Answer:
4000J
Explanation:
Given parameters:
Weight of the man = 800N
Height of ladder = 5m
Unknown:
Gravitational potential energy gained = ?
Solution:
The gravitational potential energy is due to the position of a body.
Gravitational potential energy = weight x height
Now insert the parameters;
Gravitational potential energy = 800 x 5 = 4000J
If you ride your bike at an average speed of 5 km/h and need to travel a total distance of 35 km, how long will it take you to reach your destination?
Answer:
7 hours.
Explanation:
This is because you are traveling 5 kilometers per hour so 5 multiplied by 7= 35.
PHYSICS QUESTION BRAINLIEST
Ball A with a mass of 5 kg is moving at 20 m/s, collides with ball B of unknown
mass moving at 10 m/s in the same direction. After the collision, ball
A moves at 10 m/s and ball B at 15 m/s, both still in the same
direction. What is the mass of ball B?
M = mass of the ball A = 5.0 kg
m = mass of the ball B = ?
V = initial velocity of the ball A before collision = 20 m/s
v = initial velocity of the ball B before collision = 10 m/s
V' = final velocity of the ball A after collision = 10 m/s
v' = final velocity of the ball B after collision = 15 m/s
using conservation of momentum
M V + m v = M V' + m v'
(5.0) (20) + m (10) = (5.0) (10) + m (15)
100 + 10 m= 50 + 15 m
5 m = 50
m= 10 m/s
Answer: 5
Explanation:
The unit for mass is grams. The units for molar mass is g/mol? True or false
Answer:
2.False 1.False.
Explanation:
2. I think Because it's formula is kilogram/mole
1 The si unit of mass is KG.
A brass rod and an iron rod differ in length by 28cm at 20°C. What should be the original length of the iron rod for the difference in length to remain the same when both rods are heated to 90°C? Linear expansivity of brass=1.9×10^-5. Linear expansivity of iron=1.2×10^-5.
Pls show workings.
Answer:
The original length of the iron rod is approximately 572.189 meters.
Explanation:
This is a case of linear expansion, which is defined by the following differential equation:
[tex]\alpha = \frac{1}{L}\cdot \frac{dL}{dT}[/tex] (1)
Where:
[tex]\alpha[/tex] - Linear expansion coefficient, measured in [tex]\frac{1}{^{\circ}C}[/tex].
[tex]L[/tex] - Length of the element, measured in centimeters.
[tex]\frac{dL}{dT}[/tex] - First derivative of the length of the element with respect to temperature, measured in centimeters per degree Celsius.
If we assume that thermal deformation are small regarding the length of the element, then we simplify (1) in the following form:
[tex]\alpha \approx \frac{\Delta L_{o}}{L\cdot \Delta T}[/tex]
[tex]L_{f} -L_{o} = \alpha \cdot L_{o} \cdot \Delta T[/tex]
[tex]L_{f} = L_{o}\cdot [1+\alpha\cdot (T_{f}-T_{o})][/tex] (2)
Where:
[tex]L_{o}[/tex], [tex]L_{f}[/tex] - Initial and final lengths of the element, measured in centimeters.
[tex]T_{o}, T_{f}[/tex] - Initial and final temperatures of the element, measured in degrees Celsius.
Given that brass has a higher coefficient of linear expansion, it is suppose that initial length is less than the initial length of the iron element. Then, we have the following system of linear equations:
Brass
[tex]L = L_{o}\cdot [1+\alpha_{B}\cdot (T_{f}-T_{o})][/tex] (3)
Iron
[tex]L = (L_{o}+28)\cdot [1+\alpha_{I}\cdot (T_{f}-T_{o})][/tex] (4)
Where [tex]\alpha_{B}[/tex], [tex]\alpha_{I}[/tex] are coefficients of linear expansion of brass and iron, measured in [tex]\frac{1}{^{\circ}C}[/tex].
By equalizing (3) and (4), we have the following formula:
[tex]L_{o} \cdot [1+\alpha_{B}\cdot (T_{f}-T_{o})] = (L_{o}+28)\cdot [1+\alpha_{I}\cdot (T_{f}-T_{o})][/tex]
[tex]L_{o} \cdot (\alpha_{B}-\alpha_{I})\cdot (T_{f}-T_{o}) = 28\cdot [1+\alpha_{I}\cdot (T_{f}-T_{o})][/tex]
[tex]L_{o} = \frac{28\cdot [1+\alpha_{I}\cdot (T_{f}-T_{o})]}{(\alpha_{B}-\alpha_{I})\cdot (T_{f}-T_{o} )}[/tex]
If we know that [tex]\alpha_{B} = 1.9\times 10^{-5}\,\frac{1}{^{\circ}C}[/tex], [tex]\alpha_{I} = 1.2\times 10^{-5}\,\frac{1}{^{\circ}C}[/tex], [tex]T_{o} = 20\,^{\circ}C[/tex] and [tex]T_{f} = 90\,^{\circ}C[/tex], then the initial length of the iron rod is:
[tex]L_{o} = \frac{28\cdot [1+\left(1.2\times 10^{-5}\,\frac{1}{^{\circ}C} \right)\cdot (90\,^{\circ}C-20\,^{\circ}C)]}{\left(1.9\times 10^{-5}\,\frac{1}{^{\circ}C}-1.2\times 10^{-5}\,\frac{1}{^{\circ}C} \right)\cdot (90\,^{\circ}C-20\,^{\circ}C)}[/tex]
[tex]L_{o} = 57190.857\,cm[/tex]
[tex]L_{o, I} = 57218.857\,cm[/tex]
[tex]L_{o,I} = 572.189\,m[/tex]
The original length of the iron rod is approximately 572.189 meters.
What is polarization. How can we remove it?
What factor from the choices below determines whether
a star will evolve into a white dwarf, a neutron star, or a
black hole?
A) mass
B) percentage of helium
C) percentage of carbon
D) apparent brightness
Answer:
a
Explanation:
Stars with very big masses will become neutron stars or black holes.
Smaller stars will become white dwarfs
A body with a mass of 10.0kg is assumed to be in earth's gravitational field with g=9.80m/s2. What is the acceleration?
Answer:
a = F/m
a = 9.80÷ 10.0
a 0.98 m/s²
A car starts from rest and after 20 seconds it's velocity becomes 108km find the acceleration of the car
Answer:
1.5 km/s²
Explanation:
Given that:
a car starts from rest; it means the initial velocity (u) = 0 km/hr = 0 m/s
after time (t) = 20 seconds
the final velocity = 108 km/hr = 30 m/s
The acceleration (a) of the car can be determined by using the formula:
[tex]a = \dfrac{v-u}{t}[/tex]
[tex]a = \dfrac{30\ m/s -0 \ m/s}{20 \ s}[/tex]
[tex]a = \dfrac{30 \ m/s}{20 \ s}[/tex]
a = 1.5 km/s²
A 1600 kg car is moving north with a speed of 25 m/s. An impulse of 10,000 N-s south is applied to the car. After the impulse is applied, the speed of the car is:
Answer:
The speed of the car is 18.75 m/s
Explanation:
Impulse-Momentum Change Equation
The impulse received by an object is equal to the change in momentum:
Impulse = Change in momentum
The change in momentum is:
[tex]\Delta p=m\Delta v=m(v_2-v_1)[/tex]
Where m is the mass of the object, v2 is the final speed, and v1 is the initial speed. This means the impulse J is:
[tex]J=m(v_2-v_1)[/tex]
It's given an m=1,600 kg car is moving at v1=25 m/s North and then an impulse of J=-10,000 N.s is applied South. The minus sign indicates the impulse is opposite to the original direction of the velocity.
Solving for v2:
[tex]\displaystyle v_2=\frac{J}{m}+v_1[/tex]
Substituting:
[tex]\displaystyle v_2=\frac{-10,000}{1,600}+25[/tex]
[tex]\displaystyle v_2=-6.25+25=18.75[/tex]
[tex]v_2=18.75\ m/s[/tex]
The speed of the car is 18.75 m/s
DO NOT ANSWER IF YOU DONT KNOW What is the medium for a wave:
i. At the beach?
ii. In a football stadium?
iii. Of sound?
1. How do humans get energy from food?
Answer: small intestine
Explanation:
I took a test
9.In a perfectly elastic collision, a 0.400 kg ball moving toward the east at 3.7 m/s suddenly
collides head-on with a 0.200 kg ball sitting at rest.
If the velocity of the first ball just after the collision is 4.24 m/s, what is the velocity of the second ball ?
Answer:
The velocity of the second ball is 15.88 m/s East
Explanation:
The given parameters are;
The type of collision = Perfectly elastic collision
The mass of the ball moving East, m₁ = 0.400 kg
The velocity of the ball moving East, v₁ = 3.7 m/s
The mass of the ball at rest, m₂ = 0.200 kg
The initial velocity of the ball at rest, v₂ = 0 m/s
The velocity of the first ball, 'm₁', after the collision, v₃ = 4.24 m/s (The direction is assumed as being towards the West)
The velocity, 'v₄', of the second ball, 'm₂', after the collision is given by the law of conservation of linear momentum as follows;
m₁·v₁ + m₂·v₂ = m₁·v₃ + m₂·v₄
Substituting the known values gives;
0.400 kg × 3.7 m/s + 0.200 kg × 0 m/s = 0.400 kg × (-4.24 m/s) + 0.200 kg × v₄
∴ 0.200 kg × v₄ = 0.400 kg × 3.7 m/s - 0.400 kg × (-4.24 m/s)
Therefore;
0.200 kg × v₄ = 0.400 kg × 3.7 m/s + 0.400 kg × 4.24 m/s = 3.176 kg·m/s
v₄ = 3.176 kg·m/s/(0.200 kg) = 15.88 m/s
The velocity of the second ball = v₄ = 15.88 m/s East.
How is the brightness of the star related to how quickly it pulses?
Answer:
The greater the luminosity of a star, the longer its period of oscillation.
Can anybody helppp me with this
Which is an example of a physical change?
cake baking
gasoline combusting
salt dissolving
iron rusting
Answer:
Salt dissolving
Explanation:
Dis solving salt in water doesn't change it's chemical composition
Answer:
Iron Rusting
6th grade stuff or 5th i think
Explanation:
A force of 5.0 moves a 6.0 kg object along a rough floor at a constant speed of 2.5 m/s. What power is being used? What force of friction is acting on the object?
Complete Question:
A force of 5.0 N moves a 6.0 kg object along a rough floor at a constant speed of 2.5 m/s.
(a) How much work is done in 25 s?
(b) What power is being used?
(c) What force of friction is acting on the object?
Answer:
a. Workdone = 312.5 Joules.
b. Power = 12.5 Watts.
c. Force of friction = 5 Newton.
Explanation:
Given the following data;
Force = 5N
Speed = 2.5m/s
Mass = 6kg
Time = 25 secs
a. To find the work done.
We would first solve for the distance covered by the object.
Distance = speed * time
Distance = 2.5 * 25
Distance = 62.5 meters.
Workdone = force * distance
Workdone = 5 * 62.5
Workdone = 312.5 Joules.
b. To find the amount of power being used.
Power = workdone/time
Power = 312.5/25
Power = 12.5 Watts.
c. To find the force of friction is acting on the object.
Since we know that there is no net force acting on the body. Therefore, the force of friction is equal to 5 Newton.
a) Power is being used will be 12.5 Watts.
b) Force of friction is acting on the object will be 5 Newton.
What is the friction force?It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
Mathematically it is defined as the product of the coefficient of friction and normal reaction.
The given data in the problem is;
m is the mass of object =6.0 kg
u is the speed =2.5 m/s
a) Power is being used will be 12.5 Watts.
The distance traveled by the object is;
[tex]\rm d = v \times t \\\\ \rm d = 2.5 \times 25 \\\\ \rm d = 62.5 m[/tex]
The work done is the product of force and displacement;
[tex]\rm W = F \times d \\\\ \rm W = 5 \times 62.5 \\\\ \rm W = 312.5 \ J[/tex]
The power is the ratio of work done to the time;
[tex]\rm P= \frac{W}{t} \\\\ \rm P= \frac{312.5 }{25 } \\\\\ \rm P=12.5 \ Watt[/tex]
Hence Power is being used will be 12.5 Watts.
b) Force of friction is acting on the object will be 5 Newton.
If the all unbalanced force is zero the net force is equal to the force of friction;
The force of friction= Net force
Force of friction= 5 Newton.
Hence the force of friction is acting on the object will be 5 Newton.
To learn more about the friction force refer to the link;
https://brainly.com/question/1714663
Mr. Covault gave his students an assignment to design and conduct experiments that would allow them to find the relationship between force, mass, and acceleration. Some students decided to use a setup like the one below:
The applied force in this setup is equal to the weight of the item attached to the end of the pulley. One student tested the acceleration due to this force on a wooden block (as shown above); another student used a wheeled cart instead of a wooden block.
Mr. Covault's other students decided to simply drop objects. The force in that case is the weight of the object.
All the students measured the time it took each object, starting from rest, to move a certain distance. They used the time and distance to calculate the average acceleration.
Four students' results are shown below. A successful experiment should eliminate all forces acting on the object except the force being investigated. The experiment should confirm Newton's second law:
One newton (N) is 1 kg·m/s2.
Which of the following students had a well designed and conducted experiment?
Student Mass
(kg) Force
(N) Measured
Acceleration
(m/s2)
Kira 0.21 0.098 0.0047
Sophie 0.11 0.098 0.88
Jacques 0.050 0.49 4.9
Chase 0.50 4.9 8.9
Answer:
Sophie
Explanation:
Only Sophie's results supported Newton's second law. The other students' measured accelerations were significantly lower than expected. This could indicate that their experimental designs had not sufficiently eliminated drag forces.
F=ma Rearranging when solving for acceleration gives: a=F/m
A student climbs to the top of the gym and drops a baseball to the ground below. A physics student standing nearby has a motion sensor and determines that the baseball hit the ground with a velocity of 17 m/s. if you ignore air resistance, how long was the baseball falling through the air?
Answer:
The baseball was falling during 1.733 seconds.
Explanation:
The baseball experimented a free fall, which is a particular case of uniformly accelerated motion, in which object is accelerated by gravity. In this case, we need to determine the time spent by the ball before hitting the ground ([tex]t[/tex]), measured in seconds, which is determined by the following kinematic formula:
[tex]t = \frac{v-v_{o}}{g}[/tex] (1)
Where:
[tex]v_{o}[/tex], [tex]v[/tex] - Initial and final speeds of the baseball, measured in meters per second.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v = 17\,\frac{m}{s}[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the time spent by the baseball is:
[tex]t = \frac{17\,\frac{m}{s}-0\,\frac{m}{s} }{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]t = 1.733\,s[/tex]
The baseball was falling during 1.733 seconds.
A horizontal force of 200 N is applied to move a 55 kg television set across a 10 m level surface. What is the
work done on the television set?
What does potential energy between two magnets become kinetic energy?
Answer:
HI!
Here is ur answer
100% correct Potential energy between two magnetsMagnetic PE depends on both distance and orientation in the magnetic field. ... The magnetic fields are aligned. This makes an attractive force between the two magnets. The farther apart they are, the higher their potential energy.
PLEASE MARK ME AS BRAINLISTcat walks 100 m  East, then turns around and walks 25 m west. What is the cats displacement?
a) 75 m
b) 100 m
c) 25 m
d) 125 m
Answer:
75mExplanation:
If a cat walk 100m East, this means that it is walking in the positive x direction, the distance will therefore be +100m
If it turns around and walks 25 m west, the direction of movement is in the negative x direction i.e -25m
Taking the sum;
Displacement = +100m - 25m
Displacement of the cat = 75m
hence the cats displacement is 75m
Does which object have the compacity that is the best measured in pints?
option 1: water bottle
option 2: swimming pool
option 3: trash barrel
option 4: soup spoon
:)
Answer:
1 or 2
Explanation:
describe how an electromagnet can be switched on and off.
Answer:The magnetic field around an electromagnet is just the same as the one around a bar magnet. It can, however, be reversed by turning the battery around. Unlike bar magnets, which are permanent magnets, the magnetism of electromagnets can be turned on and off just by closing or opening the switch.
a concave mirror of radius of curvature 20cm produces an inverted image three times the size of an object placed on and perpendicular to the axis. calculate the position of the object and the image
Answer:
Check attachment for your answer
Good luck
A 20 Kg box is sliding across a horizontal table at 10m/s. If the coefficient of Friction is 0.5, what is the acceleration of the box?
By Newton's second law,
• the net vertical force on the box is
∑ F = n - mg = 0
where n = magnitude of the normal force, m = mass of the box, and g = acceleration due to gravity; then
n = mg = (20 kg) (9.8 m/s²) = 196 N
• the net horizontal force on the box is
∑ F = -f = -ma
where f = mag. of kinetic friction and a = acceleration of the box. Since
f = µn
where m is the coefficient of kinetic friction, so that
f = 0.5 (196 N) = 98 N
Then the box's acceleration is
-98 N = - (20 kg) a → a = (98 N)/(20 kg) = 4.9 m/s²
How much work is required to pull a sled 5 meters if you use 60N of force?
Answer:
300Nm
Explanation:
work = force x distance = 60x5=300
The work required to pull a sled by 5 meters using 60 N of force is 300 joules.
What is Work done in physics?Work done in physics is defined as the dot product of force and displacement caused by it. Mathematically -
W = F.x
W = |F| |x| cos Ф
where Ф is the angle between force and displacement
Given is to calculate the work required to pull a sled by 5 meters by using 60 N of force.
We can write the following data -
Applied Force [F] = 60 Newtons
Displacement needed [x] = 5 meters
Angle between force and displacement [Ф] = 0°
Using the formula for work done, we can write -
W = |F| |x| cos Ф
W = 60 x 5 x cos(0)
W = 300 x 1
W = 300 Joules
Therefore, the work required to pull a sled by 5 meters using 60 N of force is 300 joules.
To solve more questions on Work, energy and power, visit the link below-
brainly.com/question/3893991
#SPJ6
A simple pendulum takes 128s to complete 40 oscillations what is the time period of the pendulum
Explanation:
Parameters
Number of oscillation = 40 oscillations
Time of oscillation = 128s
Formular
Period (T) = No of oscillation
Time taken
T = 128
40
T=3.2s
what law of motion does this picture represent?
A machine with a velocity ratio moves a load of 3000N when an effort of 200N is applied. Calculate the efficiency energy
Explanation:
A machine with a velocity ratio of 30 moves a load of 3000N when an effort of 200N is applied.the efficiency of the machine is what?
Scott needs to split a chunk of wood for the fire. There is a crack in the chunk. If he uses an axe it will use 200 Joules of energy. He receives 100 Joules of work out of the machine.
A) What type of simple machine has he used?
B) What is the efficiency of the machine?
Answer:
A) A wedge
B) The efficiency of the machine is 50%
Explanation:
The type of simple machine Scott used is the wedge
A) A wedge is a simple machine that consists of a long thick metal or wood stem that tapers to a thin (Relatively sharp) end. It is used for a splitting, tightening, and lifting.
An example of a wedge is an axe
B) The efficiency of a machine is given approximately as the ratio of the work output to the work input as follows;
The amount of input energy the axe uses = 200 Joules
The amount of output work Scott receives from the axe = 100 Joules
Therefore, the efficiency of the machine, η, is given as follows;
[tex]\eta = \dfrac{The \ out.put \ work \ Scott \ receives \ from \ the \ axe}{The \ amount \ of \ in.put \ energy \ the \ axe \ uses} \times 100[/tex]
[tex]\therefore \eta = = \dfrac{100 \ J}{200 \ J} \times 100 = 50\%[/tex]
The efficiency of the machine, η = 50%
A flashlight is shone on a mirror. As you move the flashlight to increase the angle of incidence of the light on the mirror, the angle of the reflection also changes. How does the angle of reflection change?
Answer:
the angle of reflection also increases
Explanation:
this is because the law of reflection states that the angle of incidence equals to the angle of reflection, thus, if the angle of incidence increases, angle of reflection should also increase, basically , think as if the angle of incidence is the same thing as the angle of reflection
hope this helps, if not please report it someone else can try it
