Answer:8
Explanation:
A 60 kg cyclist approaches the bottom of a gradual hill at a speed of 11 m/s. The hill is 5.0 m high, and the cyclist estimates that she is going fast enough to coast up and over it without peddling. Using conservation of energy, find the speed of the cyclist when she reaches the top of the hill (ignoring air resistance and friction). g
Answer:
9.91m/s
Explanation:
According to conservation of energy, the potential energy at the top of the hill is equal to the kinetic energy along the hill.
PE = KE
mgh = 1/2mv²
m is the mass of the cyclist
g is the acceleration due to gravity
h is the height of the hill
v is the speed of the cyclist
From the formula:
gh = v²/2
Given
g = 9.81m/s²
h = 5.0m
Substitute into the formula:
9.81(5.0) = v²/2
9.81*5*2 = v²
98.1 = v²
v = √98.1
v = 9.91m/s
Hence the speed of the cyclist when she reaches the top of the hill is 9.91m/s
Question 3: In one half hour, a car traveled 25.2 km. Its average speed is
50.4 km/min
12.6 km/hr
14 m/s
none of them
Answer:
12.6 km/hr
explanation:
25.2 x [tex]\frac{1}{2}[/tex]
A 0.1-kg meter stick is supported at both ends by strings, and there is a 200-g mass attached to it at the 70 cm mark. What is the force exerted by the string connected at the 0 cm mark, if the system is in static equilibrium
Answer:
The force exerted by the string connected at the 0 cm mark is 1.078 N
Explanation:
Given;
mass of the meter stick, m = 0.1 kg
weight of the meter stick, = 0.1 kg x 9.8 m/s² = 0.98 N
the weight attached at 70 cm mark = 0.2 kg x 9.8 m/s² = 1.96 N
0cm--------------------------50cm-----------70cm-------------100cm
↑ ↓ ↓ ↑
F₁ 0.98N 1.96N F₂
Take the moment about F₂: clockwise moment = anticlockwise moment
F₁(100 - 0) = 0.98(100 -50) + 1.96(100 -70)
100F₁ = 49 + 58.8
100F₁ = 107.8
F₁ = 107.8 / 100
F₁ = 1.078 N
Therefore, the force exerted by the string connected at the 0 cm mark is 1.078 N
A car comes to a stop in 3 s from the moment the driver applies the brakes. If the initial angular velocity of the tires was 57 rad/s, what is their angular acceleration (assumed constant) over this time interval
Answer:
-19rad/s
Explanation:
Given
Initial angular velocity w0 = 57rad/s
Time = 3s
Final angular velocity wf = 0rad/s (car comes to stop)
Required
angular acceleration (α)
Using the equation of motion
wf = w0 + αt
Substitute the given value and get α;
0 = 57 + 3α
-57 = 3α
α = -57/3
α = -19rad/s (the negative sign shows that the car decelerated over time)
Which statement is true regarding eligibility for becoming pharaoh?
People believed that the gods selected the pharaoh.
A woman could be selected as pharaoh.
The pharaoh’s title had to pass to a royal son.
The title of pharaoh was passed to a military commander.
Answer:
A woman could be selected as pharaoh.
Explanation:
Most pharaohs were men, but there were some exceptions, such as queen Hatshepsut.
Answer:
A
Explanation:
A toy cork gun contains a spring whose spring constant is 10.0 N/m. The spring is compressed 5.00 cm and then used to propel a 6.00-g cork. The cork, however, sticks to the spring for 1.00 cm beyond its unstretched length before separation occurs. The muzzle velocity of this cork is:
Answer:
2 m/s
Explanation:
We will apply the law of conservation of energy to solve this.
The initial spring potential energy = final spring potential energy + kinetic energy of cork, this means that
1/2kx² = 1/2kx'² + 1/2mv², and this is
kx² = kx'² + mv²
From the question, we are given that
k = 10N/m
x = 5cm = 0.05 m
x' = 1cm = 0.01 m
m = 6g = 0.006 kg
Substituting these values to the formula, and we get,
10(0.05)² = 10(0.01)² + 0.006v²
10 * 0.0025 = 10 * 0.0001 + 0.006 v²
0.025 = 0.001 + 0.006v², multiplying all sides by 1000, we have
25 = 1 + 6v²
6v² = 25 - 1
6v² = 24
v² = 24 / 6
v² = 4
v = √4
v = 2 m/s
⇒25 - 1 = 6v²
⇒v² = 24/6 = 4
⇒v = 2 m/s
1. Two loads are connected in parallel to a 120 V source. If the 40 W lamp is left on for 3 hr/day and the 75 W lamp is left on for 5 hr/day calculate the current in each loa
Answer:
Explanation:
Since both loads at connected in parallel, the voltage across them will be the same.
The voltage across parallel connected loads are the same.
Let the loads be R1 and R2
Power = IV
I is the current
V is the voltage
P= V²/R
R = V²/P
Get the first load R1
R1 = V²/P1
R1 = 120²/40
R1 = 360ohms
Get the second load R2
R2 = V²/P2
R2 = 120²/75
R2 = 192ohms
Current in R1 will be;
P1 = I1²R1
I1² = P/R1
I1² = 40/360
I1² = 1/9
I1 = √1/9
I1 = 1/3
I = 0.33A
Current in R2(second load) will be;
P2 = I2²R2
I2² = P/R2
I2² = 75/192
I2² = 0.390625
I2 = √0.390625
I2 = 0.625A
draw a roller coaster track that starts with one large hill, then is followed
by a valley and another, smaller hill.
Draw a cart in four positions on the track as outlined below.
i. Draw the first cart at the top of the first hill. Label it A.
ii. Draw the second cart going down the first hill into the valley. Label it B.
iii. Draw the third cart at the bottom of the valley. Assume that the height of the cart in
this position is zero. Label it C.
iv. Draw the last cart at the top of the second, smaller hill. Label it D.
okay, this is the part that I don't know lol. ill give you brianlist or whatever its called:
Type one to two paragraphs describing the changes in potential and kinetic energy of the cart.
Be sure to discuss how the potential and kinetic energy of the cart changes at each of the four
positions along the track, and explain why these changes occur.
Answer:
The cart starts out in position A with high potential energy, low kinetic energy, and some thermal energy. As the cart progresses into position B, the kinetic energy begins to increase and the potential energy begins to decrease; as the thermal energy increases as thermal energy from the track is transferred to the cart through friction. Once the cart reaches position C, it has high kinetic energy, low potential energy, and some thermal energy. At position D, the cart has high potential energy, low kinetic energy, and some thermal energy. The potential energy throughout is correlated to gravity, the kinetic energy is correlated to momentum, and the thermal energy is correlated to friction. The potential energy is at its maximum during position A, and its minimum at position C; the kinetic energy is at its maximum during position C, and its minimum at position A; the thermal energy is at its maximum during position B, and its minimum at position A.
Explanation:
At the top of the hill the cart has maximum potential energy and zero kinetic energy. At the valley the cart has maximum kinetic energy and zero potential energy.
According to the principle of conservation of mechanical energy when an object changes position, the mechanical of the object is always conserved.
An object has maximum potential energy at maximum height.When the object is at lowest point, it has maximum kinetic energy.The sum of kinetic energy and potential energy of the object at any point is the total mechanical energy.
M.A = P.E + K.E
At the top of the hill the cart has maximum potential energy and zero kinetic energy. At the valley the cart has maximum kinetic energy and zero potential energy.
Learn more here:https://brainly.com/question/19969393
A bullet is fired directly upward with a speed of 330 m/s. How far above the ground will the bullet be in 65 s
Answer:
d = 21450 m
Explanation:
Given that,
The speed of a bullet, v = 330 m/s
We need to find the distance covered by the bullet in 65 s.
We know that,
Speed of an object = distance covered divided by time taken.
It means,
d = vt
d = 330 m/s × 65 s
d = 21450 m
So, the bullet will cover a distance of 21450 m.
A ballistic pendulum is a device often used for measuring the velocity of a projectile. Assume to projectile fully embeds itself in the pendulum. If the projectile mass is 25 grams, the pendulum mass is 2.5 kg, and the vertical displacement of the pendulum after the collision is 12 cm, find the velocity of the projectile just prior to the collision.
Answer:
Explanation:
pendulum rises upto height of .12 m
velocity of projectile and pendulum after collision = √ ( 2gH )
= √ ( 2 x 9.8 x .12 )
= 1.53 m / s
Applying conservation of momentum during collision
mv = (m + M ) X V
m and M are mass of projectile and pendulum and v and V are velocity of projectile before and after collision .
.025 x v = ( 2.5 + .025 ) x 1.53
v = 154.5 m /s
as seen from above, a 73.8 kg water skier is pulled by a 105 N force at a 22.5 angle, while the water creates a 74.8 N force pulling directly backward. What is the y-component of the total force on the skier?
Answer:
40.2 N
Explanation:
I'm an Acellus student who got it correct.
The magnitude of y-component of the total force on the skier is 40.18 N.
What is force?Force is an influence which tends to set a stationary body in motion or stop a moving body, or which tends to change the speed and direction of a moving body, or which tends to change the shape and size of body.
Given parameters:
Mass of the water: M = 73.8 kg.
Magnitude of force acting at 22.5° angle is: F₁ = 105 N.
Magnitude of force acting at directly backwards is: F₂ = 74.8 N.
We have to find: the y-component of the total force on the skier = ?
As F₂ acting directly backwards, this force has no y-component.
Hence, resultant y-component of the total force on the skier
= F₁ sin22.5°
= 105 ×sin22.5° N.
= 40.18 N.
Hence, magnitude of y-component of the total force on the skier is 40.18 N.
Learn more about force here:
https://brainly.com/question/13191643
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A 27 kg child slides down a playground slide at a constant speed. The slide has a height of 4.0 m and is 7.0 m long. Part A Using the law of conservation of energy, find the magnitude of the kinetic friction force acting on the child.
Answer:
Explanation:
The forces acting along the playground are:
Kinetic friction force Fk and
Wsintheta (weight along the plane)
Since sintheta = opp/hyp
Sintheta = 4/7
Taking the sum of forces along the playground
\sumFx = max
Wsintheta - Fk = max
Since the speed is constant, ax = 0, the equation becomes
Wsin theta - Fk = 0
Fk = Wsintheta
From the question
W = mg
W = 27×9.81
W = 264.87N
Fk = 264.87 × 4/7
Fk = 1059.48/7
Fk = 151.35N
Hence the magnitude of the kinetic friction force acting on the child is 151.35N
A cyclist who is riding a bike, whose wheels have a radius of r, is moving with constant speed v, along a flat straight road. The cyclist realizes that there is a single rock stuck between the threads of his tires that is causing the bike to vibrate violently. How frequently do the vibrations caused by the rock occur
Answer:
[tex]f = \frac{v}{2\pi r }[/tex]
Explanation:
A cyclist riding a bike whose wheels have a radius of r, moving with constant speed v along a flat straight road.
The frequency of the vibrations is the same as the frequency of the rotation of the wheel and this can be represented as below
V = R(w ) where ; w = [tex]2\pi f[/tex]
[tex]f = \frac{v}{2\pi r }[/tex]
A charged particle is injected at 109 m/s into a 0.0691‑T uniform magnetic field perpendicularly to the field. The diameter of its orbit is measured and found to be 0.0427 m. What is the charge–to–mass ratio of this particle?
We know, radius of the orbit is given by :
[tex]r=\dfrac{mv}{qB}[/tex]
So, ratio is given by :
[tex]\dfrac{q}{m}=\dfrac{v}{Br}\\\\\dfrac{q}{m}=\dfrac{109\ m/s}{0.0691 \ T \times 0.0427\ m}\\\\\dfrac{q}{m}=36942.01 \ C/kg\\\\\dfrac{q}{m}=3.69\times 10^{4}\ C/kg[/tex]
Therefore, the charge–to–mass ratio of this particle is [tex]3.69\times 10^{4}\ C/kg[/tex] .
Hence, this is the required solution.
The velocity of an object as a function of time is given by v = (12.5t –7.2t2) + (4.3t3). What is its acceleration as a function of time?
Answer:
a(t) = 12.5 -10.4t + 12.9t²Explanation:
The equation is not well written:
let the velocity of an object as a function of time be given by
v = (12.5t - 7.2t²) + (4.3t³)
v(t) = 12.5t - 7.2t² + 4.3t³
acceleration is the change in velocity with respect to time:
a(t) = d(v(t))/dt
a(t) = 12.5 - 2(7.2)t^2-1 + 3(4.3)t^3-1
a(t) = 12.5 -10.4t + 12.9t²
Hence the acceleration as a function of time t is expressed as;
a(t) = 12.5 -10.4t + 12.9t²
A 3.3 kg block of copper at a temperature of 74°C is dropped into a bucket containing a mixture of ice and water whose total mass is 1.2 kg. When thermal equilibrium is reached the temperature of the water is 8°C. How much ice was in the bucket before the copper block was placed in it? (Neglect the heat capacity of the bucket.
Answer:
0.228 kilograms of ice were in the bucket before the copper block was placed in it.
Explanation:
In this case, we assume that water inside the bucket was in the form of ice at a temperature of 0 ºC and an atmospheric pressure of 101.325 kilopascals, It is also known that block of copper is cooled down whereas ice is melted and heated up until thermal equilibrium is reached. If ice-block system is an isolated system, then First Law of Thermodynamics depicts the following model:
[tex]-Q_{b} +Q_{l,w}+Q_{s, w} = 0[/tex]
[tex]Q_{b} = Q_{l,w}+Q_{s,w}[/tex] (Eq. 1)
Where:
[tex]Q_{b}[/tex] - Heat released by the block of copper, measured in kilojoules.
[tex]Q_{l,w}[/tex] - Latent heat received by water, measured in kilojoules.
[tex]Q_{s,w}[/tex] - Sensible heat received by water, measured in kilojoules.
By definitions of sensible and latent heat, we expand the equation as follows:
[tex]m_{b}\cdot c_{b}\cdot (T_{b,o}-T) = m_{w}\cdot [L_{f}+c_{w}\cdot (T-T_{w,o})][/tex] (Eq. 2)
Where:
[tex]m_{b}[/tex] - Mass of the block of copper, measured in kilograms.
[tex]c_{b}[/tex] - Specific heat of copper, measured in kilojoules per kilogram-degree Celsius.
[tex]T_{b,o}[/tex] - Initial temperature of block of copper, measured in degrees Celsius.
[tex]m_{w}[/tex] - Mass of water, measured in kilograms.
[tex]L_{f}[/tex] - Latent heat of fusion of water, measured in kilojoules per kilogram.
[tex]T_{w,o}[/tex] - Initial temperature of water, measured in degrees Celsius.
[tex]T[/tex] - Final temperature of water-block system, measured in degrees Celsius.
[tex]c_{w}[/tex] - Specific heat of water, measured in kilojoules per kilogram-degree Celsius
And we clear the mass of water of the system:
[tex]m_{w} = \frac{m_{b}\cdot c_{b}\cdot (T_{b,o}-T)}{L_{f}+c_{w}\cdot (T-T_{w,o})}[/tex]
If we know that [tex]m_{b} = 3.3\,kg[/tex], [tex]c_{b} = 0.385\,\frac{kJ}{kg\cdot ^{\circ}C}[/tex], [tex]T_{b,o} = 74\,^{\circ}C[/tex], [tex]T = 8\,^{\circ}C[/tex], [tex]L_{f} = 334\,\frac{kJ}{kg}[/tex], [tex]c_{w} = 4.186\,\frac{kJ}{kg\cdot ^{\circ}C}[/tex] and [tex]T_{w,o} = 0\,^{\circ}C[/tex], then the mass of the ice inside the bucket is:
[tex]m_{w} = \frac{(3.3\,kg)\cdot \left(0.385\,\frac{kJ}{kg\cdot ^{\circ}C} \right)\cdot (74\,^{\circ}C-8\,^{\circ}C)}{334\,\frac{kJ}{kg}+\left(4.186\,\frac{kJ}{kg\cdot ^{\circ}C} \right)\cdot (8\,^{\circ}C-0\,^{\circ}C) }[/tex]
[tex]m_{w} = 0.228\,kg[/tex]
0.228 kilograms of ice were in the bucket before the copper block was placed in it.
Help please, thank you.
Answer:
a. skin and mucous membranes
hope this helps you
❤❤❤❤❤❤❤
itz Indian heart
I need help with this science work.
Answer:
ok what is it?
Explanation:
Answer:how can I help?
Explanation:
A uniform plank 8.00 m in length with mass 30.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the left-hand end. What is the maximum additional mass you could place on the right-hand end of the plank and have the plank still be at rest
Answer:
30 kg
Explanation:
Given that a uniform plank 8.00 m in length with mass 30.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the left-hand end. What is the maximum additional mass you could place on the right-hand end of the plank and have the plank still be at rest ?
Taking the moment at the middle
Sum of Clockwise moment = sum of anticlockwise moments
Sum of clockwise moment will be:
30 × ( 4 - 1 ) = 30 × 3 = 90Nm
At equilibrium, the sum of anticlockwise moments will be equal to sum of clockwise moment.
90 = 30 × ( 5 - 4 ) + M
90 = 30 × 1 + M
90 = 30 + M
M = 90 - 30
M = 60
Moment = W × distance
Let the minimum distance = 2
60 = 4W
W = 60/2
W = 30 kg
Therefore, the maximum mass will be 30kg
A Category 1 hurricane with a radius of 70km has sustained tangential winds of about 33 m/s. What is the acceleration of a particle on th edge
of the hurricane.
Answer:
[tex]a_c=0.016 \ m/s^2[/tex]
Explanation:
Centripetal acceleration: Is the acceleration experienced by an object while in a uniform circular motion. It points toward the center of rotation and is perpendicular to the linear velocity v and has the magnitude:
[tex]\displaystyle a_c=\frac{v^2}{r}[/tex]
Where v is the linear velocity and r is the radius of the circle.
It's given a hurricane with a radius r=70 Km = 70,000 m has a tangential wind velocity of v=33 m/s. The acceleration of a particle lying on the edge of the hurricane is:
[tex]\displaystyle a_c=\frac{33^2}{70,000}[/tex]
[tex]\boxed{a_c=0.016 \ m/s^2}[/tex]
An object with a mass of 10 kg accelerates 16 m/s2 when an unknown force is applied to it what is the amount of force
Answer:
[tex] \huge{ \boxed {\bold{ \tt{160 \: N}}}}[/tex]
Explanation:
Given :Mass ( m ) = 10 kgAcceleration ( a ) = 16 m / s²To find :Force = ?We know,[tex] \boxed{ \sf{Force = Mass \: \times \: Acceleration}}[/tex]
[tex] \sf{➛ force \: = 10\: \times 16}[/tex]
[tex] \sf{➛ force = 160 \: Newton}[/tex]
The amount of force is 160 N
Hope I helped!
Best regards! :D
~TheAnimeGirl
How many different uniquecurrents will there be in this circuit? How many different "branch" are there in this circuit where the currents could differ.
Answer:
hello your question is incomplete attached below is the missing circuit diagram
answer : 3 unique currents
Explanation:
From the circuit attached below it can be seen that there will be Three(3) unique currents flowing through this circuit and the Branches where this currents could differ is at the Three(3) resistors
g First find the magnitude of the force F on a positive charge q in the case that the velocity v⃗ (of magnitude v) and the magnetic field B⃗ (of magnitude B) are perpendicular. Express your answer in terms of v, q, B, and other quantities given in the problem statement.
Answer:
The magnitude of the force F on a positive charge q in the case that the velocity v and the magnetic field B are perpendicular is [tex]F = q\cdot v\cdot B[/tex], measured in newtons.
Explanation:
From classical theory on Magnetism, the vectorial form of the magnetic force on a particle is given by:
[tex]\vec F = q\cdot \vec v \times \vec B[/tex] (Eq. 1)
Where:
[tex]\vec F[/tex] - Magnetic force, measured in newtons.
[tex]q[/tex] - Electric charge, measured in coulombs.
[tex]\vec v[/tex] - Velocity of the particle, measured in meters per second.
[tex]\vec B[/tex] - Magnetic field, measured in tesla.
By definition of cross product, we get that magnitude of magnetic force on a positive charge [tex]q[/tex] is:
[tex]F = q\cdot v\cdot B \cdot \sin \theta[/tex] (Eq. 2)
Where:
[tex]v[/tex] - Speed of the particle, measured in meters per second.
[tex]B[/tex] - Magnitude of the magnetic field, measured in tesla.
[tex]\theta[/tex] - Angle between the velocity of the particle and magnetic field, measured in sexagesimal degrees.
If velocity and magnetic field are perpendicular, then (Eq. 2) is reduced into this form: ([tex]\theta = 90^{\circ}[/tex])
[tex]F = q\cdot v\cdot B[/tex]
If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges
Complete question is;
A 13g bullet traveling 230 m/s penetrates a 2.0 kg block of wood and emerges going 170 m/s. If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?
Answer:
0.39 m/s
Explanation:
From the question, we can deduce that there is no external force acting on the
system in the horizontal direction. This means that momentum will be conserved for the system of the 2 kg block & the bullet.
Thus;
Initial momentum(p_i) = final momentum(p_f)
Mass of bullet; = 13g = 0.013 kg
Mass of Block = 2 kg
Speed of bullet = 230 m/s
Speed of Block = 170 m/s
We know that formula for momentum is; p = mv.
Thus, the initial momentum of the (bullet + block) is given as;
p_i = 0.013 × 230
Final momentum is;
p_f = (0.013 × 170) + (2 × v_f)
Where v_f is the speed at which the bullet emerges.
Since p_i = p_f from conservation of momentum, then;
(0.013 × 230) = (0.013 × 170) + (2 × v_f)
2.99 = 2.21 + 2v_f
2.99 - 2.21 = 2v_f
0.78 = 2v_f
v_f = 0.78/2
v_f = 0.39 m/s
At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball's acceleration is 810 m/s2 and the vertical or y component of its acceleration is 910 m/s2. The ball's mass is 0.40 kg. What is the magnitude of the net force acting on the soccer ball at this instant?
Answer:
487.13N
Explanation:
According to Newton's second law;
F = ma
F is the net force
m is the mass of the object
a is the acceleration
First we need to find the resultant of the acceleration
a =√810²+910²
a = √656,100+828,100
a =√1,484,200
a = 1218.28m/s²
Ball mass = 0.40kg
Next is to get the magnitude of the net force
F = 0.4 × 1218.28
F = 487.31N
Hence the magnitude of the net force acting on the soccer ball at this instant is 487.13N
2. A 60 kg crate is being lowered from a loading dock to the floor of a warehouse by a thick rope. The rope exerts a constant upward force of 500 N as it is descending to the floor. a) Draw a free-body diagram for the crate. b) Calculate the acceleration (if any) of the crate.
Answer:
Acceleration =8.33m/s^2Explanation:
Step one:
Kindly find attached a sketch of the free body diagram
Given data
mass m=60kg
Force F=500N
a=?
Step two:
From Newton's second law "An object’s acceleration equals the vector sum of the forces acting on it, divided by it mass".
mathematically
F=ma
Step three:
Substituting we have
500=60*a
a=500/60
a=8.33m/s^2
If the current density is 1.55 and the drift velocity is , how many charge carriers are present? The charge on electron is .(Enter your answer to three significant figures.)
Answer:
There are [tex]9.053\times 10^{14}[/tex] charge carriers.
Explanation:
Given that,
Current density (current per unit time), [tex]\dfrac{I}{A}=1.55\ A/cm^2=0.000155\ A/m^2[/tex]
The drift velocity of the electron, [tex]v_d=107\ cm/s=1.07\ m/s[/tex]
We need to find the no charge carriers present. The formula for drift velocity is given by :
[tex]v_d=\dfrac{I}{neA}\\\\\text{n is no of charge carriers}\\\\n=\dfrac{I}{eAv_d}\\\\n=\dfrac{0.000155}{1.6\times 10^{-19}\times 1.07}\\\\n=9.053\times 10^{14}[/tex]
So, there are [tex]9.053\times 10^{14}[/tex] charge carriers.
If I am driving down the highway going north at 50 miles per hour, and another car is driving south at 75 miles per hour. How fast is the car coming toward me?
Answer:
V₂₋₁ = 50 [miles/h]
Explanation:
In order to solve this problem we must use the concept of relative velocities.
Where:
V1 = 50 [miles/h]
V2 = 75 [miles/h]
V₂₋₁ = Relative velocity among the two cars.
V₂₋₁ = 75 - 25 = 50 [miles/h]
That is, the car that goes at 50 miles per hour, sees how the second car approaches at 50 miles per hour.
"We want to know how microgravity will
affect ____
Answer:
humans
Explanation:
mhm
What is the drift speed of the electrons when there is a voltage drop of 118.8 V across the wire? Assume there is one conduction electron per atom.
Answer:
Explanation:
If e be the charge on the electron and v be its drift speed after undergoing voltage drop of 118.8 V
eV = 1/2 m v²
1.6 x 10⁻¹⁹ x 118.8 = 1/2 x 9.1 x 10⁻³¹ v²
v² = 41.775 x 10¹²
v = 6.46 x 10⁶ m /s
drift speed of electron = 6.46 x 10⁶ m /s