If a student cycles along a level road at a speed of 5.0 m / s. The total mass of the student and bicycle is 120 kg. The student applies the brakes and stops. The braking distance is 10 m, then the average braking force would be 150 Newtons.
What is power?The rate of doing work is known as power. The Si unit of power is the watt.
Power =work / time
Work done by the braking force = change in kinetic energy
F × s = 1/2 × m × v²
F = 0.5 x 120 x 5² / 10
F = 150 Newtons
Thus, the average braking force would be 150 Newtons.
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Timothy wants to know how far his math class is from the orange tree across the street from the school. His feet are ideal feet (meaning they are 1 foot long. 1 foot is 12 inches). Timothy finds that the orange tree is 159 feet from the door of the math classroom. He wants to know that distance in kilometers (km).a. Convert from feet to inches (1 ft =12 in)b. Convert from inches to centimeters (1 in =2.54c. Conver from centimeters to meters (1m = 100cm)d. Convert from meters to kilometers (1km=1000m)
a) 1 foot = 12 inches
159 feet = 159 x 12 = 1908
The distance in inches is 1908 inches
b) 1 inch = 2.54 cm
1908 inches = 1908 x 2.54 = 4846.32
The distance in centimeters is 4846.32 cm
c) 100 cm = 1 m
4846.32 cm = 4846.32/100 = 48.4632
The distance in meters is 48.4632 m
d) 1000m = 1 km
48.4632 m = 48.4632/1000 = 0.0484632
The distance in kilometers is 0.0484632 km
Which of these is a property of an electromagnetic wave? A)magnetic and electric fields oscillate perpendicular to each other but not to the velocity of the wave B)transports energy C)has a magnetic wave but no electric wave
Electromagnetic waves :
- are transverse waves
- Can travel through a vacuum
- Tranport energy from one place to another
- can be reflected
-can be refracted
Correct options:
B)transports energy
Yea thanks thank you for the info thanks
To help us solve this problem let's plot the points given in the table:
From the graph we notice that this the position can be modeled by a sine function, we also notice that the period of this function is 8. We know that a sine function can be modeled by:
[tex]A\sin(B(x+C))+D[/tex]where A is the amplitude, C is the horizontal shift, D is the vertical shift and
[tex]\frac{2\pi}{B}[/tex]is the period.
From the graph we have we notice that we don't have any horizontal or vertical shift, then C=0 and D=0. We also notice that the amplitude is 15, then A=15. Finally, as we said, the period is 8, then:
[tex]\begin{gathered} 8=\frac{2\pi}{B} \\ B=\frac{2\pi}{8} \\ B=\frac{\pi}{4} \end{gathered}[/tex]Plugging these values in the sine function we have:
[tex]x(t)=15\sin(\frac{\pi}{4}t)[/tex]If we graph this function along the points on the table we get the following graph:
We notice that we don't get an exact fit but we get a close one.
Now, that we have a function that describes the position we can find the velocity by taking the derivative:
[tex]\begin{gathered} x^{\prime}(t)=\frac{d}{dt}\lbrack15\sin(\frac{\pi}{4}t)\rbrack \\ =\frac{15\pi}{4}\cos(\frac{\pi}{4}t) \end{gathered}[/tex]Therefore, the velocity is:
[tex]x^{\prime}(t)=\frac{15\pi}{4}\cos(\frac{\pi}{4}t)[/tex]Once we have the expression for the velocity we can find values for the times we need, they are shown in the table below:
From the table we have that:
[tex]x^{\prime}(0.5)=10.884199\text{ cm/s}[/tex]And that:
• The earliest time when the velocity is zero is 2 s.
,• The second time when the velocity is zero is 6 s.
,• The minimum velocity happens at 4 s.
,• The minimum velocity is -11.780972 cm/s
An aluminum rod and a nickel rodare both 5.00 m long at 20.0°C.The temperature of each is raisedto 70.0°C. What is the differencein length between the two rods?AluminumNickela = 23.10-6C+ B = 69.10-6 0-1a = 13.10-6C1 B = 39.10-6-1(Unit = m)Enter
The difference in length between the two rods = 0.0025m
Explanations:Linear expansivity of a material is given by the formula:
[tex]\alpha\text{ = }\frac{l_2-l_1}{l_1(\theta_2-\theta_1)}[/tex]For the Aluminium rod:
[tex]\begin{gathered} l_{A1}\text{ = 5.0m} \\ \theta_{A1}=20^0C \\ \theta_{A2}=70^0C \\ \alpha_A\text{ = }23\times10^{-6}C^{-1} \\ \alpha_A\text{ = }\frac{l_{A2}-l_{A1}}{l_{A1}(\theta_{A2}-\theta_{A1})} \\ \text{ }23\times10^{-6}\text{ = }\frac{l_{A2}-5}{5(70-20)} \\ 5\times50\times\text{ }23\times10^{-6}=\text{ }l_{A2}-5 \\ l_{A2}=\text{ (}5750\text{ }\times10^{-6})\text{ + 5} \\ l_{A2}=\text{ 0.00575+5} \\ l_{A2}=\text{ 5.00575m} \end{gathered}[/tex]For the Nickel rod:
[tex]\begin{gathered} l_{N1}\text{ = 5.0m} \\ \theta_{N1}=20^0C \\ \theta_{N2}=70^0C \\ \alpha_N=\text{ 13}\times10^{-6}C^{-1} \\ \alpha_N\text{ = }\frac{l_{N2}-l_{N1}}{l_{N1}(\theta_{N2}-\theta_{N1})} \\ \text{ 1}3\times10^{-6}\text{ = }\frac{l_{N2}-5}{5(70-20)} \\ 5\times50\times\text{ 1}3\times10^{-6}=\text{ }l_{A2}-5 \\ l_{N2}=\text{ (32}50\text{ }\times10^{-6})\text{ + 5} \\ l_{N2}=\text{ }0.00325+5 \\ l_{N2}=\text{ 5.00325m} \end{gathered}[/tex]The difference in length between the two rods will be given as:
[tex]\begin{gathered} l_{A2}-l_{N2}=\text{ 5.00575-5.00325} \\ l_{A2}-l_{N2}=0.0025m \end{gathered}[/tex]The difference in length between the two rods = 0.0025m
How much work is done on a medicine ball with a force of 29 newtons when you lift it 5 meters?
Given data
*The given force is F = 29 N
*The given distance is s = 5 m
The formula for the work is done on a medicine ball is given as
[tex]W=F\mathrm{}s[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} W=(29)(5) \\ =145\text{ J} \end{gathered}[/tex]Hence, the work is done on a medicine ball is W = 145 J
A dentist causes the bit of a high speed drill to accelerate from an angular speed of 1.76 x 10^4 rads to an angular speed of 4.61 x 10^4 rat. In the process, the bit turns through 1.97 x 10 ^4 rad. Assuming a constant angular acceleration, how long would it take the reach its maximum speed of 7.99 x 10^4 rads starting from rest?
The time taken for the bit to reach the maximum speed is 1.35 seconds.
What is the angular acceleration of the bit?The angular acceleration of the bit is determined by applying the following kinematic equation as shown below.
ωf² = ωi² + 2αθ
where;
ωf is the final angular speedωi is the initial angular speedθ is the angular displacementα is the angular accelerationα = (ωf² - ωi²)/2θ
α = (46,100² - 17,600²) / (2 x 19,700)
α = 46,077.4 rad/s²
The time taken for the bit to reach the maximum speed is calculated as follows;
ωf = ωi + αt
t = (ωf - ωi) / α
t = (79,900 - 17,600) / (46,077.4)
t = 1.35 seconds
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Assume a water strider has a roughly circular foot of radius 0.0203 mm. The surface tension of water is 0.0700 N/m.A. What is the maximum possible upward force on the foot due to surface tension of the water? NB. What is the maximum mass of this water strider so that it can keep from breaking through the water surface? The strider has six legs. mg
Part (A)
The maximum possible upward force acting on the foot is,
[tex]F=2\pi r\sigma[/tex]Substitute the known values,
[tex]\begin{gathered} F=2(3.14)(0.0203\text{ mm)(}\frac{10^{-3}\text{ m}}{1\text{ mm}})(0.0700\text{ N/m)} \\ =8.9\times10^{-6}\text{ N} \end{gathered}[/tex]Thus, the maximum possible upward force on the foot is
[tex]8.9\times10^{-6}\text{ N}[/tex]Part (B)
The maximum force due to six legs can be expressed as,
[tex]6F=mg[/tex]Substitute the known values,
[tex]\begin{gathered} 6(8.9\times10^{-6}N)=m(9.8m/s^2) \\ m=\frac{6(8.9\times10^{-6}\text{ N)}}{9.8m/s^2}(\frac{1kgm/s^2}{1\text{ N}}) \\ =(5.45\times10^{-6}\text{ kg)(}\frac{1\text{ mg}}{10^{-6}\text{ kg}}) \\ =5.45\text{ mg} \end{gathered}[/tex]Thus, the maximum mass of water strider is 5.45 mg.
A 60.0 kg skier with an initial speed of 14 m/s coasts up a 2.50 m high rise as shown in the figure.
Find her final speed right at the top, in meters per second, given that the coefficient of friction between her skis and the snow is 0.38?
The final speed of the skier at the top mountain is determined as 9.27 m/s.
What is the change in the energy of the skier?
The change in the energy of the skier due to frictional force is calculated as follows;
ΔP.E = Pi + Ef
where;
Pi is the initial potential at the topEf is the energy lost to frictionThe distance of the plane travelled is calculated as;
sin35 = 2.5/L
L = 2.5 / sin35
L = 4.36 m
ΔP.E = mghi - μmgcosθ(L)
where;
m is the masshi is the initial heightg is acceleration due to gravityμ is coefficient of frictionΔP.E = (60 x 9.8 x 2.5) - (0.38)(60)(9.8) cos(35) x (4.36)
ΔP.E = 671.98 1 J
The final speed of the skier at the top of the plane;
P.E = K.E
P.E = ¹/₂mv²
v² = 2P.E /m
v = √(2P.E /m)
v = √(2 x 671.98) / 60)
v = 4.73 m/s
Total speed = -4.73 m/s + 14 m/s = 9.27 m/s
Thus, due to frictional force opposing the upward motion of the skier, the final speed at the top will be smaller than the initial speed.
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Two equal charges q1=q2= -6uC are on the y-axis at y1=3cm and y2= -3cm. What is the magnitude and direction of the electric field on the x-axis at x=4cm. If a test charge q0=2uC is placed at x =4cm find the force the test charge experiences?
The electric field charges, q₁, and q₂, which are each -6×10⁻⁶ μC, gives:
First part:
The magnitude of the electric field at x = 4 cm is -3.456×10⁷ N/CThe direction of the electric field is towards the origin, along the x-axisSecond part:
The force experienced by the charge is 69.12 NWhat is an electric field?An electric field is the field around a particle that is electrically charged and which exerts a force on charged particles within the field.
The given information are:
The electric charges, q₁ = q₂ = -6 μC
The location of the charge q₁ = y₁ = 3 cm on the y-axis
Location of the charge q₂ = y₂ = -3 cm
First part:
The required location of the point where the electric field magnitude and direction is required is x = 4 cm
The electric field formula is: [tex]\displaystyle{E = \frac{k\cdot q}{r^2}[/tex]
Where:
k = The electrostatic constant ≈ 9 × 10⁹ N·m²/C²
The distances, r, of the charges from the required point are therefore obtained using Pythagorean theorem as follows:
r = √(3² + 4²) = 5
r = 5 cm = 0.05 m
Which gives;
[tex]\displaystyle{E = \frac{9 \times 10^9\times (-6) \times 10^{-6}}{(0.05)^2} = -2.16\times 10^{7}[/tex]
Given that the magnitude of the electric field along the y-axis cancel out, the magnitude of the electric field along the x-axis is found as follows:
[tex]E_x = 2 \times -2.16\times 10^{7}\times \dfrac{4}{5} = -3.456 \times 10^7[/tex]
The magnitude of the electric field at x = 4 is -3.456 × 10⁷ N/C
Second part: The magnitude of the test charge is q₀ = 2 × 10⁻⁶ μC
The force of an electric field, F = E × q
The force experienced by the test charge is therefore:
F = -3.456 × 10⁷ × 2 × 10⁻⁶ = -69.12
The force the test charge experiences is 69.12 N acting towards the origin from the point x = 4 cm.
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What is the mass of 12 m3 of methylated spirit whose relative density is 0.8 ? ( Hint the density of water = 1000 kgm-3 ) A. 9600 kg B.9400 kg C. 8600 kg
ANSWER
A. 9600 kg
EXPLANATION
Given:
• The volume of the substance, V = 12m³
,• The relative density of the substance, 0.8
,• The density of water, 1000 kg/m³
Unknown:
• The mass of the substance, m
The relative density of a substance is defined as,
[tex]\rho_{relative}=\frac{\rho_{substance}}{\rho_{water}}[/tex]And the density of a substance is,
[tex]\rho=\frac{m}{V}[/tex]Let's solve the first formula for the density of the substance,
[tex]\rho_{substance}=\rho_{relative}\cdot\rho_{water}=0.8\cdot1000\operatorname{kg}/m^3=800\operatorname{kg}/m^3[/tex]Then, solve the second formula for m,
[tex]m=\rho\cdot V=800\operatorname{kg}/m^3\cdot12m^3=9600\operatorname{kg}[/tex]Hence, the mass of this substance is 9600 kg
A train is traveling down a straight track at 26 m/s when the engineer applies the brakes, resulting in an acceleration of −1.0 m/s2 as long as the train is in motion. How far does the train move during a 52-s time interval starting at the instant the brakes are applied?______ m
In order to calculate the distance the train will move, we can use the formula below:
[tex]\Delta S=V_0t+\frac{at^2}{2}[/tex]Where V0 is the initial speed, t is the time and a is the acceleration.
Since the initial speed is 26 m/s and the train acceleration is -1 m/s², the train will completely stop after 26 seconds. In the remaining 26 seconds to complete the total of 52, the train will be stopped already, so there is no displacement.
Because of that, we will use a time t = 26 seconds.
So, using V0 = 26 m/s and a = -1 m/s², we have:
[tex]\begin{gathered} \Delta S=26\cdot26+\frac{(-1)26^2}{2}\\ \\ \Delta S=676-338\\ \\ \Delta S=338\text{ m} \end{gathered}[/tex]Therefore the train will move 338 meters.
How much kinetic energy does Usain Bolt (m=94kg) have when he hits his top
speed of 12 m/s?
Answer:
6768 Joules (J)
Explanation:
kinetic energy = 1/2mv^2
1/2 (94x12^2) = 6768
carts, bricks, and bands
10. Predict the acceleration that would occur if four rubber bands were used to pull a cart loaded with two bricks.
a. Approximately 0.16 m/s2
b. Approximately 0.50 m/s2
c. Approximately 0.64 m/s2
d. Approximately 1.00 m/s2
D. The acceleration that would occur if four rubber bands were used to pull a cart loaded with two bricks is approximately 1 m/s².
What is acceleration?The acceleration of an object is the rate of change of velocity of the object with time.
The acceleration that would occur when four rubber bands were used to pull a cart loaded with two bricks, is determined by applying Newton's second law of motion as follows.
a = F/m
where;
a is accelerationF is the applied forcem is the massLet the mass of a brick = mass of a band = m
the mass of a cart = 2 bricks = 2m
a = (force applied by 4 rubber) / (mass of 1 cart + mass of 2 brick)
a = (4m) / (2m + 2m)
a = (4m)/(4m)
a = 1 m/s²
Thus, the acceleration that would occur if four rubber bands were used to pull a cart loaded with two bricks is approximately 1 m/s².
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what is the general importance of water?
100 POINTS
A man pulls a crate with a rope. The crate slides along the floor in the horizontal direction (x direction). The man exerts a force of 50 N on the rope, and the rope is at an angle . Describe how the force components change as the angle increases from 0° to 90° and use your graph to explain your answer. Give a detailed explanation of the forces at . Show a sample calculation at one angle for both components.
The exerted 50 N force at an angle on the crate can be resolved into a horizontal and vertical force component in which the horizontal component, Fₓ, decreases, while the vertical force component, [tex]F_y[/tex], increases as the value of the angle formed by the rope, θ, increases from 0° to 90°.
What is a component of a force?The components of a force are the force parts acting in perpendicular directions, horizontal and vertical, that combine to give the specified force.
The direction the crate is sliding = The horizontal, x-direction
The force exerted by the man = 50 N
The angle of the rope = θ
The components of the force are therefore:
Horizontal component, Fₓ = 50 × cos(θ)
Vertical component, [tex]F_y[/tex] = 50 × sin(θ)
The value of cos(θ) and sin(θ) as the angle the rope makes with the horizontal, θ, increases from 0° to 90° are as follows:
[tex]\begin{center} \begin{tabular}{ |c|c |c |} \theta & cos(\theta) & sin(\theta) \\ 0^{\circ} & 1 & 0 \\ 30^{\circ} & \sqrt{3} /2 & 0.5 \\ 45^{\circ}&\sqrt{2}/2 &\sqrt{2}/2 \\ 60^{\circ}&0.5&\sqrt{3}/2 \\ 90^{\circ} &0&1\\ \end{tabular}[/tex]
Therefore, the horizontal component of the force exerted by the man, Fₓ has a maximum value at θ = 0, and decreases to 0, as θ increases from 0° to 90°.
The vertical component of the force exerted, [tex]F_y[/tex], has a minimum value of 0 at θ = 0°, and the value of sin(θ) and therefore [tex]F_y[/tex], increases to a maximum of (sin(90°) = 1) 50 N as increases to 90°.
Please find attached the graph showing the components of the force, Fₓ, and [tex]F_y[/tex], exerted by the man as the angle formed by the rope increases from 0° to 90°
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What is the resistance in a circuit that has a current of 2.5A and a voltage of 40v
16 ohms
Explanation
Ohm's law relates the strength of a direct current is directly proportional to the potential difference and inversely proportional to the resistance of the circuit,it is given by the expresssion
[tex]\begin{gathered} V=IR \\ if\text{ we isolate R} \\ R=\frac{V}{I} \end{gathered}[/tex]then
Step 1
a) Let
[tex]\begin{gathered} I=2.5\text{ A} \\ V=40\text{ V} \end{gathered}[/tex]b)replace in the formula
[tex]\begin{gathered} R=\frac{V}{I} \\ R=\frac{40V}{2.5A} \\ R=16\text{ ohms} \end{gathered}[/tex]therefore, the answer is 16 ohm
I hope this helps you
what happens to the state of a variable if it goes through a two series connected NOT gate.
what is the mass on grams of 0.56 moles of NaCl
Answer:
1 mole of Na = 23 g
1 mole of Cl = 35 g
1 mole of NaCl = 58 g
.56 * 58 g = 32.5 g
Whats the percent of 10 of 20
In order to determine the associated percent of 10 related to 20, proceed as follow:
If x is the percentage, then, you can write:
[tex]\frac{x}{100}\cdot20=10[/tex]which means that x percentage of 20 is equal to 10. By solving for x, you get:
[tex]\begin{gathered} x=\frac{10}{20}\cdot100 \\ x=50 \end{gathered}[/tex]Hence, 10 is the 50% of 20
Answer:2
Explanation:multiply 0.20 times 10 you get it
Suppose a 345-g kookaburra (a large kingfisher bird) picks up a 75-g snake and raises it 2.1 m from the ground to a branch.How much work, in joules, did the bird do on the snake? How much work, in joules, did it do to raise its own center of mass to the branch?
The work done in each case can be calculated with the change in potential energy of the body:
[tex]Work=m\cdot g\cdot h[/tex]The work done by the bird on the snake will use only the mass of the snake (in kg):
[tex]\begin{gathered} Work=0.075\cdot9.8\cdot2.1\\ \\ Work=1.5435\text{ J} \end{gathered}[/tex]The work done by the bird to raise its own center of mass will use only the bird mass (in kg):
[tex]\begin{gathered} Work=0.345\cdot9.8\cdot2.1\\ \\ Work=7.1\text{ J} \end{gathered}[/tex]The block of a mass 10.2 kg is sliding at an initial velocity of 3.40 m/s in the positive x-direction. The surface has a coefficient of kinetic friction of 0.153.
m = mass = 10.2 kg
vo = initial velocity = 3.40 m/s
u = coefficient of kinetic friction = 0.153
g= gravity = 9.8m/s^2
a)
Fr = force of kinetic friction = u m g
Fr = 0.153 x 10.2 x 9.8 = -15.30N
b) Block's acceleration
Newton's second law of motion:
F = m*a
a = F/m = -15.30 / 10.2 = -1.5 m/s^2
c) USe the third equation of motion:
2as = vf^2 - vo^2
Where:
Vf= final velocity= 0 m/s
s = displacement
2 * -1.5 * s = 0^2 - 3.40^2
-3s = -11.56
s= -11.56/-3
s= 3.85 m
Neptune circles the Sun at a distance of 4.50 × 1012 m once every 164 years. Saturn circles the Sun at a distance of 1.43 × 1012 m. What is the orbital period of Saturn?
The orbital time period of the Saturn is 29.6 years
We are given that,
Distance from Sun to Saturn is = a = 1.43 × 10¹²
The mass of the Sun is = M =1.99 × 10³⁰kg
The Gravitational constant = G = 6.67 × 10⁻¹¹N-m²kg⁻²
To find the orbital period of Saturn we can use the equation ,
[tex]T^{2} = \frac{4\pi }{GM}a^{3}[/tex]
Where, T is the orbital time period of the of the Saturn , M is the mass of the sun , G is the gravitational constant.
Therefore, after putting the value in above equation we can get,
[tex]T^{2} = \frac{4(\(3.14)^{2} }{(6.67*10)^{-11} )N-m^{2} kg^{-2}}(1.43*10^{12}) ^{3}m[/tex]
[tex]T^{2} = \sqrt{8.688*10^{17} } s[/tex]
[tex]T = 932094415.818s[/tex]
So that , from above to convert the orbital time period of Saturn from second into year i.e. above seconds divided by seconds (1 sec = 3.154 ×10⁷ Earth years)
Thus, the orbital time period can be ,
[tex]T = 29.6 years[/tex]
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How many moles of a gas sample are in a 5.0 L container at 251 K and 370 kPa?(The gas constant is8.31L kPamol K)Round your answer to one decimal place and enter the number only with no units.
Given:
The volume of the gas, V=5.0 L
The temperature of the gas, T=251 K
The pressure of the gas, P=370 kPa
The gas constant, R=8.31 L kPa/(mol K)
To find:
The moles of the gas sample.
Explanation:
From the ideal gas equation,
[tex]PV=\text{nRT}[/tex]Where n is the moles of the gas present.
On substituting the known values,
[tex]\begin{gathered} 370\times5.0=n8.31\times251 \\ \Rightarrow n=\frac{370\times5.0}{8.31\times251} \\ =0.9\text{ mol} \end{gathered}[/tex]Final answer:
The moles of the gas present in the sample is 0.9 mol
Writing Simple ExpressionsChoose all of the TRUE statement(s).Add 6 and 5, then multiply by 4 is the same as 4(6 + 5).4 times greater than 80 + 25 is the same as 4 x (80 + 25).Subtract 15 from 42 is the same as 15 − 42.9 times greater than 11 + 12 is the same as 9 + 11 + 12.8 times greater than 21 + 15 is the same as 8(21 + 15).
Add 6 and 5, then multiply by 4 is the same as 4(6 + 5). TRUE
4 times greater than 80 + 25 is the same as 4 x (80 + 25). TRUE
Subtract 15 from 42 is the same as 15 − 42. TRUE
9 times greater than 11 + 12 is the same as 9 + 11 + 12. FALSE
9 (11+12)
8 times greater than 21 + 15 is the same as 8(21 + 15). TRUE
Calculate the force between charges of 5.0x10^-8c and 1.0x10^-7c if they are .15m apart
Given data:
[tex]\begin{gathered} Q_1=5.0\times10^{-8}\text{ C} \\ Q_2=1.0\times10^{-7}\text{ C} \\ r=0.15\text{ m} \end{gathered}[/tex]The electrostatic force is given as,
[tex]F=\frac{KQ_1Q_2}{r^2}[/tex]Here, K is electrostatic force constant.
Substituting all known values,
[tex]\begin{gathered} F=\frac{9\times10^9\times5\times10^{-8}\times1\times10^{-7}^{}}{(0.15)^2} \\ =2\times10^{-3}\text{ N} \end{gathered}[/tex]Therefore, the force between the charges is 2×10^(-3) N. Hence, option (d) is the correct choice.
An object is placed 15cm in front of a convex mirror and an image is produced 5cm behind the mirror. Calculate the focal length of the mirror
The inverse of the optical power of an optical system, the focal length provides a measurement of how strongly the system converges or diverges light. A system's light converges when the focal length is positive, while it diverges when the focal length is negative.
A diverging mirror with the reflective surface bulging towards the direction of the light source is referred to as a convex mirror. Since they reflect light outward, they are not employed to focus light. Convex mirrors create an image that is smaller than the object and grows larger as it approaches closer to the mirror.
Given that the image of an object is generated v cm from a spherical mirror with a focal length of f, and that the object is situated u cm in front of the mirror, u, v, and f are connected by the equation 1/f=1/u + 1/v. The mirror formula is the term used to describe this equation. Both concave and convex mirrors can be used using the formula.
Therefore using, 1/f = 1/u + 1/v
1/f = 1/15 + 1/5
1/f = (1 + 3) / 15
f = 15/4 = 3.75cm
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If the electric intensity between two parallel plates, placed 1cm apart is 104 NC-1and the direction of the field of this intensity is vertically upward. Find the force on an electron in this field and compare the force on the electron with the electron’s weight
The force of a charge in an electric field is:
[tex]\vec{F}_e=q\vec{E}[/tex]In this case we know the electric field is:
[tex]\vec{E}=104\hat{j}[/tex]and that the charge is that of the electron, then we have:
[tex]\begin{gathered} \vec{F}_e=-1.6\times10^{-19}(104\hat{j}) \\ \vec{F}_e=-1.664\times10^{-17}\text{ N} \end{gathered}[/tex]Therefore, the magnitude of the force is
[tex]1.664\times10^{-17}\text{ N}[/tex]and in points down.
The weight of the electron is:
[tex]\begin{gathered} W=1.67\times10^{-27}(9.98) \\ W=1.6366\times10^{-26} \end{gathered}[/tex]Making the quotient between the force we have:
[tex]\frac{1.664\times10^{-17}}{1.6366\times10^{-26}}=1.02\times10^9[/tex]Therefore, the electric force is approximately 1e9 times the weight.
What letter from the picture below represents the position of the maximum kinetic energy?
Given that a pendulum has a mean position as C, and two extreme points A and E.
We have to find the position of kinetic energy.
Here, the total energy is conserved. So, the sum of potential energy and kinetic energy is constant.
Potential energy increases with the increase in height.
At the extreme positions, A and E, potential energy are maximum and potential energy is zero at point C.
Also, Kinetic energy is zero is at points A and E.
As energy is conserved, Kinetic energy is maximum at point C and potential energy is zero.
A small mailbag is released from a helicopter that is descending steadily at 2.82 m/s.(a) After 3.00 s, what is the speed of the mailbag?v = m/s(b) How far is it below the helicopter?d = m(c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 2.82 m/s?v = m/sd = m
a)
When the package is released from the moving helicopter, the package and the helicopter has a common speed. The package is in freefall. We would calculate the speed of the helicopter after a given time t by applying the formula,
v = vo + gt
where
vo is the initial velocity of of the package and it is equal to the speed of the helicopter
v is the final velocity of the package after time t
g is th acceleration due to gravity
From the information given
vo = 2.82
t = 3
g = 9.8
Thus,
v = 2.82 + 9.8 * 3 = 2.82 + 29.4
v = 32.22 m/s
After 3.00 s, the speed of the mailbag is 32.22 m/s
b) We want to calculate the distance covered by the mailbag in 3 s. We would apply the formula which is expressed as
s = vot + 1/2 x g x t^2
where
s is the distance
vo = 2.82
g = 9.8
t = 3
Thus,
s = 2.82 x 3 + 1/2 x 9.8 x 3^2 = 8.46 + 44.1
s = 52.56 m
Since we want to calculate the distance from the helicopter, we would calculate the diatance that the helicopter also travelled downwards in 3 s by applying the formula for calculating distance which is expressed as
distance = speed x time
Thus
distance = 2.82 x 3 = 8.46 m
Difference in distance = 52.56 - 8.46 = 44.1
The package is 44.1 m from the helicopter
c) If the helicopter is moving upwards, it would be thrown out and it would attain a certain height before it starts descending. The height is calculated by the formula,
h = vo^2/2g
By substituting the values,
h = 2.82^2/2 x 9.8 = 0.406 m
When the mail bag attains this height, it will start moving downwards. At this height, the final velocity is zero. We would calculate the time taken to attain this height by applying the formula,
v = vo - gt
v = 0
Thus,
0 = 2.82 - 9.8 x t
9.8t = 2.82
t = 2.82/9.8 = 0.288
The time left for freefall within the first 3 seconds is
3 - 0.288 = 2.712 s
The height attained by the mailbag in 2.712s is calculated by the formula,
h = gt^2/2
h = 9.8 x 2.712^2/2 = 36.04 m
Distance travelled by helicopter by ascending upward in 3 s is
distance = 2.82 x 3 = 8.46
Height of mailbag from final position after 3 seconds is
36.04 - 0.406 = 35.634
Difference in distance = 35.634 + 8.46 = 44.094
The package is 44.094 m from the helicopter
For the velocity of the mailbag after 3 seconds,
v = - vo + gt
v = - 2.82 + 9.8 x 3 = - 2.82 + 29.4
v = 26.54 m/s
A mass of 0.520 kilograms is attached to a spring and lowered into a resting position, causing the spring to stretch 18.7 centimeters. The mass is then lifted up and released. a. What is the spring constant of the spring? Include units in your answer.b. What is the frequency of its oscillation? Include units in your answer.Answer must be in 3 significant digits.
Given data
*The given mass is m = 0.520 kg
*The spring stretches at a distance is x = 18.7 cm = 0.187 m
*The value of the acceleration due to gravity is g = 9.8 m/s^2
(a)
The formula for the spring constant of the spring is given as
[tex]\begin{gathered} F=kx \\ mg=kx \\ k=\frac{mg}{x} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} k=\frac{(0.520)(9.8)}{(0.187)} \\ =27.2\text{ N/m} \end{gathered}[/tex]Hence, the spring constant of the spring is k = 27.2 N/m
(b)
The formula for the frequency of its
