Answer:
There is net loss of gravitational energy .
Explanation:
When Xanaxa is on the ground , her potential energy is assumed to be zero . When she leaps to a height of 153 m , she gains gravitational energy . When she dives and reaches the surface , she loses potential energy and on reaching the ground her potential energy becomes zero . When she further goes down inside ground to a depth of 17.5 m , she loses potential energy further . Her potential energy becomes less than zero or negative .
Ultimately her potential energy changes from zero to negative in the whole process . So there is net loss of potential energy .
A clump of soft clay is thrown horizontally from 9.80 m above the ground with a speed of 20.0 m/s. Assume it sticks in place when it hits the ground At what time will the clay hit the ground
Answer:
Explanation:
The time to hit the ground will be same as time taken to fall from the height of 9.8 m with initial vertical velocity of zero .
Considering vertical displacement
initial velocity u = 0
displacement s = 9.8 m
acceleration a = g = 9.8 m /s²
time t = ?
s = ut + 1/2 g t²
9.8 = 0 + .5 x 9.8 x t²
t² = 2
t = √2 = 1.4 s
1 example of a conductor and 1 example of a insulator in your EVERYDAY world.
Answer: Examples of conductors include metals, aqueous solutions of salts (i.e., ionic compounds dissolved in water), graphite, and the human body. Examples of insulators include plastics, Styrofoam, paper, rubber, glass and dry air.
16. An object has a gravitational potential energy 41,772.5 Jof and has a mass of 1550 kg. How high is it
above the ground?
Plz help
Answer:
2.75 m.
Explanation:
From the question given above, the following data were obtained:
Potential energy (PE) = 41772.5 J
Mass (m) of object = 1550 kg
Height (h) =?
Potential energy is the energy possess by an object due to its location. Mathematically, potential energy is expressed as shown below:
PE = mgh
Where
PE => potential energy
m => mass of the object
g => acceleration due to gravity
h => height to which the object is located.
With the above formula, we can obtain the height to which the object is located as follow:
Potential energy (PE) = 41772.5 J
Mass (m) of object = 1550 kg
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) =?
PE = mgh
41772.5 = 1550 × 9.8 × h
41772.5 = 15190 × h
Divide both side by 15190
h = 41772.5 / 15190
h = 2.75 m
Thus, the object is located at 2.75 m above the ground.
A basketball with a mass of 20 kg is accelerated with a force of 10 N. If resisting forces are ignored, what is the acceleration of the basketball?
a long solid rod 4.5 cm in radius carries a uniform volume charge density. if the electric field strength at the surface of the rod (not near either end) is 16 kn/c, what is the volume charge density
Answer:
6.29 μC/m³
Explanation:
Volume charge density is the quantity of charge per unit volume.
The direction of the electric field was not specified, therefore the volume charge density (ρ) is given by:
2πRLE = ρπR²L/ε₀
ρ = 2Eε₀ / R
Where E = electric field strength = 16 kN/C = 16 * 10³ N/C, R = radius of rod = 4.5 cm = 0.045 m, ε₀ = relative permittivity of free space = 8.85 * 10⁻¹² C² / Nm²
Therefore:
ρ = 2(16 * 10³ N/C)(8.85 * 10⁻¹² C²/Nm²) / 0.045 m = 6.29 * 10⁻⁶ C/m³
ρ = 6.29 μC/m³
If someone walkes 1000m
in 20min, what is their speed?
Answer:
Distance - 1000m
Time - 20min
Speed - ?
Use the formula of distance ÷ time = speed.
s = d/t
s = 1000m/20min
s = 50 m/min
Hope this helps, thank you !!
What must the charge (sign and magnitude) of a particle of mass 1.48 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 640 N/C
Answer:
[tex]q=-2.26\times 10^{-5}\ C[/tex]
Explanation:
Given that,
The mass of a particle, m = 1.48 g = 0.00148 kg
The electric field, E = 640 N/C
We need to find the charge of the particle when placed in a downward-directed electric field.
The force of gravity is balanced by the electric force such that,
mg = qE
Where
q is the charge of the particle
[tex]q=\dfrac{mg}{E}\\\\q=\dfrac{0.00148\times 9.8}{640}\\\\q=2.26\times 10^{-5}\ C[/tex]
q must be negative, the force must be upward (opposite direction of the electric field).
If rider A is moving 8 meters per second, and rider B is moving 3 meters per second, how far away from rider A was rider B when first observed at the start of the graph?
Answer:
the answer is 32
Explanation:
he was riding pretty far
Orion, also called the Hunter, has three stars that make up Orion's belt.
Which star is at the tip of the arrow? PLEASE HELP I NEED THIS FAST
A. Sirius
B. Betelgeuse
C. Rigel
D. Polaris
Answer - B. Betelguese.
I really hope this helps!!
15 points!!:-) Need help ASAP!
When does a compass NOT point towards magnetic north?
A.during a solar eclipse, which changes
Earth's magnetic field.
B. When there is another magnet close by.
C.When there is an unused battery close by.
D. When there is a coil of copper wire close by.
Answer:
b i think so because it makes senes
Answer: When there is another magnet close by.
Explanation: The needle of a compass is itself a magnet, and thus the north pole of the magnet always points north, except when it is near a strong magnet. ... When you take the compass away from the bar magnet, it again points north. So, we can conclude that the north end of a compass is attracted to the south end of a magnet.
Which image best illustrates diffraction
Answer:
There is no image
Explanation:
Answer: Send me image than I will be able to help
Explanation:
which newton's laws applies to a slinky?
PLEASE HELP !!
When a Slinky sits atop a staircase, gravity acts on the toy, keeping it still. Knock over the Slinky, and Newton's second law comes into play. As middle school physics class may have taught you, this law states that providing force to an object increases its acceleration.
hopes this helps uh ❣
Answer:
We know from Newton's First Law of motion that an object at rest stays at rest unless acted upon by an external force. So in the case of the slinky, that is exactly why the bottom of the slinky does not move.
Explanation:
fertilization that takes place when the union of the sex cells happens outside the body
Answer:external
Explanation:EDGE 2021
Suppose a cat climbs a tree to a height of 2 meters. If the cat doubles its height to 4 meters, its potential energy will
1)not change
2)reduce to halve
3)double
4)quadruple
1. In calculating the work W necessary to stretch a spring from its natural length of 8 centimeters to a length of 12 centimeters, we need some more information about the spring. There are two distinct kinds of information, each of which separately would provide the necessary assistance so we could find W. Describe what the two additional kinds of information are.
Answer:
* displacement from equilibrium position
* The spring constant,
Explanation:
The work done by a spring is two times the expressions
W = ∫ F. dx
in a spring the force is given by hooke's law
F = - k Δx
we substitute
W = - k ∫ x dx cos θ
In the case of the spring, the displacement is in the same direction of work, therefore the angle is zero and the cosine is equal to 1
we integrate
W = ½ k Δx²
we can see that we need two parts to calculate the work
* displacement from equilibrium position
Δx = 12 -8 = 4 cm
* The spring constant, this constant can be found from the displacement measurements as a function of the applied force.
A box of bananas weighing 51.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.46 and the coefficient of kinetic friction is 0.23. Part A If no horizontal force is applied to the box and the box is at rest, how large is the frictional force exerted on the box by the surface
static friction force = The coefficient of static friction * normal force
static friction force = 0.46 * 51 = 23.46 N
kinetic friction force = The coefficient of kinetic friction * normal force
kinetic friction force = 0.23 * 51 = 11.73 N
the applied force acting on the object must be more than 23.5 N if the object was stationary to move it and must be more than 11.7 N during the movement to keep the object moving
From the Water in each of the next move
Fusion probability is greatly enhanced when appropriate nuclei are brought close together, but mutual Coulomb repulsion must be overcome. This can be done using the kinetic energy of high-temperature gas ions or by accelerating the nuclei toward one another. Calculate the potential energy of two singly charged nuclei separated by 1.00 x 10-12 m by finding the voltage of one at that distance and multiplying by the charge of the other.
Answer:
the Potential Energy is 2.304 × 10⁻¹⁶ J
Explanation:
Given the data in the data in the question;
The expression for the electric potential energy between the charges can be expressed as follows;
PE = qV ------equ 1
where q is the charge and V is the electric potential
Also the formula for electric potential due to point a point in a field is;
V = kq / r -------equ 2
where k is the electrostatic constant and r is the distance form the charged particle
input equation 2 into 1
PE = q × kq / r
PE = kq²/r ------- equ 3
so we substitute into equation 3; 1.00×10⁻¹² for r, 9.00×10⁹ for k( constant ) and 1.60×10⁻¹⁹ for q( charge )
PE = ((9.00×10⁹) (1.60×10⁻¹⁹)²) / 1.00×10⁻¹²
PE = 2.304 × 10⁻²⁸ / 1.00×10⁻¹²
PE = 2.304 × 10⁻¹⁶ J
Therefore, the Potential Energy is 2.304 × 10⁻¹⁶ J
Two kilograms of air is contained in a rigid wellinsulated tank with a volume of 0.6 m3 . The tank is fitted with a paddle wheel (stirrer) that transfers energy to the air at a constant rate of 10 W for 1h. If no changes in kinetic or potential energy occur, determine a) The specific volume at the final state, in m3 /kg. b) The energy transfer by work, in kJ. c) The change in specific internal energy of the air, in kJ/kg.
Answer:
[tex]0.3\ \text{m}^3/\text{kg}[/tex]
[tex]36\ \text{kJ}[/tex]
[tex]18\ \text{kJ/kg}[/tex]
Explanation:
V = Volume of air = [tex]0.6\ \text{m}^3[/tex]
P = Power = 10 W
t = Time = 1 hour
m = Mass of air = 2 kg
Specific volume is given by
[tex]v=\dfrac{V}{m}\\\Rightarrow v=\dfrac{0.6}{2}\\\Rightarrow v=0.3\ \text{m}^3/\text{kg}[/tex]
The specific volume at the final state is [tex]0.3\ \text{m}^3/\text{kg}[/tex]
Work done is given by
[tex]W=Pt\\\Rightarrow W=10\times 60\times 60\\\Rightarrow W=36000\ \text{J}=36\ \text{kJ}[/tex]
The energy transfer by work, is [tex]36\ \text{kJ}[/tex]
Change in specific internal energy is given by
[tex]\Delta u=\dfrac{Q}{m}+\dfrac{W}{m}\\\Rightarrow \Delta u=0+\dfrac{36}{2}\\\Rightarrow \Delta u=18\ \text{kJ/kg}[/tex]
The change in specific internal energy of the air is [tex]18\ \text{kJ/kg}[/tex]
to what temperature it will a 30 KG of glass raise if it absorbs 4275 joules of heat in its specific heat is 0.5 J/KG degree celsius. The initial temperature of the glass is 35°C
Answer:
230° C
Explanation:
A substance's specific heat tells you how much heat much either be added or removed from 1 g of that substance in order to cause a 1∘C
A hanging wire made of an alloy of titanium with diameter 0.05 cm is initially 2.7 m long. When a 15 kg mass is hung from it, the wire stretches an amount 1.68 cm. A mole of titanium has a mass of 48 grams, and its density is 4.54 g/cm3. Based on these experimental measurements, what is Young's modulus for this alloy of titanium
Answer:
Explanation:
Young's modulus of elasticity Y = stress / strain
stress = force / cross sectional area
= weight of 15 kg / π r²
= 15 x 9.8 / 3.14 x ( .025 x 10⁻² )²
stress = 74.9 x 10⁷ N / m²
strain = Δ L / L , Δ L is change in length and L is original length
Putting the values
strain = .0168 / 2.7 =.006222
Young's modulus of elasticity Y = 74.9 x 10⁷ / .006222
= 120.88 x 10⁹ N / m² .
An electron moves from point i to point f, in the direction of a uniform electric field. During this motion:Group of answer choicesthe work done by the field is positive and the potential energy of the electron-field system increasesthe work done by the field is negative and the potential energy of the electron-field system increasesthe work done by the field is positive and the potential energy of the electron-field system decreasesthe work done by the field is negative and the potential energy of the electron-field system decreasesthe work done by the field is positive and the potential energy of the electron-field system does not change
Answer:
the work done by the field is positive and the potential energy of the electron field system decreases
Explanation:
This exercise asks to find the work and the potential energy of an electron in an electric field.
Work is defined by
W = F .d = F d cos θ
the electric force is
F_e = q E
W = q E d cos θ
since the charge of the electron is negative the force is in the opposite direction to the electric field
W = - e E d
we select the direction to the right is positive, point i is to the left of point f,
therefore the work moving from point i to point F has two possibilities
* The electric field lines go from i to f point , so that point i is on the side of the positive charges, so the electron approaches them, This movement is opposite to that indicated
* the field line reaches point i, this implies that the charges are negative, so the electrioc field is then negativeand the electron charge is negative too. The electron moves away from this point, this is in accordance with the indicated movement
In the latter case the electric field lines go from f to i point, therefore the Work is positive
Now let's examine the potential energy
ΔU = - q E .d
so we see that this definition is related to work,
ΔU = -W
Therefore, as the work is positive, the power energy must decrease
When reviewing the different answers, the correct ones are:
the work done by the field is positive and the potential energy of the electron field system decreases
The work done by the electron while moving from point [tex]i[/tex] to point [tex]f[/tex] in the direction of uniform electric field is negative and the potential energy of the electron increases.
An electron moves from point i to point f, in the direction of a uniform electric field, then the potential energy of the electron can be calculated s given below.
[tex]\Delta V=-qEd[/tex]
Where [tex]\Delta V[/tex] is the potential energy, [tex]E[/tex] is the electric field, [tex]q[/tex] is the charge and [tex]d[/tex] is the displacement of the electron.
The work done by the electron in the uniform electric field can be calculated as,
[tex]W = F\times d \times cos\theta[/tex]
Where [tex]W[/tex]is the work done by electron, [tex]F[/tex] is the electric force, [tex]d[/tex] is the displacement of the electron and for uniform electric field, the value of [tex]\theta[/tex] is zero.
Hence [tex]W=F\times d\times 1\\W=F \times d[/tex]
Electric force [tex]F = q E[/tex]
By substituting the value of electric force on the above formula,
[tex]W = qEd[/tex]
Hence, the relation between the work done the electron in an uniform electric field and potential energy of the electron can be given below.
[tex]W = -\Delta V[/tex]
The work done by the electron is negative and the potential energy of the electron increases.
For more information, follow the link given below.
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What is the correct description for kinetic energy?Immersive Reader
(1 Point)
the energy an object has because of it temperature
the energy an object has because it is moving
the energy stored in an object because of its position
the energy stored in an object when you stretch or squash it
Answer:
The energy an object has because it is moving
Explanation:
It has been a while since I have talked about kinetic energy so I can't give you an explanation why that answer is right but it is.
The driver of a 3000 lb. car, coasting down a hill, sees a red light at the bottom, and must stop. His speed when he applies the brakes is 60 mph, and he is 100 feet (vertically) above the bottom of the hill. (a)How much energy as heat must be dissipated by the brakes if we neglect wind resistance and other frictional effects
Answer:
Explanation:
60 mph = 60 x 1760 x 3 / (60 x 60) ft /s
speed of car , v = 88 ft /s
kinetic energy of car = 1/2 m v²
= .5 x 3000 x 88²
= 11616 x 10³ poundal - foot
Potential energy = mgh
= 3000 x 32 x 100
= 9600 x 10³ poundal - foot
Total energy = potential energy + kinetic energy
= ( 11616 + 9600 )x 10³
= 21216 x 10³ poundal - foot .
This energy is dissipated as heat when brakes are applied on the car to stop the car .
8) A train enters a curved horizontal section of the track at a speed of 100 km/h and slows down with constant deceleration to 50 km/h in 12 seconds. If the total horizontal acceleration of the train is 2 m/s2 when the train is 6 seconds into the curve, calculate the radius of curvature of the track for this instant.
Answer:
the radius of curvature of the track for this instant is 266 m
Explanation:
Given that;
The Initial Velocity u = 100 km/h = 100 × [tex]\frac{5}{18}[/tex] = 27.77 m/s
velocity of the train at t=12 s is;
[tex]V_{t=12}[/tex] = 50 km/h = 50 × [tex]\frac{5}{18}[/tex] = 13.89 m/s
now, we calculate the deceleration of the train
[tex]V_{t=12}[/tex] = u + at
13.89 = 27.77 + [tex]a_{t}[/tex]12
[tex]a_{t}[/tex] = (13.89 - 27.77) / 12
[tex]a_{t}[/tex] = -13.88 / 12
[tex]a_{t}[/tex] = - 1.1566 m/s²
Now, the velocity of the train at 6 seconds is;
[tex]V_{t=6}[/tex] = u + at
[tex]V_{t=6}[/tex] = 27.77 + ( - 1.1566 m/s²)6
[tex]V_{t=6}[/tex] = 27.77 - 6.9396
[tex]V_{t=6}[/tex] = 20.83 m/s
The acceleration at t=6 s is;
a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]
a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]
we substitute
2m/s² = √[ (- 1.15 )² + ([tex]a_{n}[/tex])²]
4 = (- 1.1566 )² + ([tex]a_{n}[/tex])²
4 = 1.3377 + ([tex]a_{n}[/tex])²
([tex]a_{n}[/tex])² = 4 - 1.3377
([tex]a_{n}[/tex])² = 2.6623
[tex]a_{n}[/tex] = √2.6623
[tex]a_{n}[/tex] = 1.6316 m/s²
Now the radius of curve is;
a = V² / p
[tex]p_{t=6}[/tex] = ( [tex]V_{t=6}[/tex] )² / [tex]a_{n}[/tex]
[tex]p_{t=6}[/tex] = ( 20.83 m/s )² / 1.6316 m/s²
[tex]p_{t=6}[/tex] = 433.8889 / 1.6316
[tex]p_{t=6}[/tex] = 265.9 m ≈ 266 m
Therefore; the radius of curvature of the track for this instant is 266 m
Which is the weakest of the four fundamental forces?
strong nuclear
weak nuclear
electromagnetic
gravitational
Answer:
Gravitational
Explanation:
gravitational
Answer:
Gravitational
Explanation:
In order from strongest to weakest.
Strong nuclear
Electromagnetic
Weak nuclear
Gravitational
Micro-bats use a form of radar called echolocation to navigate and find their prey such as flying insects. They locate the surrounding objects by bouncing sound wave pulses off these objects and detecting the time delay between the emitted pulses and the reflected pulses. Determine the time delay between the pulse emitted by the micro-bat and the detected pulse reflected from an insect located 10 m away from the micro-bat. Assume the approximate speed of sound waves to be 340 m-s-1
Answer:
t = 5.88 10⁻² s
Explanation:
The speed of the sound wave after it is emitted by the bat is constant, so we can use the uniform motion relationships
v = [tex]\frac{x}{t}[/tex]
t = [tex]\frac{x}{v}[/tex]
in this case the distance is that of the sound in going from the bat to the insect and back
x = 2d
x = 2 10
x = 20 m
let's calculate
t = 20/340
t = 5.88 10⁻² s
We can see that the time is very short, so the distance traveled by the two animals has little influence on the result.
At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.70 m/s2. At the same instant a truck, traveling with a constant speed of 9.50 m/s, overtakes and passes the automobile. (a) How far beyond the traffic signal will the automobile overtake the truck
Answer:
66.85 m
Explanation:
We are given that
Acceleration ,a=[tex]2.7m/s^2[/tex]
Speed of truck, v=9.5 m/s
We have to find the distance beyond which the traffic signal will the automobile overtake the truck.
Initial speed of automobile, u=0
We know that
[tex]s=ut+\frac{1}{2}at^2[/tex]
Using the formula
[tex]s=0+\frac{1}{2}(27)t^2=\frac{27}{2}t^2[/tex]
For constant speed
Acceleration, a=0
Again
[tex]s=vt+0=9.5t[/tex]
[tex]9.5t=\frac{27}{2}t^2[/tex]
[tex]t=\frac{9.5\times 2}{2.7}=7.037s[/tex]
Substitute the value of t
[tex]x=9.5(7.037)=66.85m[/tex]
Hence, the distance beyond which the traffic signal will the automobile overtake the truck=66.85 m
which is the product of cellular respiration? A. ATTP B. light C. oxygen D.sugar
a. As you coast down a hill on your bicycle, you accelerate at 0.5 m/s2. If the total mass of your body and the bicycle is 80 kilograms what is the net force pulling you down the hill (gravity - friction)?
The net force pulling you down the hill will be = 40 N
What is Newton's Second Law of motion?The Second Law of motion states that the acceleration of an object depend upon the object and the mass of the object.
F = Mass * acceleration
Given
Total mass : 80 kg
Acceleration : 0.5 m / s^2
Net Force = mass * acceleration (Second Law of motion )
Net Force = 80 * 0.5 = 40 N
The net force pulling you down the hill will be = 40 N
Learn more about Newton's Second Law of motion:
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