Answer:
Explanation:
The minimum magnitude of acceleration = 3 m /s²
displacement at t = 1
s = ut + 1 /2 at²
= -3 x 1 + .5 x 3 x 1²
= - 3 + 1.5
= - 1.5 m
position at t = 1 s
= 10 - 1.5
= 8.5 m
The maximum magnitude of acceleration = 6 m /s²
displacement at t = 1
s = ut + 1 /2 at²
= -3 x 1 + .5 x 6 x 1²
= - 3 + 3
= 0
position at t = 1 s
= 10 +0
= 10 m
So range of position is 8.5 m to 10 m .
We want to find the range of possible positions for the cart at t = 1s.
The range is:
14.5m ≤ p ≤ 16m
Finding the cart's motion equations.
First, we know that the acceleration of the cart is a, a constant (but we don't know the exact value yet) so we write the acceleration as:
a(t) = a
To get the velocity equation we integrate over time, and we know that the velocity at t = 0s is 3m/s, so that will be the constant of integration.
v(t) = a*t + 3m/s
To get the position we integrate again, the initial position is x = 10m, so that will be the constant of integration.
p(t) = (a/2)*t^2 + (3m/s)*t + 10m
Now, to get the range of possible positions for t = 1s we need to use the minimum and maximum accelerations and evaluate the position equation in t = 1s.
The minimum acceleration is a = 3m/s^2, so we have the minimum position:
p(1s) = (3m/s^2/2)*(1s)^2 + (3m/s)*1s + 10m = 14.5m
The maximum acceleration is 6m/s^2, so the maximum position is:
P(1s) = (6m/s^2/2)*(1s)^2 + (3m/s)*1s + 10m = 16m
So the range of the position at t = 1s is:
14.5m ≤ p ≤ 16m
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Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds.
This question is incomplete, the complete question is;
A student dropped a textbook from the top floor of his dorm and it fell according to the formula s(t) = -16t² + 8√t, where t is the time in seconds and s(t) is the distance in feet from the top of the building.
(a) Write a formula for the average velocity of the ball for t near 4.
(b) Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds
(c) What is your estimate for the instantaneous velocity of the ball at t = 4
Answer:
a)
Average velocity, (Vavg) of the ball for t near 4.
Vavg = [s(4) - s(0)] / (4 - 0)
Where s(4) = -16 × 4² + 8 × √4= - 240 m
s(0) = -16 × 0 + 8 * 0 = 0
b)
duration = 1 sec
Vavg = [s(5) - s(4)] / (5 - 4)
s(5) = -16 × 52 + 8 × √5 = - 382 m
s(4) = -16 × 42 + 8 √4 = - 240 m
Vavg = (-382 - (-240)) / (5 - 4)
Vavg = - 142.1 m/s
duration = 0.5 sec
Vavg = [s(4.5) - s(4)] / (4.5 - 4)
s(4.5) = -16 × 4.52 + 8 × √4.5 = - 307 m
s(4) = -16 × 42 + 8 × √4 = - 240 m
Vavg = (-307 - (-240)) / (4.5 - 4)
Vavg= - 134.1 m/s
duration = 0.05 sec
Vavg = [s(4.05) - s(4)] / (4.05 - 4)
s(4.05) = -16 × 4.052 + 8 × √4.05 = - 246 m
s(4) = -16 × 42 + 8 × √4 = - 240 m
Vavg = (-246 - (-240)) / (4.05 - 4)
Vavg= - 126.8 m/s
c)
Instantaneous velocity, v = ds/dt
= - 16 × 2 × t + 8 ×× (0.5 / √t )
= - 32 × t + 4/√t
ds/dt at t = 4 is,
v = - 32 × 4 + 4 / √4
= - 126 m/s
A hole is drilled in a metal plate. When the metal is raised to a higher temperature, what happens to the diameter of the hole?
Answer:
The diameter of the hole increases
Explanation:
Metals expand and contract with temperature. Whenever metal is heated, it usually expands in relation to its thermal expansivity. This expansion leads to a slight increase in surface area.
Once the surface area of the metal changes, this means that the dimensions of the whole metal surface changed. As a result, the diameter of the hole drilled in the metal plate will change also. In our case, the diameter of the hole will increase.
You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.0dB . So you decide to move closer to give the conversation a sound level of 80.0dB instead. How close should you come?
Answer:
The distance is [tex]r_2 = 0.24 \ m[/tex]
Explanation:
From the question we are told that
The distance from the conversation is [tex] r_1 = 24.0 \ m[/tex]
The intensity of the sound at your position is [tex]\beta _1 = 40 dB[/tex]
The intensity at the sound at the new position is [tex]\beta_2 = 80.0dB[/tex]
Generally the intensity in decibel is is mathematically represented as
[tex]\beta = 10dB log_{10}[\frac{d}{d_o} ][/tex]
The intensity is also mathematically represented as
[tex]d = \frac{P}{A}[/tex]
So
[tex]\beta = 10dB * log_{10}[\frac{P}{A* d_o} ][/tex]
=> [tex]\frac{\beta}{10} = log_{10} [\frac{P}{A (l_o)} ][/tex]
From the logarithm definition
=> [tex]\frac{P}{A * d_o} = 10^{\frac{\beta}{10} }[/tex]
=> [tex]P = A (d_o ) [10^{\frac{\beta }{ 10} } ][/tex]
Here P is the power of the sound wave
and A is the cross-sectional area of the sound wave which is generally in spherical form
Now the power of the sound wave at the first position is mathematically represented as
[tex]P_1 = A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ][/tex]
Now the power of the sound wave at the second position is mathematically represented as
[tex]P_2 = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ][/tex]
Generally power of the wave is constant at both positions so
[tex]A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ] = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ][/tex]
[tex]4 \pi r_1 ^2 [10^{\frac{\beta_1 }{ 10} } ] = 4 \pi r_2 ^2 [10^{\frac{\beta_2 }{ 10} } ][/tex]
[tex]r_2 = \sqrt{r_1 ^2 [\frac{10^{\frac{\beta_1}{10} }}{ 10^{\frac{\beta_2}{10} }} ]}[/tex]
substituting value
[tex]r_2 = \sqrt{ 24^2 [\frac{10^{\frac{ 40}{10} }}{10^{\frac{80}{10} }} ]}[/tex]
[tex]r_2 = 0.24 \ m[/tex]
Which equation describes the cosine function for a right triangle?
A. cos(O)
opposite
adjacent
B. cos(2)
adjacent
hypotenuse
C. cos(0)
opposite
hypotenuse
D. cos(e)
hypotenuse
adjacent
SUENT
Answer:
cosine is adjacent over hypotenuse
Explanation:
How I know? It's just a basic formula for triangles, and my teacher taught me this. Sorry if this didn't help.
Answer:
D.cos(e)
hypotenuse adjacent
SUENT
Explanation:
it is helpful to you
You toss a 100-g rock in the air with a kinetic energy of 5.0 J. How much is its speed as it leaves your hand
Answer:
The speed is 10 m/s
Explanation:
Th mass if the rock = 100 g = 0.1 kg
The kinetic energy = 5.0 J
speed of the rock = ?
The kinetic energy of the rock is given by the equation
E = [tex]\frac{1}{2} mv^{2}[/tex]
where
m is the mass of the rock
V is the velocity of the rock
substituting, we have
5 = [tex]\frac{1}{2} *0.1*v^{2}[/tex]
[tex]v^{2}[/tex] = 5/0.05 = 100
v = [tex]\sqrt{100}[/tex] = 10 m/s
1. An insect inside a bus flies from the back toward the front at 2 m/s. The bus is moving in a
straight line at 20 m/s. What is the total speed of the insect as seen by a person standing on the
side of the road?
Answer:
Total speed of the insect as seen by a person standing on the side of the road is 22 m/s
Explanation:
There i a composition of velocities, since the air inside the bus is moving at the same speed as the bus, and the insect flying relative to it.
So for an observer on the side of the road, the insect is flying at a speed of :
20 m/s + 2 m/s = 22 m/s
water is boiling in a pan at 100c if heat is transferred to the pan at a rate of 2 kw how much water in kg will evaporate in 10min
Answer:
The mass of water that will evaporate 2.857 kg.
Explanation:
Given;
temperature of the water, T = 100 c
rate of heat transfer, H = 2 kW = 2000 w
time, t = 10 min = 10 min x 60s = 600s
heat capacity of water is given by;
Q = McΔT
Where;
Q is the quantity of heat transfer to the pan = P x t = 2000 x 600 = 1200000 J.
c is specific heat capacity of water = 4200 J/kg.°C
m = (Q) / (cΔT)
m = (1200000) (4200 x 100)
m = 2.857 kg
Therefore, the mass of water that will evaporate 2.857 kg.
Tell uses of cancave mirror and convex mirror.
Answer:
Uses of concave mirror:
Shaving mirrors.
Head mirrors.
Ophthalmoscope.
Astronomical telescopes.
Headlights.
Solar furnaces.
Uses of convex mirror:
Convex mirrors always form images that are upright, virtual, and smaller than the actual object. They are commonly used as rear and side view mirrors in cars and as security mirrors in public buildings because they allow you to see a wider view than flat or concave mirrors.
please give me full points.
Andy took a bus and then walked from his home to downtown.
For the first 16 hour, the bus drove at an average speed of 15
km/h. For the next 0.4 hours, he walked at an average speed
of 4.5 km/h. What was the average speed for the whole
journey?
Given :
Speed of bus for first 16 hours is , u = 15 km/h .
Distance he walked for next 0.4 hour is , v = 4.5 km/h .
To Find :
The average speed for the whole journey.
Solution :
Average speed is given by :
[tex]v_{avg}=\dfrac{u_1t_1+u_2t_2}{t_1+t_2}\\\\v_{avg}=\dfrac{15\times 16 + 4.5\times 0.4}{16+0.4}\\\\v_{avg}=14.74\ km/h[/tex]
Therefore , the average speed of the journey is 14.74 km/h .
Hence , this is the required solution .
Sapphire has an index of refraction of 1.77. The critical angle for sapphire in air with an index of refraction of 1.0 is _____ degrees.
Answer:
The value is [tex]\theta = 34.40^o[/tex]
Explanation:
From the question we are told that
The index of refraction is [tex]n_s = 1.77[/tex]
The refractive index of air is [tex]n_a = 1.0[/tex]
Generally the critical angle is mathematically represented as
[tex]\theta = sin ^{-1} [\frac{1}{n_s} ][/tex]
=> [tex]\theta = sin ^{-1} [\frac{1}{1.77} ][/tex]
=> [tex]\theta = sin ^{-1} [0.565 ][/tex]
=> [tex]\theta = 34.40^o[/tex]
If a planet has an angular size of 0.29 degrees, what will be its angular size in arcminutes?
Answer:
Angular size in arc-minutes = 17.4 arc-minutes
Explanation:
Given:
Angular size = 0.29
Find:
Angular size in arc-minutes
Computation:
1 degree = 60 arc-minutes
So,
0.29 degree = 60 × 0.29 arc-minutes
0.29 degree = 17.4 arc-minutes
Angular size in arc-minutes = 17.4 arc-minutes
An electron in a region where there is an electric field experiences a force of magnitude 3.1 10-16 N. What is the magnitude of the electric field at the location of the electron?
Answer:
1.938*10^3N/C
Explanation:
Using F= qE
E= F/q
Where q= 1.6x10-19
So 3.1*10^-16/1.6*10^-19
= 1.938*10^3N/C
Explanation:
Multiple choice! To be dimensionally consistent, velocity [L/T], force [ML/T^2], mass [M], and length [L] must be related as follows. Multiple Choice A. velocity^2 = force × length/mass B. velocity^2 = force × length/mass^2 C.velocity = force × length^2/mass D. velocity = force × length/mass
Answer:
velocity^2=force*length/mass
Explanation:
velocity^2=force*length/mass is dimensionally consistent
as we know that
unit of force=Newton
unit of length=meter
unit of mass=kilogram so
velocity^2=newton*meter/kilogram
newton=kg*m/sec^2 therefore
v^2=kg*m*m/sec^2*kilogram
v^2=kg*m^2/sce^2*kg
v^2=m^2/sec^2
taking square root on both side
√ v²=√m²/sec²
v=m/s now the dimensions of V is
v=(L/T)
so option A is dimensionally consistent
It should be related like velocity^2=force*length/mass
Calculation of the relationship:
Since we know that
unit of force=Newton
unit of length=meter
And,
unit of mass=kilogram
Therefore,
velocity^2=newton*meter/kilogram
Now
newton=kg*m/sec^2
So,
v^2=kg*m*m/sec^2*kilogram
v^2=kg*m^2/sce^2*kg
v^2=m^2/sec^2
Now we take square root on both side
So,
√ v²=√m²/sec²
v=m/s
v=(L/T)
Therefore, option A is correct
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Velocity head is the mechanical energy of a fluid per unit weight of the fluid measured relative to the bottom of the channel.
a. True
b. False
Answer:
b. False
Explanation:
The Bernoulli's equation is given by;
[tex]P + \rho gh + \frac{1}{2} \rho v^2 = constat[/tex]
Divide through by the density of the fluid and acceleration due to gravity (ρg), to obtain energy per unit weight of the fluid known as head;
[tex]\frac{P}{\rho g} + h + \frac{v^2}{2g} = constant[/tex]
where;
[tex]\frac{P}{\rho g}[/tex] is pressure head
h is datum head (mechanical energy due to height of the fluid)
[tex]\frac{v^2}{2g}[/tex] is velocity head (mechanical energy due to velocity of the fluid)
Thus, the mechanical energy of a fluid per unit weight of the fluid, measured relative to the bottom of the channel is datum head.
The correct option is "b. False"
A
is a precise way to represent data, but it does not identify trends in the data.
O graph
O table
Ovariable
O diagram
Answer: table
Explanation:
Answer:Table
Explanation:
A table is a way to represent data , but does not identify trends in the data
The World Snail Racing Championships have been held in the English village of Congham, Norfolk since the 1960s. The contestants are placed at the center of a circular table covered with a damp tablecloth and race outwards 13 inches (33 cm) towards the finish line. A contestant known only as Archie set the current world record of 2 minutes 20 seconds in 1995. Determine Archie's speed in m/s and km/h.
Speed = distance / time
Speed = (33 cm) / (2min 20sec)
Speed = (0.33 m) / (140 sec)
Speed = 0.0024 m/s
Speed = (0.0024 m/s) x (3600 sec / 1 hr) x (1 km / 1000 m)
Speed = (0.0024 x 3600 / 1000) (m-sec-km / sec-hr-m)
Speed = 0.0085 km/hr
For the time of 2 min and 20 seconds, Archie's speed in m/s will be 0.0024 m/s and Archie's speed in km/hr will be 0.0085 km/hr.
What is Speed?The "speed at which an object is traveling," or speed, is a scalar quantity. The speed of an object is the rate at which it moves through space. A fast-moving object covers a considerable distance in a short amount of time while traveling at a high speed. An object moving slowly, on the other hand, moves a comparatively modest distance in the same amount of time. An object with zero speed is one that is not moving at all.
According to the question, the given values are :
Distance, d = 33 inches or,
d = 33 cm
Time, t = 2 minutes 20 seconds, or,
t = 140 seconds.
Speed is the ratio of distance to time.
s = 33/140
s = 0.0024 m/s
Now the distance in km/hr will be :
s = 0.0024 m/s × 3600 sec / 1hr × 1 km/ 1000 m
s= 0.0085 km/hr
Hence, the speed in km/hr is 0.0085 km/hr and in m/s speed is 0.0024 m/s.
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Two campers in a national park hike from their cabin to the same spot on a lake, each taking a different path, as illustrated below. The total distance traveled along Path 1 is 7.5 km, and that along Path 2 is 8.2 km. What is the final displacement of each camper?
Answer:
Their displacement is 5 Km from the cabin.
Explanation:
The displacement is define is the shortest distance between two points. It is the shortest distance of a line that joins two points together. It quantifies both the distance and direction of the net or total motion along a straight line from the initial position to the final position of the point trajectory. In this case, the straight line joining the cabin to the stream is line S which is equal to 5 Km
A 0.40 kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.5 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 63 N, what is the maximum speed the ball can have?
Answer:
The maximum speed the ball can have is 15.37 m/s
Explanation:
Given;
mass of the ball, m = 0.4 kg
radius of the chord, r = 1.5 m
maximum tension on the chord, T = 63 N
The maximum tension on the chord is given by;
[tex]T_{max} = \frac{mv_{max}^2}{r} \\\\v_{max}^2 = \frac{T_{max} *r}{m}\\\\ v_{max} = \sqrt{\frac{T_{max} *r}{m}} \\\\ v_{max} =\sqrt{\frac{63 *1.5}{0.4}}\\\\v_{max} = 15.37 \ m/s[/tex]
Therefore, the maximum speed the ball can have is 15.37 m/s
A 150 lb diver stands at the end of a rigid 8 ft long diving board. What is the magnitude of the torque the diver exerts on the diving board
Answer:
1200 lb-ft
Explanation:
Weight of diver W = 150 lb
length of board L = 8 ft
This board is pivoted at one end therefore, the perpendicular distance from the pivot = 8 ft
Torque = weight x perpendicular distance from pivot
T = WL
Torque = 150 x 8 = 1200 lb-ft
g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
Complete Question
A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.
Answer:
The value [tex]v_2 = 4 \sqrt{10} \ m/s[/tex]
Explanation:
From the question we are told that
The charge on the first sphere is [tex]q_1 = 2\mu C = 2*10^{-6} \ C[/tex]
The charge on the second sphere is [tex]q_2 = 8 \mu C = 8*10^{-6} \ C[/tex]
The mass of the second charge is [tex]m = 1.50 \ g = 1.50 *10^{-3} \ kg[/tex]
The distance apart is [tex]d = 0.4 \ m[/tex]
The speed of the second sphere is [tex]v_1 = 20 \ ms^{-1}[/tex]
Generally the total energy possessed by when [tex]q_2[/tex] and [tex]q_1[/tex] are separated by [tex]0.8 \ m[/tex] is mathematically represented
[tex]Q = KE + U[/tex]
Here KE is the kinetic energy which is mathematically represented as
[tex]KE = \frac{1 }{2} m (v_1)^2[/tex]
substituting value
[tex]KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2[/tex]
[tex]KE = 0.3 \ J[/tex]
And U is the potential energy which is mathematically represented as
[tex]U = \frac{k * q_1 * q_2 }{d }[/tex]
substituting values
[tex]U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }[/tex]
[tex]U = 0.18 \ J[/tex]
So
[tex]Q = 0.3 + 0.18[/tex]
[tex]Q = 0.48 \ J[/tex]
Generally the total energy possessed by when [tex]q_2[/tex] and [tex]q_1[/tex] are separated by [tex]0.4 \ m[/tex] is mathematically represented
[tex]Q_f = KE_f + U_f[/tex]
Here [tex]KE_f[/tex] is the kinetic energy which is mathematically represented as
[tex]KE_f = \frac{1 }{2} m (v_2^2[/tex]
substituting value
[tex]KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2[/tex]
[tex]KE_f = 7.50 *10^{ -4} (v_2 )^2[/tex]
And [tex]U_f[/tex] is the potential energy which is mathematically represented as
[tex]U_f = \frac{k * q_1 * q_2 }{d }[/tex]
substituting values
[tex]U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }[/tex]
[tex]U_f = 0.36 \ J[/tex]
From the law of energy conservation
[tex]Q = Q_f[/tex]
So
[tex]0.48 = 0.36 +(7.50 *10^{-4} v_2^2)[/tex]
[tex]v_2 = 4 \sqrt{10} \ m/s[/tex]
Calculate the radius of each ball to the nearest tenth. Use the formula r = d/2 where r is the radius and d is the diameter. Record the average in Table A of your Student Guide. What is the radius of the table tennis ball? cm What is the radius of the golf ball? cm
Answer:
What is the radius of the table tennis ball?
⇒ 2.1 cm
What is the radius of the golf ball?
2.0 cm
Explanation:
divide the radius and round it to the nearest 10th place..... but hope that help ;)
Answer:
1) What is the radius of the table tennis ball?
2.1 cm
2) What is the radius of the golf ball?
2.0 cm
Explanation:
Did this on Edge 2023 and got it right. Hope this helps :)
Two cars, P and Q, travel in the same direction on a long, straight section of a highway. car P passes car Q, and is adjacent to car Q at time t (subscript 0).
a) Suppose that car P and car Q each move with constant speed. At time t (subscript 0), is the magnitude of the instantaneous velocity of car P greater than, less than, or equal to the magnitude of the instantaneous velocity of car Q? Explain.
b) Suppose instead that car P is moving with constant speed but car Q is speeding up. At time (subscript 0), is the magnitude of the instantaneous velocity of car P greater than, less than, or equal to the magnitude of the instantaneous velocity of car Q? Explain.
Answer:
the instantaneous velocity of P is greater that Q because P obviously passes Q at that instant in time t
2. P has greater velocity at that instant (to) because P will need to be at a greater velocity to pass Q
A dolphin swims due east for 1.90 km, then swims 7.20 km in the direction south of west. What are the magnitude and direction of the vector that will take the dolphin back to its starting point? (Enter the magnitude in km and the direction in degrees north of east.)
Answer:
magnitude = 7.446 km, direction = 75.22° north of east
Explanation:
From the questions,
To get the the magnitude of the resultant vector we use Pythagoras theorem
a² = b²+c²
From the diagram,
y² = 1.9²+7.2²
y² = 55.45
y = √(55.45)
y = 7.446 km.
The direction of the dolphin is given as,
θ = tan⁻¹(7.2/1.9)
θ = tan⁻¹(3.7895)
θ = 75.22° north of east
Hence the magnitude of the resultant vector = 7.446 km, and it direction is 75.22° north of east
URGENT!!!!! I WILL MARK U AS BRAINLIEST!!!!!!! A step-down transformer converts 120 V to 5 V in a phone charger. If the primary coil has 600 turns, how many turns are in the secondary coil? SHOW ALL WORK WITH CORRECT ANSWER WITH CORRECT UNIT AS WELL.
Answer:
25 turns
Explanation:
The voltage ratio is the turns ratio:
secondary turns / primary turns = secondary voltage / primary voltage
secondary turns / 600 = 5/120
secondary turns = 600(5/120) = 25 . . . turns
The secondary coil has 25 turns.
E
Which is the downward pull on an object due to gravity?
volume
mass
weight
matter
Answer:
Weight
Explanation:
Weight is the downward pull on an object due to gravity.
For example, the moon has less gravity than Earth so we would weigh less on the moon. Our Mass and volume always stay the same but our weight could change.
Answer:
weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight
Explanation:
(a) Calculate the kinetic energy, in joules of a 1200 -kg automobile moving at 18 m/s .
(b) Convert this energy to calories.
(c) When the automobile brakes to a stop is the "lost" kinetic energy converted mostly to heat or to some form of potential energy?
Answer:
a) The kinetic energy of the 1200-kg automobile moving at 18 meters per second is 194400 joules, b) The kinetic energy of the 1200-kg automobile moving at 18 meters per second is 46440.516 calories, c) the kinetic energy is transformed into work due to friction, which is a non-conservative force. And such work is dissipated in the form of heat.
Explanation:
a) Let be the automobile considered as particle travelling on horizontal ground, so that motion is entirely translational and whose formula for kinetic energy, measured in joules, is:
[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v[/tex] - Speed of automobile, measured in meters per second.
If [tex]m = 1200\,kg[/tex] and [tex]v = 18\,\frac{m}{s}[/tex], the kinetic energy of the automobile is:
[tex]K = \frac{1}{2}\cdot (1200\,kg)\cdot \left(18\,\frac{m}{s} \right)^{2}[/tex]
[tex]K = 194400\,J[/tex]
The kinetic energy of the 1200-kg automobile moving at 18 meters per second is 194400 joules.
b) A calory equals 4.186 joules. The kinetic energy in calories is:
[tex]K = 194400\,J \times \left(\frac{1}{4.186}\,\frac{cal}{J} \right)[/tex]
[tex]K = 46440.516\,cal[/tex]
The kinetic energy of the 1200-kg automobile moving at 18 meters per second is 46440.516 calories.
c) When the automobile brakes to a stop, the kinetic energy is transformed into work due to friction, which is a non-conservative force. And such work is dissipated in the form of heat. Hence, such energy cannot be recovered. Potential energies are conservative by nature.
How does the decrease in gravitational potential energy of a falling ball compare to its increase in kinetic energy? (Ignore air friction.)
Answer:
for every decrease in potential energy, a kinetic energy proportional to 2gΔh is gained.
Explanation:
Let us consider a ball falling from its maximum height
For a body falling from its maximum height to a point p
change in height = Δh
The potential energy decrease is then proportional to
ΔPE = mgΔh
where
ΔPE is the decrease in kinetic energy
m is the mass of the ball
g is acceleration due to gravity
Δh is the change in height
For a body falling from its maximum height, the increase change in velocity
Δv = u + 2gΔh (at maximum height u = 0)
where
u is the initial kinetic energy of the ball
Δv = 0 + 2gΔh
Δv = 2gΔh
The kinetic energy increases by
ΔKE = [tex]\frac{1}{2}[/tex]m(Δv)^2
but Δv = 2gΔh
therefore
ΔKE = [tex]\frac{1}{2}[/tex]m(2gΔh)^2 = 2m(gΔh)^2
comparing the increase in kinetic energy to the decrease in potential energy, we have
(2m(gΔh)^2)/(mgΔh) = 2gΔh
This means that for every decrease in potential energy, a kinetic energy proportional to 2gΔh is gained.
Which of the following describe a condition or location for the right environment for tropical cyclone formation over the north eastern tropical Pacific Ocean?
A. light surface winds
B. a La Nina event
C. stable air
D. a deep layer of humid air
E. converging surface winds
F. ocean surface temperature greater than 80 °F
Answer:
E. converging surface winds
Explanation:
For the formation of tropical cyclones to occur in the northeast of the tropical Pacific Ocean, it is necessary to have converging surface winds that present a strong convection. This allows for the formation of vertical movements of an ascending nature that have great strength and are intense and violent. This also allows closed cyclonic circulation to occur at the lower levels which together with a low frontal pressure system over the waters of the Pacific Ocean can form the tropical cyclone.
A current of 1.70 A flows in a wire. How many electrons are flowing past any point in the wire per second
Answer:
1.0625×10¹⁹ electrons
Explanation:
From the question,
Using,
Q = It........................ Equation 1
Where Q = charge, I = current flowing in the wire, t = time (seconds)
Given: I = 1.70 A, t = 1 Seconds.
Substitute these values into equation 1
Q = 1.7(1)
Q = 1.7 C.
But,
1.6×10⁻¹⁹ C = 1 electron
Therefore,
1.7 C = 1.7/(1.6×10⁻¹⁹) electrons
= 1.0625×10¹⁹ electrons
Hence, 1.0625×10¹⁹ electrons flows through the wire per seconds.
A three-phase load has a phase voltage of 240 V and a phase current of 18 A. What is the apparent power of this load
Answer:
7482.5VA
Explanation:
We have that,
phase voltage of a three phase load is,240v
The phase current of a three phase load is, 18A
So To find apparent load we use
VA= √3*E*I
=√ 3*240*18
=7482.5VA