Answer:
10.68 kgExplanation:
The mass of the object can be found by using the formula
[tex]m = \frac{p}{v} \\ [/tex]
p is the momentum
v is the velocity
From the question we have
[tex]m = \frac{56.59}{5.3} \\ = 10.677358...[/tex]
We have the final answer as
10.68 kgHope this helps you
A man pushes a crate along a factory floor by exerting a force of 55 N. If
the crate moves a distance of 4.0 m, how much work does the man
perform?
Answer:
it's 220 J
the second one
3 QUESTIONS (SCIENCE)
First question: What can bad space weather do to earth?
Second question: What is the purpose of the United Nations Committee, and what do they do?
Third question: What protect Earth form much of the damage from space weather?
John is pushing (horizontally) on a 100 kg (Fw = 980N) bench with an unknown force. If the coefficient of static friction is 0.20 and the acceleration is 1.84m/s/s, what is the applied force?
Answer:
1176N
Explanation:
Given parameters:
Mass of bench = 100kg
Forward force = 980N
Coefficient of static friction = 0.2
Acceleration = 1.84m/s²
Unknown:
Magnitude of the applied force =?
Solution:
Frictional force is a force that opposes motion, for a body to move, the applied force must be greater than the frictional force;
The Force applied;
Force applied = Frictional force + Weight of the body
Frictional force = umg
u is the coefficient of static friction
m is the mass
g is the acceleration due to gravity
Force applied = 0.2 x 100 x 9.8 + 980 = 1176N
A box has a momentum of 38.0 kg*m/s to the right. A 88.3 N force pushes it to the right for 0.338 s. What is the final momentum of the box? (unit = kg*m/s)
Answer:
63.8454kgm/s
Explanation:
According to Newton's second law;
Force = mass ×acceleration
Fore = m(v-u)/t
Force = mv-mu/t
MV is the final momentum
mu is the initial momentum = 38kgm/s
Force = 88.3N
Time = 0.338s
Substitute and get mv
88.3 = mv-38/0.338
88.3×0.338 = mv-38
29.8454 = mv-38
MV = 29.8454+34
mv = 63.8454kgm/s
Hence the final momentum is 63.8454kgm/s
Answer:
67.8
Explanation:
im an acellus student, and this is the answer that i got when doing the problem.
A car traveling at a speed of 13 meters per second accelerates uniformly to a speed of 25 meters per second in 5.0 seconds. 11- Calculate the magnitude of the acceleration of the car during this 5.0-second time interval.
Answer:
[tex]a=2.4\ m/s^2[/tex]
Explanation:
Given that,
Initial speed of a car, u = 13 m/s
Final speed of a car, v = 25 m/s
Time, t = 5 s
We need to find the acceleration of the car during this 5.0 second time interval. Let a is the acceleration. It can be calculated as :
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{25-13}{5}\\\\=2.4\ m/s^2[/tex]
So, the acceleration of the car is [tex]2.4\ m/s^2[/tex].
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How are ribosomes different from other complex animal cell organelles?
Explanation:
Ribosomes are not quite the same as different organelles since they have no layer around them that isolates them from different organelles, they comprise of two subunits, and when they are creating sure proteins they can become film bound to the endoplasmic reticulum, however they can likewise be free drifting while at the same time performing
where would you expect light waves to move faster?
Answer:
In space
Explanation:
I took the quiz
Answer:
in space
Explanation:
If a current of 2.5 Ampere flow through a circuit in 15 minutes calculate amount of charge through a crcuit
Answer:
Q = 2250 Coulombs
Explanation:
We are given;
Current: I = 2.5 A
Time: t = 15 minutes = 15 × 60 seconds = 900 s
Now formula for the amount of charge through the circuit is given by:
Q = It
Thus;
Q = 2.5 × 900
Q = 2250 Coulombs
what are the uses of X-rays.
Explanation:
id : 237 514 2470
pass : frmCV5
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You used 100 N of force to push a shopping cart 5m down the aisle in the grocery store. How much work did you do? *
Answer:
[tex]\boxed {\boxed {\sf 500 \ Joules}}[/tex]
Explanation:
Work can be found by multiplying the force by the distance.
[tex]W=F*d[/tex]
The force is 100 Newtons and the distance is 5 meters. Therefore,
[tex]F= 100 \ N \\d= 5 \ m[/tex]
We can substitute the values into the formula.
[tex]W= 100 \ N * 5 \ m[/tex]
Multiply.
[tex]W= 500 \ N*m[/tex]
1 Newton meter is equal to 1 Joule. Thus, our answer of 500 Newton meters is equal to 500 Joules.[tex]W= 500 \ J[/tex]
You did 500 Joules of work to push a shopping cart 5 meters with a force of 100 Newtons.
Answer:
Given :-Force (F) = 100 NDisplacement (D) = 5 mTo Find :-Work done
Solution :-As we know that Work done is the product of Force and displacement
[tex] \tt \: w = force \: \times displacement[/tex]
[tex] \tt \: w = 100 \times 5[/tex]
[tex] \tt \: w = 500 \: j[/tex]
Therefore, Work done is 500 Joule
A child's first bicycle has 12-inch tires. These tires have a 6-inch radius. How fa r does the bicycle move forward each time the wheel makes one complete rotation? Give your answer in meters. (1 inch = 0.022 meters) b. A woman's ten-speed bicycle has 27-inch tires (13.5-inch radius). How far does this bicycle …
Answer:
a. s = 0.96 m
b. s = 2.15 m
Explanation:
a.
The relationship between the linear displacement and the angular displacement is given as follows:
[tex]s = r\theta[/tex]
where,
s = linear distance covered = ?
r = radius of wheel = (6 in)(0.0254 m/1 in) = 0.1524 m
θ = angular displacement = (1 rev)(2π rad/1 rev) = 2π rad
Therefore,
[tex]s = (0.1524\ m) 2\pi[/tex]
s = 0.96 m
b.
Assuming, we have to find linear displacement here, as well:
[tex]s = r\theta[/tex]
where,
s = linear distance covered = ?
r = radius of wheel = (13.5 in)(0.0254 m/1 in) = 0.3429 m
θ = angular displacement = (1 rev)(2π rad/1 rev) = 2π rad
Therefore,
[tex]s = (0.3429 \ m) 2\pi[/tex]
s = 2.15 m
a_______ is a region where the magnetic fields of a large number of atoms are lined up parallel to a magnets field.
Answer:
Magnetic Domain
Explanation:
1. Why do we need to travel?
Answer:
We're like children who move away from their parents - then back,
HUMANS LOVE to travel. Mentally or physically, they escape to other places. They crave adventure. They seek the unknown. They want to see new places, learn new things, enjoy new experiences and then return home to the familiar, the predictable, the secure, before taking off again on their travels.
This departure-return behaviour has its origins, perhaps, in the early exploratory behaviour of the child. Research has shown that children who are "securely attached" to their parents, sure of their affection and protection, and who know that their parents will respond to their needs, are more independent, adventurous and exploratory in their play and behaviour.
Observations of infants show that from the safe base of having a parent present and available, the infant dares to move away a short distance and then return to the parent, to move a little further and return, until in incremental and amazingly measurable distances it explores the environment that surrounds it and has the courage to move beyond it.
This may be the first symbolic travel: the first departure, the first expedition and the possible beginning of the travel bug for those for whom their initial independent voyaging was successful.
With these first independent steps away from the parental presence lies the atavistic or recurring wish to reach out beyond the current confines of space and relationship
and see what is further away. And yet further again to see what is not visible, what is around the next corner, in the next room and eventually on
the other side of our geographical or psychological world. This is a human need. It is why we have uncovered our world, travelled its length, hiked to the top of its mountains and the depth of its oceans. It is why we became discontent with exploring our own plant but had to move beyond it into the unknown. It is why, as poet TS Eliot reminds us, we will "not cease from exploration".
It is our nature to travel. It is our joy to travel. It is our paradoxical psychological disposition to wish for the contradictory conditions of stability and change: to stay and to go, to be and to explore, to rest and to travel unceasingly.
Travel is more than going somewhere. It is a psychological event. It requires motivation and imagination. It begins in childhood with faltering first steps and with imagination fired by the tales of "long, long ago in a faraway land".
We emerge from these fantasies of childhood determined to cross the oceans, to find those faraway lands, the characters they containand the magic they hold. Determined to find the adventures they promise and the prospect of returning home triumphant with trophies having vanquished all fear, overcome all obstacles and live happily eve after.
Answer:
to see the world and experience new things. it opens your mind and you can meet more people. you have a new view on the world after seeing it from other places, not only new views on the world but new cultures, and languages.
Explanation:
On an air track, a 4-kg cart moving at 3 m/s collides with a 2-kg cart that is
initially at rest. If the two carts stick together, what will be their velocity
after the collision? Show all calculations, including any equation(s) used,
and include correct measurement units. *
Your answer
Answer:
v₃ = 2 [m/s]
Explanation:
To solve such problems we must use the principle of conservation of momentum. That is, the linear momentum is conserved before and after the collision.
P = m*v
where:
P = linear momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]
The momentum is conserved before and after the collision, in this way we can obtain the following equation.
[tex](m_{1}*v_{1})+(m_{2}*v_{2})=(m_{1}+m_{2})*v_{3}[/tex]
where:
m₁ = mass of the cart moving = 4 [kg]
v₁ = velocity of the cart moving before the collision = 3 [m/s]
m₂ = mass of the cart initially at rest = 2 [kg]
v₂ = velocity of the cart at rest = 0
v₃ = velocity of the two carts combined (carts stick together) after the collision [m/s]
[tex](4*3)+(2*0)=(4+2)*v_{3}\\v_{3}=12/6\\v_{3}=2[m/s][/tex]
What is a gravitational field, and how does it apply to Earth? what is its formula. and what is an example of how you can use it?
Answer:
Explanation:
a gravitational field is the region of space surrounding a body in which another body experiences a force of gravitational attraction.
The formula is: weight/mass = gravitational field strength. On Earth the gravitational field strength is 10 N/kg.
an example of how you can use it if you need to find the gravitational force from earth.
What scientific law did Archimedes establish? the law of geometry the law of astronomy the law of lever and balance the law of Earth and universe
Answer:
Archimedes discovered the buoyancy laws
Explanation:
when asked by King Hiero of Syracuse to determine whether his new crown was pure gold (SG = 19.3). Archimedes measured the weight of the crown in air to be 11.8 N and its weight in water to be 10.9 N.
Answer:
The correct answer is C
Explanation:
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What is a selectively permeable membrane?????????????
Answer:
It is that kind of permeable membrane that allows certain molecules to pass through. Not every molecules can pass through it
Explanation:
Which of the four spheres is this pictures/scene?
A) hydrosphere
B) biosphere
C) geosphere
D) atmosphere
I believe the correct answer is atmosphere (D).
Answer:
D) atmosphere
Explanation:
but the correct answer is troposhare
The dead sea has a very high salt content. Explain why it is easier for a swimmer to float in the dead sea than in a swimming pool containing fresh water
Answer:
Dead sea is full of salt.
Explanation:
The water of the Dead Sea is full of salt, which makes it much denser and heavier than freshwater. If you swim in it, you float very easily.
When salt is dissolved in water, for example sea water, dissolved salt adds to the mass of the water and makes the water denser than it would be without salt. Because objects float better on a dense surface, they float better on salt water than fresh water.
A long-jumper leaves the ground at an angle of 20.0° above the horizontal and at a speed of 11.0 m/s.How far does he jump in the horizontal direction?(Assume his motion is equivalent to that of a particle.).
a. 12m
b. 7.94m
c. 10m
d. 0.384m
Answer:
The correct option is;
b. 7.94 m
Explanation:
The given parameters of the jump of the long jumper are;
The angle above the horizontal with which the long jumper leaves the ground, θ = 20.0°
The speed with which the long jumper leaves the ground, u = 11.0 m/s
The furthest horizontal distance the long jumper jumps, given that the motion is equivalent to that of a particle, is given by the formula for the range, R, of a projectile motion as follows;
[tex]R = \dfrac{u^2 \times sin \left (2 \cdot \theta \right )}{g}[/tex]
Where;
g = The acceleration due to gravity ≈ 9.8 m/s²
u = The initial velocity of the long-jumper = 11.0 m/s
θ = The angle of the direction above the horizontal the long-jumper jumps = 20.0°
Plugging in the values, gives;
[tex]R = \dfrac{(11.0 \ m/s)^2 \times sin \left (2 \times 20.0 ^{\circ} \right )}{9.8} = \dfrac{121 \ m^2/s^2 \times sin \left (40.0 ^{\circ} \right )}{9.8 \ m/s^2} \approx 7.94 \ m[/tex]
How far the long-jumper goes = The range, R, of the projectile motion ≈ 7.94 m.
help please 10 pts and quick!
Answer:
A
Explanation:
how high above the ground is a 3 kg object that has 147 J (joules) of potential energy
Answer:
4.9949
Explanation:
M = 3 kg
P.E = 147 joules
G = 9.81 ms^-2
H = ?
P.E = MGH
147 = 3 x 9.81 x H
147 = 29.43H
147/29.43 = 29.43H/29.43
4.9949 = H
:. H = 4.9949
What is the correct answer?
Answer:3
Explanation:
A 1000-kg car comes to a stop without skidding. The car's brakes do 50,000 J of work to stop the car. Which of the following was the car's velocity when the brakes were initially applied?
Answer:
10m/s
Explanation:
What is a joule? Pls only answer if you know the answer.
Answer:
the SI unit of work or energy, equal to the work done by a force of one newton when its point of application moves one meter in the direction of action of the force, equivalent to one 3600th of a watt-hour.
Explanation:
Joule, unit of work or energy in the International System of Units (SI); it is equal to the work done by a force of one newton acting through one metre. Named in honour of the English physicist James Prescott Joule, it equals 107 ergs, or approximately 0.7377 foot-pounds.
Madeline fires a bullet horizontally. The rifle is 1.9 meters above the ground. The bullet travels 200 meters horizontally before it hits the ground. What speed did Madeline's bullet have when it exited the rifle?
Answer: 321.027 m/s
Explanation:
First let´s analyze the vertical problem:
Always when an object is above the ground and nothing is holding it (like the bullet after being fired) the gravitational force will start acting on it.
Then the vertical acceleration of the bullet will be the gravitational acceleration, we can write this as:
a(t) = -g
Where g = 9.8m/s^2
and the negative sign is because this acceleration pulls the bullet downwards.
To get the vertical velocity we need to integrate over time, this will lead to:
v(t) = (-9.8m/s^2)*t + v0
where v0 is the initial vertical velocity, as the bullet is fired horizontally, there is no initial vertical velocity, then we have v0 = 0m/s
And the velocity equation is:
v(t) = (-9.8m/s^2)*t
Now for the vertical position, we need to integrate again, to get:
p(t) =(1/2)*(-9.8m/s^2)*t^2 + p0
Where p0 is the initial vertical position, in this case, is 1.9 meters above the ground, then p0 = 1.9m
And the vertical position equation will be:
p(t) = (1/2)*(-9.8m/s^2)*t^2 + 1.9 m
Now we want to find the time such that the vertical position is equal to zero, this will mean that the bullet it the ground.
p(t) = 0m = (1/2)*(-9.8m/s^2)*t^2 + 1.9 m
(1/2)*(9.8m/s^2)*t^2 = 1.9m
(4.9m/s^2)*t^2 = 1.9m
t = √(1.9m/(4.9m/s^2)) = 0.623 seconds.
This means that the bullet will travel for 0.623 seconds before hitting the ground.
Now we also can ignore the air friction for the horizontal motion, then we can assume that the horizontal speed does not change.
Then we can use the relationship:
Distance = speed*time
We know that:
time = 0.623 seconds
distance = 200m
now we can replace that in the equation to find the horizontal speed.
200m = speed*0.623s
200m/0.623s = speed = 321.027 m/s
Which term best matches the description of an earthquake?
Where all mass movements occur
Acts as a lubricant to reduce friction between soil grains
Determines how much material is available for mass movement
Occurs when a sheet of rock moves downhill on a sliding surface
Can trigger a sudden mass movement
A force that works to pull material downslope
can tigger sudden mass movement
At which point is the kinetic energy the greatest?
Answer:
Kinetic energy is energy an object has because of its motion and is equal to one-half multiplied by the mass of an object multiplied by its velocity squared (KE = 1/2 mv2). Kinetic energy is greatest at the lowest point of a roller coaster and least at the highest point.
Explanation:
here is an example Where is the apple when it has the most kinetic energy? When the apple is falling and is right before it hits the ground. As an object falls under the influence of gravity, kinetic energy and potential energy are equal everywhere/at the halfway point only. 10. As an object falls under the influence of gravity, potential energy is greater than kinetic energy after halfway point/ before the halfway point
In the roller coaster, the kinetic energy is greatest at point X.
The kinetic energy of a body is given by 1/2mv^2.
Where;
m = mass of the object
v =velocity of the object.
For a roller coaster, the velocity is maximum at the lowest points of the roller coaster. If the velocity is maximum at the lowest point, it also follows that the kinetic energy is also maximum at the lowest point. Hence point X has the highest kinetic energy.
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A hunter shoots a tranquilizer dart straight at a monkey who is
hanging on a tree branch that is 1.5 meters high. The tree is 16 meters
away from the hunter. The man shot the dart at the monkey with an
initial sideways velocity of 40 m/s, and at the same time, the monkey
let go of the branch. what time will the monkey reach the same height
of the arrow?
t= _S
Answer:
The donkey might be dead
a beach ball is left in the bed of a pickup truck. Describe what happens to the ball when the truck accelerates forward