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:
Work is being done when
Answer:
To express this concept mathematically, the work W is equal to the force f times the distanced, or W = fd. If the force is being exerted at an angle θ to the displacement, the work done is W = fd cos θ.
The SI unit of work is the joule (J). It is defined as the work done by a force of one newton through a distance of one meter.
Explanation:
Work is done when a force is applied to an object through a distance. This means that when a force is applied to an object through a distance, the object's total energy will be affected.
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
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How much force must be applied for a 150.W motor to keep a package moving at 3.00m/s?
The force must be applied for motor to keep the package moving is 50 N.
What is force?Force is the action of push or pull the object in order to make it move or stop.
Power is related to force and velocity as
P = F x v
150 W = F x 3 m/s
F =50 N
Thus, the force must be applied for motor to keep the package moving is 50 N.
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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
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
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.
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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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
how can we change the frequency of the wave
Answer:
.Also we can change the frequency by decreasing wavelength , or increasing velocity
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
Which rocket has the highest acceleration?
A.Rocket 3
B.Rocket 1
C.Rocket 4
D.Rocket 2
Answer:
Rocket 4
Explanation:
since it has the smallest mass it is going to the bigger acceleration.
Answer:
C
Explanation:
In theory, rocket 4 has the least mass and highest force output in comparison.
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
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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In a given chemical reaction, the energy of the products is greater than the energy of the reactants. Which statement is true for this reaction?
A.
Energy is absorbed in the reaction.
B.
Energy is released in the reaction.
C.
No energy is transferred in the reaction.
D.
Energy is created in the reaction.
E.
Energy is lost in the reaction.
Answer:
A - energy is absorbed in the reaction
Explanation:
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
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
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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
If viewed at the correct time and in the right conditions, asteroids, comets, and meteors are all visible to the nked eye.
true or false
Answer:
True
Explanation:
If it is at right conditions and correct time (night time most likely) they will all be visible
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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2 neutral metal spheres, A and B, are touching each other.
A negatively charged rod is brought near to B.
(a) Draw the charges induced on the 2 spheres.
The negatively charged rod is now brought closer to touch B.
(b) Explain why A momentarily moves away from B.
i need help with b
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!
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
Pulling up on a rope, you lift a 9.27-kg bucket of water from a well with an acceleration of 1.20 m/s2. Part A What is the tension in the rope
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
A boat moves 3.0 km due north and then moves 1.0 km due south. The time for this trip is half an hour. What is the average velocity of the boat
The average velocity of the boat given the data from the question is 6.32 Km/h
What is velocity?Velocity is defined as the rate of change of displacement with time. Mathematically, it is expressed as
Velocity = displacement / time
How to determine the resultant displacementDisplacement 1 (d₁) = 3 KmDisplacement 2 (d₂) = 1 KmResultant displacement (D) =?
D² = d₁² + d₂²
D² = 3² + 1²
D² = 9 + 1
D² = 10
Take the square root of both sides
D = √10
D = 3.16 Km
How to determine the average velocityResultant displacement = 3.16 KmTime = 1/2 hour = 0.5 hourAverage velocity =?
Average velocity = resultant displacement / time
Average velocity = 3.16 / 0.5
Average velocity = 6.32 Km/h
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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)75mPlease answer these! Will mark brainiest if correct!
Answer:
11 is d. 12 is c. 13 is b. 14 is b. 15 is b.
Explanation:
A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with
an initial speed of 20.0 m/s. The point of release is 45.0 m above the ground. Neglect air resistance.
(i) How long does it take for the ball to hit the ground?
(ii) Find the ball’s speed at impact.
(iii) Find the horizontal range of the stone.
Answer:
See below
Explanation:
Vertical position = 45 + 20 sin (30) t - 4.9 t^2
when it hits ground this = 0
0 = -4.9t^2 + 20 sin (30 ) t + 45
0 = -4.9t^2 + 10 t +45 = 0 solve for t =4.22 sec
max height is at t= - b/2a = 10/9.8 =1.02
use this value of 't' in the equation to calculate max height = 50.1 m
it has 4.22 - 1.02 to free fall = 3.2 seconds free fall
v = at = 9.81 * 3.2 = 31.39 m/s VERTICAL
it will also still have horizontal velocity = 20 cos 30 = 17.32 m/s
total velocity will be sqrt ( 31.39^2 + 17.32^2) = 35.85 m/s
Horizontal range = 20 cos 30 * t = 20 * cos 30 * 4.22 = 73.1 m