Answer:
-71 V
Explanation:
Given that,
An electron starts from rest at point A and has a speed of 5.0 × 10⁶ m/s at point B.
We need to find the electric potential difference [tex]V_A-V_B[/tex]. It can be calculated by using the conservation of charges as follows :
[tex]qV=\dfrac{1}{2}mv^2[/tex]
m and q are mass and charge on electrons
[tex]V=\dfrac{mv^2}{2(-q)}\\\\V=-\dfrac{9.1\times 10^{-31}\times (5\times 10^6)^2}{2\times 1.6\times 10^{-19}}\\\\V=-71.09\ V[/tex]
So, the electric potential difference is (-71 V). So, the correct option is (a).
1. Synthesize Information You push your
younger sister on a swing in a park. Then you
give her a harder push. Explain what happens
in each case, in terms of the second and third
laws of motion
Answer:separate
Explanation:
A car travels at a constant speed up a ramp making an angle of 28 degrees with the horizontal component of velocity is 40 kmh^-1, find the vertical and horizontal component of velocity.
Answer:
Vx = 35.31 [km/h]
Vy = 18.77 [km/h]
Explanation:
In order to solve this problem, we must decompose the velocity component by means of the angle of 28° using the cosine function of the angle.
[tex]v_{x} = 40*cos(28)\\V_{x} = 35.31 [km/h][/tex]
In order to find the vertical component, we must use the sine function of the angle.
[tex]V_{y}=40*sin(28)\\V_{y} = 18.77 [km/h][/tex]
List four safety measures one should take when there is a cyclone warning.
Answer:
I'm gonna give you more than 4 So u can choose.
Turn off all electricity, gas and water.
Unplug all electrical appliances.
Keep your emergency kit close at hand.
If the building starts to collapse, protect yourself with mattresses, rugs.
Stay tuned to the radio for updates.
Stay inside until told it is safe to go outside.
Superman strikes a golf ball on the ground at a 38 degree angle above the horizontal at 147 m/s. What is the maximum height the golf ball reaches?
255 m
418 m
687 m
1103 m
The ball is hit with an initial vertical velocity of
(147 m/s) sin(38º) ≈ 90.5 m/s
Recall that
v ² - u ² = 2 a ∆y
where u and v are initial and final velocities, respectively; a is acceleration; and ∆y is displacement.
Vertically, the ball is in freefall, so it is only subject to acceleration due to gravity, with magnitude g = 9.80 m/s² in the downward direction. At its maximum height, the ball has zero vertical velocity (v = 0) and the displacement is equal to the maximum height, so
0² - (90.5 m/s)² = 2 (-g) ∆y
∆y = (90.5 m/s)² / (2g)
∆y ≈ 418 m
A small truck has a mass of 2085 kg. How much work is required to decrease the speed of the vehicle from 22.0 m/s to 13.0 m/s on a level road?
Answer:
Work required is 328387.5 Joules.
Explanation:
Given the following data;
Mass = 2085kg
Initial velocity, Vi = 13m/s
Final velocity, Vf =22m/s
To find the workdone;
We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.
Mathematically, it is given by the equation;
W = Kf - Ki
Where;
W is the work required.
Kf is the final kinetic energy possessed by the object.
Ki is the initial kinetic energy possessed by the object.
But Kinetic energy = ½MV²
W = ½MVf² - ½MVi²
Substituting into the equation, we have;
W = ½(2085)*22² - ½(2085)*13²
Simplifying the equation, we have;
W = 1042.5 * 484 - 1042.5 * 169
W = 504570 - 176182.5
W = 328387.5J
Therefore, the work required to decrease the speed of the vehicle is 328387.5 Joules.
A barge is pulled by two tugboats. If the resultant of the forces exerted by the
tugboats is 5kN force directed along the axis of the barge, determine the tension in
each of the ropes for α = 45
Answer:
Approximately [tex]3.5\; \rm kN[/tex] in each of the two ropes.
Explanation:
Let [tex]F_1[/tex] and [tex]F_2[/tex] denote the tension in each of the two ropes.
Refer to the diagram attached. The tension force in each rope may be decomposed in two directions that are normal to one another.
The first direction is parallel to resultant force on the barge.
The component of [tex]F_1[/tex] in that direction would be [tex]\displaystyle F_1\cdot \cos(45^\circ) = \frac{F_1}{\sqrt{2}}[/tex].Similarly, the component of [tex]F_2[/tex] in that direction would be [tex]\displaystyle F_2\cdot \cos(45^\circ) = \frac{F_2}{\sqrt{2}}[/tex].These two components would be in the same direction. The resultant force in that direction would be the sum of these two components: [tex]\displaystyle \frac{F_1 + F_2}{\sqrt{2}}[/tex]. That force should be equal to [tex]5\; \rm kN[/tex].
The second direction is perpendicular to the resultant force on the barge.
The component of [tex]F_1[/tex] in that direction would be [tex]\displaystyle F_1\cdot \sin(45^\circ) = \frac{F_1}{\sqrt{2}}[/tex].Similarly, the component of [tex]F_2[/tex] in that direction would be [tex]\displaystyle F_2\cdot \sin(45^\circ) = \frac{F_2}{\sqrt{2}}[/tex].These two components would be in opposite directions. The resultant force in that direction would be the difference of these two components: [tex]\displaystyle \frac{F_1 - F_2}{\sqrt{2}}[/tex]. However, the net force on the barge is normal to this direction. Therefore, the resultant force should be equal to zero.
That gives a system of two equations and two unknowns:
[tex]\displaystyle \frac{F_1 + F_2}{\sqrt{2}} = 5\; \rm kN[/tex], and[tex]\displaystyle \frac{F_1 - F_2}{\sqrt{2}} = 0[/tex].The second equation suggests that [tex]F_1 = F_2[/tex]. Hence, replace the [tex]F_2[/tex] in the first equation with [tex]F_1[/tex] and solve for [tex]F_1\![/tex]:
[tex]F_1 = \displaystyle \frac{5\; \rm kN}{2\, \sqrt{2}} \approx 3.5\; \rm kN[/tex].
Because [tex]F_1 = F_2[/tex] (as seen in the second equation,) [tex]F_2 = F_1 \approx 3.5\; \rm kN[/tex].
In other words, the tension in each of the two ropes is approximately [tex]3.5\; \rm kN[/tex].
The tension in each of the rope is 3,535.5 N
The given expression:
the resultant force, R = 5 kN = 5000 N
the angle in between the forces, α = 45
To find:
the tension in each of the ropesThe tension in each of the ropes is calculated as follows;
The tension in vertical direction
[tex]F_y = F \times sin(\alpha)\\\\F_y = 5000 \times sin(45)\\\\F_y = 5000 \times 0.7071\\\\F_y = 3,535.5 \ N[/tex]
The tension in horizontal direction;
[tex]F_x = F \times cos(\alpha)\\\\F_x = 5000 \times cos(45)\\\\F_x = 5000 \times 0.7071\\\\F_x = 3,535.5 \ N[/tex]
Thus, the tension in each of the rope is 3,535.5 N
Learn more here:https://brainly.com/question/11194858
Lauren walks 100m in half a minute. What must her speed have been to travel this distance?
Answer:
3.33 m/s
Explanation:
Using the formula: s = d / t (where s is speed, d is distance, and t is time)
[Convert minutes to seconds then solve]
Half a minute is 30 seconds. Therefore:
s = 100 / 30 = 3.33 m/s
A piano string having a mass per unit length equal to 4.70 10-3 kg/m is under a tension of 1 400 N. Find the speed with which a wave travels on this string.
Answer:
The speed of the sound wave on the string is 545.78 m/s.
Explanation:
Given;
mass per unit length of the string, μ = 4.7 x 10⁻³ kg/m
tension of the string, T = 1400 N
The speed of the sound wave on the string is given by;
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
where;
v is the speed of the sound wave on the string
Substitute the given values and solve for speed,v,
[tex]v = \sqrt{\frac{T}{\mu} }\\\\v = \sqrt{\frac{1400}{4.7*10^{-3}} }\\\\v = \sqrt{297872.34}\\\\v = 545.78 \ m/s[/tex]
Therefore, the speed of the sound wave on the string is 545.78 m/s.
can someone help im not sure of my answer
Answer:
Yes
Explanation:
But I feel also 2 is correct but your answer is right
А A pool of water of refractive index
4/3! is 60cm deep. Find its apparent
depth when viewed vertically through
air.
Answer:
Apparent depth = 45 cm
Explanation:
The refractive index of water in a pool, n = 4/3
Real depth, d = 60 cm
We need to find its apparent depth when viewed vertically through air. The ratio of real depth to the apparent depth is equal to the refractive index of the material. Let the apparent depth is d'. So,
[tex]n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{60}{\dfrac{4}{3}}\\\\d'=45\ cm[/tex]
So, the apparent depth is 45 cm.
as an object falls:
a) both velocity and acceleration increase
b) velocity increases and acceleration decreases
c) velocity increases and acceleration is unchanged
Answer:
c it is not accelerating on it's on but gravity pulls it there for velocity increases.
As an object falls, velocity increases and acceleration is unchanged. Therefore, option C is correct.
What do you mean by velocity ?The term velocity is defined as the direction of the movement of the body or the object.
Speed is fundamentally a scalar quantity. Velocity is, in essence, a vector quantity. It is the rate at which distance changes. It is the displacement rate of change.
The velocity of something is the rate at which it moves in a specific direction. For example, the speed of a car driving north on a highway or the speed of a rocket after launch. The scalar denotes that the absolute magnitude of the velocity vector is always the speed of motion.
Thus, it is not accelerating on it's on, but gravity pulls it there as velocity increases.
To learn more about velocity, follow the link:
https://brainly.com/question/18084516
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how much heat is necessary to warm 500g of water from 20°c to 65°c
Explanation:
so sorry
don't know but please mark me as brainliest please