Answer:
Option (c) [tex]2\times 10^{-3}\ m=2\ millimeters[/tex]
Explanation:
Here, it is required to describe the given length in a particular unit. Firstly, we need to see the following conversions as :
1 meter = 0.001 kilometers
1 meter = 10⁻⁶ megameters
1 meter = 1000 millimeters
1 meter = 1000000 micrometers
From the given option, the correct one is (c) because, 1 meter = 1000 millimeters
So, [tex]2\times 10^{-3}\ m=2\ millimeters[/tex]
Hence, the correct option is (c). Hence, this is the required solution.
2.0×10^-3 of a meter is equivalent to 2.0 millimeters, hence option c is the correct answer.
Explanation:The given length, 2.0×10^-3 of a meter, corresponds to a unit of length commonly used in the metric system. In this system, 1 meter is equal to 10^3 millimeters. Therefore, 2.0×10^-3 of a meter equates to 2.0 millimeters. This means the best answer from the provided options would be option c: 2.0 millimeters.
Learn more about Metric System here:https://brainly.com/question/25966695
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A car moves with constant velocity along a straight road. Its position is x1 = 0 m at t1 = 0 s and is x2 = 80 m at t2 = 5.0 s . Answer the following by considering ratios, without computing the car's velocity. Part A What is the car's position at
t = 2.5 s ?
Part B What will be its position at
t = 15 s ?
Answer:
Part A) x=40m at t= 2.5s
Part B) x=240m at t=15 s
Explanation:
Data:
x₁=0 , t₁=0
x₂=80m, t₂=5 s
t₃= 2.5 s
t₄= 15 s
problem development
We set the ratios x / t:
Part A)
[tex]\frac{x_{3} }{t_{3} } =\frac{x_{2} }{t_{2} }[/tex]
[tex]\frac{x_{3} }{2.5} =\frac{80}{5 }[/tex]
x₃=16*2.5
x₃=40m
Part B)
[tex]\frac{x_{4} }{t_{4} } =\frac{80 }{5 }[/tex]
[tex]\frac{x_{4} }{15} =\frac{80}{5 }[/tex]
x₄=16*15
x₄=240m
Final answer:
The car's position at t = 2.5 s will be 40 m as it is half of the interval to reach 80 m. At t = 15 s, three times the interval, the car will be at 240 m assuming constant velocity.
Explanation:
Given that the car moves from x1 = 0 m at t1 = 0 s to x2 = 80 m at t2 = 5.0 s, we can determine its position at other times under the assumption of constant velocity. Because the velocity is constant, the ratios of times to positions are constant as well.
Part A: Car's Position at t = 2.5 s
The time t = 2.5 s is exactly half of the time interval given (5.0 s), so the position should be half of 80 m, which is 40 m.
Part B: Car's Position at t = 15 s
To find the position at t = 15 s, let's first determine the time ratio. The ratio of 15 s to 5.0 s is 3:1. Since the velocity is constant, the positions scale with time. So, the position x at t = 15 s is three times 80 m, giving us 240 m.
You drop a rock into a deep well and hear the sounds of it hitting the bottom 5.50 s later. If the speed of sound is 340 m/s, determine the depth of the well.
Answer:
The depth of well equals 120.47 meters.
Explanation:
The time it takes the sound to reach our ears is the sum of:
1) Time taken by the rock to reach the well floor.
2) Time taken by the sound to reach our ears.
The Time taken by the rock to reach the well floor can be calculated using second equation of kinematics as:
[tex]h=ut+\frac{1}{2}gt^2[/tex]
where,
'u' is initial velocity
't' is the time to cover a distance 'h' which in our case shall be the depth of well.
'g' is acceleration due to gravity
Since the rock is dropped from rest hence we infer that the initial velocity of the rock =0 m/s.
Thus the time to reach the well base equals
[tex]h=\frac{1}{2}gt_1^2\\\\\therefore t_{1}=\sqrt\frac{2h}{g}[/tex]
Since the propagation of the sound back to our ears takes place at constant speed hence the time taken in the part 2 is calculated as
[tex]t_{2}=\frac{h}{Speed}=\frac{h}{340}[/tex]
Now since it is given that
[tex]t_{1}+t_{2}=5.50[/tex]
Upon solving we get
[tex]\sqrt{\frac{2h}{g}}+\frac{h}{340}=5.5\\\\\sqrt{\frac{2h}{g}}=5.5-\frac{h}{340}\\\\(\sqrt{\frac{2h}{g}})^{2}=(5.5-\frac{h}{340})^{2}\\\\\frac{2h}{g}=5.5^2+\frac{h^2}{(340^2)}-2\times 5.5\times \frac{h}{340}[/tex]
Solving the quadratic equation for 'h' we get
h = 120.47 meters.
A ball is thrown straight upward and rises to a maximum
heightof 16 m aboves its lauch point. At what height above
itslaunch point has the speed of the ball decreased to one-half of
itsinitial value?
Answer:
Explanation:
As we know that the ball is projected upwards so that it will reach to maximum height of 16 m
so we have
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
here we know that
[tex]v_f = 0[/tex]
also we have
[tex]a = -9.81 m/s^2[/tex]
so we have
[tex]0 - v_i^2 = 2(-9.81)(16)[/tex]
[tex]v_i = 17.72 m/s[/tex]
Now we need to find the height where its speed becomes half of initial value
so we have
[tex]v_f = 0.5 v_i[/tex]
now we have
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex](0.5v_i)^2 - v_i^2 = 2(-9.81)h[/tex]
[tex]-0.75v_i^2 = -19.62 h[/tex]
[tex]0.75(17.72)^2 = 19.62 h[/tex]
[tex]h = 12 m[/tex]
A physics professor is trying to flee a class of angry students by falling onto a moving freight train from the Howard and 26th St overpass. The train is y = −5 m below the overpass. It is moving at a velocity of v = 10 m/s. Taking the acceleration due to gravity to be g = −10 m/s^2, how far away should the train be before the professor falls from the overpass (and escapes the wrath of the students)?
Answer:
The train should be a 10 meters away before the professor falls
Explanation:
Calculation of the teacher's fall time:
[tex]y(t)=v_{o}t-1/2*g*t^{2}[/tex]
in this case Vo=0 and y=-5:
[tex]-5=-1/2*10*t^{2}[/tex]
[tex]t=1s[/tex]
Calculation how far away is the train when the professor falls:
The distance the train travel while the professor is falling:
[tex]d=v_{train}*t=10m/s*1s=10m[/tex]
So the train should be a 10 meters away before the professor falls
The mass of a muon is 207 times the electron mass; the average lifetime of muons at rest is 2.21 μs. In a certain experiment, muons moving through a laboratory are measured to have an average lifetime of 7.04 μs. For the moving muons, what are (a) β (b) K, and (c) p? The rest energy of the electron is 0.511 MeV.
Answer:
Given, average lifetime of muons at rest is To = 2.21μs
Average lifetime of moving muons, T = 7.04μs
mass of muon = 207 mass of electron
a)
rest mass energy of an electron is 0.511 MeV.
[tex] T =\frac{T_o}{\sqrt{1-\beta^2}}[/tex]
Substitute the values:
[tex]7.04 =\frac{2.21}{\sqrt{1-\beta^2}}\\ \beta^2 = 0.90146\\ \beta = 0.95[/tex]
b) K = (γ-1) m₀c²
[tex]\gamma = \frac{1}{\sqrt{1-\beta^2}} = 3.185[/tex]
K = (3.185-1) (207 ×9.1×10⁻³¹) (3×10⁸)² = 3.7×10⁻¹¹J
c) p = γmv = (3.185)((207 ×9.1×10⁻³¹)(0.95×3×10⁸)= 1.71×10⁻¹⁹ N/s
A charge if 2 nC is placed 10cm to the right of a conducting sphere with a diameter of 2 cm. A charge of 5 nC is placed 10 nC to the left of the same sphere. determine the charge at the center of the sphere.
Answer:
The charge at the center of the conducting sphere is zero.
Explanation:
A principle of conductor materials is that the electric field inside a conductor in electrostatic state is always zero. The gauss law says that the flux of a electric field in a closed surface is proportional to the charge enclosed by the surface. Then if the Electric field inside of a conductor is zero, imperatively the charge anywhere inside the conductor is zero too, so the charge at the center of the sphere is zero.
A car traveling at 45 m/s can brake to a stop within 10m. Assume that the braking force is effective for stop and constant with speed. If the car speeds up to 90 m/s, what is its stopping distance? Provide your explanation using principle of work and energy.
Answer:
40 m
Explanation:
A car travelling at 45 ms has a kinetic energy of
Ec = 1/2 * m * v^2
We do not know the mass, so we can use specific kinetic energy:
ec = 1/2 * v^2
In this case
ec = 1/2 * 45^2 = 1012 J/kg
If it stops in 10 m, the braking force performed a sppecific work of 1012 J/kg in 10 m
L = F * d
F = L / d
F = 1012 / 10 = 101.2 N/kg
If the car is running at 90 m/s, the specific kinetic energy of:
ec2 = 1/2 * 90^2 = 4050 J/kg
With the same braking force the braking distance is:
d2 = L2 / F
d2 = 4050 / 101.2 = 40 m
A road goes down a slope. For every 12 km measured along the ground, the road drops 500 m in elevation. a) What is the angle of slope on the road?
b) What is the map distance along the road for every 12 km actually traveled? (measured in km)
c) What is the map distance along the road for every 1 mile actually traveled? (measured in miles)
Answer:
a) 2.24°
b) 11.99 km
c) 0.9992 miles
Explanation:
We can think of the road as a triangle with a 12 km hypotenuse and a 0.5 km side. Then:
l = 12
h = 0.5
and
sin(a) = h / l
a = arcsin(h / l)
a = arcsin(0.5 / 12)
a = 2.24°
That is the slope.
The map distance travelled would be the other side of the triange.
cos(a) = d / l
d = l* cos(a)
d = 12 * cos(2.24) = 11.99 km
And for miles
d = 1 mile * cos(2.24) = 0.9992 miles
A boy uses a slingshot to launch a pebble straight up into the air. The pebble reaches a height of 37.0 m above the launch point 2.3 seconds later. Assume air resistance is negligible (a) What was the pebble's initial speed (just after leaving the slingshot)? m/s (b) How much time did it take for the pebble to first reach a height of 18.5 m above its launch point? s
The pebble's initial speed is approximately 22 m/s. It takes approximately 1.45 seconds for the pebble to first reach a height of 18.5 m.
Explanation:To determine the pebble's initial speed, we can use the equation for projectile motion:
h = v0yt - (1/2)gt2
Where h is the height, v0y is the initial vertical speed, t is the time, and g is the acceleration due to gravity.
Since the pebble is launched straight up, the final height is equal to the initial height. Plugging in the given values, we have:
37 m = v0y(2.3 s) - (1/2)(9.8 m/s2)(2.3 s)2
Simplifying this equation gives us the value of v0y, the initial vertical speed. To find the pebble's initial speed, we can use the Pythagorean theorem:
v0 = √(v0x2 + v0y2)
Where v0 is the initial speed and v0x is the initial horizontal speed. Since the pebble is launched straight up, v0x = 0. Plugging in the calculated value of v0y, we can solve for v0.
(a) The pebble's initial speed is approximately 22 m/s.
(b) To find the time it takes for the pebble to first reach a height of 18.5 m, we can use the equation for height:
18.5 m = v0yt - (1/2)gt2
Solving for t gives the time it takes for the pebble to reach the desired height.
(b) It takes approximately 1.45 seconds for the pebble to first reach a height of 18.5 m.
Two rockets are flying in the same direction and are side by side at the instant their retrorockets fire. Rocket A has an initial velocity of +5300 m/s, while rocket B has an initial velocity of +8700 m/s. After a time t both rockets are again side by side, the displacement of each being zero. The acceleration of rocket A is -11 m/s^2. What is the acceleration of rocket B?
Answer:[tex]-18.05 m/s^2[/tex]
Explanation:
Given
Initial velocity of rocket A ([tex]v_A[/tex]) 5300m/s
Initial velocity of rocket B([tex]v_B[/tex]) 8700 m/s
after time t they have displacement=0
[tex]s_A=5300\times t+\frac{1}{2}\times \left ( -11\right )t^2[/tex]
[tex]0=5300\times t+\frac{1}{2}\times \left ( -11\right )t^2[/tex]
[tex]t=\frac{5300\times 2}{11}=963.63 s[/tex]
for Rocket B
[tex]s_B=8700\times t+\frac{1}{2}\times \left ( a_B\right )t^2[/tex]
time is same for A & B
[tex]0=8700\times 963.63+\frac{1}{2}\times \left ( a_B\right )963.63^2[/tex]
[tex]a_B=\frac{2\times 8700}{963.63}=-18.05 m/s^2[/tex]
What is gravity at north pole, South pole and at different point on the equatorial regions. Give reasoning for your answers why do you think it is different or same. Can you imagine same concept for the electric charge, yes or No
Answer:
The gravity at Equator is 9.780 m/s2 and the gravity at poles is 9.832 m/s2. The gravity at poles are bigger than at equator, principally because the Earth is not totally round. The gravity is inversely proportional to the square of the radius, that is the reason for the difference of gravity (The radius at Poles are smaller than at Equator).
If Earths would have a net charge Q. The Electric field of Earth would be inversely proportional to the square of the radius of Earth (Electric field definition for a charge), the same case as for gravity. So there would be a difference between the electric field at poles and equator, too.
A baseball is thrown straight up from a building that is 25 meters tall with an initial velocity v = 10 m/s. How fast is it going when it hits the ground?
Answer:-24,5m/s
Explanation: what we have here is a UALM with these gravity as acceleration (-9.8 m/s^2). The initial position is 25 m and initial speed is 10m/s.
Speed and gravity are increasing in the opposite direction, speed upwards and gravity downwards, while the position is also upwards, depending on your reference system.
The first thing I need to know is the maximum high it will reach.
Hmax=- S(0)^2/2g=
S= speed.
0= initial
G= gravity
Hm= 100/19,6= 5.1 m
So, the ball will go 5,1 m higher than the initial position, and from there it will fall free.
Then, I need to know how long it takes to fall. For that we use UALM equation:
X(t)= X(0) + S(0)*t + (A*t^2)/2.
X: position
S: speed
A: acceleration
T:time
0: initial
0 = 25m +10*t -(9.8 * t^2)/2
Solving the quadratic equation we get
T= 3,5 sec. ( Negative value for time is impossible)
So now we know that the ball to go up and then fall needs 3,5 sec.
Let's see how long it takes to go up:
30,1=25+10*t-4,9*t^2
0=-5,1+10*t-4,9*t^2
T= 1 sec. So it will take 1 sec to the ball to reach the maximum high and 0=speed and then it'll fall during the resting 2,5 sec
Finally, to know the speed just before it touches the ground, we use the following formula:
A= (St-S0)/t
-9.8m/s^2 = (St- 0m/s)/ 2,5s
-24,5 m/s= St
-24,5 m/s is the speed at 3,5 sec, which is the time just before falling
The proton with initial v1 = 3.00 x 10^5 m/s enters a region 1.00 cm long where it is accelerated. Its final velocity is v = 6.40 x 10^6 m/s . What was its constant acceleration. Answer in 10^15 m/s^2
Answer:
The constant acceleration, [tex]a_{c} = 2.044\times 10^15 m/s^2[/tex]
Solution:
As per the question:
[tex]v_{1} = 3 x 10^5 m/s[/tex]
[tex]v_{2} = 6.40 x 10^6 m/s[/tex]
length of the re region, l = 1 cm = 0.01 m
Now,
Using the third equation of motion:
[tex]v_{2}^2 = v_{1}^2 + 2a_{c}l[/tex]
[tex]a_{c} = \frac{v_{2}^2 - v_{1}^2}{2l}[/tex]
[tex]a_{c} = \frac{(6.40 x 10^6)^2 - (3 x 10^5)^2}{2\times 0.01}[/tex]
[tex]a_{c} = 2.044\times 10^15 m/s^2[/tex]
An alert physics student stands beside the tracks as a train rolls slowly past. He notes that the frequency of the train whistle is 496 Hz when the train is approaching him and 478 Hz when the train is receding from him. Using these frequencies, he calculates the speed of the train. What value does he find? (Assume the speed of sound in air is 343 m/s.)
Answer:
6.455 m/s
Explanation:
Given;
Frequency of train whistle, f = 496 Hz
Frequency observed, f' = 478 Hz
Speed of sound in air = 343 m/s
Now,
From the instant Doppler's effect, we have the relation
[tex]f = \frac{v+v_s}{v-v_s}\times f'[/tex]
here,
v is the speed of the sound
[tex]v_s[/tex] is the speed of the train
[tex]496 = \frac{343+v_s}{343-v_s}\times478[/tex]
or
1.037 × ( 343 - [tex]v_s[/tex] ) = 343 + [tex]v_s[/tex]
or
355.91 - [tex]v_s[/tex] = 343 + [tex]v_s[/tex]
or
12.91 = [tex]2v_s[/tex]
or
[tex]v_s[/tex] = 6.455 m/s
An echo is heard from a building 0.485 s after you shout "hello." How many feet away is the building if the air temperature is 30.6°C?
Answer: 84.63 m
Explanation:
The speed of sound in the air, at [tex]0 \°C[/tex] is [tex]331.5 m/s[/tex], and for each degree Celsius the temperature rises, the speed of the sound increases by [tex]0.6 m/s[/tex]. So, if we estimate the speed of sound at [tex]30 \°C[/tex] it will be [tex]349 m/s[/tex].
On the other hand, the speed of sound [tex]V[/tex] is defined as the distance traveled [tex]d[/tex] in a especific time [tex]t[/tex]:
[tex]V=\frac{d}{t}[/tex]
Where:
[tex]V=349 m/s[/tex] is the speed of sound
[tex]t=\frac{0.485 s}{2}=0.2425 s[/tex] is half the time the sound wave travels since you say "hello", the sound wave hits the building and then returns to you
[tex]d[/tex] is the distance between you and the building
[tex]349 m/s=\frac{d}{0.2425 s}[/tex]
Finding [tex]d[/tex]:
[tex]d=(349 m/s)(0.2425 s)[/tex]
Finally:
[tex]d=84.63 m[/tex]
To find the distance to the building, calculate the speed of sound at 30.6°C, multiply it by the echo time of 0.485 seconds, and divide by 2 to account for the round trip of the sound. Then, convert the distance from meters to feet to get approximately 280.95 feet.
Explanation:To calculate the distance of the building when an echo is heard 0.485 seconds after shouting, we need to consider the speed of sound at the given air temperature of 30.6°C. Since the speed of sound varies with temperature, we use the formula v = v_0 + 0.6t, where v_0 is the speed of sound at 0°C (approximately 331.5 m/s), t is the temperature in Celsius, and v is the speed of sound at t degrees Celsius. At 30.6°C, the speed of sound is v = 331.5 m/s + 0.6(30.6°C), which calculates to roughly 349.86 m/s. Since the echo travels the distance to the building and back, the actual distance to the building is half the total distance the sound wave traveled.
Therefore, the one-way distance to the building is 0.485 s × 349.86 m/s ÷ 2. Converting meters to feet (1 meter = 3.28084 feet), we find that the building is approximately 280.95 feet away.
If an object falls twice as far, how do you think the energy acquired changes? halves
doesn't change
doubles
quadruples
Answer:
Doubles
Explanation:
Before starting to fall, the object will have energy in the form of potential energy. This will be the energy that will be transformed into kinetic energy, once the object starts its fall. And the potential energy that the object had before starting to fall, will be equal to the kinetic energy acquired during the fall. The potential energy is given by:
[tex]E_p = m*g*h[/tex]
Where m is the mass of the object, g is the gravity, and h is the distance to the floor. As you can see, The potential energy is proportional to the height of the fall, so if this height doubles, the potential energy and therefore, the kinetic energy acquired will doubles.
A 100kg cannon at rest contains a 10kgcannonball. when fired,
the cannonball leaves the cannon with avelocity of 90m/s. what is
the magnitude of the recoil speed of thecannon.
Answer:
9 m/s
Explanation:
mass of cannon, M = 100 kg
mass of cannon ball, m = 10 kg
velocity of cannon ball, v = 90 m/s
Let the recoil velocity of cannon is V.
Us ethe conservation of linear momentum, as no external force is acting on the system, so the linear momentum of the system is conserved.
Momentum before the firing = momentum after the firing
M x 0 + m x 0 = M x V + m x v
0 = 100 x V + 10 x 90
V = - 9 m/s
Thus, the recoil velocity of cannon is 9 m/s.
In reaching her destination, a backpacker walks with an average velocity of 1.32 m/s, due west. This average velocity results, because she hikes for 5.21 km with an average velocity of 3.49 m/s due west, turns around, and hikes with an average velocity of 0.687 m/s due east. How far east did she walk (in kilometers)?
Answer:
distance in east is 1273.78 m
Explanation:
given data
average velocity = 1.32 m/s west =
hike = 5.21 km = 5.21 × 10³ m
average velocity = 3.49 m/s west
average velocity = 0.687 m/s east
to find out
distance in east
solution
we consider here distance in east is = x
so distance from starting point = 5.21 × 10³ - x ...................1
and we can say time required to reach end
time required = distance / speed
time required = [tex]\frac{5.21 *10^3 - x}{1.32}[/tex] ................2
and
time required for 6.44 km west
time required = [tex]\frac{5.21 *10^3 - x}{3.49}[/tex] ................3
and time required for distance x
time required = [tex]\frac{x}{0.687}[/tex] ................4
so from equation 2 , 3 and 4
[tex]\frac{5.21 *10^3 - x}{1.32}[/tex] = [tex]\frac{5.21 *10^3 - x}{3.49}[/tex] + [tex]\frac{x}{0.687}[/tex]
x = 1273.78 m
so distance in east is 1273.78 m
A cart is given an initial velocity of 5.0 m/s and
experiencesa constant acceleration of 2.0 m/s ^2. What is the
magnitude of thecart's displacement during the first 6.0 s of its
motion?
Answer:
cart displacement is 66 m
Explanation:
given data
velocity = 5 m/s
acceleration = 2 m/s²
time = 6 s
to find out
What is the
magnitude of cart displacement
solution
we will apply here equation of motion to find displacement that is
s = ut + 0.5×at² .............1
here s id displacement and u is velocity and a is acceleration and time is t here
put all value in equation 1
s = ut + 0.5×at²
s = 5(6) + 0.5×(2)×6²
s = 66
so cart displacement is 66 m
If an object moves twice as fast how do you think its energy changes? halves
doesn't change
doubles
quadruples
Answer:
If an object moves twice as fast its kinetic energy quadruples.
Explanation:
The kinetic energy (K₁) of a body of mass (m) that moves with speed (v) is:
K₁= 1/2 * m* v²
If we double the speed of the body, its kinetic energy (K₂) will be:
K₂= 1/2 * m*( 2v)²
K₂= 1/2 * m* 4 *v²
K₂= 4(1/2 * m *v²)
K₂= 4*K₁
Two 2.4 cm -diameter disks face each other, 1.0 mm apart. They are charged to ±11nC. What is the electric field strength between the disks?
Answer:
[tex]\rm 1.374\times 10^6\ N/C.[/tex]
Explanation:
Given:
Diameter of each disc, D = 2.4 cm.Distance between the discs, d = 1.0 mm.Charges on the discs, q = ±11 nC = [tex]\rm \pm 11\times 10^{-9}\ C.[/tex]The surface area of each of the disc is given by
[tex]\rm A=\pi \times Radius^2.\\\\Radius = \dfrac D2=\dfrac{2.4\ cm}{2} = 1.2\ cm = 1.2\times 10^{-2}\ m.\\A = \pi \times (1.2\times 10^{-2} )^2=4.524\times 10^{-4}\ m^2.[/tex]
For the case, when d<<D, the strength of the electric field at a point due to a charged sheet is given by
[tex]\rm E=\dfrac{|\sigma|}{2\epsilon_o}[/tex]
where,
[tex]\sigma[/tex] = surface charge density of the disc = [tex]\rm \dfrac qA[/tex].[tex]\epsilon_o[/tex] = electrical permittivity of free space = [tex\rm ]9\tiimes 10^9\ Nm^2/C^2[/tex].The electric field strength between the discs due to negative disc is given by
[tex]\rm E_1 = \dfrac{|\sigma|}{2\epsilon_o}=\dfrac{|q|}{2A\epsilon_o}.[/tex]
Since, the electric field is directed from positive charge to negative charge, therefore, the direction of this electric field is towards the negative disc.
The electric field strength between the discs due to positive disc is given by
[tex]\rm E_2 = \dfrac{|\sigma|}{2\epsilon_o}=\dfrac{|q|}{2A\epsilon_o}.[/tex]
The direction of this electric field is away from the positive disc, i.e., towards negative disc.
The electric field between the discs due to both the disc is towards the negative disc, therefore the total electric field strength between the discs is given by
[tex]\rm E=E_1+E_2\\=\dfrac{|q|}{2A\epsilon_o}+\dfrac{|q|}{2A\epsilon_o}\\=\dfrac{|q|}{A\epsilon_o}\\=\dfrac{11\times 10^{-9}}{2\times (4.524\times 10^{-4})\times (8.85\times 10^{-12})}\\=1.374\times 10^6\ N/C.[/tex]
The cockroach Periplaneta americana can detect a static electric field of magnitude 8.50 kN/C using their long antennae. If the excess static charge on a cockroach is modeled as point charges located at the end of each antenna, what magnitude of charge q would each antenna possess in order for each antennae to experience a force of magnitude 2.00 μN from the external electric field? Calculate q in units of nanocoulombs (nC) .
Answer:
0.235 nC
Explanation:
Given:
[tex]E[/tex] = the magnitude of electric field = [tex]8.50\ kN/C =8.50\times 10^{3}\ N/C[/tex][tex]F[/tex] = the magnitude of electric force on each antenna = [tex]2.00\ \mu N =2.00\times 10^{-6}\ N[/tex][tex]q[/tex] = The magnitude of charge on each antennaSince the electric field is the electric force applied on a charged body of unit charge.
[tex]\therefore E = \dfrac{F}{q}\\\Rightarrow q =\dfrac{F}{E}\\\Rightarrow q =\dfrac{2.00\times 10^{-6}\ N}{8.50\times 10^{3}\ N/C}\\\Rightarrow q =0.235\times 10^{-9}\ C\\\Rightarrow q =0.235\ nC[/tex]
Hence, the value of q is 0.235 nC.
What should a graph of the velocity vs time look like if it shows an object moving toward the motion detector at a constant speed? a. An upwards curved parabola.
b. A downwards curved parabola.
c. A straight line with zero slope.
d. A straight line sloping upward.
Answer:
option C
Explanation:
the correct answer is option C.
Graph between velocity time which shows the constant velocity is straight line with slope zero.
constant velocity means velocity is not changing with respect to time hence this condition will be only when graph is straight line with slope zero.
area under velocity time graph shows the displacement of the object.
And where as slope of velocity time graph show acceleration of the object.
Final answer:
The correct option is (d) A straight line sloping upward. A graph of velocity vs time that shows an object moving toward a motion detector at a constant speed is represented as a straight line sloping upward, indicating a constant positive velocity.
Explanation:
When considering an object moving toward a motion detector at a constant speed, the velocity vs. time graph should depict a constant velocity. This is visualized as a straight line on such a graph, since the speed of the object does not change over time.
The correct choice to represent an object moving at a constant speed towards a motion detector is (d) A straight line sloping upward.
The key concept here is that when velocity is constant, the slope of the velocity vs. time graph remains constant. If the object is moving toward the motion detector, the velocity is positive, therefore, the graph slopes upwards, indicating a positive velocity that does not change.
This scenario clearly shows the relationship between velocity and time when an object moves at a constant speed.
Nerve impulses in a human body travel at a speed of about 100 m/s. Suppose a man accidentally stubs his toe. About how much time does it take the nerve impulse to travel from the foot to the brain (in s)? Assume the man is 1.80 m tall and the nerve impulse travels at uniform speed.
Answer:
t = 0.018 s
Explanation:
given,
nerve impulse in human body travel at a speed of = 100 m/s
height of the man = 1.80 m
time taken by the impulse to travel from foot to brain = ?
distance = speed × time
1.80 = 100 × t
t =[tex]\dfrac{1.80}{100}[/tex]
t = 0.018 s
hence, the time taken by the nerve to reach brain from toe is 0.018 s
A person wants to lose weight by "pumping iron". The person lifts an 80 kg weight 1 meter. How many times must this weight be lifted if one pound of fat contains approximately 3500 Calories. If the weight can be lifted once every two seconds, how long would this activity take? 1 Cal=4184 J.
Answer:
37357 sec
or 622 min
or 10.4 hrs
Explanation:
GIVEN DATA:
Lifting weight 80 kg
1 cal = 4184 J
from information given in question we have
one lb fat consist of 3500 calories = 3500 x 4184 J
= 14.644 x 10^6 J
Energy burns in 1 lift = m g h
= 80 x 9.8 x 1 = 784 J
lifts required [tex]= \frac{(14.644 x 10^6)}{784}[/tex]
= 18679
from the question,
1 lift in 2 sec.
so, total time = 18679 x 2 = 37357 sec
or 622 min
or 10.4 hrs
You have been working on a new, strong, lightweight ceramic which will be used to replace steel bearing balls. One cubic meter of the steel alloy currently in use has a mass of 8.08 ⨯ 10^3 kg, whereas a cubic meter of your new material has a mass of 3.14 ⨯ 10^3 kg. If the balls currently in use have a radius of 1.9 cm and for this application to keep balls of the same mass instead of the same size, what would the radius replacement ball of the new alloy be?
Answer:
r=2.6 cm
Explanation:
Hi!
Lets call steel material 1, and the new alloy material 2. You know their densities:
[tex]\rho_1=8.08*10^3\frac{kq}{m^3}\\\rho_2=3.14*10^3\frac{kq}{m^3}[/tex]
The volume of a sphere with radius r is given by:
[tex]V=\frac{4}{3}\pi r^3[/tex]
Then the masses of the bearings are:
[tex]m_1=\rho_1V_1=\frac{4}{3}\pi r_1^3 \\m_1=\rho_2V_2=\frac{4}{3}\pi r_2^3[/tex]
For the masses to be the same:
[tex]\rho_1 V_1 = \rho_2 V_2\\\frac{V_2}{V_1} =\frac{\rho_1}{\rho_2}\\(\frac{r_2}{r_1})^3 = \frac{8.08}{3.14}=2.57\\r_2=\sqrt[3]{2.57}\;r1 =1.37r_1=1.37*1.9cm=2.6cm[/tex]
Boyle’s Law states that when a sample of gas is compressed at a constant temperature, the pressure P of the gas is inversely proportional to the volume V of the gas. (a) Suppose that the pressure of a sample of air that occupies 0.106 m3 at 25°C is 50 kPa. Write V as a function of P. (b) Calculate dVydP when P − 50 kPa. What is the meaning of the derivative? What are its units?
(a) V = [tex]\frac{8.3}{P}[/tex]
(b) (i) the value of [tex]\frac{dV}{dP}[/tex] when P = 50kPa is - 0.00332 [tex]\frac{m^{3} }{kPa}[/tex]
(ii) the meaning of the derivative [tex]\frac{dV}{dP}[/tex] is the rate of change of volume with pressure.
(iii) and the units are [tex]\frac{m^{3} }{kPa}[/tex]
Explanation:Boyle's law states that at constant temperature;
P ∝ 1 / V
=> P = k / V
=> PV = k -------------------------(i)
Where;
P = pressure
V = volume
k = constant of proportionality
According to the question;
When;
V = 0.106m³, P = 50kPa
Substitute these values into equation (i) as follows;
50 x 0.106 = k
Solve for k;
k = 5.3 kPa m³
(a) To write V as a function of P, substitute the value of k into equation (i) as follows;
PV = k
PV = 8.3
Make V subject of the formula in the above equation as follows;
V = 8.3/P
=> V = [tex]\frac{8.3}{P}[/tex] -------------------(ii)
(b) Find the derivative of equation (ii) with respect to V to get dV/dP as follows;
V = [tex]\frac{8.3}{P}[/tex]
V = 8.3P⁻¹
[tex]\frac{dV}{dP}[/tex] = -8.3P⁻²
[tex]\frac{dV}{dP}[/tex] = [tex]\frac{-8.3}{P^{2} }[/tex]
Now substitute P = 50kPa into the equation as follows;
[tex]\frac{dV}{dP}[/tex] = [tex]\frac{-8.3}{50^{2} }[/tex] [[tex]\frac{kPam^{3} }{(kPa)^{2} }[/tex]] ----- Write and evaluate the units alongside
[tex]\frac{dV}{dP}[/tex] = [tex]\frac{-8.3}{2500 }[/tex] [[tex]\frac{kPam^{3} }{k^{2} Pa^{2} }[/tex]]
[tex]\frac{dV}{dP}[/tex] = - 0.00332 [[tex]\frac{m^{3} }{kPa}[/tex]]
Therefore,
(i) the value of [tex]\frac{dV}{dP}[/tex] when P = 50kPa is - 0.00332 [tex]\frac{m^{3} }{kPa}[/tex]
(ii) the meaning of the derivative [tex]\frac{dV}{dP}[/tex] is the rate of change of volume with pressure.
(iii) and the units are [tex]\frac{m^{3} }{kPa}[/tex]
Boyle’s Law describes the inverse relationship between the pressure and volume of a gas at constant temperature. To express volume as a function of pressure, use the formula V = k/P. The derivative of volume with respect to pressure indicates the rate at which volume changes as pressure changes, measured in cubic meters per kilopascal (m3/kPa).
Explanation:Boyle’s Law states that for a given amount of gas at a constant temperature, the volume V is inversely proportional to the pressure P. This relationship can be mathematically represented as PV = k, where k is a constant.
(a) Given a sample of air at 25°C with a volume of 0.106 m3 and a pressure of 50 kPa, we can write the volume as a function of pressure by rearranging the formula to V = k/P. Using the initial conditions, we find that k = P × V = 50 kPa × 0.106 m3 = 5.3 kPa · m3, so V(P) = 5.3 kPa · m3 / P.
(b) To calculate dV/dP when P = 50 kPa, we differentiate the function V(P) with respect to P, resulting in dV/dP = -5.3 kPa · m3 / P2. At P = 50 kPa, this becomes dV/dP = -5.3 / (502) = -0.00212 m3/kPa. The derivative represents the rate of change of volume with respect to pressure, which in this case means that for each increase of 1 kPa in pressure, the volume decreases by 0.00212 m3. The units of the derivative are m3/kPa.
Two charges q1 and q2 have a total charge of 10 mu or micro CC. When they are separated by 4.1 m, the force exerted by one charge on the other has a magnitude of 8 mN. Find q1 and q2 if both are positive so that they repel each other, and q1 is the smaller of the two. (For the universal constant k use the value 8.99 times 109 N m2/C2.)
Answer:
q₂ = 6.8 x 10⁻⁶C
q₁ = 3.2 x 10⁻⁶ C
Explanation:
First charge is q₂ and second charge will be ( 10 x10⁻⁶ - q₂ )
Force between them F = 8 X 10⁻³ N distance between them d = 4.1 m
F = k ( q₁ xq₂ ) / d²
8 x 10⁻³ = [tex]\frac{8.99\times10^9\times q_2\times (10^{-5}-q_2)}{(4.1)^2}[/tex]
14.95 x 10⁻¹² = 10⁻⁵q₂ - q₂²
q₂²-10⁻⁵q₂ + 14.95 x 10⁻¹² =0
This is a quadratic equation .The solution of it gives the value of q₂
q₂ = 6.8 x 10⁻⁶C
q₁ = 3.2 x 10⁻⁶ C
q₁ = 3.2 x 10⁻⁶ C
q₂ = 6.8 x 10⁻⁶C
What is Charge in Physics?Charge is a characteristic of a unit of matter that expresses the extent to which it has more or fewer electrons than protons.
It is also known as electric charge, electrical charge, or electrostatic charge and symbolized q.
Given,
First charge is q₂ and second charge will be ( 10 x10⁻⁶ - q₂ )
F = 8 X 10⁻³ N
d = 4.1 m
By using the Formula,
[tex]F =\frac{k ( q_{1} * q_{2} ) }{d^{2} }[/tex]
8 x 10⁻³ = [tex]\frac{8.99*10^{9}*q_{2} *10^{-5} -q_{2} }{(4.1)^{2} }[/tex]
14.95 x 10⁻¹²
= 10⁻⁵q₂ - q₂²
= q₂²-10⁻⁵q₂ + 14.95 x 10⁻¹²=0
Therefore,
The above equation is a quadratic equation , it gives the value of q₂
q₂ = 6.8 x 10⁻⁶C
q₁ = 3.2 x 10⁻⁶ C
Learn more about Charges here:https://brainly.com/question/14467306
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Two sinusoidal waves traveling in opposite directions interfere to produce a standing wave with the wave function y = (2.00) sin(0.500x) cos(300t) where x and y are in meters and t is in seconds. (a) Determine the wavelength of the interfering waves. m (b) What is the frequency of the interfering wave? Hz (c) Find the speed of the interfering waves.
Answer:
Part a)
[tex]\lambda = 4\pi[/tex]
Part b)
[tex]f = 47.7 Hz[/tex]
Part c)
[tex]v = 600 m/s[/tex]
Explanation:
Part a)
As we know that angular wave number is given as
[tex]k = \frac{2\pi}{\lambda}[/tex]
[tex]k = 0.500[/tex]
[tex]0.500 = \frac{2\pi}{\lambda}[/tex]
[tex]\lambda = \frac{2\pi}{0.500}[/tex]
[tex]\lambda = 4\pi[/tex]
Part b)
As we know that angular frequency is given as
[tex]\omega = 300 rad/s[/tex]
[tex]\omega = 2\pi f[/tex]
[tex]300 = 2\pi f[/tex]
[tex]f = \frac{300}{2\pi}[/tex]
[tex]f = 47.7 Hz[/tex]
Part c)
Speed of the wave is given as
[tex]v = \lambda \times frequency[/tex]
so we have
[tex]v = 4\pi \times 47.7[/tex]
[tex]v = 600 m/s[/tex]
An electron is travelling in a straight line with a kinetic energy K = 1.60 x 10^-17J. What are (a) the magnitude and (b) the direction of the electric field that will stop the electron in a distance of 10.0 cm? For part (b), make a drawing showing the direction of motion of the electron, the direction of the electric field and the electric force on the electron
Answer:
(a) 1000 N/C
Explanation:
Kinetic energy of electron, K = 1.6 x 10^-17 J
distance, d = 10 cm = 0.1 m
Let the potential difference is V and the electric field is E.
(a) The relation between the kinetic energy and the potential difference is
K = e V
V = K / e
Where, e be the electronic charge = 1.6 x 10^-19 C
V = [tex]\frac{1.6\times 10^{-17}}{1.6\times 10^{-19}}[/tex]
V = 100 V
The relation between the electric field and the potential difference is given by
V = E x d
100 = E x 0.1
E = 1000 N/C
(b) The force acting on the electron, F = q E
where q be the charge on electron
So, F = -e x E
It means the direction of electric field and the force are both opposite to each other.
The direction of electric field and the force on electron is shown in the diagram.