Answer:
12.3 m/s
Explanation:
To calculate the final velocity of Ben, we can use the equation KE = 1/2mv^2, where KE is the kinetic energy and m is the mass of Ben. Inserting the given values, we get:
7100 = 1/2 (63.2 kg) v^2
Therefore, v^2 = 7100/31.6
v = √(7100/31.6)
v = 12.3 m/s
5. The picture shows a mass that is attached to a spring with constant, K = 500 N/m(20 points)
The mass is of 2 kg and the values of the heights are:h1= 5 meters,h2= 10 meters andh3= 2 meters.•The spring is compressed a length of 80 cm.
What is the speed of the mass when it reaches a height of h2?
•What is the speed of the mass when it reaches a heighth3?
Answer:
[tex]\text{(a)}\\\text{Given that current energy at starting point }h_{1} \text{ is}:\\E_{h1}=mgh+\frac{1}{2}kx^2\\\text{Plug in height and mass: }E_{h1}=10g+250(0.8)^2=10g+160=258J\\\text{Using Conservation of Energy: }E_{h1}=E_{h2} \\\therefore 258J=mg(h_{2}-h_{1})+\frac{1}{2}mv^2\\=258J=10g+v^2\\=160=v^2\\\therefore v=40m/s \text{ at }h_{2}\\[/tex][tex]\text{(b)}\\\text{Again, apply conservation of energy: } 258J=mg(h_{3}-h_{2})+\frac{1}{2}mv^2\\\text{Plugging in: }258J=16g+v^2 \\\text{ Simplify: }101.2=v^2\\\therefore v=10.05m/s \text{ at } h_{3}[/tex]
A 1.0-kg cart and a 0.50-kg cart sit at different positions on a low-friction track. You push on the 1.0-kg cart with a constant 4.0-N force for 0.20 m. You then remove your hand, and the cart slides 0.35 m and strikes the 0.50-kg cart. What is the work done by you on the two-cart system? How far does the system's center of mass move while you are pushing the 1.0-kg cart? By what amount does your force change the kinetic energy of the system's center of mass?
Answer:
a)[tex]W=0.8J[/tex]
b) [tex]d_t=0.20m[/tex]
c) [tex]\triangle K.E=0.267J[/tex]
Explanation:
From the question we are told that:
Mass of cart 1 [tex]M_1=1.0kg[/tex]
Mass of cart 1 [tex]M_2=0.05kg[/tex]
Force on cart 1 [tex]F_1=4.0N[/tex]
Push Distance of cart 1 [tex]d_1=0.20m[/tex]
Slide Distance of cart 1 [tex]d_1'=0.35m[/tex]
a)
Generally the equation for work-done is mathematically given by
[tex]W=f*d\\W=4*0.20\\W=0.8J \\[/tex]
b)
The systems center of mass moved a net totally of (while being pushed)
Mass 1 =0.20m
Mass 2=0
Therefore
[tex]d_t=d_1+d_2[/tex]
[tex]d_t=0.20+0[/tex]
[tex]d_t=0.20m[/tex]
c)
Since work-done is equal to K.E energy of cart 1
Therefore
[tex]W=1/2mv^2[/tex]
[tex]V_1=\sqrt{\frac{W}{1/2m}}[/tex]
[tex]V_1=\sqrt{\frac{0.8}{1/2(1)}}[/tex]
[tex]V_1=1.264[/tex]
Therefore Kinetic energy before collision is
[tex]K.E_1=1/2mv^2[/tex]
[tex]K.E_1=1/2*1*1.264^2[/tex]
[tex]K.E_1=0.768[/tex]
Generally from the equation for conservation of momentum the Velocity of cart 2 is mathematically given by
[tex]v_2=\frac{m_1V_1}{m_1+m_2}[/tex]
[tex]v_2=\frac{1*1.264}{1+0.5}[/tex]
[tex]V_2=0.842m/s[/tex]
Therefore the final K.E is mathematically given by
[tex]K.E_2=(1/2)(m_1+m_2)V_2^2[/tex]
[tex]K.E_2=1/2*(1.5)(0.842)^2[/tex]
[tex]K.E_2=0.531J[/tex]
Generally the Change in K.E is mathematically given by
[tex]\triangle K.E=K.E_1-K.E_2[/tex]
[tex]\triangle K.E=0.798-0.531[/tex]
[tex]\triangle K.E=0.267J[/tex]
Therefore the will force change the kinetic energy of the system's center of mass by
[tex]\triangle K.E=0.267J[/tex]
(a) The work done by you when you push the cart at a constant force is 0.8 J.
(b) The distance moved by the center mass of the two cart system is 0.23 m.
(c) The change in kinetic energy of the system center of mass is 0.271 J.
Work done by you
The work done by you when you push the cart at a constant force is calculated as follows;
W = Fd
W = 4 x 0.2
W = 0.8 J
Distance moved by the center mass of the two cart systemlet the 0.5 kg mass be the reference mass at zero (0).Xcm = (m₁x₁ + m₂x₂)/(m₁ + m₂)
Xcm = (0.5(0) + 1(0.35)) / (1 + 0.5)
Xcm = (0.35)/(1.5)
Xcm = 0.23 m
Initial velocity of the 1.0 kg massF = ma
a = F/m
a = (4)/1 = 4 m/s²
v² = u² + 2as
v² = 0 + 2(4)(0.2)
v² = 1.6
v = √1.6
v = 1.265 m/s
Final velocity of the massesm₁u₁ + m₂u₂ = v(m₁ + m₂)
1(1.265) + 0 = v(1 + 0.5)
1.265 = 1.5v
v = 0.84 m/s
Change in kinetic energyK.E(initial) = ¹/₂m₁u₁² + ¹/₂m₂u₂²
K.E(initial) = ¹/₂(1)(1.265)² + ¹/₂(0.5)(0) = 0.8 J
K.E(final) = ¹/₂(m₁ + m₂)v²
K.E(final) = ¹/₂(1 + 0.5)(0.84)² = 0.529 J
Δ K.E = 0.8 J - 0.529 J = 0.271 J
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Directions: Unscramble each word. Then write each term on the line before its definition.
1. antstonc
2. desep
3. ticlovey
4. onlacietrace
5. emincedsplat
6. tomoni
7. cenferree nitop
8. spooniti
9. measure of change in velocity
10. distance an object moves in a unit of time
11. change of position
12. distance and direction from a reference point
13. not changing
14. speed of an object in a given direction
15. distance between the initial and final position
16. starting point used to describe the location of an object
Answer:
constantspeedvelocityaccelerationdisplacementmotionreference pointpositionExplanation:
Above is the unscrambled words. Hope this helps!
What is the atomic composition of methane
Answer:
As in chemical formula? It's CH4.
d. The mass of the boy is the same in all places because mass does not depend
on gravity.
True or False
(Science)
Explanation:
Remember, even if you weigh less because of a change in gravity's force on your body, your body's mass is still the same. As your body grows, you will have more mass, which also means you will weigh more. That's because when you're on the earth, the amount of gravity that pulls on you stays the same.Thank you for marking me Brainlist ☺️Answer:
true . it depends tho weight depends on gravity .
please help AFAP........................
Answer:
afap
Explanation:
HELP When the forces are applied in the same direction, how do you determine net force?
Answer:
Explanation:
If two forces act on an object in the same direction, the net force is equal to the sum of the two forces.
The first scientist misidentified the fossil as that of a big cat. What body structure did he use to make his observations?
Two students are balancing on a 10m seesaw. The seesaw is designed so that each side of the seesaw is 5m long. The student on the left weighs 60kg and is sitting three meters away from the fulcrum at the center. The student on the right weighs 45kg. The seesaw is parallel to the ground. The mass of the board is evenly distributed so that its center of mass is over the fulcrum. What distance from the center should the student on the right be if they want the seesaw to stay parallel to the ground?
a. 4m
b. 5m
c. 2m
d. 3m
NO LINKS.
Answer:
B.5m
Explanation:
not sure if i'm right;)
True or False question! ⬇⬇ Will give BRAINLIEST!
No matter the medium, light will always bend at the same angle when it passes from one medium to a different medium.
-True
-False
Answer:
Explanation:true
What does newton first law states
Answer:
Newton's first law, also known as the law of inertia, states that an object will remain at rest or in motion at a constant velocity unless acted upon by an external force. In other words, an object will not change its velocity on its own, it will only do so if something else pushes or pulls on it. This law is a fundamental principle of classical mechanics, and it helps to explain the behavior of objects under various conditions.
Explanation:
Answer: Newton first law states that an object at rest tends to stay at rest. an object in motion tends to stay in motion in a straight line at a constant speed until another force acts on the object. It is also called the law of inertia
Explanation: Newton’s first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.
Hope this was helful
A smoke detector is an electronic fire-protection device that automatically senses the presence of smoke as a key indicator of fire and creates a loud sound as a warning. In 3-5 sentences, describe what is happening inside a photoelectric smoke detector as soon as smoke enters it
Answer:
Photoelectric-type alarms aim a light source into a sensing chamber at an angle away from the sensor. Smoke enters the chamber, reflecting light onto the light sensor; triggering the alarm.
Explanation:
nfpa.org is the website with theanswer
During a hard drive crash the read/write head scrapes against the disk with a coefficient of kinetic friction of µk and normal force of N. Assume that just before the head crash the disk is rotating at ω0 rad/s, and the distance of the head from the disk axis is r. You can ignore any friction at the bearing (rotational axis of the disk). Assume that the disk is uniform, and has radius R and mass M. What is the angular acceleration associated with the torque from the crashed disk head?
Answer:
α = [tex]\frac{2 \mu \ N}{m \ r}[/tex]
Explanation:
For this exercise we use Newton's equation for rotational motion
∑ τ = I α
the troque is
α = Fr .r
the moment of inertia of a cylinder is
I = ½ m r²
we substitute
fr r = (½ m r²) α
the expression friction is
fr = μ N
we substitute
μ N r = ½ m r² α
α = [tex]\frac{2 \mu \ N}{m \ r}[/tex]
The system in the figure below, If atmospheric pressure is 101.3 KPa and the absolute pressure at the bottom of the tank is 231.3 KPa, what is the specific gravity of the olive oil?
If atmospheric pressure is 101.3 KPa and the absolute pressure at the bottom of the tank is 231.3 KPa, the specific gravity of the olive oil is 1.30
How do you calculate the gravity?To calculate the specific gravity, deduct the absolute pressure at the bottom of the tank (231.3 KPa) from the atmospheric pressure (101.3 KPa). This will give you a ratio of 130KPa, then divide it by the atmospheric pressure which is the specific gravity of olive oil.
Use the formula below:
specific gravity = absolute pressure - atmospheric pressure / atmospheric pressure
Specific gravity = (231.3 KPa - 101.3 KPa) / (101.3 KPa) = 1.30
Therefore, the specific gravity of the olive oil is 1.30.
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Which of the following describes a negative ion?
Which of the following best explains the cause and effect relationship between convection currents in the mantle and the movement of tectonic plates?
A. The tectonic plates will heat up the mantle due to being closer to the surface of the earth and sun. The less dense particles in the mantle fall and cool toward the core, driving the convection cycle.
B. The tectonic plates drive the convection cycles in the mantle. As the plates move apart or push together, particles in the mantle will be thrust toward the core.
C. The heat from the core will make the particles in the mantle more dense and rise, causing plates in the
lithosphere to move in the opposite direction of the convection current.
D. The heat from the core will make particles in the mantle less dense and rise, causing plates in the
lithosphere to move in the direction of the convection current.
C) The mantle's particles will become more dense and rise as a result of the core heat, causing the lithosphere's plates to move in the opposite direction of the convection current.
How does the mantle's density change as it heats up?Convection Currents The lithosphere becomes denser than the surrounding magma and sinks back toward the core as the heated mantle transfers heat energy to it.
How does heat travel from the core to the crust in the mantle?Conduction from the core directly heats the lower mantle. As atoms collide, heat is transferred in conduction. Convection is the movement of warm and cool materials upward and downward. Material that is hot rises to the surface through conduction.
When the mantle rises, what takes place?“hot spots," which are non-plate tectonic volcanic regions, are probably caused by mantle plumes. A diapir is formed when a mantle plume reaches the upper mantle. Volcanic eruptions are sparked by the heat generated by this molten material in the asthenosphere and lithosphere.
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Both formal and informal
Explanation:
is this a question????...
The pressure of a sample of gas is measured as 49 torrent. Convert this to atmosphere
Answer:
P = 0.0644 atm
Explanation:
Given that,
The pressure of a sample of gas is measured as 49 torr.
We need to convert this temperature to atmosphere.
The relation between torr and atmosphere is as follow :
1 atm = 760 torr
1 torr = (1/760) atm
49 torr = (49/760) atm
= 0.0644 atm
Hence, the presssure of the sample of gas is equal to 0.0644 atm.
a horizontal of 5 n is required to maintain a velocity of 2m/s for a block of 10kg
Complete Question:
A horizontal force of 5N is required to maintain a velocity of 2m/s for a block of 10kg mass sliding over a rough horizontal surface. The work done by this force in one minute is:
Answer:
Workdone = 600 Joules.
Explanation:
Given the following data;
Force = 5N
Velocity = 2 m/s
Time = 1 minute to seconds = 60 seconds.
Mass = 10 kg
To find the work done;
First of all, we would determine the distance covered by the block.
Distance = speed * time
Distance = 2 * 60
Distance = 120 seconds
Mathematically, workdone is given by the formula;
[tex] Workdone = force * distance[/tex]
Substituting into the formula, we have;
[tex] Workdone = 5 * 120 [/tex]
Workdone = 600 Joules.
Therefore, the work done by this force in one minute is 600 Joules.
The sum of the x components of vectors A and B in (Figure 1 ) is
-5
4
14
6
Vectors A and B are shown in a tip to tail arrangement. Vector A begins at the origin (0,0) and ends at the x y coordinate (9,3). Vector B begins at the tip of vector A and ends at the x y coordinate (4,6).
The resultant (x,y) coordinate of vector B is (- 5, 3) and that of A is (9,3). Then, the sum of x components of vector A and B is 4.
What are vectors ?Vectors are physical quantities having both magnitude and direction. For example force, velocity, displacement etc .are vector quantities. Coordinates are used to describe the position of these vectors.
The coordinates of tail of the vector A is (0, 0) and the head is (9,3)
Then the coordinate of A is (9 -0, 3-0) = (9,3)
Similarly, the tail coordinate of B is (9,3) and head is (4,6). The coordinates of B vector is (4-9, 6-3) = (-5, 3)
Now the sum of x components of A and B is = - 5 + 9 = 4.
Therefore, option b is correct.
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A rocket is launched from the surface of Earth with a speed v0 that will allow the rocket to escape the gravitational field of Earth. The same rocket is now launched from a different planet with the same mass as Earth and four times the radius of Earth. Which of the following is a correct expression for the escape speed from the surface of the new planet?
a. 0.5V.
b. 2V0
c. 0.25v.
d. 0.71V.
e. Vo
Answer:
option ( a ) is correct .
Explanation:
Escape velocity on the earth = √ ( 2 GM / R )
where G is universal gravitational constant , M is mass of the earth and R is radius .
V₀ = √ ( 2 GM / R )
escape velocity on the planet where mass is equal is earth's mass and radius is 4 times that of the earth
Radius of the planet = 4 R
escape velocity of planet = √ ( 2 GM / 4R )
= .5 x √ ( 2 GM / R )
= .5 V₀
option ( a ) is correct .
Find the Input force, if the Mechanical Advantage of the simple machine used is 5 and Output force is 50 N.
P.S PLEASE DO NOT POST A LINK THAT I HAVE TO DOWNLOAD EVERY TIME I ASK A QUESTION PEOPLE POST THAT JUST POST A WRITEN ANSWER WITH AN EXPLINATION THAT IS ALL
Answer:
6N
Explanation:
You have a Coke bottle with an air cavity
that is 0.220 m deep. You add water to
the bottle, shortening the column, so
that when you blow across the top, it
plays a 528 Hz fundamental. How tall is
the water column in the bottle?
Answer: 0.0576
Explanation:
The equation for fundamental frequency for closed pipes is as follows:
f(1) = v/4L
In which v is the speed of sound (343 m/s) and L is the length of the pipe. Therefore:
528 = 343/4L
x = 0.1624
This is not the answer as you still need to subtract it from the original length.
0.220 - 0.1624 = 0.0576
"Rubble pile" asteroids have low average densities due to being up to 50% empty space inside.
True
False
An object with a weigh of 170 N on Earth is brought to planet Oobla. Where it has a weight of 360 N. What is the gravitational field strength on Oobla
The gravitational field strength on Oobla is 20.75 m/s².
What is gravitational field strength?The gravitational field is the gravitational force exerted per unit mass on a small mass at a point in the field.
To calculate the gravitational field strength on Ooobla, we use the formula below
Formula:
W/g = W'/g'................. Equation 1Where:
W = Weight of the object on Earthg = Gravitational field strength on the Earthg' = Gravitational field strength on the OoblaW' = Weight of the object on OoblaFrom the question,
Given;
W = 170 Ng = 9.8 m/s²W' = 360 NSubstitute these values into equation 1 and solve for g'
170/9.8 = 360/g'g' = 360×9.8/170g' = 20.75 m/s²Hence, the gravitational field strength is 20.75 m/s².
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As a warehouse worker pushes a crate across a concrete floor, the force he
applies is not perfectly horizontal, as shown in the image below. If the
coefficient of kinetic friction between the crate and concrete floor is 0.5, what
is the net force on the crate?
Pushing Force
350 N
Weight
460 N
OA. 95 N
OB. 115 N
O C. 130 N
OD. 145 N
10⁰
-Friction Force
2
The crate will be under a net force of 95 N. Resolving the force in the x-direction yields the net force on the crate: F = 300 cos 10° - N F= 295.44-196.45, F= 95.
How do you calculate the frictional force?The resistive force of friction (Fr) divided by the normal or perpendicular force (N) pushing the objects together yields the coefficient of friction (fr), which is a numerical value. The formula fr = Fr/N serves as a representation of it.
What are the friction laws?The friction of a moving item is inversely proportional to the normal force and runs perpendicular to it. The type of surface the thing comes into touch with determines the amount of friction it experiences. As long as there is a point of contact, friction does not depend on the area of contact.
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what electrical devices that belong to physiological effect?
Answer:
Most of us have experienced some form of electric “shock,” where electricity causes our body to experience pain or trauma. If we are fortunate, the extent of that experience is limited to tingles or jolts of pain from static electricity buildup discharging through our bodies.
When we are working around electric circuits capable of delivering high power to loads, electric shock becomes a much more serious issue, and pain is the least significant result of shock.
As electric current is conducted through a material, any opposition to the current (resistance) results in a dissipation of energy, usually in the form of heat. This is the most basic and easy-to-understand effect of electricity on living tissue: current makes it heat up. If the amount of heat generated is sufficient, the tissue may be burnt.
The effect is physiologically the same as damage caused by an open flame or other high-temperature source of heat, except that electricity has the ability to burn tissue well beneath the skin of a victim, even burning internal organs.
A woman of mass 55 kg stands on the rim of a frictionless merry-go-round of radius 2.0m and rotational inertia 1250 kg-m2 that is not moving. She throws a rock of mass 350g horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is 2.0m/s. Afterward, what are (a) the angular speed of the merry-go-round and (b) the linear speed of the woman
Answer:
a) [tex]\omega=9.10*10^{-4}rad/sec[/tex]
b) [tex]V=1.8217*10^{-3}[/tex]
Explanation:
From the question we are told that:
Mass of woman [tex]M_w=55kg[/tex]
Radius of merry go round [tex]r=2.0m[/tex]
Rotational inertia [tex]i= 1250 kg-m2[/tex]
Mass of rock [tex]M_r=350g \approx 0.350kg[/tex]
Speed of rock [tex]V_r=2.0m/s[/tex]
Tangent angle to the outer edge [tex]\theta=1[/tex]
a)
Generally the equation for conservation of momentum is mathematically given by
[tex]M_r(ucos\theta)r=(I+M_wr^2)\omega[/tex]
[tex]0.350(2.0cos1)(2.0)=(1250+(55)(2.0)^2)\omega[/tex]
[tex]1.3998=1470\omega\\\omega=\frac{1.339}{1470}[/tex]
[tex]\omega=9.10*10^{-4}rad/sec[/tex]
b)
Generally the equation for linear speed V is mathematically given by
[tex]V=r\omega\\V=2.0*9.10*10^{-4}[/tex]
[tex]V=1.8217*10^{-3}[/tex]
A 30-N iPad is dropped from a height of 10 m and strikes the ground with a speed of 13 m/s.
What average force of air friction acted on the iPad as it fell?
Answer:
Initially the PE of the object was W * h = 30 * 10 = 300 Joules
The KE of the object when it struck the ground was 1/2 M v^2
KE = 1/2 * 30/9.8 * 13^2 = 259 J
So the object lost 41 J to friction during the fall
Since Work = Force * distance
Force = 41 J / 10 m = 4.1 N (the average force of friction)
Q8. An alluminum object has
a mass of 27.0 Kg and a dan
[sity of 2.70x10² Kgm–³ The object
is attached to a string and
immersed in a tank of water . determine the volume of the object and the tension force
Answer:
volume=0.1m^3 and force=270N
Explanation:
Solution
Given,
mass=27kg
density=2.70×10^2=270kg/m^3
volume =?
force=?
We have,
force=m×a=27×10=270N
density =m÷v
v×d=m
v=m÷d
=27÷270
=0.1m^3