Answer:
0.7432m²
1152in²
Explanation:
a) Usng the conversion
1ft² = 0.0929m²
8ft² = y
y = 8 × 0.0929
y = 0.7432m²
Hence 8ft² in m² is 0.7432m²
For ft² to in²
1ft² = 144in²
8ft² = x
X =8×144
x = 1152in²
Hence 8ft² expressed in in² is 1152in²
A small car of mass 833 kg is parked behind a small truck of mass 1767 kg on a level road. The brakes of both the car and the truck are off so that they are free to roll with negligible friction. A 41 kg woman sitting on the tailgate of the truck shoves the car away by exerting a constant force on the car with her feet. The car accelerates at 1.5 m/s 2 . What is the acceleration of the truck
Answer: the acceleration of the truck a_t is 0.6911 m/s²
Explanation:
Given the data in the question;
There is no external force on the system; net force on the system is 0
Mass of the truck with the woman M = 1767 kg + 41 kg = 1808 kg
mass of the car m = 833 kg
car acceleration a_c = 1.5 m/s²
now let a_t be the acceleration of the truck in opposite direction
Action force on the car = Reaction force on the car
ma_c = Ma_t
a_t = ma_c / M
we substitute
a_t = (833 × 1.5) / 1808
a_t = 1249.5 / 1808
a_t = 0.6911 m/s²
Therefore, the acceleration of the truck a_t is 0.6911 m/s²
a proton travelling along is x-axis is slowed by a uniform electric field E. at x = 20.0 cm, the proton has a speed of 3.5x10^6 m/s and at 80.0 cm the speed is zero. Determine the magnitude and direction of e,
Answer:
[tex]E=101955.8volt/m[/tex]
The electric field direction is toward the right
Explanation:
From the question we are told that
Initial X-co-ordinate of proton [tex]X_1=20.0cm => \frac{20}{100}m[/tex]
Initial speed of Proton [tex]V_1= 3.5*10^6 m/s[/tex]
Final X-co-ordinate of proton [tex]X_2=80.0cm => \frac{80.0}{100}m[/tex]
Final speed of Proton [tex]V_2=0[/tex]
Generally the mass of Proton is given by
[tex]m_P=1.67*10^-27[/tex]
Generally the kinetic energy of the proton is mathematically given by
[tex]K.E_p=1/2mv^2[/tex]
[tex]K.E_p=1/2*1.6*10^-^2^7*(3.5*10^6)^2[/tex]
[tex]K.E_p=9.8*10^-^1^5[/tex]
Generally the change in electric potential [tex]\triangle V[/tex] is mathematically given by
[tex]\triangle V =\frac{K.E_p}{q}[/tex]
Charge on a proton [tex]q=1.602*10^-^1^9[/tex]
[tex]\triangle V =\frac{9.8*10^-^1^5}{1.602*10^-^1^9}[/tex]
[tex]\triangle V =61173.5volts[/tex]
Generally the equation for magnitude of an electric field is mathematically given by
[tex]E=\frac{\triangle V}{\triangle d}[/tex]
Where
[tex]d=0.8m-0.2m\\d=0.6m[/tex]
Therefore
[tex]E=\frac{61173.5}{0.8-0.2}[/tex]
[tex]E=\frac{61173.5}{0.6}[/tex]
[tex]E=101955.8volt/m[/tex]
The direction of the charge is towards the right
Water runs out of a horizontal drainpipe at the rate of 135 kg/min. It falls 3.1 m to the ground. Assuming the water doesn't splash up, what average force does the water exert on the ground
Answer:
The average force exerted by the water on the ground is 17.53 N.
Explanation:
Given;
mass flow rate of the water, m' = 135 kg/min
height of fall of the water, h = 3.1 m
the time taken for the water to fall to the ground;
[tex]h = ut + \frac{1}{2}gt^2\\\\h = 0 + \frac{1}{2}gt^2\\\\t = \sqrt{\frac{2\times 3.1}{9.8} } \\\\t = 0.795 \ s[/tex]
mass of the water;
[tex]m = m't\\\\m = 135 \ \frac{kg}{min} \ \times \ 0.795 \ s \ \times \ \frac{1 \ \min}{60 \ s} \ = 1.789 \ kg[/tex]
the average force exerted by the water on the ground;
F = mg
F = 1.789 x 9.8
F = 17.53 N
Therefore, the average force exerted by the water on the ground is 17.53 N.
A small mass is released from rest at a very great distance from a larger stationary mass. Draw a graphs best represents the gravitational potential energy U of the system of the two masses as a function of time T.
Answer:
attached below
Explanation:
Gravitational potential
energy = [tex]- \frac{GmM}{r}[/tex] ,
V ∝ [tex]- \frac{1}{r}[/tex]
attached below is a graph that represents the gravitational potential energy U
Gravitational potential energy is the energy held in an entity as a result of its vertical position or length. This gravitational pull of a planet on an object causes the information to be stored.
This is connected with gravitational pull since it takes work to lift anything against the gravity of the Earth.
Freshwater in a raised lake or kept behind a dam demonstrates gravitational force.
According to the formula:
Gravitational potential energy [tex]\bold{=-\frac{GMm}{r^2}}[/tex]
Therefore
[tex]\to \bold{V \propto -\frac{1}{r}}[/tex]
Therefore, the answer is "Option d"
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Suppose the maximum safe intensity of microwaves for human exposure is taken to be 1.00 W/m2 . (a) If a radar unit leaks 10.0 W of microwaves (other than those sent by its antenna) uniformly in all directions, how far away must you be to be exposed to an intensity considered to be safe
Answer:
We must be approximately at least 1.337 meters away to be exposed to an intensity considered to be safe.
Explanation:
Let suppose that intensity is distributed uniformly in a spherical configuration. By dimensional analysis, we get that intensity is defined by:
[tex]I = \frac{\dot W}{\frac{4\pi}{3}\cdot r^{3}}[/tex] (1)
Where:
[tex]I[/tex] - Intensity, measured in watts per square meter.
[tex]r[/tex] - Radius, measured in meters.
If we know that [tex]\dot W = 10\,W[/tex] and [tex]I = 1\,\frac{W}{m^{2}}[/tex], then the radius is:
[tex]r^{3} = \frac{\dot W}{\frac{4\pi}{3}\cdot I }[/tex]
[tex]r = \sqrt[3]{\frac{3\cdot \dot W}{4\pi\cdot I} }[/tex]
[tex]r = \sqrt[3]{\frac{3\cdot (10\,W)}{4\pi\cdot \left(1\,\frac{W}{m^{2}} \right)} }[/tex]
[tex]r \approx 1.337\,m[/tex]
We must be approximately at least 1.337 meters away to be exposed to an intensity considered to be safe.
Why could it be argued that the respiratory system is most critical to sustaining life?
Explanation:
Energy is the most important ingredient for life. Organisms use energy in diverse ways. Scientifically, energy is defined as the ability to do work. Without this ability, organisms would not exist.
So, the most important process is one that can furnish the body with energy.
The respiratory system happens to be the one that furnishes the body with energy. During respiration, the energy needs of the body is met by series of processes. Oxygen is taken in and use to liberate calories from chemical substances packed with energy. So, without respiration, the bodily energy demands will not be met.A 9.0 kg test rocket is fired vertically from Cape Canaveral. Its fuel gives it a kinetic energy of 1905 J by the time the rocket engine burns all of the fuel. What additional height will the rocket rise
Answer:
21.6m
Explanation:
Since the rocket engine burns all the fuel hence the kinetic energy will be converted to potential energy
Potential Energy = mass × acceleration due to gravity × height
Given
PE = 1905J
Mass = 9.0kg
Acceleration due to gravity =9.8m/s²
Required
Height h
Substitute into the formula
1905 = 9(9.8)h
1905 = 88.2h
h =1905/88.3
h = 21.6m
Hence the required height is 21.6m
An object with velocity 141 ft/s has a kinetic energy of 1558.71 ft∙lbf, on a planet whose gravity is 31.5 ft/s2. What is its mass in pounds
Answer:
The mass of the object is 5.045 lbm.
Explanation:
Given;
kinetic energy of the object, K.E = 1558.71 ft.lbf
velocity of the object, V = 141 ft/s
The kinetic energy of the object is calculated as;
[tex]K.E = \frac{1}{2} mV^2\\\\mV^2 = 2K.E\\\\m = \frac{2K.E}{V^2} \\\\1 \ lbf = 32.174 \ lbm.ft/s^2\\\\m = \frac{2 \ \times \ 1558.71 \ ft.lbf \ \times \ 32.174 \ lbm.ft/s^2 }{(141 \ ft/s)2 \ \ \times \ \ \ \ 1 \ lbf\ }[/tex]
[tex]m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm.ft^2/s^2 }{(141 )^2\ ft^2/s^2 }\\\\m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm }{(141 )^2 }\\\\m = 5.045 \ lbm[/tex]
Therefore, the mass of the object is 5.045 lbm.
A block weighing 9.3 N requires a force of 3.7 N to push it along at constant velocity. What is the coefficient of friction for the surface
Answer:
0.398
Explanation:
According to friction, the frictional force is directly proportional to the normal reaction
Ff = nR
Ff is the frictional force
n is the coefficient of friction
R is the reaction
Reaction is equal to the weight
R= W = 9.3N
Fm = Ff = 3.7N
Fm is the moving force
Get the coefficient of friction
n = Ff/R
n = 3.7/9.3
n = 0.398
Hence the coefficient of friction for the surface is 0.398
is college football playoffs the best way to determine the national champion?
Answer:
No it is Not because win ratio is differs say Ohio only won 7 out of 7 games and e
was in the playoffs while Alabama won 11 out of 11 games so it is harder to win 11 out of 1a games and it is easier to win 7 out of 7
Explanation: would like brainliest but u don’t have to
Which items in this image are electrically conductive?
Check all that apply
the power lines themselves
the wooden pole that supports the lines
the rubber soles on the worker's boots
the metal tools the worker uses
the wooden ladder leaning against the lines
all except the rubber boots.
The answers should be The power lines themselves and The metal tools the worker uses (the 1st and 4th choices).
(For anyone curious, the image I attached to this answer is the image given for this problem.)
A 146N force is needed to pull a 350 N block across a horizontal surface at a constant speed by a rope making an angle of 50 degrees with the floor. Find the coefficient of friction.
Answer:
F = force
f = friction
u = coefficient of friction
R = normal reaction force
a = Acceleration
m = mass of block
g = gravity
f = uR
F = Ma
Say the block is moving to the right.
The 146N force thus acts to the right, and the friction force to the left, since it resists movement.
The 146N force acts to the right, but the horizontal component of it is 146 cos 50 = 93.84: So this is the force to the right.
Since F = uR and we're trying to find u, we need both F and R. R is easy to get since it is just m x g. This is in fact already given as the weight 350N. So R = 350.
The block is moving at a constant speed, so the force to the right must = the force to the left.
F = ma, so 93.84 - f = (350/g) x 0
This means f must be 93.84 also.
so we have f = uR,
93.84 = u x 350
so u = 0.268 or
0.27 to 2dp.
Hope you understand this.
Explanation:
The coefficient of friction is 0.26 if the 46N force is needed to pull a 350 N block across a horizontal surface at a constant speed by a rope making an angle of 50 degrees with the floor.
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.
We know:
f = uR
Where f is the friction force and u is the coefficient of friction, R is the normal reaction force.
The horizontal component of 146N is:
F' = 146cos50°
F' = 93.84 N
Since F = mass×acceleration
Because the block is traveling at a constant speed, the right-hand force must equal the left-hand force.
F' = f = 93.84 N
93.84 = u x 350 (R = 350N)
u = 0.26
Thus, the coefficient of friction is 0.26 if the 46N force is needed to pull a 350 N block across a horizontal surface at a constant speed by a rope making an angle of 50 degrees with the floor.
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The resistance of a wire depends on its length i and on its cross sectional area A the resistance is
Answer:
The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area
Explanation:
What are the three longest wavelengths for standing waves on a270-cm-long string that is fixed at both ends
Answer:
The answer is below
Explanation:
a) What are the three longest wavelengths for standing waves on a 270-cm-long string that is fixed at both ends? b. If the frequency of the second-largest wavelength is 50.0 Hz, what is the frequency of the third-longest wave length?
Solution:
a) The wavelengths (λ) for standing waves is given by the formula:
[tex]\lambda_m=\frac{2*length\ of\ string}{m}\\\\Where\ m=1,2,3,.\ .\ .\\\\Given\ that\ length\ of\ string = 270\ cm=2.7\ m,\ m=1,2,3(three\ longest\ wavelengths)\\\\Hence:\\\\\lambda_1=\frac{2(2.7)}{1}=5.4\ m\\\\\lambda_2=\frac{2(2.7)}{2}=2.7\ m \\\\\lambda_3=\frac{2(2.7)}{3}=1.8\ m[/tex]
b) The frequency (f) and wavelength (λ) is given by:
fλ = constant
Hence:
[tex]f_2\lambda_2=f_3\lambda_3\\\\f_2=50\ Hz\\\\2.7*50=f_3(1.8)\\\\f_3=\frac{2.7*50}{1.8} \\\\f_3=75\ Hz[/tex]
The three longest wavelengths for the standing waves on a 270-cm long string that is fixed at both ends are:
1. 5.4 meters.
2. 2.7 meters.
3. 1.8 meters.
Given the following data:
Length of string = 270 cm to m = [tex]\frac{270}{100} =2.7\;m[/tex]To determine the three (3) longest wavelengths for these standing waves:
Mathematically, the wavelength for standing waves is given by the formula:
[tex]\lambda_n = \frac{2L}{n}[/tex]
Where:
[tex]\lambda_n[/tex] is the wavelength for standing waves.L is the length of string.Note: n = 1, 2, and 3.
When n = 1:
[tex]\lambda_1 = \frac{2\times 2.7}{1} \\\\\lambda_1 = 5.4 \;meters[/tex]
When n = 2:
[tex]\lambda_2 = \frac{2\times 2.7}{2} \\\\\lambda_2 = 2.7 \;meters[/tex]
When n = 3:
[tex]\lambda_3 = \frac{2\times 2.7}{3} \\\\\lambda_3 =\frac{5.4}{3} \\\\\lambda_3 = 1.8 \;meters[/tex]
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All of the following are ways in which sports psychologists help athletes except __________.
A.
staying motivated
B.
managing fear of failure
C.
improving performance
D.
enhancing memory
Please select the best answer from the choices provided
A
B
C
D
Answer:
D-Enhancing memory
Explanation:
if an object is projected upward with an initial velocity of 80 ft per second, what is t at maximum height
Answer:
The time at maximum height is 2.49 s.
Explanation:
The time (t) at the maximum height can be found using the following equation:
[tex] v_{f} = v_{0} - gt [/tex]
Where:
[tex]v_{f}[/tex]: is the final velocity = 0 (at the maximum height)
[tex]v_{0}[/tex]: is the initial velocity = 80 ft/s
g: is the gravity = 9.81 m/s²
Hence, the time is:
[tex]t = \frac{v_{0}}{g} = \frac{80 \frac{ft}{s}*\frac{1 m}{3.281 ft}}{9.81 m/s^{2}} = 2.49 s[/tex]
Therefore, the time at maximum height is 2.49 s.
I hope it helps you!
PLEASE PLEASE HELP!!!!
In a particular crash test, an automobile of mass 1577 kg collides with a wall and bounces back off the wall. The x components of the initial and final speeds of the automobile are 17 m/s and 1.5 m/s, respectively. If the collision lasts for 0.18 s, find the magnitude of the impulse due to the collision. Answer in units of kg · m/s.
Answer:
Ft=24,443.5 kgm/s
Explanation:
Step one
Given data
Mass of automobile m=1577kg
Initial Velocity u=17m/s
Final Velocity v=1.5m/s
Time t=0.18s
Step two
From the impulse and momentum equation
Ft=mΔv
Substitute
Ft=1577*(17-1.5)
Ft=1577*15.5
Ft=24,443.5 kgm/s
A light ray in glass (n=1.5) hits the air-glass interface at an angle of 10 degrees from the normal. What angle from the normal is the light ray in the air (n=1.0)? (You can use the small angle approximation.)
Answer:
The angle from the normal is 15.1°.
Explanation:
We can find the angle by using Snell's law:
[tex] n_{1}sin(\theta_{1}) = n_{2}sin(\theta_{2}) [/tex]
Where:
n₁: is the first medium (glass) = 1.5
n₂: is the second medium (air) = 1.0
θ₁: is the first angle (in the glass) = 10°
θ₂: is the second angle (in the air) =?
[tex] \theta_{2} = arcsin(\frac{n_{1}sin(\theta_{1})}{n_{2}}) = arcsin(\frac{1.5*sin(10)}{1.0}) = 15.1 ^{\circ} [/tex]
Therefore, the angle from the normal is 15.1°.
I hope it helps you!
A small lead ball, attached to a 1.10-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled at a constant rate of three revolutions per second and is released on the upward part of the circular motion when it is 1.3 m above the ground. The ball travels straight upward. In the absence of air resistance, to what maximum height above the ground does the ball rise
Answer:
1.84 m
Explanation:
For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.
So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,
a = g + rω²
= 9.8 m/s² + 1.10 m × (18.85 rad/s)²
= 9.8 m/s² + 390.85 m/s²
= 400.65 m/s²
Now, using v² = u² + 2a(h₂ - h₁) where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.
v² = u² + 2a(h₂ - h₁)
So, v² - u² = 2a(h₂ - h₁)
h₂ - h₁ = (v² - u²)/2a
h₂ = h₁ + (v² - u²)/2a
substituting the values of the variables into the equation, we have
h₂ = 1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)
h₂ = 1.3 m + [-430.15 (m/s)²]/-801.3 m/s²
h₂ = 1.3 m + 0.54 m
h₂ = 1.84 m
A motorcycle moving 18.8 m/s has
57800 J of KE. What is its mass?
Answer:
m = 327.07 kg
Explanation:
Given that,
Kinetic energy of a motorcycle, E = 57800 J
Velocity of the motorcycle, v = 18.8 m/s
We need to find the mass of the motorcycle. The kinetic energy of an object is given by :
[tex]E=\dfrac{1}{2}mv^2[/tex]
m is mass
[tex]m=\dfrac{2E}{v^2}\\\\m=\dfrac{2\times 57800 }{(18.8)^2}\\\\m=327.07\ kg[/tex]
So, the mass of the motorcycle is 327.07 kg.
A 20 kg sled is pulled up a 10m tall hill. What work is done against gravity?
Answer:
1962
Explanation:
w = f × d
= 20×9.81 × 10
= 1962
Which example best matches the term refraction? (17 Points)
O A. Light spreads out after it travels through a keyhole
O B. A straw in a glass of water looks bent
O C. Seeing your image in a lake
O D. An orchestra of different sounds coming together to make a larger sound
Answer:
B
Explanation:
Just answerd that question!!!
Hope it helped!!!
.
A finch rides on the back of a Galapagos tortoise, which walks at the stately pace of 0.060 m>s. After 1.5 minutes the finch tires of
Complete Question:
A finch rides on the back of a Galapagos tortoise, which walks
at the stately pace of 0.060 m/s. After 1.5 minutes the finch tires of
the tortoise’s slow pace, and takes flight in the same direction for
another 1.5 minutes at 11 m/s.
What was the average speed of the finch for this 3.0-minute interval?
Answer:
[tex]Speed = 5.53 m/s[/tex]
Explanation:
Distance is calculated as:
[tex]Distance = Speed * Time[/tex]
First, we calculate the distance for the first 1.5 minutes
For the first 1.5 minutes, we have:
[tex]Speed = 0.060m/s[/tex]
[tex]Time = 1.5\ mins[/tex]
[tex]D_2= 0.060m/s * 1.5\ mins[/tex]
Convert 1.5 mins to seconds
[tex]D_2= 0.060m/s * 1.5 * 60s[/tex]
[tex]D_2= 5.4m[/tex]
Next, we calculate the distance for the next 1.5 minutes
[tex]Speed = 11m/s[/tex]
[tex]Time = 1.5\ mins[/tex]
[tex]D_2= 11m/s * 1.5\ mins[/tex]
Convert 1.5 mins to seconds
[tex]D_2 = 11m/s * 1.5 * 60s[/tex]
[tex]D_2= 990m[/tex]
Total distance is:
[tex]Distance = 990m + 5.4m[/tex]
[tex]Distance = 995.4m[/tex]
The average speed for the 3.0 minute interval is:
[tex]Speed = \frac{Distance}{Time}[/tex]
[tex]Speed = \frac{995.4\ m}{3.0\ mins}[/tex]
Convert 3.0 minutes to seconds
[tex]Speed = \frac{995.4\ m}{3.0 * 60 secs}[/tex]
[tex]Speed = \frac{995.4\ m}{180 secs}[/tex]
[tex]Speed = 5.53 m/s[/tex]
A person is standing on and facing the front of a stationary skateboard while holding a construction brick. The mass of the person is 67.0 kg, the mass of the skateboard is 4.10 kg, and the mass of the brick is 2.50 kg. If the person throws the brick forward (in the direction they are facing) with a speed of 23.0 m/s relative to the skateboard and we ignore friction, determine the recoil speed of the person and the skateboard, relative to the ground.
Answer:
v₁ = -0.8087 m / s
Explanation:
To solve this problem we can use the conservation of momentum, for this we define a system formed by the man, the skateboard and the brick, therefore the force during the separation is internal and the momentum is conserved
Initial instant. When they are united
p₀ = 0
Final moment. After throwing the brick
[tex]p_{f}[/tex] = (m_man + m_skate) v1 + m_brick v2
the moment is preserved
p₀ = p_{f}
0 = (m_man + m_skate) v₁ + m_brick v₂
v₁ = - [tex]\frac{ m_{brick} }{m_{man} + m_{skate} } v_{2}[/tex]
the negative sign indicates that the two speeds are in the opposite direction
let's calculate
v₁ = - [tex]\frac{2.5}{67 + 4.10} 23.0[/tex]
v₁ = -0.8087 m / s
Two floors in a building are separated by 4.1 m. People move between the two floors on a set of stairs. (a) Determine the change in potential energy of a 3.0 kg backpack carried up the stairs. (b) Determine the change in potential energy of a person with weight 650 N that descends the stairs.
Answer:
a) The change in potential energy of a 3.0 kilograms backpack carried up the stairs.
b) The change in potential energy of a persona with weight 650 newtons that descends the stairs is -2665 joules.
Explanation:
Let consider the bottom of the first floor in a building as the zero reference ([tex]z = 0\,m[/tex]). The change in potential energy experimented by a particle ([tex]\Delta U_{g}[/tex]), measured in joules, is:
[tex]\Delta U_{g} = m\cdot g\cdot (z_{f}-z_{o})[/tex] (1)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]z_{o}[/tex], [tex]z_{f}[/tex] - Initial and final height with respect to zero reference, measured in meters.
Please notice that [tex]m\cdot g[/tex] is the weight of the particle, measured in newtons.
a) If we know that [tex]m = 3\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{o} = 0\,m[/tex] and [tex]z_{f} = 4.1\,m[/tex], then the change in potential energy is:
[tex]\Delta U_{g} = (3\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.1\,m-0\,m)[/tex]
[tex]\Delta U_{g} = 120.626\,J[/tex]
The change in potential energy of a 3.0 kilograms backpack carried up the stairs.
b) If we know that [tex]m\cdot g = 650\,N[/tex], [tex]z_{o} = 4.1\,m[/tex] and [tex]z_{f} = 0\,m[/tex], then the change in potential energy is:
[tex]\Delta U_{g} = (650\,N)\cdot (0\,m-4.1\,m)[/tex]
[tex]\Delta U_{g} = -2665\,J[/tex]
The change in potential energy of a persona with weight 650 newtons that descends the stairs is -2665 joules.
How much work would be needed to lift the ball from the 2-m shelf to the 5-m shelf, and how much potential energy would it have on the 5-m shelf?
Answer:
a) 2.94 N
b) 4.90 N
Explanation:
Let us assume that the weight of the ball is 0.98 N
Solution:
a) An object’s gravitational potential energy depends on two factors which are height and its weight (or mass). The equation for gravitational potential energy (PE) is given as:
Potential energy = weight (w) * height (h)
PE = wh
Potential energy at 2 m shelf = weight * height = 0.98 N * 2 m = 1.96 N
Potential energy at 5 m shelf = weight * height = 0.98 N * 5 m = 4.90 N
The work needed to lift the ball from the 2-m shelf to the 5-m shelf = Potential energy at 5 m shelf - Potential energy at 2 m shelf
The work needed to lift the ball from the 2-m shelf to the 5-m shelf = 4.90 N - 1.96 N = 2.94 N
b) Potential energy at 5 m shelf = weight * height = 0.98 N * 5 m = 4.90 N
A 450-N rightward force is used to drag a large box across the floor with a constant velocity of 1.2 m/s. The coefficient of friction between the box and the floor is 0.795. Determine the mass of the box.
Answer:
the mass of the box is 51.98 kg.
Explanation:
Given;
applied horizontal force, F = 450 N
coefficient of friction, μ = 0.795
constant velocity, v = 1.2 m/s
At constant velocity, the acceleration of the object is zero and the net force will be zero.
[tex]F_{Net} = F - F_k\\\\0 = F - F_k\\\\F_k = F\\\\\mu \ N = F\\\\\mu (mg) = F\\\\m = \frac{F}{\mu g} \\\\m = \frac{405}{0.795 \ \times \ 9.8} \\\\m = 51.98 \ kg[/tex]
Therefore, the mass of the box is 51.98 kg.
A cart's initial velocity is +3.0 meters per second. What is its final velocity after accelerating at a rate of 1.5 m/s2 for 8.0 seconds?
A. 36 m/s
B. 9 m/s
C. 72 m/s
D. 15 m/s
Answer:
V = 15m/s
Explanation:
Given the following data;
Initial velocity = 3m/s
Time = 8secs
Acceleration = 1.5m/s²
To find the final velocity, we would use the first equation of motion;
V = U + at
Substituting into the equation, we have
V = 3 + 1.5*8
V = 3 + 12
V = 15m/s
help mmmemememeeee I'M BEING TIMED
Answer:
it's c. or A. not b.
Explanation:
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