Answer:
Speed and velocity are related in much the same way that distance and displacement are related. Speed is a scalar and velocity is a vector.
What kind of waves are light waves and radio waves
Answer:
Electromagnetic waves
Explanation:
I don’t know what to do help me
Answer:
oooh thats hard
Explanation:
well i would probaly search the page number or anthing on the page that should help
I really dont know physics
Answer:
3
Explanation:
I had this question before
A wagon having a mass of 42 kg is accelerated across a level road at 1.50 m/s2. What net force acts on the wagon horizontally?
Answer:
The net force acting on the wagon is 63 N
Explanation:
Net Force
According to the second Newton's law, the net force exerted by an external agent on an object of mass (m) is:
F = m.a
Where (a) is the acceleration of the object.
The wagon has a mass of m=42 Kg is accelerated at [tex]a=1.5\ m/s^2[/tex]. The net force is:
F = 42*1.5
F = 63 N
The net force acting on the wagon is 63 N
A force of 30 N stretches a very light ideal spring 0.73 m from equilibrium. What is the force constant (spring constant) of the spring
The forces constant (spring constant) of the spring will be 41.09 N/m.
What is spring force?The force required to extend or compress a spring by some distance scales linearly concerning that distance is known as the spring force. Its formula is;
F = kx
The given data in the problem is;
F is the spring force = 30 N
K is the spring constant= ?
x is the displacement of spring = 0.73 m
The spring constant is;
K =F/x
K=30/0.73
K=41.09 N/m
Hence the force constant (spring constant) of the spring will be 41.09 N/m.
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pls help me with this
Answer:
A screw
Explanation:
1. A book is at rest on a tabletop. Diagram the forces acting on the book.
The free-body diagram of the forces acting on a book resting on a table is shown in the figure.
When a body is at rest, the net force on the body is zero. So, no acceleration is produced in it. If a body is in contact with another body or surface, a normal contact force acts on it perpendicular to the surface in contact.
So, the forces acting on a book resting on a tabletop are:
1. Weight of the book, W(downward)
2. Normal force (N) perpendicular to the surface in contact.
Therefore, the free-body diagram of the forces acting on a book resting on a table is shown in the figure.
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Calculate the speed with which you would have to throw a rock to put it into orbit around the asteroid 234 Ida near its surface, assuming 234 Ida is spherical. Ida’s mass is 4 × 1016 kg and its radius is 16 km.
Answer:
The speed with which a rock would have to be thrown to put it in 234 Ida's orbit, near its surface is approximately 12.917 m/s
Explanation:
The given parameters are;
The mass of Ida, M = 4 × 10¹⁶ kg
The radius of 234 Ida, r = 16 km = 16,000 m
The speed, v, required to put a rock in 234 Ida's orbit near its surface is given by the orbital velocity equation as follows;
[tex]v = \sqrt{{\dfrac{G \times M}{r} } }[/tex]
Where;
G = The universal gravitational constant = 6.67408 × 10⁻¹¹ m³·kg⁻¹·s⁻²
Substituting the known values gives;
[tex]v = \sqrt{{\dfrac{6.67408 \times 10^{-11} \times 4 \times 10^{16}}{16,000} } } \approx 12.917[/tex]
Therefore, the speed required to put a rock in 234 Ida's orbit near its surface = v ≈ 12.917 m.
what does PE + KE = __
Answer:
Energy that is dependant on height is called gravitational potential energy.
The formula for potential energy is:
G.P.E.=mgh
Weight (mass x gravity) determines the amount of potential energy.
Is there potential energy in the 5 forms of energy?
Yes, potential energy can be found in fossil fuels, within the foods you eat, and the batteries that you use.
Conservation of Energy:
It is important to understand that each form of energy does not exist separately but undergoes change from one to another without a net loss in energy. Consider this example:
As a projectile is launched into the air KE is at its maximum. As the projectile gains altitude PE becomes greater than KE. At the top of its arc, PE is at its maximum. The whole cycle reverses itself on the way down.
As you can see, the loss in KE is a gain in PE. Energy is not lost but conserved.
Law of conservation of energy.
Energy cannot be created nor destroyed. It can be transformed from one form to another, but the total amount of energy never changes.
A washing machine is being dragged with a starting force of 380.8 N to the West, but this is reduced by friction. If the mass of the washing machine is 37 kg
and the coefficient of friction is 0.54, what was the Net acceleration experienced by the washing machine?
A 10.27 m/s^2 South
B 5 m/s^2 South
C 19.01 m/s^2 South
D 7.6 m/s^2 South
Answer:
B. [tex]5\,\frac{m}{s^{2}}[/tex] South
Explanation:
The correct statement is:
A washing machine is being dragged with a starting force of 380.8 N to the South, but this is reduced by friction. If the mass of the washing machine is 37 kg and the coefficient of friction is 0.54, what was the Net acceleration experienced by the washing machine?
The net force of the washing machine moving on the horizontal ground is equal to the external force minus the friction force, where the normal force from the ground on the washing machine is equal to its weight. The free body diagram of the washing machine is presented below. The equation of equilibrium associated with the acceleration of the washing machine is:
[tex]\Sigma F = P -\mu_{k}\cdot m \cdot g = m\cdot a[/tex] (1)
Where:
[tex]P[/tex] - External force exerted on the washing machine, measured in newtons.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, dimensionless.
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]a[/tex] - Net acceleration of the washing machine.
If we know that [tex]P = 380.8\,N[/tex], [tex]m = 37\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\mu_{k} = 0.54[/tex], then the net acceleration experienced by the washing machine is:
[tex]a = \frac{P}{m}-\mu_{k}\cdot g[/tex]
[tex]a = \frac{380.8\,N}{37\,kg}-(0.54)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]a = 4.996\,\frac{m}{s^{2}}[/tex]
The correct answer is B.
A skier skids to a stop at the bottom of a mountain.
What is the equal and opposite force for the force of the skier's friction pushing on the mountain as
described by Newton's third law?
Answer:
Friction of the mountain on the skier in the opposite direction
Explanation:
answered it on khan academy
Friction of the mountain on the skier in the opposite direction.
What is Newton's third law?A force is a push or a pull that acts upon an object as a results of its interaction with another object. Forces result from interactions.
Forces result from contact interactions (normal, frictional, tensional, and applied forces are examples of contact forces) and other forces are the result of action-at-a-distance interactions (gravitational, electrical, and magnetic forces).
There are two forces resulting from this interaction - a force on the chair and a force on your body. These two forces are called action and reaction forces and are the subject of Newton's third law of motion.
Therefore, Friction of the mountain on the skier in the opposite direction.
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‼️can someone help me‼️
• What are the other two ways that you can rewrite the
formula F=ma?
• If I push a wheelbarrow, with a force of 240 N, with an
acceleration of 4 m/s2, how much mass does the
wheelbarrow have? (Solve using formula F=ma)
Answer:
Part A:
The other two ways we can rewrite F = m·a are;
1) F = dp/dt
2) F = m × d(v)/dt
Part B:
The mass of the wheelbarrow is 60 kg
Explanation:
Part A:
The other two ways of rewriting the equation are;
1) Force as the rate of change of momentum, F = dp/dt = d(m·v)/dt and
2) From the above equation, where, m is constant, we have;
F = d(m·v)/dt = m × d(v)/dt
∴ F = m × d(v)/dt
Part B:
The force with which we push the wheelbarrow = 240 N
The acceleration, a, given to the wheelbarrow (by the applied force) = 4 m/s²
From F = m × a
Where;
m = The mass of the wheelbarrow
m = F/a = 240 N/(4 m/s²) = 60 kg
The mass of the wheelbarrow = m = 60 kg.
A child kicks a ball horizontally with a speed of 4.8 m/s off a deck 3.5 m off the
ground. How long will it take, in seconds, for the ball to hit the ground?
Answer:
0.73 sExplanation:
The time taken can be found by using the formula
[tex] t = \frac{d}{v} \\ [/tex]
d is the distance
v is the velocity
From the question we have
[tex]t = \frac{3.5}{4.8} \\ = 0.72916...[/tex]
We have the final answer as
0.73 sHope this helps you
Which of the following examples of energy transfer CANNOT happen?
A toy car rolling down a hill increases speed and then rolls to a stop.
A bunger jumper jumps from a bridge and goes up higher with each recoil.
A child playing on a swing moves fastest at the bottom of the swing's path.
A rock dropped from the top of a bridge increases in speed until it hits the water.
The energy transfer which cannot be happened is option c , at the bottom of the swing, the kinetic energy gets converted completely into its potential energy thus, it have no speed when it reached the bottom.
What is kinetic energy ?Kinetic energy of a body is generated by virtue of its motion. Kinetic energy increases as its mass and velocity increases. Potential energy of a body is the energy stored on it when it comes to rest.
When an object is falling freely to the ground its kinetic energy is converting to the potential energy up to the point it reaches the ground. When it touches the ground, the kinetic energy converts back to potential energy.
Therefore, when the child in the swing comes down from a height its speed is increasing and have greater kinetic energy but this is converted to potential energy when it reach the bottom. Hence no speed it have at the bottom.
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A weightlifter takes 0.5s to raise a barbell with 2 free-weights from the ground to a height of 0.9m. If each free-weight has a mass of 30kg and the barbell has a mass of 20kg, calculate how much power the weightlifter produces when lifting the barbell?
Answer:
P = 1412.82 [Watt]
Explanation:
First we must calculate the total mass that makes up the barbell and the two free weights at each end of the barbell.
[tex]m=30+30+20\\m= 80 [kg][/tex]
Weight is now defined as the product of mass by gravitational acceleration.
[tex]w=m*g[/tex]
where:
m = mass = 80 [kg]
g = gravity acceleration = 9.81 [m/s²]
[tex]w=80*9.81\\w=784.8[N][/tex]
Now using the product of the weight by the distance traveled we can calculate the work.
[tex]W=w*d[/tex]
where:
W = work [J]
w = weight = 784.8 [N]
d = distance = 0.9 [m]
[tex]W=784.8*0.9\\W=706.41[J][/tex]
And power is defined as the relationship of work at a certain time. The potency is expressed by means of the following formula.
[tex]P=W/t\\P=706.41/0.5\\P=1412.82[W][/tex]
A pinball bangs against a bumper of a pinball machine with a speed of 0.46 m/s. If the ball has a mass of 0.058 kg, what is the ball's kinetic energy?
This needs to be done in 3 mins
Answer:
Okay I might be wrong and sorry if i am either its 12 or 0.402
Explanation:
* I recommend the second one cause its more logical
Be aware I might be wrong
A friend is coming to Tim’s house to study after school. What directions would Tim give for reaching his house from the entrance of the school?
(I WILL GIVE BRAINLIEST)
Answer:
go up the street as you exit the house and make a right and keep going up for 3 blocks and you should see the school
s. The resistan Rav/I., where va 1005 and
1 - 100..2. Find the percentage
emor in R.
Complete question :
the resistance r =v/i where v = (100±5 )v and I = (10±0.2) a.find percentage error in r
Answer:
7%
Explanation:
Given that :
R = V/I
V = 100 ± 5
I = 10 ± 0.2
Percentage error in R can be obtained thus :
[(dV / V) + (dI / I)] * 100%
dV = ± 5
dI = ± 0.2
[(5 / 100) + (0.2 / 10)] * 100%
(0.05 + 0.02) * 100%
= 0.07 * 100%
= 7%
show your work. john uses a 25N force to push a boulder off a cliff that is 312m tall. What is the work done on the boulder?
Answer:7800
work=force x distance
Force in Newtons
Distance in Meters
Work in Joules
Answer the following
1) A piece of gold weighis 240 g and has a volume of 20 cm^3. Find the density of gold.
2) What is the perpendicular distance between the force and the turning point if a force
of 25 N produces a moment of force of 12.5 Nm
pls answer fast I will mark as brainlist
Answer:
1. 12 g/cm³
2. 0.5 m
Explanation:
1.
The weight of the gold is 240 g
The volume is 20 cm³
Density = Mass/ volume
Density = 240 / 20 = 12 g/cm³
2.
The formula for moment of a force is ;
Force * distance perpendicular between force and the turning point of force
Given that the moment of force is 12.5 Nm and force is 25 N then the perpendicular distance between the force and the turning point will be;
12.5 = 25 * d
12.5/25 = d
0.5 = d
d= 0.5 m
A bus accelerates forward. If an apple were on the floor of the bus it would move forward.
A: True
B: False
Answer:
False
Explanation:
Since it is on the bus, it would not move forward because the outside acceleration cannot be considered.
A 200-gram baseball traveling at 48 m/s is hit by a bat and rebounds in the opposite direction at 52 m/s. Find the average force of the bat on the ball if the time of contact is 2.0x10-3 s.
Answer:
400N
Explanation:
Given the following parameters
Mass m = 200g = 0.2kg
initial velocity u = 48m/s
Final velocity v = 52m/s
Time t = 2.0x10-3 s.
Impulse is expressed as I = Ft = m(v-u)
Ft = m(v-u)
F = m(v-u)/t
Substitute the given values into the formula
F = 0.2(52-48)/0.002
F = 0.2(4)/0.002
F = 0.8/0.002
F = 400N
Hence the average force of the bat on the ball is 400N
A boy throws a stone vertically in the air with an initial speed of 40m/sec.At the instant the stone is thrown,a monkey at the top of a tree h m high falls freely.How long will it take for the stone to hit the monkey if they meet at a point mid way between the ground and the top of tree?
∞infinaty stones∞
because why not
At a sports event, the car starts from iosi, in 5.0 siis acceleration is 5.
Calculate the distance travelled by ban
100 m
25 m
63 m
O bag 6.
The Score Tower hingga laglomery 4
werissa 13
Question:
At a sports event, the car starts from rest. in 5.0 s its acceleration is 5.0 m/s2. Calculate the distance travelled by car.
Answer:
62.5 metres
Explanation:
Given
[tex]u = 0[/tex] -- Initial Velocity
[tex]a = 5.0m/s^2[/tex] --- acceleration
[tex]t =5.0s[/tex] -- time
Required
Calculate the distance
This question will be solved using the following equation of motion
[tex]S = ut + \frac{1}{2}at^2[/tex]
Where S represents the distance
Substitute values for u, t and a
[tex]S = 0 * 5.0 + \frac{1}{2} * 5.0 * 5.0^2[/tex]
[tex]S = 0 + \frac{1}{2} * 5.0 * 25.0[/tex]
[tex]S = \frac{1}{2} * 5.0 * 25.0[/tex]
[tex]S = \frac{1}{2} * 125.0[/tex]
[tex]S = 62.5m[/tex]
Hence, the distance travelled is 62.5m
Which is heavier: 30 kilogram or 300 milligrams?
Answer:
30 killograms is 85.9803
and 300 milligrams is 0.000661387 pounds
Explanation:
Which one of the following is not a member of the solar family: (a) Morning Star (b) Evening Star (c) Pole Star (d) Tailed Star plzzz answer this...
Answer:
Tailed star
Explanation:
Tailed star
A 75 kg ball carrier is running to the right at 6.5 m/s. An 80 kg defender is chasing the ball carrier running at 7.0 m/s. The defender catches the ball carrier in a completely inelastic collision. What was the kinetic energy of the ball carrier/defender system BEFORE the tackle?
Answer:
3,544.375Joules
Explanation:
Kinetic energy is the energy possessed by a body by virtue of its motion, It is expressed as;
Kinetic energy = 1/2mv²
m is the mass of the body
v is the velocity
For the ball carrier;
KE = 1/2(75)(6.5)²
KE = 3168.75/2
KE = 1584.375Joules
For the defender;
KE = 1/2(80)(7)²
KE = 3920/2
KE = 1960Joules
The kinetic energy of the ball carrier/defender system BEFORE the tackle = KE for the carrier + KE for the defender
kinetic energy of the ball carrier/defender system BEFORE the tackle= 1584.375+1960 = 3,544.375Joules
i) The energy stored in wound watch spring is
a) K.E
b) P.E
c) Heat Energy
d) Chemical Energy
Answer:
Springs when non compressed has potential energy.
How high can a 40 N force move a load, when 395 J of work is done? (Show work)
A 40 N force would move a load 9.875 m high when 395 J of work is done on the load.
What is work done?The total amount of energy transferred when a force is applied to move an object through some distance
The work done is the multiplication of applied force with displacement.
The SI unit of work done by an object is Joules.
IJ is the work done when a force of 1 N displaces an object by 1 m .
Work Done = Force * Displacement
For the given problem Force =40N and work done = 395 J
By using the above formula
Work Done = Force * Displacement
395 j = 40 N × height
height =9.875 m
Thus a 40 N force supported by a work done of 395 J moves the load up to 9.875 m height
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A machine that uses 200 w of power moves an object a distance of 15 m in 25 sec. Find
the force needed and the work done by this machine,
Answer:dummy
Explanation:
Dum dum