Answer:
In the inner vote part of the earth
Which of the following is most likely to be a secondary source
Answer:
analyze, assess or interpret an historical event, era, or phenomenon,.
Explanation:
Secondary sources are works that analyze, assess or interpret an historical event, era, or phenomenon, generally utilizing primary sources to do so. Secondary sources often offer a review or a critique. Secondary sources can include books, journal articles, speeches, reviews, research reports, and more.
A 21 newton force keeps a 3 kg object in uniform circular motion. The speed of the object is 9 m/s. The magnitude of the centripetal acceleration is
3 m/s^2
7 m/s^2
27 m/s^2
30 m/s^2
Answer:
Explanation:
F = ma
a = F/m = 21/3 = 7 m/s²
Michelle recently started selling her invention: A bed that looks like it floats in mid-air. The bed is actually suspended by magnetic forces. Michelle is a(n)
Answer:
Explanation:
designer
illusionist
engineer
entrepreneur
salesperson
human
inventor
1 point
Kinetic friction is defined as a force that acts between moving surfaces. A
body moving on the surface experiences a force in the opposite direction
of its movement. A student is investigating the motion of a block sliding
down a ramp onto the floor. The diagram below shows the block at five
points during the investigation. The block is at rest at point V. The student
releases the block so that it slides down the ramp and stops at point Z.
Which of the following best explains where kinetic friction is acting on the
block?
Answer:
Explanation:
Not sure what your options are but anything that says something like
"at the block surface in contact with the ramp along the line from V to Z" is probably a good shot.
A 1.95 kg box sits on an incline of 24 ° with the horizontal. If the box accelerates down the incline at 0.245 m/s 2 , what is the coefficient of kinetic friction between the box and the inclined plane?
Answer:
Explanation:
F = ma
mgsinθ - μmgcosθ = ma
gsinθ - μgcosθ = a
μgcosθ = gsinθ - a
μ = (gsinθ - a) / gcosθ
μ = (9.81sin24 - 0.245) / 9.81cos24
μ = 0.4178906...
μ = 0.418
The coefficient of kinetic friction will be equal to 0.418.
What is friction?Friction is the force that prevents solid surfaces, fluid layers, and material elements from sliding against each other. There are various kinds of friction: Dry friction is the force that opposes the relative lateral motion of two in-touch solid surfaces.
Given that a 1.95 kg box sits on an incline of 24 ° with the horizontal. If the box accelerates down the incline at 0.245 m/s 2
The coefficient of kinetic friction will be calculated as,
F = ma
mgsinθ - μmgcosθ = ma
gsinθ - μgcosθ = a
μgcosθ = gsinθ - a
Solve for the value of the coefficient of friction,
μ = (gsinθ - a) / gcosθ
μ = (9.81sin24 - 0.245) / 9.81cos24
μ = 0.4178906...
μ = 0.418
Therefore, the coefficient of kinetic friction will be equal to 0.418.
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A cyclist rides in a circle with speed 8.1 m/s. What is his centripetal
acceleration if the circle has a radius of 27 m?
Explanation:
We know that the tangent velocity is 8.1 m/s. We also know that the tangent velocity can be written in the following way:
Vt = ωr with ω being the angular velocity.
We now calculate ω:
ω = Vt/r = 8.1 m/s / 27m = 0.3 rad/s
Now that we have ω we can calculate the centripetal aceleration:
a = ω^2 * r = ( 0.3 )^2 * 27 = 2.43 m/s^2
In a Little League baseball game, the 145 g ball enters the strike zone with a speed of 17.0 m/s . The batter hits the ball, and it leaves his bat with a speed of 20.0 m/s in exactly the opposite direction. Part A What is the magnitude of the impulse delivered by the bat to the ball
Hi there!
Impulse = Change in momentum
I = Δp = mΔv = m(vf - vi)
Where:
m = mass of object (kg)
vf = final velocity (m/s)
vi = initial velocity (m/s)
Begin by converting grams to kilograms:
1 kg = 1000g ⇒ 145g = .145kg
Now, plug in the given values. Remember to assign directions since velocity is a vector. Let the initial direction be positive and the opposite be negative.
I = (.145)(-20 - 17) = -5.365 Ns
The magnitude is the absolute value, so:
|-5.365| = 5.365 Ns
During an experiment, your teacher gives you two objects: tissue paper and a balloon. You observe that the tissue paper repels the balloon. What does this most likely tell you about the charges of the two objects?
Both objects have negative charges.
The tissue has a positive charge, and the balloon has a negative charge.
The tissue has a negative charge, and the balloon has a positive charge.
The objects have no interactive with each other.
Answer:
i think your answer is this: the objects have no interactive with each other.
Explanation:
if you think about it tissue paper doesn't really have a static electrical charge if it does it is very weak so therefore cannot really attract or repel anything.
2) A traffic light of weight 100 N is supported by two ropes as shown. Let T1 and T2 are the tensions.
a. Resolve the vectors into its components and find their values [ 4 marks]
b. Find sum of the x-components [1 mark]
c. Find sum of the y-components [1mark]
d. Use the above equations to find the tensions in the ropes?
the two angles on x and y are both 37 I cant upload the diagram
Hi there!
a.
We know that:
[tex]\Sigma F_y = 0 \\\\\Sigma F_x = 0[/tex]
Begin by determining the forces in the vertical direction:
W = weight of traffic light
T₁sinθ = vertical component of T₁
T₂sinθ = vertical component of T₂
b.
The ropes provide a horizontal force:
T₁cosθ = Horizontal component of T1
T₂cosθ = Horizontal component of T2
Thus:
0 = T₁cosθ - T₂cosθ
T₁cosθ = T₂cosθ
T₁ = T₂
c.
Since the angles for both ropes are the same, we can say that:
T₁ = T₂
Sum the forces:
ΣFy = T₁sinθ + T₁sinθ - W = 0
2T₁sinθ = W
d.
Now, we can begin by solving for the tensions:
2T₁sinθ = W
[tex]T_1 = T_2 = \frac{W}{2sin\theta} = \frac{100}{2sin(37)} = \boxed{83.08 N}[/tex]
A disgruntled physics student, frustrated with
finals, releases his tensions by bombarding the
adjacent building, 13.5 m away, with water
balloons. He fires one at 38◦
from the horizontal with an initial speed of 23.6 m/s.
The acceleration of gravity is 9.8 m/s
2
.
For how long is the balloon in the air?
Answer:
Explanation:
The balloon would require a time of
t = d/v = 13.5/ (23.6cos38) = 0.7259...s
to travel the horizontal distance.
the vertical position relative to the throw point at that time is
h = 0 + (23.6sin38)(0.7259) + ½(-9.8)(0.7259²)
h = 7.9652...
so as long as the adjacent building is at least 8.0 m higher than the student position, the balloon is in the air for 0.726 s.
If the building is shorter than 8.0 m above the student, the balloon will land on the building roof and will be in the air for a longer period of time
Fireworks have a very high reaction rate. Describe and explain what could be done to slow down their reaction.
Answer:
Don't light up the firework or put less fire
Explanation:
Using Pharaoh's serpent or "black snake" firework could be done to slow down their reaction.
What are fireworks chemically?Gunpowder is typically made using three reagents: potassium nitrate, carbon, and Sulphur. These kinds of elements are used in a combustion reaction that results in this detonation explosion. You have solid reagents reacting to produce gases in the form of solid potassium carbonate, solid potassium sulphate, nitrogen gas, and carbon dioxide gas.
The numerous metal salts that are added to produce the colours spread out all of that material, which is in a super-heated state. In such extremely intense environment, the metal salts heat up and get "excited," emitting light as a result.
If you want an explosion, you need the chemical reaction to go quickly in order to release a large volume of gas in a brief period of time. You might have a chemical reaction that happens rather slowly, like the Pharaoh's serpent or "black snake" firework.
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An object accelerates at a rate of -3.2 m/s^2 to a velocity of 8 m/s over a time of 14 s. What was its initial velocity?
[tex]\text{Given that,}\\\\\text{Acceleration, a = -3.2 m s}^{-2}\\\\\text{Time, t = 14 sec}\\\\\text{Velocity, v = 8 m s}^{-1}\\\\\text{Initial velocity, v}_0 = ?\\\\\\v = v_0 +at\\\\\implies v_0 = v -at = 8 -(-3.2)(14) = 8 + 44.8 =52.8~ \text{m s}^{-1}[/tex]
48.36
g.
MgSO4 to motes
Answer:120.3676
Explanation: using the molecular calculator and molar mass of MgSO4. hope this helps!
A fan blade Spins at 3,000 revolutions per minute.
How
many degrees does it rotate in one second?
18,000 degrees in one second i believe
2 examples of non fossil fuels ?
Answer:
-> Hydropower
-> Solar power
Explanation:
-> Hydropower
[] The power of water! It is the use of falling or fast-running water to produce electricity for power. Impoundments or da*ms are mainly used in this type of power source.
-> Solar power
[] The power of the sun! It is the use of sunlight, or solar energy, to produce electricity for power. You have probably heard of solar panels, and this is the main way to collect it.
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)
- Heather
Answer:
Wind energy and solar power
Explanation:
they do not use fossil fuels
A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. Its speed is 38 m/s when it reaches ground level. What was its speed when its height was half that of its starting point
Explanation:
look !
speed= 38m/s
start from rest= 0
the turns ratio for a transformer with 225 turns of wire in its primary winding and 675 turns in the secondary is: n
The ratio of the primary turns to the secondary turns is 1/3
The correct answer to the question is Option A. 1/3
From the question given above, the following data were obtained:
Primary turn (Nₚ) = 225 turnsSecondary turn (Nᵣ) = 675 turns Ratio of primary to secondary =?Ratio = Nₚ/Nᵣ
Nₚ/Nᵣ = 225 / 675
Nₚ/Nᵣ = 1/3
Therefore, the ratio of the primary turns to the secondary turns is 1/3
Complete question:
See attached photo
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Which of these is a push or a pull? Acceleration Force Mass Inertia
Answer:
the answer is force . force is applied as a push or pull
Shorter the vibrating part more will be the pitch. How?
Answer:When the length of a string is changed, it will vibrate with a different frequency.Shorter strings have higher frequency and therefore higher pitch.
Two rocks with different masses fall side by side, pulled downward by gravity. Why are they able to experience the same motion?(1 point)
The rock with more mass will also have more force pulling it down. Even though the accelerations are different, the rocks will have the same speed.
The rock with more mass will also have more force pulling it down. Even though the accelerations are different, the rocks will have the same speed.
The rock with more mass will also have more force pulling it down, so the accelerations can be the same.
The rock with more mass will also have more force pulling it down, so the accelerations can be the same.
Gravity will pull the rocks down equally, so the accelerations can be the same.
Gravity will pull the rocks down equally, so the accelerations can be the same.
Gravity will pull the rocks down equally. Even though the accelerations are different, the rocks will have the same speed.
Answer:
The rock with more mass will also have more force pulling it down, so the accelerations can be the same.
Explanation:
You are driving your car around a roundabout when you get a flat tire and you decelerate at a constant rate to a stop. The diameter of the roundabout is 100m. It takes you 20 sec to come to a complete stop. While slowing down, you continue to drive in a circle and you stop halfway around the loop. What must your speed have been before the pop?
Answer:
2.5 meters per second
Explanation:
stops half way which is 50m and if its at a constant speed of 2.5 meters multiply that by the seconds and you get 50m
The speed depends on the distance and time. The speed before the pop is 2.5 m/s.
What is the speed?The speed of an object is defined as the total distance traveled by the object within a given time interval.
Given that the diameter of the roundabout is 100m. It takes you 20 sec to come to a complete stop.
While slowing down, you continue to drive in a circle and you stop halfway around the loop. It means that the half distance is 50 m. The speed is calculated as given below.
[tex]s = \dfrac {D}{t}[/tex]
[tex]s = \dfrac {50}{20}[/tex]
[tex]s = 2.5 \;\rm m/s[/tex]
Hence we can conclude that the speed before the pop is 2.5 m/s.
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As the amount of charge on either of two charged objects increases, the electric force between the objects decreases.
O True O False
Answer:
False.
Explanation:
provided the distance between the forces remains the same, the force will increase with increased charge, whether attractive or repulsive.
4. The winding ridge of a screw
Answer:
Thread.
Explanation:
The most common form consists of a cylindrical shaft with helical grooves or ridges called threads around the outside.
Two a.c V1=100sin(wt) and V2 = 150cos(wt) are fed into one circuit, determine the combined output of the two as a single a.c
Answer:
Explanation:
V = 100sin(ωt) + 150cos(ωt)
let x = ωt
V = 100sin(x) + 150cos(x)
a maximum or minimum will occur when the derivative is zero
V' = 100cos(x) - 150sin(x)
0 = 100cos(x) - 150sin(x)
100cos(x) = 150sin(x)
100/150 = sin(x)/cos(x)
0.6667 = tan(x)
x = 0.588 rad
V = 100sin(0.588) + 150cos(0.588)
V = 180.27756
as the maximum will not occur until ωt = 0.588 radians, for a cosine function we subtract that amount as a phase angle φ
V = 180.3 cos(ωt - 0.588)
or as a sine function, the phase angle lags the cosine by a difference of π/2
V = 180.3sin(ωt - (0.588 - π/2)
V = 180.3sin(ωt + 0.983)
You will create your own circuit , discuss how electricity flows and describe how circuits are made in
Answer:
An electric current flows in a loop,powering bulbs or other electric COMPONENTS. The loop is an electric circuit. A circuit is made up of various components linked together by wires. The current is driven around the circuit by a power source, such as a BATTERY.
Voltage: is the energy given to each unit of charge that flows in a circuit
Current: is the amount of electric charge flowing past a point in a circuit each second
Wattage: is the amount of electrical energy a circuit uses each second
What is the difference between speed and velocity?
Hope you could get an idea from here.
Doubt clarification - use comment section.
How does an emergency action plan benefit your workplace
2. The system shown is accelerated by applying a tension Ti to the right-most cable. Assume the system is
frictionless. Solve for the tension in the cable between the blocks, T2, in terms of T. (not a).
The tension in the cable between the blocks, T₂, is [tex]\frac{2T_i}{7}[/tex]
The given parameters:
Mass of the first block, = 2 kgMass of the second block, = 5 kgThe net force on the block system is calculated by applying Newton's second law of motion as follows;
Total mass of the block-system = 2 kg + 5 kg = 7 kg
The acceleration of the block-system;
[tex]a = \frac{T_i}{7}[/tex]
The tension in the cable between the blocks, T₂, is calculated as;
[tex]T_2 = m_2 a\\\\T_2 = 2a\\\\T_2 = 2 \times \frac{T_i}{7} \\\\T_2 = \frac{2T_i}{7}[/tex]
Thus, the tension in the cable between the blocks, T₂, is [tex]\frac{2T_i}{7}[/tex].
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Which statement about diffraction is correct?
A. Sound waves bend around the corners of various obstacles.
B. Sound waves can only travel in straight lines.
C. The amplitudes of two waves combine to appear as a wave smaller than the individual waves.
D. The amplitudes of two waves combine to appear as one big wave.
Answer:
I think A
Explanation:
Which statement about diffraction is correct?
A. Sound waves bend around the corners of various obstacles.
B. Sound waves can only travel in straight lines.
C. The amplitudes of two waves combine to appear as a wave smaller than the individual waves.
D. The amplitudes of two waves combine to appear as one big wave.
I can learn new things but I cannot change how good I am at math. A strongly agree B. agree C disagree D. strongly disagree
Mathematics is a broad subject that anyone who constantly practices by learning from first principle and from example will grow in perfection of the subject as time goes by. Hence, I strongly disagree
During my schools days, I struggled learning mathematics, at the time it was difficult, I began practising and learning from text examples, with time I got a hang of it and my grade and confidence level in the subject increased.
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