance between crests or trough in the adjacent cycle of the waveform.
Wavelength is denoted by (lambda).
It is measured in meter, or centimeter, or millimeters.
Mathematically,
Wavelength is equal to velocity divided by frequency,
So,
Wavelength (lambda) = Velocity/frequency.
Velocity is in meter per second.
Frequency is in 1/second.
Prism is a transparent object in which if sunlight is passed then it will be split in a VIBGYOR
where V = vilot, I = indigo, B = blue, G = green, Y = Yellow, O = orange, R = red.
In this series wavelength of red is higher and vilot is smaller and frequency of violet is higher and red is smaller.
Therefore, A light ray in medium 1 striking a boundary at an angle of theta Subscript 1 Baseline. In medium 2 is a second light ray that is at a larger angle from the normal with angle theta Subscript 2 Baseline. Thus, 1 is the correct option.
Learn more about wavelength here:
https://brainly.com/question/13533093
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outs each inning?
Each team gets (how many).
o 2
3
(Basketball )
Help!!
?
lubricants reduce friction by replacing sliding friction with fluid friction
Parallel incident rays appear to bounce like they have all originated from the same point. What is this point called?
A. cross point
B. midpoint
C. bounce point
D. focal point