Answer:
The position function is [tex]s_{t}=2t^3+5t^2-5t+9[/tex].
Explanation:
Given that,
Acceleration [tex]a =12t+10[/tex]
Initial velocity [tex]v_{0} = -5\ cm/s[/tex]
Initial displacement [tex]s_{0}=9\ cm[/tex]
We know that,
The acceleration is the rate of change of velocity of the particle.
[tex]a = \dfrac{dv}{dt}[/tex]
The velocity is the rate of change of position of the particle
[tex]v=\dfrac{dx}{dt}[/tex]
We need to calculate the the position
The acceleration is
[tex]a_{t} = 12t+10[/tex]
[tex]\dfrac{dv}{dt} = 12t+10[/tex]
[tex]a_{t}=dv=(12t+10)dt[/tex]
On integration both side
[tex]\int{dv}=\int{(12t+10)}dt[/tex]
[tex]v_{t}=6t^2+10t+C[/tex]
At t = 0
[tex]v_{0}=0+0+C[/tex]
[tex]C=-5[/tex]
Now, On integration again both side
[tex]v_{t}=\int{ds_{t}}=\int{(6t^2+10t-5)}dt[/tex]
[tex]s_{t}=2t^{3}+5t^2-5t+C[/tex]
At t = 0
[tex]s_{0}=0+0+0+C[/tex]
[tex]C=9[/tex]
[tex]s_{t}=2t^3+5t^2-5t+9[/tex]
Hence, The position function is [tex]s_{t}=2t^3+5t^2-5t+9[/tex].
To find the position function of a particle given acceleration a(t) = 12t + 10, one integrates twice. The first integral gives the velocity function, the second gives the position or displacement function. These are determined to be v(t) = 6t^2 + 10t - 5 and s(t) = 2t^3 +5t^2 - 5t + 9, respectively.
Explanation:The subject of this problem involves calculating the position function, or displacement, of a particle given an acceleration function, initial velocity, and initial position. In this case, acceleration is given by a(t) = 12t + 10.
To find the velocity function, v(t), you integrate the acceleration function: ∫a(t) dt = ∫(12t + 10) dt = 6t^2 + 10t + C1, where C1 is the constant of integration. We know that v(0) = −5 cm/s, so C1 is -5: the full velocity function is v(t) = 6t^2 + 10t - 5.
We then integrate the velocity function to find the position function, s(t): ∫v(t) dt = ∫(6t^2 + 10t - 5) dt = 2t^3 +5t^2 - 5t + C2. Given s(0) = 9 cm, we find C2 = 9. The position function s(t) is therefore s(t) = 2t^3 +5t^2 - 5t + 9.
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What net external force is exerted on a 1100-kg artillery shell fired from a battleship if the shell is accelerated at 2.40×104 m/s2? What is the magnitude of the force exerted on the ship by the artillery shell?
Answer:
Force exerted, F = 2.64 × 10⁷ Newton
Explanation:
It is given that,
Mass of the artillery shell, m = 1100 kg
It is accelerated at, [tex]a=2.4\times 10^4\ m/s^2[/tex]
We need to find the magnitude of force exerted on the ship by the artillery shell. It can be determined using Newton's second law of motion :
F = ma
[tex]F=1100\ kg\times 2.4\times 10^4\ m/s^2[/tex]
F = 26400000 Newton
or
F = 2.64 × 10⁷ Newton
So, the force exerted on the ship by the artillery shell is 2.64 × 10⁷ Newton.
Answer: The force exerted on the artillery shell is [tex]2.64\times 10^6N[/tex] and the magnitude of force exerted on the ship by artillery shell is [tex]2.64\times 10^6N[/tex]
Explanation:
Force is defined as the push or pull on an object with some mass that causes change in its velocity.
It is also defined as the mass multiplied by the acceleration of the object.
Mathematically,
[tex]F=ma[/tex]
where,
F = force exerted on the artillery shell
m = mass of the artillery shell = 1100 kg
a = acceleration of the artillery shell = [tex]2.40\times 10^4m/s^2[/tex]
Putting values in above equation, we get:
[tex]F=1100kg\times 2.40\times 10^4m/s^2\\\\F=2.64\times 10^6N[/tex]
Now, according to Newton's third law, every action has an equal and opposite reaction.
So, the force exerted on the artillery shell will be equal to the force exerted on the ship by artillery shell acting in opposite direction.
Hence, the force exerted on the artillery shell is [tex]2.64\times 10^6N[/tex] and the magnitude of force exerted on the ship by artillery shell is [tex]2.64\times 10^6N[/tex]
A spring has a natural length of 8 m. If a 12-N force is required to keep it stretched to a length of 10 m, how much work W is required to stretch it from 8 m to 16 m? (Round your answer to two decimal places.)
Answer:
The work is required to stretch it from 8 m to 16 m is 192 N-m
Explanation:
Given that,
Natural length = 8 m
Force F = 12 N
After stretched,
length = 10 m
We need to calculate the elongation
[tex]x = 10-8=2\ m[/tex]
Using hook's law
The restoring force is directly proportional to the displacement.
[tex]F\propto (-x)[/tex]
[tex]F = -kx[/tex]
Where, k = spring constant
Negative sign shows the displacement in opposite direction
Now, The value of k is
[tex]k = \dfrac{F}{x}[/tex]
[tex]k = \dfrac{12}{2}[/tex]
[tex]k = 6[/tex]
When stretch the string from 8 m to 16 m.
Then the elongation is
[tex]x=16-8=8\ m[/tex]
Now, The work is required to stretch it from 8 m to 16 m
[tex]W = \dfrac{1}{2}kx^2[/tex]
Where, k = spring constant
x = elongation
[tex]W=\dfrac{1}{2}\times6\times8\times8[/tex]
[tex]W=192\ N-m[/tex]
Hence, The work is required to stretch it from 8 m to 16 m is 192 N-m
An experiment is designed to test the relationship between the initial height of a basketball before it is dropped to the height of its rebound bounce. The height of the rebound bounce is measured using a scale positioned behind the ball. In the above experiment, which condition would not be controlled?
A). the starting height of the ball
B). the surface the ball bounces on
C). the type of ball
D). the method to measure the rebound height of the ball
Answer:
A. The starting height of the ball
Explanation:
When we talk about controlled variables, we refer to the variable that should be kept the same throughout the experiment. The reason why we do this, is to limit anything else that is not being tested, that may affect the results of the experiment.
In the scenario given, the experiment is to see the relationship between the initial height of a basketball and the height of its rebound bounce.
So you the starting height of the ball should vary, meaning it is NOT controlled.
Answer:
Starting height of the ball
Explanation:
An intercontinental ballistic missile goes from rest to a speed of 6.50 km/s in 60.0 s. In multiples of g, what is its acceleration?
Answer:
The acceleration is [tex]11.1g\ m/s^2[/tex]
Explanation:
Given that,
Speed [tex]v= 6.50\ km/s=6.5\times10^{-3}\ m/s[/tex]
Time t = 60.0 sec
We need to calculate the acceleration
Using formula off acceleration
[tex]a = \dfrac{\Delta v}{t}[/tex]
[tex]a=\dfrac{v_{f}-v_{i}}{t}[/tex]
We know that,
Missile goes from rest
So, Initial velocity =0
Put the value into the formula
[tex]a =\dfrac{6.50\times10^{3}}{60.0}[/tex]
[tex]a=108.33\ m/s^2[/tex]
On right hand side multiplying and dividing by g = 9.8m/s²
[tex]a=108.33\times\dfrac{g}{g}[/tex]
Put the value of g
[tex]a = \dfrac{108.33}{9.8}g\ m/s^2[/tex]
[tex]a = 11.1g\ m/s^2[/tex]
Hence, The acceleration is [tex]11.1g\ m/s^2[/tex]
A locomotive enters a station with an initial velocity of 19 m/s and slows down at a rate of .8m/s^2 as it goes through. If the station is 175 m long, how fast is it going when the nose leaves the station?
Answer:
Final velocity, v = 25.3 m/s
Explanation:
Initial velocity of a locomotive, u = 19 m/s
Acceleration of the locomotive, a = 0.8 m/s²
Length of station, d = 175 m
We need to find its final velocity (v) when the nose leaves the station. It can be calculated using the third law of motion :
[tex]v^2-u^2=2ad[/tex]
[tex]v^2=2ad+u^2[/tex]
[tex]v^2=2\times 0.8\ m/s^2\times 175\ m+(19\ m/s)^2[/tex]
[tex]v^2=(641)\ m^2[/tex]
v = 25.31 m/s
v = 25.3 m/s
When the nose leaves the station, it will move with a velocity of 25.3 m/s. Hence, this is the required solution.
An 0.80-kilogram object tied to the end of a 2.0-meter string swings as a pendulum. At the lowest point of its swing, the object has a kinetic energy of 10.0 joules. What is the speed of the object at the instant that the string makes an angle of 50.0° with the vertical?
Answer:
3.3 m/s
Explanation:
As the object rises above the lowest point, some of the kinetic energy is converted to potential energy. From the diagram, we can see that at angle θ, the object rises to height h:
h = L - L cos θ
Conservation of energy:
KE₀ = KE + PE
KE₀ = 1/2 mv² + mgh
Substituting:
KE₀ = 1/2 mv² + mg(L - L cos θ)
Given KE₀ = 10.0 J, m = 0.80 kg, g = 9.8 m/s², L = 2.0 m, and θ = 50.0°:
10.0 = 1/2 (0.80) v² + (0.80) (9.8) (2.0 - 2.0 cos 50.0)
v = 3.32 m/s
Rounding to 2 sig-figs, the speed of the object is 3.3 m/s.
A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 3°C, the resistance of the carbon resistor is 217.6 2 Ω. What is the temperature on a spring day when the resistance is 215.3 2 Ω? (Take the temperature coefficient of resistivity for carbon to be a-5.00 x 10^-4°C^-1) (Give your answer in decimal using "degreeC" °C) as unit)
Answer:
24.14 ⁰C
Explanation:
T₀ = initial temperature = 3 °C
R₀ = initial resistance of thermometer at initial temperature = 217.62 Ω
R = Final resistance of thermometer at final temperature = 215.32 Ω
T = final temperature = ?
α = temperature coefficient of resistivity for carbon = - 5.00 x 10⁻⁴ C⁻¹
Final resistance of thermometer at final temperature is given as
R = R₀ (1 + α (T - T₀ ))
Inserting the values
215.32 = 217.62 (1 + (- 5.00 x 10⁻⁴) (T - 3))
T = 24.14 ⁰C
An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.03 m/s^2. Determine the orbital period of the satellite.
Answer:
Orbital period, T = 2.02 hours
Explanation:
It is given that, an artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.03 m/s². We have to find the orbital period (T) of the satellite.
Firstly, calculating the distance between Earth and satellite. The acceleration due to gravity is given by :
[tex]a=\dfrac{GM}{r^2}[/tex]
G = universal gravitational constant
M = mass of earth
[tex]r=\sqrt{\dfrac{GM}{a}}[/tex]
[tex]r=\sqrt{\dfrac{6.67\times 10^{-11}\times 5.97\times 10^{24}}{6.03\ m/s^2}}[/tex]
r = 8126273.3 m..........(1)
Now, according to Kepler's third law :
[tex]T^2=\dfrac{4\pi^2}{GM}r^3[/tex]
Putting the value of r from equation (1) in above equation as :
[tex]T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 5.97\times 10^{24}}\times (8126273.3)^3[/tex]
[tex]T^2=53202721.01\ s[/tex]
T = 7294.01 seconds
Since, 1 hour = 3600 seconds
Converting seconds to hour we get :
So, T = 2.02 hour
So, the orbital period of the satellite is 2.02 hours.
A spherical surface completely surrounds a collection of charges. Find the electric flux (with its sign) through the surface if the collection consists of (a) a single +6.60 × 10-6 C charge, (b) a single -1.30 × 10-6 C charge, and (c) both of the charges in (a) and (b).
Answer:
(a) 6.8 x 10^5 Nm^2/C
(b) 1.47 x 10^5 Nm^2/C
(c) 5.3 x 10^5 Nm^2/C
Explanation:
According to the Gauss's theorem
Electric flux = Charge enclosed / ∈0
(a) Charge enclosed = 6 x 10^-6 C
So, Electric flux = (6 x 10^-6) / (8.854 x 10^-12) = 6.8 x 10^5 Nm^2/C
(b) Charge enclosed = -1.3 x 10^-6 C
So, Electric flux = (1.3 x 10^-6) / (8.854 x 10^-12) = 1.47 x 10^5 Nm^2/C
(c) Charge enclosed = 6 x 10^-6 + (-1.3 x 10^-6) = 4.7 x 10^-6 C
So, Electric flux = (4.7 x 10^-6) / (8.854 x 10^-12) = 5.3 x 10^5 Nm^2/C
The electric flux through a spherical surface due to enclosed charges can be computed using Gauss's Law. The flux for a +6.60 x 10^-6 C charge is outward-directed, for a -1.30 x 10^-6 C charge it is inward-directed, and with both charges, the net flux is the sum of the individual fluxes.
Explanation:The student is asking about the concept of electric flux through a spherical surface that surrounds a collection of charges, which falls under the subject of Physics (specifically electromagnetism), and it is a high school- or introductory college-level question. According to Gauss's Law, the electric flux through a closed surface is directly proportional to the enclosed electric charge. This can be calculated using the formula Φ = q/ε0, where Φ is the electric flux, q is the electric charge, and ε0 is the permittivity of free space (approximately 8.85 x 10^-12 C2/N⋅m2).
For part (a), a spherical surface surrounding a single +6.60 × 10-6 C charge would result in an outward-directed electric flux Φ = +6.60 × 10^-6 C / 8.85 × 10^-12 C2/N⋅m2.
For part (b), a spherical surface surrounding a single -1.30 × 10-6 C charge would have an inward-directed electric flux Φ = -1.30 × 10^-6 C / 8.85 × 10^-12 C2/N⋅m2.
For part (c), when both charges are enclosed, their net flux through the surface is the sum of the individual fluxes. Therefore the net electric flux is Φ = (+6.60 × 10^-6 C - 1.30 × 10^-6 C) / 8.85 × 10^-12 C2/N⋅m2, which simplifies to the sum of the charges divided by the permittivity of free space.
A ramp leading to the entrance of a building is inclied upward at an angle of 7 degree. A suitcase is to be pulled up the ramp by a handle that makes an angle of 35 degree with the horizontal. How mch force must be applied in the direction of the handle so that the component of the force parallel to the ramp is 50 lbs?
Explanation:
The angle of the handle relative to the horizontal is 35°. The angle of the ramp to the horizontal is 7°. So the angle of the handle relative to the ramp is 28°.
cos 28° = 50 / F
F = 50 / cos 28°
F = 56.6 lbs
56.6 lbs force must be applied in the direction of the handle so that the component of the force parallel to the ramp is 50 lbs.
What is Newton's second law?Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change in momentum.
As given in the problem if a ramp leading to the entrance of a building is inclined upward at an angle of 7 degrees. A suitcase is to be pulled up the ramp by a handle that makes an angle of 35 degrees horizontal.
The angle of the handle relative to the horizontal is 35°. The angle of the ramp to the horizontal is 7°. So the angle of the handle relative to the ramp is 28°.
cos 28° = 50 / Force
Force = 50 / cos 28°
Force = 56.6 lbs
Thus, the 56.6 lbs force must be applied in the direction of the handle so that the component of the force parallel to the ramp is 50 lbs
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A piece of plastic has a net charge of +9.9 μC. How many more protons than electrons does this piece of plastic have? (e = 1.60 x 10^-19 C). (Divide your answer by 10^13, and give the result to the nearest 0.01
Answer:
The number of protons 6.19 more than electron.
Explanation:
Given that,
Charge [tex]Q=+9.9\times10^{-6}\ C[/tex]
We know that,
formula of charge
[tex]Q = nc[/tex]
Where,
Q = total charge
n = number of protons
e = charge of electron
Put the value into the formula
[tex]n = \dfrac{Q}{e}[/tex]
[tex]n=\dfrac{9.9\times10^{-6}}{1.6\times10^{-19}}[/tex]
[tex]n=6.1875\times10^{13}\ C[/tex]
According to statement of question
Divide the answer by [tex]10^{13}[/tex]
[tex]n=\dfrac{6.1875\times10^{13}}{10^{13}}[/tex]
[tex]n=6.19\ C[/tex]
Hence, The number of protons 6.19 more than electron.
To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 310 m/s at 50.0° above the horizontal. It explodes on the mountainside 39.0 s after firing. What are the x and y coordinates of the shell where it explodes, relative to its firing point?
The shell has horizontal position [tex]x[/tex] and vertical position [tex]y[/tex] at time [tex]t[/tex] according to
[tex]x=\left(310\dfrac{\rm m}{\rm s}\right)\cos50.0^\circ t[/tex]
[tex]y=\left(310\dfrac{\rm m}{\rm s}\right)\sin50.0^\circ t-\dfrac g2t^2[/tex]
where [tex]g=9.80\dfrac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity.
The shell explodes after 39.0 s, at which point its coordinates are
[tex]x=7770\,\mathrm m[/tex]
[tex]y=1810\,\mathrm m[/tex]
A ball is launched at an angle of 39.8 degrees up from the horizontal, with a muzzle velocity of 6.6 meters per second, from a launch point which is 1 meters above the floor. How high will the ball be above the floor (in meters), when it is a horizontal distance of 2.7 meters away? Use 9.82 meters per second for "g".
Answer:
1.85 m
Explanation:
The horizontal velocity of the ball is
[tex]v_x = v cos \theta = (6.6 m/s) cos 39.8^{\circ}=5.1 m/s[/tex]
The horizontal distance travelled is
d = 2.7 m
And since the motion along the horizontal direction is a uniform motion, the time taken is
[tex]t= \frac{d}{v_x}=\frac{2.7 m}{5.1 m/s}=0.53 s[/tex]
The vertical position of the ball is given by
[tex]y= h + u_y t - \frac{1}{2}gt^2[/tex]
where
h = 1 m is the initial heigth
[tex]u_y = v sin \theta = (6.6 m/s) sin 39.8^{\circ}=4.2 m/s[/tex] is the initial vertical velocity
g = 9.82 m/s^2 is the acceleration due to gravity
Substituting t = 0.53 s, we find the height of the ball at this time:
[tex]y=1 m + (4.2 m/s)(0.53 s) - \frac{1}{2}(9.82 m/s^2)(0.53 s)^2=1.85 m[/tex]
Answer:
To convert kilometers per hour to meters per second, perform dimensional analysis. Remember that:
1 km = 1000 m
1 hr = 3600 seconds
Using these conversion, perform dimensional analysis:
16.2 km/ hr (1000m/1km) (1hr/60 sec) = 4.5 m/s
The analysis basically just uses the conversion factors and canceling of units. The final answer is 4.5 m/s.
_________________________________________________________
Correction: That should be *(1 hr/3600 sec). The answer is still 4.5 m/s.
___________________________________________________________
Hope this helps, i did the test and this answer was right, oh and brainliest, Good luck.
The capacitor in the flash of a disposable camera has a value of 165 μF. 1) What is the resistance of the filament in the bulb if it takes 10 s to charge the capacitor to 80% of its maximum charge? (Express your answer to two significant figures.)
Answer:
3.8 x 10⁴ Ω
Explanation:
C = Capacitance = 165 μF
R = resistance of the filament = ?
t = time taken to charge the capacitor = 10 s
Q₀ = maximum charge stored by capacitor
Q = charge stored by capacitor at time "t" = 0.80 Q₀
T = Time constant
Charge stored by capacitor at any time is given as
[tex]Q = Q_{o}(1 - e^{\frac{-t}{T}})[/tex]
[tex]0.80 Q_{o} = Q_{o}(1 - e^{\frac{-10}{T}})[/tex]
T = 6.21 s
Time constant is given as
T = RC
6.21 = R (165 x 10⁻⁶)
R = 3.8 x 10⁴ Ω
The resistance of the filament in the bulb R = 3.8 x 10⁴ Ω
What will be the resistance in the filament of the bulb?It is given that:-
Capacitance C= 165 μF
Resistance of the filament = R=?
Time taken to charge the capacitor = t = 10 s
Now,
Q₀ = maximum charge which can be stored by the capacitor
Since the capacitor to 80% of its maximum charge
Then,
Q = charge stored by capacitor at time "t" = 0.80 Q₀
T = Time constant
The charge stored by the capacitor at any time is given as
[tex]Q=Q_{0} (1-e^{\dfrac{-t}{T} } )[/tex]
[tex]0.80Q_{0} =Q_{0} (1-e^{\dfrac{-10}{T} } )[/tex]
[tex]T=6.21 sec[/tex]
Now the Time constant is given as
[tex]T=R\times C[/tex]
[tex]6.21= R\times (165\times 10^{-6} )[/tex]
[tex]R=3.8\times 10^{4}[/tex]Ω
The resistance of the filament in the bulb R = 3.8 x 10⁴ Ω
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A potato is shot out of cylinder at an angle of 17 degrees above the horizontal with an initial speed of 20 m/s. What is its maximum height?
Answer:
Maximum height, h = 1.74 meters
Explanation:
It is given that,
A potato is shot out of the cylinder. It is a case of projectile motion. The potato makes an angle of 17 degrees above the horizontal.
Initial speed with which the potato is shot out, u = 20 m/s
We have to find the maximum height of the potato. The maximum height of a projectile (h) is given by the following formula as :
[tex]h=\dfrac{u^2sin^2\theta}{2g}[/tex]
Where
[tex]\theta[/tex] = angle between the projectile and the surface
g = acceleration due to gravity
[tex]h=\dfrac{(20\ m/s)^2sin^2(17)}{2\times 9.8\ m/s^2}[/tex]
h = 1.74 m
or h = 1.74 meters
Hence, this is the required solution.
Formula One race cars are capable of remarkable accelerations when speeding up, slowing down, and turning corners. At one track, cars round a corner that is a segment of a circle of radius 95 m at a speed of 68 m/s. What is the approximate magnitude of the centripetal acceleration, in units of g?
Answer:
Centripetal acceleration of the car is (4.96 g) m/s²
Explanation:
It is given that,
Radius of circle, r = 95 m
Speed of the car, v = 68 m/s
We need to find the centripetal acceleration. It is given by :
[tex]a_c=\dfrac{v^2}{r}[/tex]
So, [tex]a_c=\dfrac{(68\ m/s)^2}{95\ m}[/tex]
[tex]a_c=48.67\ m/s^2[/tex]
Since, g = 9.8 m/s²
So,
[tex]a_c=(4.96\ g)\ m/s^2[/tex]
So, the magnitude of the centripetal acceleration is (4.96 g) m/s². Hence, this is the required solution.
The potential difference between two parallel plates is 227 V. If the plates are 6.8 mm apart, what is the electric field between them? O S.0x10 N/C O 28 x 10* N/C O 4.1 x 10 N/C O 3.3 x 10 N/C
Answer:
[tex]E=3.3\times 10^4N/C[/tex]
Option D is the correct answer.
Explanation:
Electric field, E is the ratio of potential difference and distance between them.
Potential difference, V = 227 V
Distance between plates = 6.8 mm = 0.0068 m
Substituting,
[tex]E=\frac{V}{d}=\frac{227}{0.0068}=3.3\times 10^4N/C[/tex]
Option D is the correct answer.
Suppose a disk with constant angular velocity has rotational kinetic energy 1280 J. If the moment of inertia of the disk is 35 kg-m^2, then what is its angular velocity? (a) 7.604 rad/s (b) 8.552 rad/s (c) 10.12 rad/s (d) 6.818 rad/s (e) 9.952 rad/s (f) 8.935 rad/s f
Answer:
Angular velocity of the disk is 8.552 rad/s
Explanation:
It is given that,
Rotational kinetic energy, KE = 1280 J
The moment of inertia of the disk, I = 35 kg m²
We have to find the angular velocity of the disk. In rotational mechanics the kinetic energy of the disk is given by :
[tex]KE=\dfrac{1}{2}I\omega^2[/tex]
[tex]\omega=\sqrt{\dfrac{2KE}{I}}[/tex]
[tex]\omega=\sqrt{\dfrac{2\times 1280\ J}{35\ kgm^2}}[/tex]
[tex]\omega=8.552\ rad/s[/tex]
Hence, the angular velocity of the disk is 8.552 rad/s.
please explain vector addition, triangle method and parallelogram method
Start with 2 arbitrary vectors, [tex]\vec v_1[/tex] and [tex]\vec v_2[/tex]. (pic 1)
Vectors are determined by their lengths and direction. This means that translating the vector (i.e. sliding it left/right and up/down in the plane) doesn't fundamentally change that vector. To this end, we could just as easily represent [tex]\vec v_2[/tex] as if it had originated from the tip of [tex]\vec v_1[/tex]. This "new" [tex]\vec v_2[/tex] and the "old" [tex]\vec v_2[/tex] are the same vector. (pic 2)
If we connect the origin of [tex]\vec v_1[/tex] with the tip of "new" [tex]\vec v_2[/tex], we get a new vector, and this we define as the vector sum [tex]\vec v_1+\vec v_2[/tex]. (pic 3)
We can do this other way, by first traslating [tex]\vec v_1[/tex] to the tip of [tex]\vec v_2[/tex], then connecting the origin of [tex]\vec v_2[/tex] with the tip of "new" [tex]\vec v_1[/tex]. This demonstrates that vector addition is commutative (order of the vectors being added doesn't matter - you always end up at the same terminus). The "parallelogram method" refers to how a parallelogram is traced out. (pic 4)
Multiplying a vector by -1 reverses its direction. (pic 5)
Adding [tex]\vec v_1[/tex] and [tex]-\vec v_2[/tex] works the same way as standard vector addition, giving us the new vector [tex]\vec v_1-\vec v_2[/tex]. (pic 6)
We can do the same in the reverse order, but now we get a different vector, [tex]\vec v_2-\vec v_1[/tex]. (pic 7)
These vectors have the same length but point in opposite directions. (pic 8)
But notice that we can translate the vectors [tex]\vec v_1-\vec v_2[/tex] and [tex]\vec v_2-\vec v_1[/tex] so that we get a vector that either starts at the tip of [tex]\vec v_2[/tex] and ends at the tip of [tex]\vec v_1[/tex] (pic 9), or starts at the tip of [tex]\vec v_1[/tex] and ends at the tip of [tex]\vec v_2[/tex] (pic 10). The "triangle method" refers to the triangles that are traced out by either vector sum [tex]\vec v_1-\vec v_2[/tex] and [tex]\vec v_2-\vec v_1[/tex] together with [tex]\vec v_1[/tex] and [tex]\vec v_2[/tex].
Calculate the magnitude of the linear momentum for the following cases. (a) a proton with mass 1.67 10-27 kg, moving with a speed of 5.10 106 m/s kg · m/s (b) a 15.0-g bullet moving with a speed of 480 m/s kg · m/s (c) a 72.5-kg sprinter running with a speed of 10.5 m/s kg · m/s (d) the Earth (mass = 5.98 1024 kg) moving with an orbital speed equal to 2.98 104 m/s. kg · m/s
The magnitude of linear momentum can be calculated using the equation p=mv. The momentum for the different objects given are: (a) Proton: 8.517 * 10^-21 kg*m/s, (b) Bullet: 7.2 kg*m/s, (c) Sprinter: 761.25 kg*m/s, and (d) Earth: 1.78324 * 10^29 kg*m/s.
Explanation:The magnitude of the linear momentum of an object is calculated using the equation p=mv, where 'p' is the momentum, 'm' is the mass of the object, and 'v' is its velocity.
(a) Using the given values, m = 1.67 * 10^-27 kg and v = 5.10 * 10^6 m/s, the momentum of the proton is p = m*v = (1.67 * 10^-27 kg)*(5.10 * 10^6 m/s) = 8.517 * 10^-21 kg*m/s.
(b) The bullet's mass is 15.0 g which is 0.015 kg. Its velocity is 480 m/s. Thus, its momentum is p = m*v = (0.015kg)*(480m/s) = 7.2 kg*m/s.
(c) The sprinter's mass is 72.5 kg and velocity is 10.5 m/s. So, his momentum is p = m*v = (72.5kg)*(10.5m/s) = 761.25 kg*m/s.
(d) The Earth's mass is 5.98 * 10^24 kg and its orbital speed is 2.98 * 10^4 m/s. Hence, its momentum is p = m*v = (5.98 * 10^24 kg)*(2.98 * 10^4 m/s) = 1.78324 * 10^29 kg*m/s.
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The magnitude of the linear momentum for a proton, a bullet, a sprinter, and the Earth is approximately 8.52 * 10^-21 kg*m/s, 7.2 kg*m/s, 760.5 kg*m/s, and 1.78 * 10^29 kg*m/s, respectively.
Explanation:To calculate the magnitude of the linear momentum, we need to multiply the mass of the object by its velocity, which follows the formula: momentum = mass * velocity.
For the given cases:
Proton (1.67 * 10^-27 kg, 5.10 * 10^6 m/s): Using the momentum formula, you multiply these values together to get an approximate momentum of 8.52 * 10^-21 kg*m/s.Bullet (15.0 * 10^-3 kg, 480 m/s): Similarly, multiplying these entities gives us a momentum of approximately 7.2 kg*m/s.Sprinter (72.5 kg, 10.5 m/s): For the sprinter, the momentum will be approximately 760.5 kg*m/s.Earth (5.98 * 10^24 kg, 2.98 * 10^4 m/s): The Earth's momentum is a whopping 1.78 * 10^29 kg*m/s!Learn more about Linear Momentum here:https://brainly.com/question/34645079
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A 75 cm straight wire moves straight up through a 0.53 T magnetic field with a velocity of 16 m/s. What is the induced emf in the wire? O 3.18 V O 6.36 v O 4.78 V O 12.3V
Answer:
Induced emf of the wire is 6.36 Volts.
Explanation:
It is given that,
Length of the wire, l = 75 cm = 0.75 m
Magnetic field, B = 0.53 T
Velocity, v = 16 m/s
The wire is moving straight up in the magnetic field. So, an emf is induced in the wire. It is given by :
[tex]\epsilon=Blv[/tex]
[tex]\epsilon=0.53\ T\times 0.75\ m\times 16\ m/s[/tex]
[tex]\epsilon=6.36\ V[/tex]
So, the induced emf of the wire is 6.36 V. Hence, the correct option is (b) "6.36 V".
Answer:
6.36
Explanation:
If a 75 cm straight wire moves straight up through a 0.53 T magnetic field with a velocity of 16 m/s, the induced emf in the wire is 6.36.
Consider two metallic rods mounted on insulated supports. One is neutral, the other positively charged. You bring the two rods close to each, but without contact, and briefly ground the the neutral rod by touching it with your hand. show answer Correct Answer What would be resulting charge (if any) on the initially neutral rod
Answer:
I think it will be half of the initial charge
Explanation:
Because we know, the resulting charge will be q1+q2/2, since one is neutral so the charge will be half q/2
A person walks across a 60 m wide room with a constant speed of 2 m/s, and they run back with a constant speed of 6 m/s. What is their average speed for the trip? a. 2 m/s b 3m/s c. 4 m/s d. 5 m/s 2 e 6 m/s
Answer:
Option B is the correct answer.
Explanation:
Total distance traveled = 60 + 60 = 120 m
Time taken to walk across [tex]=\frac{60}{2}=30s[/tex]
Time taken to run back [tex]=\frac{60}{6}=10s[/tex]
Total time taken = 30 + 10 = 40 s
Average speed = Total distance traveled / Total time taken
Average speed [tex]=\frac{120}{40}=3m/s[/tex]
Option B is the correct answer.
15. Give an example for some particles or waves that are moving faster than light in everyday life 16. What is a laser? 17. What is an oscilloscope? 18. What does it means practically that nothing is faster than light in vacuum? 19. What is vacuum?
Answer:
15.Radiowave
16.laser is device that generates an intense beam of other electromagnetic radiation by emission of photons from excited atoms.
17.this is a laboratory instrument commonly used to display and analyse the waveformof electronic signals.
19. this is a space entirely devoid of matter.
Julie is walking around a track at a 2m/s for some exercise. She then decides to start jogging so she accelerates at a rate of 0.5m/s² for 3 seconds. How far did Julie travel from the time she started to accelerate to the end of the 3 seconds?
Answer:
The distance is 8.25 m.
Explanation:
Given that,
Speed = 2 m/s
Acceleration = 0.5 m/s^2
Time = 3 sec
We need to calculate the distance
Using equation of motion
[tex]s = ut+\dfrac{1}{2}at^2[/tex]
Where, u = initial velocity
a = acceleration
t = time
Put the value in the equation
[tex]s=2\times3+\dfrac{1}{2}\times0.5\times3^2[/tex]
[tex]s=8.25\ m[/tex]
Hence, The distance is 8.25 m.
Julie's distance at the end of the 3 seconds is 8.25 m.
To calculate the distance traveled by Julie at the end ]of 3 seconds, we use the formula below.
Formula:
s = ut+at²/2.............. Equation 1Where:
s = distance traveled by Julie after 3 secondsu = Initial velocityt = timea = acceleration.From the question,
Given:
u = 2 m/sa = 0.5 m/s²t = 3 secondsSubstitute these given values into equation 1
s = 2(3)+(0.5×3²)/2s = 6+2.25s = 8.25 mHence, Julie's distance at the end of the 3 seconds is 8.25 m.
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A pitcher claims he can throw a 0.146-kg baseball with as much momentum as a 2.70-g bullet moving with a speed of 1.50 ✕ 103 m/s. (a) What must the baseball's speed be if the pitcher's claim is valid? (b) Which has greater kinetic energy, the ball or the bullet? a. The bullet has greater kinetic energy. b. The ball has greater kinetic energy. c. Both have the same kinetic energy.
(a) The pitcher must throw the ball at 27.7 m/s
The momentum of an object is given by:
[tex]p=mv[/tex]
where
m is the mass of the object
v is the object's velocity
Let's calculate the momentum of the bullet, which has a mass of
m = 2.70 g = 0.0027 kg
and a velocity of
[tex]v=1.50\cdot 10^3 m/s[/tex]
Its momentum is:
[tex]p=mv=(0.0027 kg)(1.50\cdot 10^{3} m/s)=4.05 kg m/s[/tex]
The pitcher must throw the baseball with this same momentum. The mass of the ball is
m = 0.146 kg
So the velocity of the ball must be
[tex]v=\frac{p}{m}=\frac{4.05 kg m/s}{0.146 kg}=27.7 m/s[/tex]
So, the pitcher must throw the ball at 27.7 m/s.
(b) a. The bullet has greater kinetic energy
The kinetic energy of an object is given by
[tex]K=\frac{1}{2}mv^2[/tex]
where m is the mass of the object and v is its speed.
For the bullet, we have:
[tex]K=\frac{1}{2}(0.0027 kg)(1.50\cdot 10^3 m/s)^2=3037.5 J[/tex]
For the ball:
[tex]K=\frac{1}{2}(0.146 kg)(27.7 m/s)^2=56.0 J[/tex]
So, the bullet has greater kinetic energy.
Final answer:
The baseball's speed must be 27.74 m/s to match the momentum of the bullet, and the bullet has greater kinetic energy than the baseball.
Explanation:
Calculating Baseball's Speed and Comparing Kinetic Energies
To validate the pitcher's claim that a 0.146-kg baseball can have the same momentum as a 2.70-g bullet traveling at 1.50 x 10³ m/s, we must use the formula for momentum (p = mv), which gives us:
Momentum of bullet = (2.70 g) x (1.50 x 10³ m/s) = (0.0027 kg) x (1500 m/s) = 4.05 kg m/s.
Therefore, the baseball's speed v needed to have the same momentum is calculated by rearranging the formula to v = p/m, which gives us:
Baseball's speed v = 4.05 kg m/s / 0.146 kg = 27.74 m/s.
To determine which has the greater kinetic energy, we use the kinetic energy formula KE = (1/2)mv². Calculating the kinetic energy of both the bullet and the baseball:
KE of bullet = (1/2)(0.0027 kg)(1500 m/s)² = 3037.5 J.
KE of baseball = (1/2)(0.146 kg)(27.74 m/s)² = 55.90 J.
Comparing the results shows that the bullet has greater kinetic energy than the baseball.
On January 22, 1943, in Spearfish, South Dakota, the temperature rose from −4.00°F to 45.0°F over the course of two minutes (the current world record for the fastest recorded temperature change). By how much did the temperature change on the Kelvin scale? HINT
Answer:
27.22 K
Explanation:
T₁ = initial temperature in fahrenheit = - 4.00 ⁰F
T₂ = final temperature in fahrenheit = 45.0 ⁰F
To convert the temperature from fahrenheit to kelvin, we can use the relation
[tex]K = \frac{F - 32}{1.8} + 273.15[/tex]
where F = Temperature in fahrenheit and K = temperature in kelvin
T'₁ = initial temperature in kelvin = (- 4.00 - 32)/1.8 + 273.15 = 253.15 K
T'₂ = final temperature in kelvin = (45.0 - 32)/1.8 + 273.15 = 280.37 K
ΔT = Change in temperature
Change in temperature on kelvin scale is given as
ΔT = T'₂ - T'₁
ΔT = 280.37 - 253.15
ΔT = 27.22 K
To find the temperature change on the Kelvin scale, convert the given temperatures from Fahrenheit to Kelvin and subtract them.
Explanation:The temperature change on the Kelvin scale can be determined by converting the given temperatures from Fahrenheit to Kelvin and then finding the difference between them.
First, convert -4.00°F to Kelvin:
273.15 K + (-4.00°F + 459.67 °F) × (5/9)
Next, convert 45.0°F to Kelvin:
273.15 K + (45.0°F + 459.67 °F) × (5/9)
Finally, subtract the two Kelvin temperatures to find the temperature change.
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For some applications, it is important that the value of a resistance not change with temperature. For example, suppose you made a 3.80?k? resistor from a carbon resistor and a Nichrome wire-wound resistor connected together so the total resistance is the sum of their separate resistances.What value should each of these resistors have (at 0 ?C) so that the combination is temperature independent?(two answers)
To achieve a temperature independent resistance of 3.80 kΩ, you need to use a carbon resistor and Nichrome wire-wound resistor that counterbalance each other. This is possible because carbon and Nichrome have opposite temperature coefficients of resistance.
Explanation:In order for the resistance to remain constant with temperature, the carbon resistor and the Nichrome wire-wound resistor must counterbalance each other. Meaning, when one's resistance increases with temperature, the other's resistance decreases, keeping the total resistance the same. Given that carbon and Nichrome have opposite temperature coefficients of resistance, they can accomplish this task.
Generally, the resistance R of a resistor is given by the formula R = R0(1 + α(T-T0)), where α is the temperature coefficient, T is the temperature and R0 is the resistance at reference temperature T0. As the temperature increases, a positive α will increase the resistance while a negative α will decrease it.
To make the combined resistance temperature independent, the sum of the change in resistance of the carbon resistor and the Nichrome resistor should be zero. Therefore, you would set up the equation where the increase of the carbon resistance equals the decrease of the Nichrome resistance. Solving this equation will give you the exact values required for the resistances of carbon and Nichrome at 0ºC in order to have a total resistance of 3.80 kΩ.
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A wire with a weight per unit length of 0.071 N/m is suspended directly above a second wire. The top wire carries a current of 72.8 A, and the bottom wire carries a current of 72.7 A. The permeability of free space is 4π × 10−7 T · m/A . Find the distance of separation between the wires so that the top wire will be held in place by magnetic repulsion. Answer in units of mm.
Answer:
d = 15 mm
Explanation:
Force of repulsion between two current carrying wire is given by
[tex]F = \frac{\mu_0 i_1 i_2 L}{2\pi d}[/tex]
now this force of repulsion is counterbalanced by the weight of the wire
so we have
[tex]mg = \frac{\mu_0 i_1 i_2 L}{2\pi d}[/tex]
now we have
[tex]d = \frac{\mu_0 i_1 i_2 L}{2\pi mg}[/tex]
so here we can say that
[tex]d = \frac{\mu_0 i_1 i_2}{2\pi (m/L)g}[/tex]
now plug in all values in it
[tex]d = \frac{4\pi \times 10^{-7} (72.8)(72.7)}{2\pi (0.071)}[/tex]
[tex]d = 0.015 m[/tex]
d = 15 mm
A 1.0-kg block moving to the right at speed 3.0 m/s collides with an identical block also moving to the right at a speed 1.0 m/s. Both blocks stick together and move to the right. What is their speed after collision?
Answer:
Speed of both blocks after collision is 2 m/s
Explanation:
It is given that,
Mass of both blocks, m₁ = m₂ = 1 kg
Velocity of first block, u₁ = 3 m/s
Velocity of other block, u₂ = 1 m/s
Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :
[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]
[tex]v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]
[tex]v=\dfrac{1\ kg\times 3\ m/s+1\ kg\times 1\ m/s}{2\ kg}[/tex]
v = 2 m/s
Hence, their speed after collision is 2 m/s.