Answer:
5.68 meters
Explanation:
hope this helps!
Answer:
5.68
Explanation:
to convert cm to m you move the decimal point 2x to the left
A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 4.55 m/s. If the roof is pitched at 22.0° below the horizon and the roof edge is 2.90 m above the ground, find the time the baseball spends in the air and the horizontal distance from the roof edge to the point where the baseball lands on the ground. HINT Solve for the time of flight using vy2 = v0y2 − 2gΔy and then vy = v0y − gt. Solve for the horizontal distance using the horizontal displacement equation. Click the hint button again to remove this hint. (a) the time the baseball spends in the air (in s) 0.434 Incorrect: Your answer is incorrect. s (b) the horizontal distance from the roof edge to the point where the baseball lands on the ground (in m) m
Answer:
Explanation:
The time that the baseball spends in the air is known as time of flight and it is expressed as shown;
T = Using(theta)/g where;
U is the velocity of the baseball
g is the acceleration due to gravity.
Given parameters
U = 4.55m/s
theta = 22.0°
g = 9.81m/s²
Substituting the values in the formula;
T = 4.55sin22°/9.81
T = 4.55(0.3746)/9.81
T = 1.7045/9.81
T = 0.1738second
Hence the time flight of the baseball is 0.1738second
b) The horizontal distance covered by the ball is called the RANGE in projectile.
Range = U√2H/g
U = 4.55m/s
H is the maximum height = 2.90m
g = 9.81m/s²
Substitute the given parameters into the formula
Range = 4.55√2(2.90)/9.81
Range = 4.55√5.8/9.81
Range = 4.55√0.5912
Range = 4.55(0.7689)
Range = 3.4986m
Hence the horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.4986m
If we want to find the size of the force necessary to just barely overcome static friction (in which case fs=μsNfs=μsN), we use the condition that the sum of the forces in both directions must be 0. Using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as:
Complete Question
Two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. The block in ( The diagram is shown on the first uploaded image ) has a mass m= 10kg and is being pulled by a force F on a table with coefficient of static friction μs=0.3. Four forces act on it:
The applied force F (directed θ=30∘ above the horizontal).
The force of gravity Fg=mg (directly down, where g=9.8m/s2).
The normal force N (directly up).
The force of static friction fs (directly left, opposing any potential motion).
If we want to find the size of the force necessary to just barely overcome static friction (in which case fs=μsN), we use the condition that the sum of the forces in both directions must be 0. Using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as:
Fcosθ−μsN=0
Fsinθ+N−mg=0
In order to find the magnitude of force F, we have to solve a system of two equations with both F and the normal force N unknown. Use the methods we have learned to find an expression for F in terms of m, g, θ, and μs (no N)
Note the diagram is shown on the first uploaded image
Answer:
The expression for F is [tex]F = \frac{\mu_s * m* g}{[\mu_s * sin (\theta )] * cos (\theta)}[/tex]
Explanation:
Generally from the diagram we see that
[tex]Fcos\theta -f_s = 0[/tex]
From the question we are told that
[tex]f_s = \mu_s * N[/tex]
So
[tex]Fcos\theta - \mu_s * N = 0[/tex]
=> [tex] N = \frac{Fcos(\theta)}{\mu_s}[/tex]
Also from the diagram
[tex]Fsin(\theta )+N - F_g = 0[/tex]
Here [tex]F_g = m * g[/tex]
So
=> [tex]Fsin(\theta )+ \frac{Fcos(\theta)}{\mu_s} - m* g = 0 [/tex]
=> [tex]F = \frac{\mu_s * m* g}{[\mu_s * sin (\theta )] * cos (\theta)}[/tex]
Two vectors A and B are at right angles to each other. The magnitude of A is 3.00. What should be the length of B, so that the magnitude of their vector sum is 5.00?
Answer: length of B =4.00
Explanation:
for the vectors A and B and the angle between them as x.
Magnitude of the sum of A and B is given as = √(A²+B²+2ABcosx
where
Magnitude of A = 3.00
Magnitude of the sum of A and B is 5.00
5.00=√(A²+B²+2ABcos90°
5.00= √3² +b² +0
5²= 3² +b²
25=9+b²
b²= 25-9
b² = 16
b= √16
b= 4
g A very early, simple satellite consisted of an inflated spherical aluminum balloon 38 m in diameter and of mass 22 kg. Suppose a meteor having a mass of 21 kg passes within 8.0 m of the surface of the satellite. What is the magnitude of the gravitational force on the meteor from the satellite at the closest approach?
Answer:
6.37*10^-13 N
Explanation:
Given that
We r = 38/2 = 14 m
m1 = 22kg
m2 = 21 kg
so we can say that distance between center of the satellite and meteor,
d = r + h
= 14 + 8
= 22m
So the gravitational force on the meteor, is
F = G x m1 x m2/d²
= 6.67*10^-11 x 21 *22/22²
= 6.37*10^-13 N
Suppose the two carts have the same mass m. In the initial state, these two carts are moving toward each other with the same initial speed, vi, along a frictionless track (implying no net external forces acting on the two carts). These carts collide and the result is some final state. The three parts of this question are concerned with three different final states. A. Assume that the carts hit each other and stop (both carts are not moving). Draw a momentum chart for this situation: make a separate row for each cart. B) Assume that the carts bounce off each other so that the final state of the system has each cart moving oppositely to its initial motion but with the same speed. Draw a momentum chart for this situation. C) As in (B), assume that the carts bounce off each other with equal speeds and in opposite directions, but now assume that the final speeds are smaller than the initial speeds. Draw a momentum chart. D) For each case does the total momentum of the two cars change? How do the momentum charts tell you this? E) Is the total kinetic energy constant for all three cases? How do you know?
Answer:
in all three cases the total moment is zero
cases A and B the kinetic energy is conserved.
In case C the velocity decreases so the kinetic energy decreases
Explanation:
This is a momentum conservation exercise
p = mv
In order for the moment to be preserved, we must define a system formed by the two cars, so that the forces during coke have been internal.
Before crash
car 1 p₀₁ = m v₀
car 2 p₀₂ = - mv₀
pose us several situations, we analyze each one
A) After the crash the cars stop
[tex]p_{f}[/tex] = 0
p₀₁ m v₀
p₀₂ -m v₀
p_{f} 0
B) After the collision, each vehicle reverses its direction
p₀₁ m v₀
p₀₂ -m v₀
p_{f1} -m v₀
p_{f2} m v₀
C) In this case some of the kinetic energy is lost which is converted into internal energy, for example, deformation, heat, friction.
Consequently the speed of the cars is
v < v₀
p₀₁ m v₀
p₀₂ - m v₀
p_{f1} -m v
p_{f2} m v
D) in cases A and B the momentum is maintained, but in case C the total momentum is maintained, even when the speed of the cars decreases, this is pf_total = 0
In all cases the total impulse is zero
p₀ = p₀₁ + p₀₂ = m v₀ - mv₀
p₀_total = 0
in all three cases the total moment is zero
E) The total kinetic energy is the sum of the kinetic energy of each car
K_total = K₀₁ + K₀₂
K_total = ½ m v₀² + ½ m (-v₀)²
K_total = m v₀²
we see that because it is squared, the sign of the velocity does not matter, therefore in cases A and B the kinetic energy is conserved.
In case C the velocity decreases so the kinetic energy decreases
Kf_total < K₀_total
the missing energy is transformed into internal energy during sackcloth.
In the attachment we can see a vector diagram of the momentum in each case
What is the control group used for
Answer:
What is a Control Group? The control group (sometimes called a comparison group) is used in an experiment as a way to ensure that your experiment actually works. It's a way to make sure that the treatment you are giving is causing the experimental results, and not something outside the experiment
Explanation:
A runner is jogging in a straight line at a
steady vr= 5.9 km/hr. When the runner is
L= 6.6 km from the finish line, a bird begins
flying straight from the runner to the finish
line at vb= 29.5 km/hr (5 times as fast as
the runner). When the bird reaches the finish
line, it turns around and flies directly back to
the runner.
What cumulative distance does the bird
travel? Even though the bird is a dodo, assume that it occupies only one point in space
(a "zero" length bird), travels in a straight
line, and that it can turn without loss of
speed.
Answer in units of km.
Answer:
11.88 km
Explanation:
Given that the bird begins flying straight from the runner to the finish
line at vb= 29.5 km/hr (5 times as fast as the runner). When the bird reaches the finish line, it turns around and flies directly back to the runner.
Then the first distance covered by the bird is 6.6 km.
Since the speed of the bird is five times the speed of the man, the man must have covered the distance one - fifth of the 6.6 km. That is,
1/5 × 6.6 = 1.32
Take 1.32 away from 6.6 you will get
6.6 - 1.32 = 5.28
The cumulative distance the bird
travel will be:
Cumulative distance = 6.6 + 5.28 = 11.88km
Therefore, the cumulative distance the bird travelled is 11.88 km
An air filter can remove dust particles from air but will reach capacity (saturation) at 50.0 mg. Of air containing 225 µg dust particles per m3 air is passed through the filter at 400 ft3 /min and air leaving the filter has 15.0 µg dust/m3 air, how much time, in days, is required for the filter to reach saturation
Answer:
Time to Reach Saturation = 0.0146 day
Explanation:
In order to solve this problem, we first need to calculate the dust filtered by the filter per cubic meter of air:
Filtered Dust per m³ = Dust Particles Entering per m³ - Dust Particles Leaving per m³
Filtered Dust per m³ = 225 μg/m³ - 15 μg/m³
Filtered Dust per m³ = 210 μg/m³ = 210 x 10⁻³ mg/m³
Now, we find volume flow rate of air through filter:
Volume Flow Rate of Air = (400 ft³/min)(0.3048 m/1 ft)³
Volume Flow Rate of Air = 11.33 m³/min
Now, we calculate rate of dust filtered:
Rate of Dust Filtered = (Filtered Dust per m³)(Volume Flow Rate of Dust)
Rate of Dust Filtered = (210 x 10⁻³ mg/m³)(11.33 m³/min)
Rate of Dust Filtered = 2.38 mg/min
Now, for the time required to reach saturation:
Time to Reach Saturation = (Saturation Capacity)/(Rate of Dust Filtered)
Time to Reach Saturation = (50 mg)/(2.38 mg/min)
Time to Reach Saturation = (21.02 min)(1 day/24 h)(1 h/60 min)
Time to Reach Saturation = 0.0146 day
what kind of energy is called mechanical energy?
Answer:
Mechanical energy is the energy that is possessed by an object due to its motion or due to its position. Mechanical energy can be either kinetic energy (energy of motion) or potential energy (stored energy of position).
Hope this helps! :)
Explanation:
Two masses m1m1 and m2m2 exert a gravitational force of 12 N onto each other when they are 6 m apart. What will the gravitational force be if the masses are moved closer to be 3 m apart?
Answer:
48N
Explanation:
because as the distance is halfed. the force on the two objects are quadrupled.
a total journey of 170 miles, in 3 hours, what would be your average velocity? (in miles/hour) *
Answer:
Velocity = 56.67 mi/hr
Explanation:
Distance = 170 miles
Time = 3 hours
Velocity =?
[tex]Velocity = \frac{Distance}{Time} \\\\V = \frac{170}{3} \\\\V =56.67[/tex]
What was the velocity of a dog that ran 50 meters in 15 seconds?
Answer:
5m/s
Explanation:
there are 3,600 seconds per hour so 5•3600 18,000 hours and 18 km/h
Answer:
velocity = 3.33 m/sec
Explanation:
velocity = distance / time
where
distance = 50 meters i
time = 15 seconds
plugin values into the formula
velocity = 50 m
15 sec
velocity = 3.33 m/sec
Two forces are applied to a tree stump to pull it out of the ground. Force A has a magnitude of 2410 newtons (N) and points 36.0 ° south of east, while force B has a magnitude of 4470 N and points due south. Using the component method, find the (a) magnitude and (b) direction of the resultant force A + B that is applied to the stump. Specify the direction as a positive angle with respect to due east.
Answer:
The magnitude of the resultant force is 6201 N
The direction of the resultant force is 72°
Explanation:
Please find the attached file for explanation;
Can an object with constant acceleration reverse its direction of travel? Can it reverse its direction twice? Explain.
Answer:yes
Explanation:The constan acceleration means that it wont stop moving but if you kick it a different direction then it will change direction
A lounging leopard decides to come down out of her tree and hunt for her lunch in the savanna. The graph above represents the motion of the leopard as a function of time.
What is the distance traveled by the leopard during the entire 11 seconds? Please include a number and uni
Which three statements make up the law of gravity
A) The strength of the gravitational force between two objects decreases as the distance between their centers increases.
B) The strength of the gravitational force between two objects increases as the total mass of the objects increases
C) matter attracts all other matter in the universe
D) no matter where you are in the universe you will always have the same weight.
Answer:
the answer is A, B, and C
Explanation:
D is wrong because your weight does change...it is your mass that doesn't change
One afternoon, a couple walks three-fourths of the way around a circular lake, the radius of which is 2.08 km. They start at the west side of the lake and head due south to begin with. (a) What is the distance they travel? (b) What is the magnitude of the couple’s displacement? (c) What is the direction (relative to due east) of the couple’s displacement?
Answer:
(a). The distance is 9.80 km.
(b). The magnitude of the couple’s displacement is 2.94 km.
(c). The direction of the couple’s displacement is 45°.
Explanation:
Given that,
Radius of lake = 2.08 km
(a). We need to calculate the total distance
Using formula of distance
[tex]d=\dfrac{3}{4}(2\pi R)[/tex]
Put the value into the formula
[tex]d=\dfrac{3}{4}(2\pi\times2.08)[/tex]
[tex]d=9.80\ km[/tex]
(b). We need to calculate the magnitude of the couple’s displacement
Using formula of displacement
[tex]D=\sqrt{R^2+R^2}[/tex]
[tex]D=\sqrt{2R^2}[/tex]
[tex]D=\sqrt{2\times(2.08)^2}[/tex]
[tex]D=2.94\ km[/tex]
(c), The direction of the displacement is given by
Using formula of direction
[tex]\tan\theta=\dfrac{R}{R}[/tex]
[tex]\theta=\tan^{-1}(1)[/tex]
[tex]\theta=45^{\circ}[/tex]
Hence, (a). The distance is 9.80 km.
(b). The magnitude of the couple’s displacement is 2.94 km.
(c). The direction of the couple’s displacement is 45° relative to the east.
The distance they travel is approximately 6.54644 km. The magnitude of the couple's displacement is 4.16 km. The direction of their displacement is opposite to due east.
(a) The distance they travel is:
Distance = (3/4) × 2 × 3.14159 × 2.08
Distance = 3.14159 × 2.08 km
Distance = 6.54644 km
So, the distance they travel is approximately 6.54644 km.
(b) The magnitude of the couple's displacement is:
Displacement = 2 × 2.08
Displacement = 4.16 km
So, the magnitude of the couple's displacement is 4.16 km.
(c) The direction of their displacement is opposite to due east.
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The energy stored in a wood log is transformed when the log is burned. Which of the following explanations best describes how the
chemical energy stored in the log compares to the heat and light energy produced by burning?
O A The chemical energy is equal to the amount of heat, light, and other applicable energies.
O B The chemical energy is less than the amount of heat, light, and other applicable energies.
OCThe chemical energy is more than the amount of heat, light, and other applicable energies.
D. The chemical energy stays the same; additional energy is produced as heat, light, and other applicable energies.
Answer:
D. The chemical energy stays the same; additional energy is produced as heat, light, and other applicable energies.
Hope this helps!
The chemical energy stays the same; additional energy is produced as heat, light, and other applicable energies. Thus option D is correct.
What is chemical energy ?Energy is the ability to do work It can be movement of a body to do some physical activity.
If the energy is stored in the form of chemical bonds of a complex molecule then the energy is called as chemical energy.
The energy released in the chemical reaction and produced as as a by-product, that process is known as an exothermic reaction.
For instance, chemical energy sored in biomass, batteries, natural gas, petroleum, and coal.
Dry wood also the storage of chemical energy, as it burns the chemical energy is released and converted into light energy and thermal energy
The food we eat is also an example of chemical energy storage as it is liberated during digestion process.
Thus option D is correct.
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The classic Millikan oil drop experiment was the first to obtain an accurate measurement of the charge on an electron. In it, oil drops were suspended against the gravitational force by a vertical electric field. Consider an oil drop with a weight of 2.9 x 10-14N, if the drop has a single excess electron, find the magnitude (in N/C) of the electric field needed to balance its weight. Your should round your answer to an integer, indicate only the number, do not include the unit.
Answer:
[tex]E=1.81\times 10^5\ N/C[/tex]
Explanation:
In Millikan oil drop experiment, oil drops were suspended against the gravitational force by a vertical electric field such that its weight is balanced by the electron force i.e.
W = qE,
W is weight, W = mg
q is charge,
E is electric field
⇒ [tex]2.9\times 10^{-14}\ N=qE[/tex]
or
[tex]E=\dfrac{2.9\times 10^{-14}\ N}{1.6\times 10^{-19}\ C}\\\\E=181250\ N/C\\\\\text{or}\\\\E=1.81\times 10^5\ N/C[/tex]
So, the magnitude of the electric field is [tex]1.81\times 10^5\ N/C[/tex].
The water is wide character descriptions:__________
Answer:
Explanation:
I can't name the whole characters, I will try to name as much as I can remember, thanks.
Pat Conroy (Conrack) or (C'roy)
C'roy happens to be the male teacher on Yamacraw Island.
Dr. Henry Piedmont
Dr. Henry is the Superintendent for the school district in Beaufort, South Carolina.
Ezra Bennington
Ezra is the Deputy Superintendent of Beaufort County schools, in South Carolina
Mrs. Brown
Mrs Brown is a teacher on the Yamacraw Island. A racist teacher, if I might add.
Barbara Bolling Jones Conroy
Barbara lives with a Yamacraw Island teacher.
Ted Stone
Ted is the man with an awful lot of responsibilities on Yamacraw Island.
Lou Stone
Lou is the Yamacraw Island postmaster, in addition to being the school bus driver.
Zeke Skimberry
Zeke is the Yamacraw Island maintenance man who eventually becomes friends with one of the teachers.
Babies
This is a nickname is for the students of Yamacraw Island.
That's the lot I can mention. I hope that's enough.
which best describes the big bang theory ? A. matter was compressed into a single point and then exploded outward to form the universe. B. stars burn off their fuel, expand, and shed their shells. C. a black hole exploded, creating stars.
Answer:
A. matter was compressed into a single point and then exploded outward to form the universe
Explanation:
The big bang is how astronomers explain the way the universe began. It is the idea that the universe began as just a single point, then expanded and stretched to grow as large as it is right now (and it could still be stretching).
A. matter was compressed into a single point and then exploded outward to form the universe.
this option describes the big bang theory
What is the Big Bang theory?The Big Bang hypothesis states that all of the current and past matter in the Universe came into existence at the same time, roughly 13.8 billion years ago. At this time, all matter was compacted into a very small ball with infinite density and intense heat called Singularity.
How did the universe begin?The Big Bang was the moment 13.8 billion years ago when the universe began as a tiny, dense, fireball that exploded. Most astronomers use the Big Bang theory to explain how the universe began. But what caused this explosion in the first place is still a mystery.
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You attach a meter stick to an oak tree, such that the top of the meter stick is 1.471.47 meters above the ground. Later, an acorn falls from somewhere higher up in the tree. If the acorn takes 0.2810.281 seconds to pass the length of the meter stick, how high h0h0 above the ground was the acorn before it fell, assuming that the acorn did not run into any branches or leaves on the way down?
Answer:
The value is [tex]h_a = 1.712 \ m [/tex]
Explanation:
From the question we are told that
The height of the top meter stick above the ground is [tex]h_1 = 1.47 \ m[/tex]
The time taken for the acorn to pass the length of the stick is [tex]t = 0.281 \ s[/tex]
Generally the height of the acorn at the point it is the same height with the metered stick is mathematically represented as
[tex]h = h_m = u_a * t + \frac{1}{g} t^2[/tex]
Here [tex]h_m[/tex] is height of the meter stick and the value is 1 m (This because we are told in the question that the stick is 1 meter in length ( a meter stick))
So
[tex]1 = u_a * 0.281 + \frac{1}{9.8} (0.281)^2[/tex]
=> [tex]u_a = -2.2 \ m/s[/tex]
Generally the velocity of the acorn just before passing the top of the meter stick is mathematically represented by a kinematic equation as
[tex]u^2_a = u^2 + 2gs[/tex]
here u is zero since the acorn started from rest
So
[tex] (-2.2) = 0 + 2 * 9.8 * s[/tex]
[tex]s = 0.242 \ m[/tex]
Generally the height of the acorn is
[tex]h_a = h_1 + s[/tex]
[tex]h_a = 0.242 + 1.47[/tex]\
[tex]h_a = 1.712 \ m [/tex]
A bus accelerated at 1.8 m/s2 from rest for 15 s. It then traveled at constant speed for 25 s, after which it slowed to a stop with an acceleration of 1.8 m/s2. The distance traveled by the bus was:______.
Answer:
The distance traveled by the bus was 1080 m
Explanation:
Please find the attached file for explanation
A projectile is launched on earth at an angle θ, relative to horizontal direction. At half of its maximum height the speed of the projectile is 1.00 m/s, and at its maximum height the speed of the projectile is 0.50 m/s. What is the angle θ ?
Answer:
the angle is about 67.79 degrees
Explanation:
We know that at its maximum height, the vertical component of the projectile's launching (initial) velocity (Vyi) is zero, so at that point it total velocity equals the horizontal component of the initial velocity (Vxi = 0.5 m/s)
We also know that the maximum height of the projectile is given by the square of its initial vertical component of the velocity (Vyi) divided by 2g, therefore half of such distance is :
[tex]half\,\,max-height = \frac{v_{yi}^2}{4\,g}[/tex]
we can use this information to find the y component of the velocity at that height via the formula:
[tex]v_{yf}^2-v_{yi}^2=-2\,g\,\Delta y\\\\v_{yf}^2-v_{yi}^2=-2\,g\,(\frac{v_{yi}^2}{4\,g} )\\v_{yf}^2=v_{yi}^2-\frac{v_{yi}^2}{2} \\v_{yf}^2=\frac{v_{yi}^2}{2}[/tex]
Now we use the information that tells us the speed of the projectile at this height to be 1 m/s. That should be the result of the vector addition of the vertical and horizontal components:
[tex]1=\sqrt{v_{yf}^2+0.5^2} \\1=\sqrt{\frac{v_{yi}^2}{2} +0.5^2}\\1^2=\frac{v_{yi}^2}{2} +0.5^2}\\1-0.5^2=\frac{v_{yi}^2}{2} \\2(1-0.5^2)=v_{yi}^2\\1.5 = v_{yi}^2\\v_{yi}=\sqrt{1.5} \\[/tex]
Now we can use the arc-tangent to calculate the launching angle, since we know the two initial component of the velocity vector:
[tex]tan(\theta)=\frac{v_{yi}}{v_{xi}} =\frac{\sqrt{1.5} }{0.5} \\\theta= arctan(\frac{\sqrt{1.5} }{0.5})=67.79^o[/tex]
A proton (mass=1.67x10^-27 kg, charge= 1.60x10^-19 C) moves from point A to point under the influence of an electrostatic force only. At point A the proton moves with a speed of 50 m/s. At point B the speed of the proton is 80 km/s. Determine the potential difference VB-VA in volts.
Answer:
[tex]VB - VA = - 33.4[/tex]
Explanation:
Generally the workdone in moving the proton is mathematically represented as
[tex]W = KE_f - KE_i[/tex]
Where [tex]KE_i \ and \ KE_f \ are\ the\ initial \ and \ final \ kinetic \ energy [/tex]
So
[tex]KE_i = \frac{1}{2} m v_a^2[/tex]
Here [tex]v_a[/tex] is the velocity at A with value 50 m/s
So
[tex]KE_i = \frac{1}{2} (1.67*10^{-27}) * 50^2 [/tex]
[tex]KE_i = 2.09 *10^{-24} \ J [/tex]
Also
[tex]KE_f = \frac{1}{2} m v_b^2[/tex]
Here [tex]v_a[/tex] is the velocity at A with value [tex]80 km/s = 80000 m/s [/tex]
=> [tex]KE_f = \frac{1}{2} (1.67*10^{-27}) * 80000^2 [/tex]
=> [tex]KE_f = 5.34 *10^{-18} \ J[/tex]
So
[tex]W = 5.34 *10^{-18} - 2.09 *10^{-24}[/tex]
[tex]W = 5.34 *10^{-18} m/s[/tex]
Now this workdone is also mathematically represented as
[tex]W = q * V[/tex]
So
[tex] q * V = 5.34 *10^{-18} [/tex]
Here [tex]q = 1.60*10^{-19} C[/tex]
So
[tex] V = \frac{5.34 *10^{-18} }{1.60*10^{-19}}[/tex]
[tex] V = 33.4 \ V [/tex]
Generally proton movement is in the direction of the electric field it means that [tex] VA>VB [/tex]
So
[tex]VB - VA = - 33.4[/tex]
What is the effect on the gas’ pressure when compressing a gas to 1/3 of its volume? Explain
Explanation:
It's pressure become three times larger because according to Boyles Law the pressure of fixed mass of gas is inversely proportional to it's volume provided that temp remains constant. That means a reduction in volume, will result in an increase in pressure and vice versa.
The resultant velocity with respect to the ground of a fighter
plane flying at 100 km/hr air speed will be
encounters a 45 km/hr tailwind.
Answer:145 km/hr
Explanation: v1+v2 100+45 =145 km/hr
how many key reasons are there for time getting away from you?
Answer:
4
Explanation: the problem with this one is 75 the problem the problem with
is weight included in free body diagram
Answer:
Yes
Explanation:
I took physics last year that that was a requirement.
Grant sprints 50 meters to the right with an average velocity of 3.0 m/s.
How many seconds did Grant sprint?
Answer:
The Answer Is 17s
Explanation:
Displacement trianglex = 50m
Time t = ?
Average velocity v = 3.0 m/s
You can rearrange the equation v = trianglex/t to solve for time t.
v = tianglex/t
t = trianglex/v
= 50m/3.0m/s
= 17s