Answer: 6 J
Explanation:
Total force applied = 47 N Assuming that direction of movement of pencil and applied force is same. Work done by force in moving the pencil W = Force × Distance through which force moves ⇒ W = 47 × 0.25 = 11.75 J .-.-.-.-.-.-.-.-. In case we are asked useful work done then we calculate net force used for pushing the pencil: Net force used for pushing the pencil = 47 − 23 = 24 N Assuming that direction of movement of pencil and net force is same. Useful Work done by force in moving the pencil W u = Force × Distance ⇒ W u = 24 × 0.25 = 6 J
The work done the pushing force is required.
The work done by the pushing force is [tex]6\ \text{J}[/tex]
[tex]F_1[/tex] = Pushing force = 47 N
[tex]F_2[/tex] = Opposing force = 23 N
m = Mass of object = 0.025 kg
s = Displacement = 0.25 m
Work done is given by
[tex]W=F_ns[/tex]
[tex]\Rightarrow W=(47-23)\times 0.25[/tex]
[tex]\Rightarrow W=6\ \text{J}[/tex]
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The potential energy stored in the compressed spring of a dart gun, with a spring constant of 32.50 N/m, is 0.640 J. Find by how much is the spring is compressed. Submit Answer Tries 0/20 A 0.050 kg dart is fired straight up. Find the vertical distance the dart travels from its position when the spring is compressed to its highest position. Submit Answer Tries 0/20 The same dart is now fired horizontally from a height of 3.90 m. The dart remains in contact until the spring reaches its equilibrium position. Find the horizontal velocity of the dart at that time. Submit Answer Tries 0/20 Find the horizontal distance from the equilibrium position at which the dart hits the ground.
Answer:
A
[tex]x = 0.198456 \ m [/tex]
B
[tex]h = 1.3061 \ m [/tex]
C
[tex] v = 5.06 \ m/s [/tex]
D
[tex]d = 4.0273 \ m [/tex]
Explanation:
Considering the first question
From the question we are told that
The spring constant is [tex]k = 32.50 N/m[/tex]
The potential energy is [tex]PE = 0.640 \ J[/tex]
Generally the potential energy stored in spring is mathematically represented as [tex]PE = \frac{1}{2} * k * x^2[/tex]
=> [tex]0.640= \frac{1}{2} * 32.50 * x^2[/tex]
=> [tex]x = \sqrt{0.03938}[/tex]
=> [tex]x = 0.198456 \ m [/tex]
Considering the second question
From the question we are told that
The mass of the dart is m = 0.050 kg
Generally from the law of energy conservation
[tex]PE = mgh[/tex]
=> [tex]0.640 = 0.050 * 9.8 * h[/tex]
=> [tex]h = 1.3061 \ m [/tex]
Considering the third question
The height at which the dart was fired horizontally is [tex]H = 3.90\ m[/tex]
Generally from the law of energy conservation
[tex]PE = KE [/tex]
Here KE is kinetic energy of the dart which is mathematical represented as
[tex]KE = \frac{1}{2} * mv^2[/tex]
=> [tex]0.640 = \frac{1}{2} * 0.050 * v^2 [/tex]
=> [tex] v^2 = 25.6 [/tex]
=> [tex] v = 5.06 \ m/s [/tex]
Considering the fourth question
Generally the total time of flight of the dart is mathematically represented as
[tex]t = \frac{ 2 * H }{g}[/tex]
=> [tex]t = \frac{ 2 * 3.90 }{9.8 }[/tex]
=> [tex]t = 0.7959 \ s [/tex]
Generally the horizontal distance from the equilibrium position to the ground is mathematically represented as
[tex]d = v * t[/tex]
=> [tex]d = 5.06 * 0.7959[/tex]
=> [tex]d = 4.0273 \ m [/tex]
Why is a very small fraction (0.001%) of the water found in the atmosphere?
There is a very small fraction (0.001%) of the water found in the atmosphere due to evaporation of water.
What is Atmosphere?This is defined as mixture of gases surrounding the Earth or other celestial body, held in place by gravity.
A very small fraction (0.001%) of the water in the form of water vapor gas exist in the atmosphere due to evaporation by heat from the Sun.
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Define kinetic Energy and thermal energy. Describe what happens to each as the temperature of a substance is increases.
Explanation:
Kinetic energy is the energy by virtue of
object's motion whereas Thermal energy is
the internal energy of an object due to the
kinetic energy of its atoms.
On Increasing temperature, they both
increases
Lana's class is studying the various kinds of energy found on a playground
Which of the following would be the BEST way to test for an equal transformation of energy on a playground?
СА
Using the slide, Lana could measure her kinetic energy when she was climbing up the ladder, measure her potential energy before she slid down, and her total energy used at the bottom of the slide. If the amount of potential
energy and kinetic energy added up to equal the total amount of energy used, it would show an equal transformation of energy
OB
Using the slido, Lana could measure her potential and kinetic energy sitting at the top of the slide, and then she could measure her kinetic and potential energy at various points while sliding to the bottom of the total value of
her potential energy and her kinetic energy were equal at every measuremont, it would show that there had been an equal transformation
C
Using the swing, Lana could measure her potential energy when she was sitting still on the swing, and she could measure her kinetic energy when the swing was in motion. If the kinetic energy was twice the amount of her
potential energy, it would show there had boon an equal transformation of potential energy to kinetic energy
D
Using the soesaw, Lana could measure her potential energy when she was in motion, and she could measure her kinetic energy when she was resting. If her kinetic and potential energy were equal, it would show that there
had been an equal transformation
If the amount of potential energy and kinetic energy added up to equal the total amount of energy used, it would show an equal transformation of energy.
What is Energy?This can be defined as the ability or capacity to do work. The potential and kinetic energy equals the total amount of energy involved when Lana was on the playground.
This depicts an equal transformation of energy which is why the most appropriate choice in this scenario is option A.
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The ratio of carbon-14 to nitrogen-14 is an artifact is 1:3. Given that half-life of carbon-14 is 5730years, how old is the artifact?
Answer:
9155 years old
Explanation:
We use the following expression for the decay of a substance:
[tex]N = N_0\,\,e^{-k*t}[/tex]
So we first estimate the value of k knowing that the half-life of the C14 is 5730 years:
[tex]N = N_0\,\,e^{-k*t}\\N_0/2=N_0\,\,e^{-k*5730}\\1/2 = e^{-k*5730}\\ln(1/2)=-k*5730\\k= 0.00012[/tex]
so, now we can estimate the age of the artifact by solving for"t" in the equation:
[tex]1/3=e^{-0.00012*t}\\ln(1/3)= -0.00012*t\\t=9155. 102[/tex]
which we can round to 9155 years old.
Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper, with a circular cross section of radius 1.53 mm. The other wire is composed of 19 strands of thin copper wire bundled together. Each strand has a circular cross section of radius 0.306 mm. The current density ???? in each wire is the same. ????=1750 A/m2 Two circles. One circle is solid, whereas the other contains 19 tightly packed smaller circles How much current does each wire carry? solid wire current: A stranded wire current: A The resistivity of copper is ????=1.69×10−8 Ω·m. What is the resistance of a 3.25 m length of each wire? solid wire resistance: Ω stranded wire resistance: Ω
Answer:
a
Solid Wire [tex]I = 0.01237 \ A [/tex]
Stranded Wire [tex]I_2 = 0.00978 \ A [/tex]
b
Solid Wire [tex]R = 0.0149 \ \Omega [/tex]
Stranded Wire [tex]R_1 = 0.0189 \ \Omega [/tex]
Explanation:
Considering the first question
From the question we are told that
The radius of the first wire is [tex]r_1 = 1.53 mm = 0.0015 \ m[/tex]
The radius of each strand is [tex]r_0 = 0.306 \ mm = 0.000306 \ m[/tex]
The current density in both wires is [tex]J = 1750 \ A/m^2[/tex]
Considering the first wire
The cross-sectional area of the first wire is
[tex]A = \pi r^2[/tex]
= > [tex]A = 3.142 * (0.0015)^2[/tex]
= > [tex]A = 7.0695 *10^{-6} \ m^2 [/tex]
Generally the current in the first wire is
[tex]I = J*A[/tex]
=> [tex]I = 1750*7.0695 *10^{-6}[/tex]
=> [tex]I = 0.01237 \ A [/tex]
Considering the second wire wire
The cross-sectional area of the second wire is
[tex]A_1 = 19 * \pi r^2[/tex]
=> [tex]A_1 = 19 *3.142 * (0.000306)^2[/tex]
=> [tex]A_1 = 5.5899 *10^{-6} \ m^2[/tex]
Generally the current is
[tex]I_2 = J * A_1[/tex]
=> [tex]I_2 = 1750 * 5.5899 *10^{-6} [/tex]
=> [tex]I_2 = 0.00978 \ A [/tex]
Considering question two
From the question we are told that
Resistivity is [tex]\rho = 1.69* 10^{-8} \Omega \cdot m[/tex]
The length of each wire is [tex]l = 6.25 \ m[/tex]
Generally the resistance of the first wire is mathematically represented as
[tex]R = \frac{\rho * l }{A}[/tex]
=> [tex]R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }[/tex]
=> [tex]R = 0.0149 \ \Omega [/tex]
Generally the resistance of the first wire is mathematically represented as
[tex]R_1 = \frac{\rho * l }{A_1}[/tex]
=> [tex]R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }[/tex]
=> [tex]R_1 = 0.0189 \ \Omega [/tex]
describe an example that shows thermal energy cannot be stored for a long time
Answer: heat always flows from higher temperature to lower temperature
Explanation: there:)
Heat always flows from higher temperature to lower temperature so can not be stored.
What is thermal energy?The energy present in a system that determines its temperature is referred to as thermal energy. Thermal energy flows as heat. Thermodynamics is a whole field of physics that studies how heat is transmitted across various systems and how work is performed in the process.
Heat dissipation always happen decrease in temperature so thermal energy can not be stored for long time.
Heat always flows from higher temperature to lower temperature so can not be stored
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calculate the monentun pf 75 kg bicycle and boy who has ghe velocitg of 3m/s
Explanation:
n=75 kg
v=3m/s
m=n×v
m=75x3
m= 225n
SAVE MEEEE WILL MARK BRAINLY
How much time would it take for a 0.17 kg ice hockey puck to decrease its speed by 9.0 m/s if the coefficient of kinetic friction between the ice and the puck is 0.05
How much time would it take for a 0.17 kg ice hockey puck to decrease its speed by 9.0 m/s if the coefficient of kinetic friction between the ice and the puck is 0.05
Given:- mass = 0.17 kg change in speed= 9 m/scoefficient of kinetic friction = 0.05 To Find :-Time taken to decrease the speed
Answer:-Equation :-
[tex]f {\tiny{k} }= u {\tiny{k}}.f \tiny{N}[/tex]
[tex]f {\tiny{N} }= mg = 0.17 \times 10 \\f {\tiny{N} }= 1.7 {N}[/tex]
[tex]u{ \tiny{N}} = 0.05[/tex]
[tex]mg = f {\tiny{k} }= u {\tiny{k}}.f {\tiny{N}} \\ 0.17a = 0.05 \times 1.7 \\ a = 0.05 \times \frac{ \cancel{1.7} {}^{ \: \: 10} }{ \cancel{0.17}} \\ a = 0.5m {s}^{ - 2} [/tex]
change in speed = 9 ms-1 (Given)
[tex]change \: in \: speed = at \\ 9 = 0.5 \times t \\ \frac{9 \times 10}{5} = t \\ \frac{9 \times \cancel{10} {}^{ \: \: 2} }{ \cancel{5}} = t \\ 18 \: sec = t \: [/tex]
How far away is the light away in meters
299,792,458 is the right answer
) The square plates of a 5000-pF parallel-plate capacitor measure 50 mm by 50 mm and are separated by a dielectric that is 0.23 mm thickand totally fills the region between the plates. The voltage rating (the maximum safe voltage) of the capacitor is 400 V. What is the maximum energy that can be stored in this capacitor without damaging it
Answer:
4 x 10⁻⁴ J
Explanation:
C = 5000 pF, V = 400 V
Energy = CV²/2 = 5000 x 10⁻¹² x 400²/2 = 4 x 10⁻⁴ J
2.27 A gas is compressed from V1= 0.3 m3, p1=1 bar to V2= 0.1 m3, p2 =3 bar. The pressure and
the volume are related linearly during the process. For the gas, find the
work, in kJ.
2.29 A gas expands from an initial state where p1=500kPa and V1=0.1m3 to a state
end where p2= 100 kPa. The relationship between pressure and volume during the process is
pV=constant. Schematize the process in a p-V diagram and determine the work, in
kJ.
Answer:
-40 kJ
80 kJ
Explanation:
Work is equal to the area under the pressure vs volume graph.
W = ∫ᵥ₁ᵛ² P dV
2.27) Pressure and volume are linearly related. When we graph P vs V, the area under the line is a trapezoid. So the work is:
W = ½ (P₁ + P₂) (V₂ − V₁)
W = ½ (100 kPa + 300 kPa) (0.1 m³ − 0.3 m³)
W = -40 kJ
2.29) Pressure and volume are inversely proportional:
pV = k
The initial pressure and volume are 500 kPa and 0.1 m³. So the constant is:
(500) (0.1) = k
k = 50
The final pressure is 100 kPa. So the final volume is:
(100) V = 50
V = 0.5
The work is therefore:
W = ∫ᵥ₁ᵛ² P dV
W = ∫₀₁⁰⁵ (50/V) dV
W = 50 ln(V) |₀₁⁰⁵
W = 50 (ln 0.5 − ln 0.1)
W ≈ 80 kJ
1. Allen is driving North on Highway 69 at 90 km/h and sees a large moose on the road. He
quickly slams on his brakes, but his reaction time is 0.85 s (as he sees the moose, thinks
about his response, and then presses the brake pedal). He presses the brake for 3.5 s and
comes to a stop just in time.
a) Find the distance travelled after seeing the moose and before pressing the brake.
b) Find the total distance he travelled before coming to a stop.
c) Find the average acceleration once he presses the brake.
Take the moment Allen sees the moose to be the origin.
First, convert his speed to m/s.
90 km/h = (90 km/h) • (1000 m/km) • (1/3600 h/s) = 25 m/s
(a) For the time it takes him to react (0.85 s), Allen is moving at a constant speed of 25 m/s, so that before he actually does anything, he covers a distance of
(25 m/s) • (0.85 s) = 21.25 m
(b) Once he presses the brakes, Allen's vehicle covers a distance x in time t of
x = 21.25 m + (25 m/s) t + 1/2 a t²
and has a speed v of
v = 25 m/s + a t
It takes him 3.5 s to come to a full stop. Use this to find the acceleration:
0 = 25 m/s + a (3.5 s)
a = - (25 m/s) / (3.5 s)
a ≈ - 7.1 m/s²
After 3.5 s, he will have traveled a total distance of
x = 21.25 m + (25 m/s) (3.5 s) + 1/2 (- 7.1 m/s²) (3.5 s)²
x = 152.5 m ≈ 150 m
(c) This one is worded a bit strangely, specifically "once he presses the brake" seems to suggest instantaneous acceleration, not average. Average acceleration is defined for some duration of time. You're probably expected to report the acceleration of the car as it comes to a stop, which we found earlier to be
a ≈ - 7.1 m/s²
Assignment 1. Define Couple?
If electrons are added to an object the object becomes
charge.
negative
positive
neutral
nothing happens
Answer:
It can be either, depending on how much and how you add the electrons
a strong inductive argument must have true premises
True
False
we can see objecf in bright room because
Answer:
because there is a reflection due to light in the the room.
Explanation:
if if it's helped you please mark as brainliest and like and follow please
We can see objects in a bright room because, the object reflect the light falling on them
A force of 3N acts on 90degree to a force of 4N.find the magnitude and direction of the resultant R.
R=A2+B2+2ABcosβ−−−−−−−−−−−−−−−−√R=A2+B2+2ABcosβ
A=4NA=4N , B=3NB=3N , β=90°β=90° , cosβ=0cosβ=0
R=A2+B2−−−−−−−√R=A2+B2
R=42+32−−−−−−√=25−−√=5NR=42+32=25=5N
tanα=Bsin90°A+Bcos90°=34tanα=Bsin90°A+Bcos90°=34
α=37°α=37°
Therefore the resultant of the two forces has a magnitude of 5N5N and is at an angle of 37°37° with respect to
e.) calculate the total heat required to change 10kg of ice at 0°c into water at 50°c.
Answer:
the heat change to 10kg of ice to water 0
so quantity of heat required is answer: 5460 J.
hope its helps!
Answer:
Q=ml+mc∆+ml'
=10*80+10*1*(100-0)+10*540
=800+1000+5400
=7200cal.7.2kcel
L=heat of fusion of ice
L'=heat of vapourisation of water
Amoving object is in equilibrium. Which best describes the motion of the object if no forces change?
a. It will change directions.
b. It will slow down and stop.
c. It will maintain its state of motion.
d. It will speed up and then slow down.
22/22 is an even number greater than 20?
Answer:
yes
Explanation:
becuse it is an even that is 2 number then 20
Answer:
Yes
Explanation:
It is because 22 is greater than 20
what is the diffrence between scalar and vector physical quantites write all examples that you know about each of them
Answer:
Scalar quantity is defined as the physical quantity with magnitude and no direction.
There are plenty of scalar quantity examples, some of the common examples are:
Mass
Speed
Distance
Time
Area
Volume
Density
Temperature
A vector quantity is defined as the physical quantity that has both direction as well as magnitude.
Vector quantity examples are many, some of them are given below:
Linear momentum
Acceleration
Displacement
Momentum
Angular velocity
Force
Electric field
Polarization
A 6.0 kg object is moving at 5.0 m/s along the x axis in the positive direction. It collides
with and sticks to a 2.0 kg object moving also along the x axis. After the collision the
composite object is moving 2.0 m/s along the x axis in the negative direction.
Determine the velocity (magnitude and direction) of the 2.0 kg object before the collision.
Answer:
V2 = 23 [m/s] to the left.
Explanation:
In order to solve this problem, we must use the definition of conservation of linear momentum. That is, the momentum is conserved before and after the collision. The values before the collision will be taken to the left of the equality, and the values after the collision will be taken to the right of the equality, in this way we have:
Σbefore = Σafter
ΣPbefore = ΣPafter
where:
P = m*v
The positive momentum will be taken to the right and the negative momentum is to the left in this way we formulate the following equation:
[tex](m_{1}*v_{1}) + (m_{2}*v_{2})=-(m_{1} +m_{2})*v_{3}\\[/tex]
where:
m1 = mass of the first object = 6 [kg]
v1 = velocity of the first object = 5 [m/s]
m2 = mass of the stick = 2 [kg]
v2 = velocity of the stick [m/s]
v3 = velocity of the composite object = - 2 [m/s]
(6*5) + (2*V2) = - (6 + 2)*2
30 + (8*2) = - 2*V2
46 = - 2*V2
V2 = - 23 [m/s]
Note: the negative sign means the stick moves to the left
metal sphere having an excess of +5 elementary charges has a net electric charge of
Answer:
[tex]q=+8.01\cdot 10^{-19}\ coulombs[/tex]
Explanation:
Elementary charge
The elementary charge, denoted by the symbol e is the electric charge carried by a proton or, equivalently, the magnitude of a negative electric charge carried by an electron, which has charge −e.
The value of the elementary charge is a fundamental constant in physics:
[tex]\mathbf{e}=1.60217662 \cdot 10^{-19}\ coulombs[/tex]
If a metal sphere has an excess of +5 elementary charge, then it has a net charge of:
[tex]q=5*\mathbf{e}=+5*1.60217662 \cdot 10^{-19}\ coulombs[/tex]
[tex]\boxed{q=+8.01\cdot 10^{-19}\ coulombs}[/tex]
what are the applications of measurement?
If object A exerts a force on another body B, then body B exerts an equal and opposite for on body A. Which newton is this?
Answer:
newton 's 3rd law which states that to every action there's equal but opposite reaction
If you were to pour 4 different liquids into a glass and the liquids separated into 4 layers, which liquid would have the highest density?
A. the layer at the bottom of the glass
B. the third layer from the bottom of the glass
C. the layer at the top of the glass
D. the second layer from the bottom of the glass
Answer:
A.
Explanation:
The liquid with the higher density will be found at the bottom of the glass because is the heaviest of all. The liquids which have a lower density than the liquid at the bottom of the glass will be found at the top of the glass. We assume that the liquids are not soluble in each other because otherwise they will be mixed.
what is E=mc^2?
please explain..
Answer:
In physics, mass–energy equivalence is the principle that anything having mass has an equivalent amount of energy and vice ve
5
Find the resultant of the following four forces.
A= 25 N due north
B = 50 N due east
C= 30 N due west
D=40 N due south
25+50-30-40 = 0
north and east are positive directions, south and west are negative
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A block is 10cm long, 5cm wide and 2cm high and weighs 100g. What is the volume of the block? What is the density?
Answer:
1gm/cm^3
Explanation:
its the answer