Answer:
When an external force acts upon them
15C of charge flow through the filament of a light bulb in 22 seconds. What is the strength of the current in the filament?
Answer:
Current(I)= charge(q)/time
I=15/22
I=0.680.68A
Explanation:
Current is flow of charge per unit time. Always make sure the time is in seconds. istthe
give an example of the 4 steps of the scientific method
Answer:
1) asking a question about something you observe, 2) doing background research to learn what is already known about the topic, 3) constructing a hypothesis, 4) experimenting to test the hypothesis
Answer:
1) make an observation that describes a problem - see something you can fix or improve and try to describe the problem like which type of vegetables does rabbit like
2) create a hypothesis - What do you think will happen - If the rabbit eat the lettuce than .........
3) test the hypothesis - Do your experiment with the variables and follow the procedures
4) draw conclusions and refine the hypothesis - see if your hypothesis was correct - In conclusion my hypothesis was not correct because.......
(a) In Coulomb scattering of 7.50-MeV protons by a target of 'Li, what is the energy of the elastically scattered protons at 90°? (b) What is the energy of the inelastically scattered protons at 90° when the 'Li is left in its first excited state (0.477 MeV)?
Answer:
First the charge is given 7.50×10^-6
Explanation:
so that we have
[tex]90 [/tex]
so that their is no cross sectional area of this anglethen the direction is one is left and other one is right so thats my hint
Mrs. dela Vega had 27 students in her class last year and this year she had 30 students in her math class. What is the percent of difference?
a boat travels 500 m down a strait river. if starts from the rest accelerates uniformly to a velocity of 5m/s. how long does this take?
PLSSSS HELPPPPP NOWWW!!!!!
A man is pushing a large box with a force of 44 newtons. He slides over a floor that has a sliding frictional force of 12 Newtons. What is the acceleration?
Two horses pull horizontally on ropes attached to a stump. The two forces Fu and F2 that they apply to the stump are such that the net (resultant) force R has a magnitude equal to that of Fu and makes an angle of 90° with Fu. Let F1 = 1300 N and R = 1300 N also. Find the magnitude of F and its direction (relative to Fi)
For two horses pull horizontally on ropes attached to a stump, the magnitude of F and its direction is mathematically given as
F=1711 N
dF= 135 degrees
What is the magnitude of F and its direction?
Generally, the equation for the components of F1 is mathematically given as
For The x axis
1210 - F2x = 0
F2x = 1210
For the y axis
F2y = R
F2y = 1210
In conclusion
F = sqrt(2)*1210
F=1711 N
Fopt the direction of F2
dF= 90 + 45
dF= 135 degrees
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Difine Ripple Factor
-,- ty!!
Answer:
The answer is in the picture above.
Conceptual Example 14 provides useful background for this problem. A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carousel itself (without riders) has a moment of inertia of 105 kg·m2. When one person is standing at a distance of 1.57 m from the center, the carousel has an angular velocity of 0.512 rad/s. However, as this person moves inward to a point located 0.507 m from the center, the angular velocity increases to 0.795 rad/s. What is the person's mass?
For one person standing at a distance of 1.57 m from the center, the carousel has an angular velocity of 0.512 rad/s, the person's mass is mathematically given as
m = 24.2 kg
What is the person's mass?Generally, the equation for the conservation angular momentum is mathematically given as
(I + m * r0^2) * w0 = (I + m * r^2) * w^2
Therefore
(105 + m * 1.78^2) * 0.517 = (105 + m * 0.524^2) * 0.841
m = 24.2 kg
In conclusion, the mass
m = 24.2 kg
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Two children are playing tetherball, in which a ball at the end of a cord spins around a pole. After a really good hit, the ball makes three complete revolutions in 2.0 s. What is the angular speed of the ball?
Answer:
3 pi R / s or 9.42 R/s or 540 degrees/sec
Explanation:
Ball covers 2 pi radians each rotation
2 pi * 3 R / 2 s = 3 pi R / s
the angular speed of the ball is 3 pi R / s or 9.42 R/s or 540 degrees/sec
What is angular speed ?Angular speed can be defined as the measures of speed of how fast the central angle of a rotating body changes with respect to time
Angular speed mainly implies how quickly the rotation of an object occur means it is described as the change in the angle of the object per unit of time.
To calculate the speed of a rotational motion the angular speed value need to be known, hence The angular speeds formula is used to calculate the distance traveled by an object in terms of rotation and revolutions per unit of time.
The unit of angular speed of an object is radian per second and Both angular speed, angular velocity are determined by using the same formula, angular velocity is vector quantity which describes both magnitude and direction.
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2. Box A has 2x the mass of Box B. Compare the inertia of the 2 bodies.
Box A has 2x the mass of Box B so Box A has more inertia as compared to Box B.
Relationship between inertia and massMore the mass of an object, there is more inertia of the body. Thus, if mass of two bodies is same their inertia will also be same. Hence, inertia of A and B will be same as mass of both A and B are same.
So we can conclude that Box A has 2x the mass of Box B so Box A has more inertia as compared to Box B.
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A positive charge +Q is moving to the right and experiences a vertical (upward) magnetic force. In which direction is the magnetic field?
O into the screen
O upward
O out of the screen
O to the left
O to the right
Answer:
out of the screen
Explanation:
because charge +Q is already moving upward
What season would Texas be having at point D?
Answer:
Almost winter so fall
Explanation:
Hope this helps! Please let me know if you want me to elaborate more or think my answer is incorrect. Brainliest would be MUCH appreciated. Have a wonderful day!
What is refractive index?
[tex] \huge \mathbb \red{HEY \: THERE ♡}[/tex]
Refractive Index:The ratio of the sine of angle of incidence to the sine of angle of refraction in case of lens for light of a given colour and given pair of media is constant. Note** It is also called Snell's Law of Refraction. [tex] \mathsf \orange{\frac{sine \: i}{sine \: r} = constant}[/tex][tex] \huge \mathbb\pink{HOPE \: IT \: HELPS}[/tex]
A 75-W light bulb is turned on. It has an operating voltage of 120 V. (A)How much current flows through the bulb? (B)What is the resistance of the bulb? (C)How much energy is used each second?
Required Answer:
Given:
Power (P) = 75 W Voltage (V) = 120 V(A)
As we know that,
I = P/V↠ Current = 75/120
↠ Current = 0.625 A
(B)
R = V/IHere,
R is resistanceV is Voltage, and I is current↠ Resistance = 120/0.625
↠ Resistance = 192 Ω
(C)
Energy = P × tHere,
P is power T is time↠ Energy = 75 × 1
↠ Energy = 75 J
A karate expert breaks a stack of bricks with his bare hand. If the force applied is 520 newtons and the impact time is 5.0 × 10 seconds, what is the value of impulse
A.
0.26 newton seconds
B.
0.58 newton seconds
C.
1.82 newton seconds
D.
2.20 newton seconds
A karate expert breaks a stack of bricks with his bare hand. If the force applied is 520 newtons and the impact time is [tex] \sf{5.0 \times 10^{-4}} [/tex] seconds, what is the value of impulse ?
So, the value of impulse is (A). 0.26 newton seconds.
IntroductionHi ! Here, I will help you about the impulse. Impulse is described as a quantity which expresses the integral of the amount force respect to time. The amount of impulse will be proportional to the amount of force or time (the greater value of the force or the value of time, the value of impuls is greater too). The relationship between impulse, force, and time is expressed in this equation :
[tex] \boxed{sf{\bold{I = F \times \Delta t}}} [/tex]
With the following condition:
I = impulse that occurs (N.s)F = force that given (N)[tex] \sf{\Delta t} [/tex] = interval of the time (s) Problem SolvingWe know that :F = force that given = 520 N[tex] \sf{\Delta t} [/tex] = interval of the time = [tex] \sf{5.0 \times 10^{-4} \: s}[/tex]What was asked :
I = impulse that occurs = ... N.sStep by step :
[tex] \sf{I = F \times \Delta t} [/tex]
[tex] \sf{I = 520 \times (5.0 \times 10^{-4}} [/tex]
[tex] \sf{I = 2,600 \times 10^{-4}} [/tex]
[tex] \sf{I = 2.60 \times 10^3 \times 10^{-4}} [/tex]
[tex] \sf{I = 2.60 \times 10^{3 + (-4)}} [/tex]
[tex] \sf{I = 2.60 \times 10^{-1} = \boxed{0.26 \: N.s}} [/tex]
Conclusion :So, the value of impulse is (A). 0.26 newton seconds
What is the magnetic flux going through a coil with 0.065 m2 area if the intensity of the magnetic field is 0.114 T.
Hi there!
[tex]\Phi _B = \int B \cdot dA[/tex]
φ = Magnetic Flux (Wb or Tm²)
B = Magnetic field strength (T)
A = Area (m²)
This equation includes a dot product, so this can be rewritten as:
[tex]\Phi _B = B A cos\theta[/tex]
In this instance, the field is parallel to the area, so cos(0) = 1.
Calculate the magnetic flux by plugging in the given values.
[tex]\Phi _B = BA = 0.114 * 0.065 = \boxed{0.00741 Wb}[/tex]
-What is the potential energy at point A?
-What is the kinetic energy at point A?
-What is the kinetic energy at point B?
-What is the potential energy at point D?
-What is the kinetic energy at point D?
Answer:
1560520156Explanation:
Assuming there is no friction or other force involved, recall that energy is conserved in a system as long as no external force acts on the system.
Using the data from point C, we can find out that the total energy of the system is 156 because [tex]E = K+Pe[/tex].
Since at point A the object doesn't move, it has Kinetic energy of 0, because [tex]K=\frac{1}{2} mv^2[/tex], therefore [tex]0=\frac{1}{2} m0^2[/tex]. However at point A it has maximum Potential energy, because [tex]Pe=mgh[/tex].
At point B, we can find the Kinetic energy by using [tex]E = K+Pe[/tex]. Substitute values:
[tex]156=104+K\\52=K[/tex]
At point D, the object has maximum kinetic energy and no potential energy, therefore it's the opposite of point A.
A 2 kg solid disk with a radius of 0.22 m has a tangential force of 300N applied to it.
a) What is the torque acting on the disk?
b) What is the moment of inertia of the disk?
c) What angular acceleration is produced by the torque?
d) If the disk starts from rest and the acceleration is constant for 3.0s, what is the angular velocity of the disk at the end of 3.0s?
e) Through what angle in radians has the disk rotated during this time?
(a) The torque acting on the disk is 66 Nm.
(b) The moment of inertia of the disk is 0.05 kgm².
(c) The angular acceleration is produced by the torque is 1,320 rad/s².
(d) The final angular velocity of the disk is 3,960 rad/s.
(e) The angle of rotation of the disk is 5,940 rad.
Torque acting on the diskThe torque acting on the disk is calculated as follows;
τ = Fr
τ = 300 x 0.22
τ = 66 Nm
Moment of inertiaThe moment of inertia of a solid disk is calculated as follows;
I = ¹/₂MR²
I = ¹/₂ x 2 x (0.22)²
I = 0.05 kgm²
Angular acceleration of the diskThe angular acceleration of the disk is calculated as follows;
τ = Iα
[tex]\alpha = \frac{\tau }{I} \\\\\alpha = \frac{66}{0.05} \\\\\alpha = 1,320 \ rad/s^2[/tex]
Angular velocity of the disk after 3 sωf = ωi + αt
ωf = 0 + (1320 x 3)
ωf = 3,960 rad/s
Angle of rotation of the diskωf² = ωi²+ 2αθ
(3,960)² = 0 + 2(1320)θ
θ = (3,960²) / (2 x 1320)
θ = 5,940 rad
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A uniformly charged insulating rod is bent into the shape of a semicircle of radius R = 5 cm. If the rod has a total charge of Q = 3.10-9C, find the magnitude and direction of the electric field at O, the center of the circle.
Hi there!
We can begin by using Coulomb's Law:
[tex]E = \frac{kq}{r^2}[/tex]
k = Coulomb's Constant (8.99 × 10⁹ Nm²/C²)
E = Electric field strength (N/C)
r = distance from point (m)
q = charge (C)
Since this is a continuous charge, we must use calculus.
We can express this as the following:
[tex]q = \lambda L[/tex]
λ = Linear charge density (C/m)
L = Length of rod (m)
Now, since this is an arc, L = s (arc length). Additionally, we must find the differential elements of each:
[tex]dq = \lambda ds\\\\dq = \lambda rd\theta[/tex]
Our new equation is:
[tex]dE = \frac{kdq}{r^2}\\\\dE = \frac{k\lambda rd\theta}{r^2}[/tex]
However, we will only take the cosine component of the electric field since the vertical components will cancel out. (Electric fields are a vector). Therefore:
[tex]dE = \frac{k\lambda rd\theta}{r^2}cos\theta\\\\dE = \frac{k\lambda}{r}cos\theta d\theta[/tex]
Integrate. For a semicircle, the bounds will be from -π/2 to π/2.
[tex]E = \frac{k\lambda}{r}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}} {cos\theta} \, d\theta\\\\E = \frac{k\lambda}{r}sin\theta\left \|{\frac{\pi}{2}} \atop {-\frac{\pi}{2}}} \right. \\\\E = \frac{k\lambda}{r}(1 - (-1)) = \frac{2k\lambda}{r}[/tex]
We need to solve for λ, which is Q/ L:
[tex]\lambda = \frac{3.10 \times 10^{-9} C}{\pi (0.05)} = 1.9735 \times 10^{-8} \frac{C}{m}[/tex]
Now, plug and solve for the electric field strength:
[tex]E = \frac{2(8.99\times 10^9)(1.9735\times 10^{-8})}{0.05} = \boxed{7096.783 \frac{N}{C}}[/tex]
**A diagram was not provided, but if the hemisphere's focus was to the right, the electric field would be to the right, and etcetera.
A single coil of wire, with a radius of 0.13 m is rotated in a uniform magnetic field such that the angle between the field vector and the area vector obeys θ=ωt. If the strength of the field is 3.746 T, and the angular frequency is 524.7 rad/s, what is the induced emf in the loop at t=1.16 s?
The magnitude of emf induced in the single coil of wire rotated in the uniform magnetic field is 0.171 V.
The emf induced in the loop is determined by applying Faraday's law of electromagnetic induction.
emf = N(dФ/dt)
where;
N is number of turns of the wireФ is magnetic fluxФ = BA
where;
B is magnetic field strengthA is the area of the loopemf = NBA/t
A = πr²
A = π x (0.13)²
A = 0.053 m²
emf = NBA/t
emf = (1 x 3.746 x 0.053)/(1.16)
emf = 0.171 V
Thus, the magnitude of emf induced in the single coil of wire is 0.171 V.
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a true statement of
kinetic theory
Answer:
real kinetic theory means that kinetic energy
The surface of a lake has an area of 15.5 km2. What is the area of the lake in m2?
Answer:
1.55 x 10⁷ m²
Explanation:
Unit conversion
km² → m² 1 km² = 10⁶ m²Solving
15.5 km² = 10⁶ x 15.5 m²155 x 10⁵ m²1.55 x 10⁷ m²1. A drag racer accelerates from rest at 18ft/sec^2. How long does it take to acquire a speed of 60mph? What is required?
2. A contestant ran a 100-m dash in 10.6sec. What was his speed a) In feet per second and b) in miles per hr?
(1) The time taken for the drag racer to accelerate is 4.89 s
(2) The speed of the contestant (a) in feet per second is 30.89 ft/s (b) in miles per hr is 21.12 miles per hr.
(1) To calculate the time required to accelerate 18 ft/sec² from rest to a velocity of 60 mph, we use the formula below.
Formula:
t = (v-u)/a........... Equation 1Where:
t = timev = Final velocityu = initial velocitya = acceleration.From the question,
Given:
a = 18 ft/sec² = (18×0.3048) = 5.4864 m/s²v = 60 mph = (60×0.44704) = 26.82 m/su = 0 m/s ( from rest)Substitute these values into equation 1
t = (26.82-0)/5.4864t = 4.89 seconds(2) To calculate the speed of the contestant, we use the formula below
Formula:
s = d/t............ Equation 1Where:
s = speed of the contestantd = distancet = time.From the question,
Given:
d = 100 mt = 10.6 sSubstitute these values into equation 1
s = 100/10.6s = 9.43 m/s(a) In feet = (9.43/0.3048) = 30.94 ft/s
(b) in miles per hr = (9.43×2.24) = 21.12 miles per hr
Hence, (1) The time taken for the drag racer to accelerate is 4.89 s (2) The speed of the contestant (a) in feet per second is 30.89 ft/s (b) in miles per hr is 21.12 miles per hr.
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Yesterday you walked 20 meters to the right from your house to the bus stop, which took you 20 seconds. You waited at the bus stop for 1 minute before realizing you forgot your Mathematics homework at home. You ran back to your house in 5 seconds. It took you 10 seconds to find your homework, and then you ran back to the bus stop in 5 seconds just in time to catch the bus. What was your average velocity for the entire period of motion?
a.
0.2 m/s
b.
1 m/s
c.
100 m/s
d.
20 m/s
average velocity= tot displacement/total time
avg v = 20/(20+60+5+10+5)=20/100=0.2 m/s
What are the essential components of successfully competing in the field events:
Power, technique, and timing
Power, speed, and strength
Technique, flexibility, and speed
None of the above
Answer:
Power, technique, and timing
Explanation:
I just did the quiz and it was correct!! Hope this helps!
Spring
The pulling force acting on an object is 350.0 N. It also is being affected by a 50.0 N
kinetic friction force. As a result, its acceleration is 3.0 m/s?, what is its mass?
A. 67 kg
B. 83 kg
C. 100 kg
D. 600 kg
E. 0.0 kg
If the gravitational force of the Farth nulling you"down" in th
Answer:
100 kg
Explanation:
Net force = 350 - 50 = 300 N
F= ma
300 N = m * 3 m/s^2
100 kg = mass
Calculate the frequency of a sound wave produced when a tennis racquet string is plucked. The tension of the string is 274 N, the mass of the string is 28 kg and the length of the string is 0.74 m.
Answer:
Explanation:
The velocity of sound from the plucked string in the tennis racquet is:
[tex]v=\sqrt{\frac{FL}{m}}=\sqrt{\frac{274(0.74)}{28}}\approx 2.69 ms^{-1}[/tex]
Then the frequency will be:
[tex]f=\frac{v}{2L}=\frac{2.69}{2(0.74)}\approx 1.81 Hz[/tex]
Why is there two π in the formula of a mathematical pendulum (T=2π √l/g)?
Answer:
The π is from the initial formula of period T=2π/omega
If electromagnetic radiation acted like particles in the double-slit experiment, what would be observed? (1 point)
O A series of light and dark bands would appear on the screen.
O Two bright bands would appear on the screen in line with the slits.
O The screen would remain dark because no radiation would reach the screen.
O One bright band would appear in the center of the screen.
If electromagnetic radiation acted like particles in the double-slit experiment, we would observe one bright band would appear in the center of the screen.
Bahavior of particles in double-slit experimentIn a double-slit experiment, single particles, such as photons, pass one at a time through a screen containing two slits.
The photons behave like wave and the constructive interfernce of the waves of these photons will generate a high amplitude wave seen as a bright band in the center of the screen.
Thus, if electromagnetic radiation acted like particles in the double-slit experiment, we would observe one bright band would appear in the center of the screen.
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