we can use 3 formula for inclined plane
W=-ΔPE
W=ΔKE
W=mg sinθ. d
let's use the third formula
sinθ = 1.2 : 3
sinθ = = 0.4
W = m.g. sinθ. d
W = 110 x 10 x 0.4 x 3
W = 1320 J
Match the term to its best description.
Covalent
Electronegativity
Anion
Metallic
?
Determines what type of bond will form
?
Electronegativity of two atoms are similar
?
A type of covalent bond
?
Negative ion
3. Classify the different types of spectra by how they are created. Physical science
Answer:
Most light sources can be classified into three main types: continuous, absorption, and emission. A hot, opaque object, like the filament in an incandescent light bulb, emits a continuous spectrum, having light of all wavelengths. A hot, dense gas is another example of an object that emits a continuous spectrum.
Explanation:
1. Use the picture on page 3 of the lesson in Connexus to help you label each part of the wave below:
a.
b.
c.
d.
Explain the difference between the direction of force and the plane of application of force. Use a diagram to help you.
The direction of force refers to the path along which the force acts. The plane of application of force, on the other hand, is the surface or area where the force is applied
What is the difference between the direction of force and the plane of application of force?The direction of force refers to the path along which the force acts. It is the straight line in which the force is being applied. For example, if you push a box on the ground, the direction of force is the line of action of the force vector that points in the direction of the push.
The plane of application of force is the surface or area where the force is applied.
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How far did an object travel if they rode a bike at 15m/s for 5 seconds?
A. 75m
B. 3 m/s
C.20 m
Answer:
A) 75m
Explanation:
distance = speed x time taken
Here given:
speed: 15 m/s
time: 5 second
So, distance covered:
15 x 5
75 m
Distance
Speed×Time15(5)75mA car moving with a velocity of 30 ms-1 is stopped by the application of brakes which imparts a radiation of 6 ms-2 to the car.
1. how long does it take for the car to come to stop
2. how far does the car travel during the time the brakes are applied
somebody please help me solve this and do explain..
[tex]\huge{\color{red}{\fbox{\textsf{\textbf{Answer}}}}} [/tex]
1) 5 seconds
2) 75 meters
Explanation:
Refer to the attachment
A Ball A and a Ball B collide elastically. The initial momentum of Ball A is 14.0kgm/s and the initial momentum of Ball B is -9.00kgm/s. Ball A has a mass of 2.00kg and is traveling at -5.00 m/s after the collision. What is the velocity of ball B if it has a mass of 3.00kg
The total momentum before collision is
14.0 kg•m/s + (-9.00 kg•m/s) = 5.00 kg•m/s
If v is the velocity of ball B after colliding with ball A, the total momentum after collision is
(2.00 kg) (-5.00 m/s) + (3.00 kg) v
Momentum is conserved, so these quantities are equal. Solve for v :
5.00 kg•m/s = (2.00 kg) (-5.0 m/s) + (3.00 kg) v
5.00 kg•m/s = -10.0 kg•m/s + (3.00 kg) v
15.0 kg•m/s = (3.00 kg) v
v = (15.0 kg•m/s) / (3.00 kg) = 5.00 m/s
7,293 ÷ 8 what is this
Answer: 911.625
Explanation:
Answer:
911.625 is the answer of your question
Ronnie kicks a playground ball with an initial velocity of 16 m/s at an angle of 40° relative to the ground. what is the approximate horizontal component of the initial velocity? 0.64 m/s 0.77 m/s 10.3 m/s 12.3 m/s
The horizontal component of the velocity with which Ronnie kicks a playground ball is 12.3 m/s.
Horizontal component of the velocity
The horizontal component of the velocity with which Ronnie kicks a playground ball is calculated as follows;
Vx = Vcosθ
where;
V is the initial velocityθ is the angle of projectionVx = 16 x cos(40)
Vx = 12.3 m/s
Thus, the horizontal component of the velocity with which Ronnie kicks a playground ball is 12.3 m/s.
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Answer:
D) 12.3
Explanation:
took the test
A motorcycle starts 20m west of a reference point. It moves at a constant velocity for 5s, reaching a position of 10m east of the reference point.
What is the motorcycle’s average velocity?
The motorcycle's average velocity will be 6 m/sec. The ratio of the overall distance traveled to the total time taken by the body is the average velocity.
What is average velocity?The average velocity is the ratio of the total distance traveled to the total time taken by the body. Its unit is m/sec.
Total distance travelled by the motorcycle;
d=20 m+10 m
d=30 m
The average velocity is found as;
[tex]\rm v_{avg}=\frac{d}{t} \\\\ \rm v_{avg}=\frac{30}{5} \\\\ v_{avg}= 6 \ m /sec[/tex]
Hence,the motorcycle’s average velocity will be 6 m/sec.
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how can we change the frequency of the wave
Answer:
.Also we can change the frequency by decreasing wavelength , or increasing velocity
Explanation:
Please answer these! Will mark brainiest if correct!
Answer:
11 is d. 12 is c. 13 is b. 14 is b. 15 is b.
Explanation:
if it takes you 10 n of force to push your bike 30 m how much work did you do>
Answer:
300 J
Explanation:
GIVEN:
Force (F) = 10 N
Displacement (s) = 30 m
or displacement (d) = 30m
Work done (W) = ?
Therefore,
W = F×s
or W= F × d
= 10×30
= 300 J
Hope it helps
Work is done when:______.a. you apply a force to an object and the object moves in any direction. b. you apply a force to an object and it moves in the same direction.c. you apply a force to an object and it moves in the opposite direction. d. You apply a force to an object and it doesn't move.
Answer:
b
Explanation:
The formula for work done (W) is:
W = Force x Distance moved in the direction of force
This means that the object needs to move in the direction of the force, otherwise no work is done in that direction.
Hope this helps!
1. The distance between a trough and a crest on a transverse wave is 12.0cm. If cycles of the wave pass a fixed point in one second, calculate the speed of the wave.
Answer:
0.12m/s
Explanation:
v=λf
Given that, λ = 12cm = 0.12m
T = 1second
(A period T is the time required for one complete cycle of vibration to pass a given point)
frequency 'f' is unknown but we can get frequency from f = 1/T = 1/1 = 1Hz
therefore, v= 0.12 × 1 = 0.12m/s
calculate the depth of water of density 1020 kg/m³ where the pressure is 3.0 x 10⁶ Pa
Answer:
Here is your answer
Explanation:
Given:
Given:Density of sea water (d) = 1020 kgm^{3}3
Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2
Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:
Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.
Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:
Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.
Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)
Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7
Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7= 102,00,000 Nm^{–2}–2
Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7= 102,00,000 Nm^{–2}–2We know that,
Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7= 102,00,000 Nm^{–2}–2We know that,P = h × d × g
Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7= 102,00,000 Nm^{–2}–2We know that,P = h × d × g\therefore∴ h = P/dg
Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7= 102,00,000 Nm^{–2}–2We know that,P = h × d × g\therefore∴ h = P/dg\begin{gathered} = \frac{10200000}{1020 \times 10} \\ \\ = 1000 \: m \\ \\ = 1 \: km.\end{gathered}=1020×1010200000=1000m=1km.
Given:Density of sea water (d) = 1020 kgm^{3}3Acceleration due to gravity (g) = 10 ms^{–2}–2To find:Depth of sea.Solution:In order to find depth of sea water first we need to find pressure of sea water.Pressure of sea water (P)= 1.02 × 10^{7}7= 102,00,000 Nm^{–2}–2We know that,P = h × d × g\therefore∴ h = P/dg\begin{gathered} = \frac{10200000}{1020 \times 10} \\ \\ = 1000 \: m \\ \\ = 1 \: km.\end{gathered}=1020×1010200000=1000m=1km.Hence, depth of sea is 1 km
Amber likes riding her bicycle through her neighborhood where there are lots of hills. she rides down a very steep hill, and then goes up a small hill. she doesn't need to pedal as her bicycle carries her up the small hill. how do amber's kinetic energy and potential energy change as she goes uphill
Answer:
Her kinetic energy decreases and her potential energy increases.
Explanation:
what temperature change is required to change the length of a 5.5-m steel beam by 0.0012m?
Answer:
Length of steel = 5.5 m
Increase in length = 0.0012 m
α of steel = 12 x 10-6/K
To find:
Temperature change needed ?
Calculation:
Let the required temperature change be ∆T. (in the photo)
So, temperature change of 18.1 Kelvin is required
I hope I have helped you and go to my profile to help me with my question, I really need It, thanks.
Please help me guys please help please
Answer:
C = frequency × wavelength, where C is the speed of the photons which is the speed of light ~3×10^8
frequency = C/wavelength
f = 3×10^8/400×10^-9m
f = 3/4 × 10^15Hz
f = 0.75×10^15Hz
f = 7.5×10^14Hz
Which pair of atoms will form a covalent bond? a. One atom of Fluorine and two atoms of Oxygen
b. two atoms of lithium and one atom of chlorine
c. one atom of magnesium and two atoms of sulfur
d. three atoms of potassium and one atom of phosphorus
Answer:
a.one atom of fluorine and two atoms of oxygen
Explanation:
hope it help
Question 21 of 25
If an electromagnetic wave has a frequency of 8 x 1014 Hz, what is its
wavelength? Use a = . The speed of light is 3 x 108 m/s.
?
3 x .
V
f
O A. 2.7 x 10-6 m
O B. 3.8 x 1021 m
O C. 3.8 x 10-7m
O D. 2.7 x 106 m
SUBMIT
Write down at least 3 equations as you solve them
Answer:
3+7= 7x3= 21➗7=
Explanation:
3+7= 10 7 8 9 10
7x3= 21 7 14 21
21➗7=3 21 14 7
mark me brainliest please
A driver traveling at 39.2 m/s spots a cop are and
hits the break. If he is able to brake for 2.1 seconds
and he accelerates at a rate of -7.6 m/s?, what is
his final speed?
Answer:
23.24 m/s
Explanation:
2.1 s * (-7.6 m/s^2) = - 15.96 m/s from original speed
39.2 - 15.96 = 23.24 m/s
One evening, a student tells her sister that the Sun only appears brighter than other stars because the Sun and stars are different distances from Earth. Her sister disagrees, so the student supports her argument by asking her sister to observe the brightness of streetlights throughout their neighborhood. Which observation would support the student’s argument?
Comparing and analyzing the brightness of near and farther streetlight would support the student’s argument.
Which observation would support the student’s argument?The brightness of nearer streetlight is more as compared to the brightness of the farther light because brightness decreases with the increase in distance.
So we can conclude that comparing and analyzing the brightness of near and farther streetlight would support the student’s argument.
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QUESTION 1
1.1 Define the term acceleration in words
Answer:
To be better at something
Explanation:
I have Accelerated at cooking
Answer:
Acceleration means to speed up
Explanation:
You want to reduce your energy bills.
A) calculate the power of a lightbulb that transfers 6000 joules every minute.
B) calculate the power of a different lightbulb that has a current of 0.05 amps through it when there is a potential difference of 240 volts across it.
C) both bulbs are equally bright. State and explain which lightbulb you should use.
Answer:
A) P = W / t = 6000 / 60 = 100 J/s == 100 watts
B) P = I V = .05 * 240 = 12 watts
C) Bulb B) uses much less power
A body which is uniformly decelerated comes to rest in 5s after travelling of 10m. What is its initial velocity a distance
Answer:
4 m/s
Explanation:
v=u+ at and v^2=u^2+2as.
It is given that v=0,t=5 s and s=10 m.
⇒0=u+5a and 0=u^2+20a.
⇒a=−u/5 and a=−u^2/20.
⇒u/5 = u^2/20⇒u=4 m/s.
⇒ The initial velocity is 4 m/s.
A company is creating an advertisement for colored contact lenses. In the ad, the company wants to include the
phrase "Each contact lens is like a second
Which part of the eye would be most accurate in completing the phrase?
retina
sclera
iris
pupil
Answer:
option C) the iris
Explanation:
2023
lookup the composition for air. what is the percentage of nitrogen and oxygen present in the air we breathe
78 percent nitrogen and 21 percent oxygen
A 9 newtwon charge is Felt on a (1x10^-4) C charge from another charge that is 3 meters away. What is the magnitude of that other charge
The magnitude of that other charge will be 9×10⁻⁵ C. The force on the charge is inverse of the distance.
What is Columb's law?The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
The magnitude of that other charge is found as;
[tex]\rm F = \frac{Kq_1q_2}{r^2} \\\\ \rm 9 = \frac{9 \times 10^9 \times 1\times 10^{-4}\times q_2}{(3)^2} \\\\ q_2=9\times 10^{-5} \ C[/tex]
Hence, the magnitude of that other charge will be 9×10⁻⁵ C.
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