Answer:
Energy stored in food is consider as carbohydrate,protein,fats,etc.
A car drives 84 meters forward. It's displacement and distance would be the same?
Answer:
Yes.
Explanation:
Displacement is a vector quantity, meaning it has to be a straight graph jointing from starting point to final point and has a direction as well as magnitude and it can also be negative as well as positive and zero.
Displacement is also changes in position (points) of distance. A distance is scalar quantity meaning it only has magnitude and does not have direction. Since distance is scalar quantity, it cannot be negative. Also graph of distance can be a curve graph or any graph.
So when does displacement equal to distance? If an object is moving in straight path or line in one/fixed direction, both displacement and distance are same.
Because if a distance is a straight path and the displacement is simply a straight line (or ray) jointing from starting point to end point, both distance and displacement are both straight path or rays. Since distance and displacement have same graph, we can conclude that both are same in values.
If we go by calculus, given x = 2t as example of straight graph where x stands for position and t stands for time. We can find the displacement from b to a by using the following formulas:
Displacement
[tex]\displaystyle \large{s=\int\limits^b_a {v} \, dt}[/tex]
Distance
[tex]\displaystyle \large{l=\int\limits^b_a {|v|} \, dt}[/tex]
v stands for velocity which we can find from:
Velocity
[tex]\displaystyle \large{v=\frac{dx}{dt}}[/tex]
If you have learnt calculus, first, differentiate x = 2t with respect to t.
[tex]\displaystyle \large{v=2}[/tex]
Then substitute v = 2 in s and l to find find if both displacement and distance are equal in straight path.
Displacement
[tex]\displaystyle \large{s=\int\limits^b_a {2} \, dt}\\\displaystyle \large{s=[2t]\limits^b_a}\\\displaystyle \large{s=2b-2a = 2(b-a)}[/tex]
Distance
[tex]\displaystyle \large{l=\int\limits^b_a {|2|} \, dt}\\\displaystyle \large{l=\int\limits^b_a {2} \, dt}\\\displaystyle \large{l=[2t]\limits^b_a}\\\displaystyle \large{l=2b-2a=2(b-a)}[/tex]
Since both displacement and distance are equal when integrating on straight graph, we can conclude that an object moving in straight fixed point has same distance and displacement.
In an action-reaction pair
a. action force is exerted first
b. action force and the reaction force are equal in magnitude and act in the same direction.
c. action force and the reaction force are contact forces only.
d. action force and the reaction force act on two different objects.
Answer:
d. action force and the reaction force act on two different objects.
Hope you could understand.
If you have any query, feel free to ask.
In an action-reaction pair, action force and the reaction force act on two different objects.
What is Newton's Third Law of Motion?Newton's Third Law of Motion states that "For every action, there is an equal and opposite reaction, action and reaction act on two different bodies."
It means that equal force are act on reaction force that is it is also known as reciprocal law of motion.
When considering the options,
The first option is incorrect because action and reaction forces act simultaneously.The second option is also false because they act in opposite direction.The third option is wrong because for action-reaction force, physical contact is not necessary.So, In an action-reaction pair, action force and the reaction force act on two different objects. Thus, Option B is the correct answer.
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#SPJ2
If a 35 kg box collides with a stationary 120 kg box with a force of 90 N, what must
be true of the magnitude of the reaction force?
Answer:
yes , true the magnitude of the reaction force.
Calculate Kylie’s weight at 3 radii away from the center of the earth. Kylie weighs 565.4N on the earths surface
Answer:
Earth's mass is 5.98 x 1024 kg and has a radius of 6.37 x 106 m. 8.06 N/kg ... what would be the weight of a 70.0-kg student on the surface of Jupiter?Explanation:
i do believe that i need help
A train starts at rest and ends up with 1500 J of kinetic energy. How much work was done on the train?
Answer:
1500 J
Explanation:
Apparently, we're to assume that no energy is lost to friction or other effects. Then all of the energy the train has is due to work being done on it.
1500 J of work was done on the train
How long does it take a car traveling at 45km/h to travel 100.0 m?
Answer:
8 seconds
Explanation:
Since the carspeed is in km/h, we need equal units, so we will make 100.0m 0.1000km.
Then we need to find how long it takes the car to travel 0.1km
We can use the formula distance=speed * time and get
0.1=45 * time
Therefore we get .002222... hours
Multiplying this by 3600 (to get seconds, 60x60), we get 8 seconds
Answer:
8m/s
Explanation:
speed = 45km/h
Distance = 100m
we have to find time = ?
Formula for speed is = Distance/ Time
Here Distance is given in 'm' so we need to convert speed value in 'm'
So to convert km/h in m formula is divide 45km/h by 3.6 then the km/h value gets converted in m so now the value is 12.5 m/s
So now,
speed = Distance/Time
we have to find Time
Then,
Time = Distance/ Speed
= 100/12.5
= 8m/s
A car travels 120 miles in 12 hours. What is
the car's speed?
Select one:
O a. 10 m/s
b. 10 mph
c. 1200 mph
O d. 0.1 m/s
Answer:
I hope it will help you...
#CARRYONLERANING
The magnitude, M, of an earthquake is represented by the equation M=23logEE0 where E is the amount of energy released by the earthquake in joules and E0=104. 4 is the assigned minimal measure released by an earthquake. In scientific notation rounded to the nearest tenth, what is the amount of energy released by an earthquake with a magnitude of 5. 5?.
An earthquake with a magnitude of 5.5 releases 4.5 × 10¹² J of energy.
What is the magnitude of an earthquake?The magnitude of an earthquake is a number that characterizes the relative size of an earthquake and is based on the measurement of the maximum motion recorded by a seismograph.
If the magnitude of the earthquake is M = 5.5, we can calculate the amount of energy released (E) using the following expression.
[tex]M = \frac{2}{3} log(\frac{E}{E_0} )\\\\E = E_0 \times 10^{\frac{3}{2}M } = 10^{4.4} \times 10^{\frac{3}{2}(5.5) } = 4.5 \times 10^{12} J[/tex]
where,
E₀ is the assigned minimal measure released by an earthquake.An earthquake with a magnitude of 5.5 releases 4.5 × 10¹² J of energy.
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What is the height of the Golden Bridge?
Answer:
hi your answer to this question is 220'
Explanation:
hope this helped you
An astronomical unit (A.U.) is 1 point A) a term for defining the luminosity of a star B) the average distance from the Earth to the sun C) the average distance of any given planet to the sun D) equal to a light year
Answer:
B) the average distance from the Earth to the Sun
Explanation:
what is a velocity of an object with a momentum of 300 kg m/s and the mass of 100 kg
Answer:
v = 3 m/s
Explanation:
P is the momentum
m is the mass
v is the velocity
P = m•v
300 = 100•v Divide both sides per 100
300/100 = v
v = 3 m/s
[tex]\text{Given that,}\\\\\text{Momentum, p = 300 kg m s}^{-1}\\\\\text{Mass, m = 100 kg}\\\\\text{We know that,}\\\\p = mv\\\\\implies v = \dfrac pm = \dfrac{300}{100} = 3 ~ \text{m s}^{-1}[/tex]
a car travels at a constant speed of 20m/s and yet it does not have a constant velocity. explain how this could be
Answer:
HOPE THIS HELPS!!
Explanation:
Velocity is speed as well as distance mentioned.
The car is traveling at a constant speed of 20m/s but it may not be traveling in any particular direction. It would be taking many turns (left and right) or it would be moving in a circular manner.
So,it is at constant speed but not constant velocity.
A 2.6 kg mass attached to a light string rotates on a horizontal,
frictionless table. The radius of the circle is 0.525 m, and the
string can support a mass of 17.9 kg before breaking. The
acceleration of gravity is 9.8m/s2. What maximum speed can
the mass have before the string breaks?
The maximum speed the mass can have before it breaks is 2.27 m/s.
The given parameters:
maximum mass the string can support before breaking, m = 17.9 kgradius of the circle, r = 0.525 mThe maximum speed the mass can have before it breaks is calculated as follows;
[tex]T = ma_c\\\\Mg = \frac{Mv^2}{r} \\\\v^2 = rg\\\\v = \sqrt{rg} \\\\v_{max} = \sqrt{0.525 \times 9.8} \\\\v_{max} = 2.27 \ m/s[/tex]
Thus, the maximum speed the mass can have before it breaks is 2.27 m/s.
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ASAP!!!!!!!! Parts of a Microscope: Use the following word bank to identify the following parts of the microscope.
Body Tube
Ocular Lens ( eye piece)
Revolving Nose piece Arm
Objectives (three of these) Diaphragm
Stage clips
Stage
Light Source
Base
Course Adjustment Knob
Fine Adjustment Knob
Answer:
1. Body Tube
2.Revolving Nose Piece
3.Low Objective
4.Med Objective
5.High Objective
6.Stage clips
7.Diaphragm
8.Light source
9.Ocular lens
10. Arm
11.Stage
12.Coarse Adjustment knob
13.Fine Adjustment knob
14. base
If the force to stretch a spring is given as k = (100N/m), then what is the potential energy of the spring if it is stretched 1 meters from rest?
Answer:
50 Joules
Explanation:
Find the heat energy is required to change 2Kg of ice at 0 C to water at 20 C ( specific latent heat of fusion of water = 336000 J/Kg and specific heat of capacity of water = 4200 J/Kg C
We want to find the energy that we need to transform 2kg of ice at 0°C to water at 20°C.
We will find that we must give 840,000 Joules.
First, we must change of phase from ice to water.
We use the specific latent heat of fusion to do this, this quantity tells us the amount of energy that we need to transform 1 kg of ice into water.
So we need 336,000 J of energy to transform 1kg of ice into water, and there are 2kg of ice, then we need twice that amount of energy:
2*336,000 J = 672,000 J
Now we have 2kg of water at 0°C, and we need to increase its temperature to 20°C.
Here we use the specific heat, it tell us the amount of energy that we need to increase the temperature per mass of water by 1°C.
We know that:
specific heat of capacity of water = 4200 J/kg°C
This means that we need to give 4,200 Joules of energy to increase the temperature by 1°C of 1kg of water.
Then to increase 1°C of 2kg of water we need twice that amount:
2*4,200 J = 8,400 J
And that is for 1°C, we need to give that amount 20 times (to increase 20°C) this is:
20*8,400 J = 168,000 J
Then the total amount of energy that we must give is:
E = 672,000 J + 168,000 J = 840,000 J
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19. A beach ball is rolling in a straight line toward you at a speed of 9 m/sec. Its momentum is 3 kg'm/sec.
What is the mass of the beach ball? moss - 0.33
aroto from 10mle to speed of 14 m/s The mass of the bicycle
The mass of the beach ball is 0.33 kg
From the question,
We are to determine the mass of the beach ball.
Using the formula
ρ = mv
Where ρ is the momentum
m is the mass
and v is the velocity
From the given information,
ρ = 3 kg m/sec
v = 9 m/sec
Putting the parameters into the formula, we get
3 = m × 9
∴ m = 3 ÷ 9
m = 0.33 kg
Hence, the mass of the beach ball is 0.33 kg
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If a 75 kg box collides with a stationary 35 kg box with a force of 110 N, what must be true of the magnitude of the reaction force?
According to Newton third law of motion, what must be true of the magnitude of the reaction force is that the reaction force will be equal and opposite to the action force of 110 newtons
From Newton third law of motion which state that in every action of force, there will be equal and opposite reaction.
If a 75 kg box collides with a stationary 35 kg box with a force of 110 N, the mass 35 kg will produce a reaction force which is equal and opposite to the action force 110 newton received from mass 75 kg.
Therefore, what must be true of the magnitude of the reaction force from mass 35 kg is that the mass will produce an equal and opposite force equal to 110 Newtons.
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Mass can be measured with a triple
beam balance in units of
Resources
A. grams
B. Newtons
C. cubic centimeters
D. millimeters
Answer:
A
Explanation:
A projectile is fired with a velocity of 320 ms at an angle of 30 degree to a horizontal.1 Find the time to reach the maximum height. 2 it's horizontal range.
(a) The time taken for the projectile to reach the maximum height is 32.65 s.
(b) The horizontal range of the projectile is 9,049.1 m.
The given parameters:
Initial velocity of the projectile, u = 320 m/sAngle of projection, = 30 degreesThe time taken for the projectile to reach the maximum height is calculated as follows;
[tex]v_f = u- gt\\\\0 = 320 - 9.8t\\\\9.8t = 320\\\\t = \frac{320}{9.8} \\\\t = 32.65 \ s[/tex]
The horizontal range of the projectile is calculated as follows;
[tex]R = \frac{u^2 sin(2\theta)}{g} \\\\R = \frac{320^2 \times sin(2\times 30)}{9.8} \\\\R = 9,049.1 \ m[/tex]
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PLZZZ HELPPP ASAP
I really need help as soon as possible
Answer:
Friction
Explanation:
As the toy cars rolls away, more friction is created. The more friction there is, the more friction on surface rubs against another which creates friction which in-term slows it down. Hope this helps.
Two boats race in the water. One turns around to head back in the other direction to complete the race, but it collides into the boat that was now directly in front. What forces may have been a part of the collision? Describe them in detail.
The two forces involved in this collision are the colliding force and the reaction force which are equal in magnitude but opposite in direction.
According to Newton's third law, action and reaction are equal and opposite. This law is very important in describing head - on collisions between objects and is particularly apt in describing the collision between the two boats.
In this collision, two forces are important these are the action and the reaction. The action is the colliding force which the boat that was hit experiences. The reaction force is the force that this boat experiences as a result of hitting the boat in front of it. The both forces are equal in magnitude but opposite in direction.
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A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates. A 73.5-kg water-skier has an initial speed of 5.2 m/s. Later, the speed increases to 12.2 m/s. Determine the work done by the net external force acting on the skier.
ANSWER FOR BRAINLIEST IF ANSWER IS GOOD
Ignore my comment, that information is unnecessary.
Assuming no friction between the skis and the water, the total work performed on the skier is due only to the pulling force (T, for tension). By Newton's second law, the net force exerted on the skier is
∑ F = T = ma
where m = 73.5 kg is the mass of the skier, and a is their acceleration.
Under constant acceleration, the skier is pulled a distance x such that
(12.2 m/s)² - (5.2 m/s)² = 2ax
and solving for x gives
x = (60.9 m²/s²) / a
Then the total work performed on the skier is
W = Tx
W = ma • (60.9 m²/s²) / a
W = (73.5 kg) (60.9 m²/s²)
W ≈ 4480 J
Explain the relationship between pitch and frequency.
Answer:
Though pitch and frequency are not equivalent, they are correlated. This means that as one goes up, the other does as well. A higher frequency produces a higher pitch, and a lower frequency produces a lower pitch.
Explanation:
A Person is carrying 6kg in one hand and 5kg in another.Calculate the resultant force applied by him
8. Placing your vehicle between the pilot/escort vehicle and an oversize/overweight vehicle can be dangerous.A. TrueB. False
Answer:
A because that's the answer
electromagnetic radiation at its maximum wavelength is:
Answer:
the radio wave. - The frequency ranges between 300 gigahertz (GHz) to 30 hertz (Hz).
Explanation:
A bullet of mass M1, is fired towards a block
of mass m2 initially at rest at the edge of a
frictionless table of height h as in the figure.
The initial speed of the bullet is vi. Consider
two cases. a соmрlеtеlу inсlаѕtiс one and an
elastic one where the bullet bounces off the block. What is the flight ratio time?
The ratio of time of flight for inelastic collision to elastic collision [tex](t_A :t_B)[/tex] is 1:2
The given parameters;
mass of the bullet, = m₁mass of the block, = m₂initial velocity of the bullet, = u₁initial velocity of the block, = u₂ = 0Considering inelastic collision, the final velocity of the system is calculated as;
[tex]m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\m_1u_1 + 0 = v(m_1 + m_2)\\\\v= \frac{m_1u_1}{m_1 + m_2} \ -- (1)\\\\[/tex]
The time of motion of the system form top of the table is calculated as;
[tex]v = u + gt\\\\v = 0 + gt\\\\v = gt\\\\t= \frac{v}{g} \\\\t_A = \frac{m_1u_1}{g(m_1 + m_2)} \ \ ---(2)[/tex]
Considering elastic collision, the final velocity of the system is calculated as;
[tex]m_1u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\m_1u_1 + 0 = m_1v_1 + m_2v_2\\\\m_1 u_1 = m_1v_1 + m_2v_2[/tex]
Apply one-directional velocity
[tex]u_1 + (-v_1) = u_2 + v_2\\\\u_1 -v_1 = 0 + v_2\\\\v_1 = v_2 -u_1[/tex]
Substitute the value of [tex]v_1[/tex] into the above equation;
[tex]m_1u_1 = m_1(v_2 - u_1) + m_2 v_2\\\\m_1u_1 = m_1v_2 -m_1u_1 + m_2v_2\\\\2m_1u_1 = m_1v_2 + m_2v_2\\\\2m_1u_1= v_2(m_1 + m_2)\\\\v_2 = \frac{2m_1u_1}{m_1+ m_2} \ --(3)[/tex]
where;
[tex]v_2[/tex] is the final velocity of the block after collision
Since the bullet bounces off, we assume that only the block fell to the ground from the table.
The time of motion of the block is calculated as follows;
[tex]v_2 = v_0 + gt\\\\v_2 = 0 + gt\\\\t = \frac{v_2}{g} \\\\t_B = \frac{v_2}{g} \\\\ t_B = \frac{2m_1u_1}{g(m_1 + m_2)} \ \ ---(4)[/tex]
The ratio of time of flight for inelastic collision to elastic collision is calculated as follows;
[tex]\frac{t_A}{t_B} = \frac{m_1u_1}{g(m_1 + m_2)} \times \frac{g(m_1 + m_2)}{2m_1u_1} \\\\\frac{t_A}{t_B} = \frac{1}{2} \\\\t_A:t_B = 1: 2[/tex]
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Question 4 of 10
When you go along with a group because you don't want to cause trouble, it is
called:
A. informational social influence.
B. normative social influence.
C. attribution
.
D. fundamental social influence.
Answer:
Normative
Explanation:
It is normal occurance but not unnatural