Answer:
349 J
Explanation:
Length L of baton = 1.0 m
Mass m of baton = 1.2 kg
Weight W of baton = 1.2 kg x 9.81 m/[tex]s^{2}[/tex] = 11.772 N
Height h reached = 5.0 m
Angular speed ω = 3.5 rev/s = 2π x 3.5 (rad/s) = 21.99 rad/s
Total work done on baton will be the work done in taking it to a height of 5.0 m and the kinetic energy with which the baton rolls.
Work done to bringing it to the height of 5.0 m = weight x height above ground
W x h = 11.772 x 5 = 58.86 J
Velocity v of spinning baton = ω x L = 21.99 x 1 = 21.99 m/s
Kinetic energy = [tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex] =
Total work done on baton = 58.86 + 290.14 = 349 J
Horizontal beam AB is 200 kg, 2.4 m long, and is welded at point A. The man is 80 kg and applies a tension of 300 N on the cable. Diameter of the pulley is 300 mm and BC = 300 mm. Determine:
(a) horizontal and vertical components of force at A,
(b) magnitude and direction of the moment supported at A.
Answer:
[tex]a)-3346.8\;N \\ b)-4937.04\;N-m[/tex]
Explanation:
a) - In Free-body diagram :
At point D, the free body diagram of a man :
[tex]D = Fn\\Fn=T+mg\\put\;values\;in\;it\\ .\;\;\;\;=300+(80)(9.81)=1084.8\;N[/tex]
[tex]Mg=200\times9.81=1962\;N[/tex]
[tex]\sum Fx=0\; where\; Ax=0[/tex]
[tex]\sum Fy=0\; where\; Ay-Mg-Fn-T=0[/tex]
Then, put the value in the equation.
[tex]Ay=3346.8\;N[/tex]
b)-
[tex]Ma=Mg(AE)+Fn(AD)+T=4937.04\;N-m[/tex]
The 5-lb collar is released from rest at A and travels along the frictionless guide. Determine the speed of the collar when it strikes the stop B. The spring has an unstretched length of 0.5 ft.
Answer:
Explanation:
Stiffness of spring k equal to 4 lb/ft.
Unstretched length of the spring L is equal to 0.5 feet.
Weight of the collar W is 15lb
Radius of curvature of curve guide is 1 feet
length of vertical rod is 1.5 feet
Initial speed of collar when released from rest at A is 0 feet per seconds
use the energy conservation equation
[tex]P_A +K_A=P_B+K_B[/tex]
Estimate the potential energy , component as position B as below
[tex]P_A=Wh_1+\frac{1}{2} ks^2_1\\\\=5\times(1.5+1)+\frac{1}{2} \times 4 \times(1.5+1-0.5)^2\\\\=20.5lb.ft[/tex]
Estimate the kinetic energy , component as position A as below
[tex]K_A=\frac{1}{2} \frac{W}{g} V^2_1\\\\=\frac{1}{2} \frac{5}{32.2} \times0^2\\\\=0lb.ft[/tex]
Estimate the kinetic energy , component as position B as below
[tex]K_A=\frac{1}{2} \frac{W}{g} V^2_2\\\\=\frac{1}{2} \frac{5}{32.2} \times V^2_2\\\\=00777V^2_2[/tex]
Substitute 20.5lb- ft for [tex]P_A[/tex]
0.5lb-ft for [tex]P_B[/tex]
0lb -ft for [tex]K_A[/tex]
[tex]0.0777V_2^2 for K_B[/tex]
[tex]20.5+0=0.5+0.777V^2_2\\\\V^2_2=257.6\\\\V_2=16.05[/tex]
= 16.05ft/sec
A glass rod that has been charged to +12.0 nC touches a metal sphere. Afterward, the rod's charge is +8.0 nC. a) What kind of charged particle was transferred between the rod and the sphere and in which direction? That is, did it move from rod to the sphere or from sphere to the rod? b) How many charged particles were transferred?
Answer:
Explanation:
charge transferred to sphere = 12 - 8 = 4 nC
a )
The positive charge is decreased on glass rod , that means electrons must have entered the glass rod from the sphere which reduces its positive charge.
b )
charge transferred = 4 nC = 4 x 10⁻⁹ C
charge on one electron = 1.6 x 10⁻¹⁹ C
No of electrons transferred
= Total charge transferred / charge on one electron
= 4 x 10⁻⁹ / 1.6 x 10⁻¹⁹
= 2.5 x 10¹⁰ .
This question involves the concepts of charges and charged particles.
a) "Electrons" were transferred "from the sphere to the rod"
b) "2.5 x 10¹⁰" charged particles were transferred.
a)
Since the charge on the rod decreases by 4 nC. Therefore, negatively charged particles, that is electrons must have moved from the sphere to the rod. Because electrons are the mobile particles that move from one atom to the other.
b)
The number of charged particles transferred can be found using the following formula:
q = ne
where,
q = charge transferred = 4 nC = 4 x 10⁻⁹ C
n = no. of charged particles (electrons) transferred = ?
e = charge on one electron = 1.6 x 10⁻¹⁹ C
Therefore,
[tex]n=\frac{4\ x\ 10^{-9}\ C}{1.6\ x\ 10^{-19}\ C}[/tex]
n = 2.5 x 10¹⁰
Learn more about charged particles here:
https://brainly.com/question/1936492?referrer=searchResults
For the different values given for the radius of curvature RRR and speed vvv, rank the magnitude of the force of the roller-coaster track on the cart at the bottom of each dip. Rank from largest to smallest. To rank items as equivalent, overlap them.
A) R = 60m v = 16 m/s
B) R = 15m v = 8m/s
C) R = 30m v = 4m/s
D) R = 45m v = 4m/s
E) 30m v = 16 m/s
F) R = 15m v =12 m/s
Explanation:
The force of the roller-coaster track on the cart at the bottom is given by :
[tex]F=\dfrac{mv^2}{R}[/tex], m is mass of roller coaster
Case 1.
R = 60 m v = 16 m/s
[tex]F=\dfrac{(16)^2m}{60}=4.26m\ N[/tex]
Case 2.
R = 15 m v = 8 m/s
[tex]F=\dfrac{(8)^2m}{15}=4.26m\ N[/tex]
Case 3.
R = 30 m v = 4 m/s
[tex]F=\dfrac{(4)^2m}{30}=0.54m\ N[/tex]
Case 4.
R = 45 m v = 4 m/s
[tex]F=\dfrac{(4)^2m}{45}=0.36m\ N[/tex]
Case 5.
R = 30 m v = 16 m/s
[tex]F=\dfrac{(16)^2m}{30}=8.54m\ N[/tex]
Case 6.
R = 15 m v =12 m/s
[tex]F=\dfrac{(12)^2m}{15}=9.6m\ N[/tex]
Ranking from largest to smallest is given by :
F>E>A=B>C>D
A 80-kg base runner begins his slide into second base when he is moving at a speed of 3.7 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.A 45-kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. Neglect air resistance, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.
Answer:
the vaulter has a height of 6.112 meters
Explanation:
Base runner: no question content.
__
Pole vaulter:
Initial kinetic energy* is ...
KE = (1/2)mv^2 = (1/2)(45 kg)(11 m/s)^2 = 2722.5 J
Above the bar, her kinetic energy is ...
KE = (1/2)mv^2 = (1/2)(45 kg)(1.1 m/s)^2 = 27.225 J
Then the amount of kinetic energy converted to potential energy is ...
2722.5 J -27.225 J = 2695.275 J
This corresponds to a change in height of ...
PE = mgh
2695.275 J = (45 kg)(9.8 m/s^2)h
h = 2695.275/(45·9.8) m = 6.112 m
Her altitude above the bar is 6.112 meters.
_____
* Here, we use the traditional equations for kinetic and potential energy, which use "m" to represent mass. When we fill in the numbers, we attach units to the numbers in which "m" represents "meters". We trust you can sort this without confusion.
A leaf floating down from a tree is an example of an object in free fall.
A. True
B. False
Answer:
its false
Explanation:
because if an leaf floats down from a tree it is not considered an object for a free-fall
A model rocket is fired straight up from the top of a 45-m-tall building. The rocket has only enough fuel to burn for 4.0 s. But while the rocket engine is burning fuel, it produces an upward acceleration of 55 m/s2. After the fuel supply is exhausted, the rocket is in free fall and just misses the edge of the building as it falls back to the ground. Ignoring air resistance, calculate (a) the height above the ground and the velocity of the rocket when its fuel runs out; (b) the maximum height of the rocket; (c) the time the rocket is in the air; and (d) the rocket's velocity the moment before it hits the ground.
Answer:
a)y = 485 m , v = 220 m / s , b) y = 2954.39 m , c) t_total = 51 s ,
d) v = 240.59 m / s
Explanation:
a) We can use vertical launch ratios for this exercise
the speed of the rocket the run out the fuel is
v = v₀ + a t
the rocket departs with initial velocity v₀ = 0
v = a t
v = 55 4
v = 220 m / s
the height at this point is
y = y₀ + v₀t + ½ a t²
y = y₀ + 1/2 a t²
y = 45 + ½ 55 4²
y = 485 m
b) the maximum height of the rocket is when its speed is zero
for this part we will use as the initial speed the speed at the end of the fuel (v₀´ = 220 m / s) and the height of y₀´ = 485 m
v² = v₀´² + 2 g (y-y₀´ )
0 = v₀´² + 2 g (y-y₀´ )
y = y₀´ + v₀´² / 2g
y = 485 + 220 2/2 9.8
y = 2954.39 m
c) the time that the rocket is in the air is the acceleration time t₁ = 4 s, plus the rise time (t₂) plus the time to reach the ground (t₃)
let's calculate the rise time
v = v₀´- g t
v = 0
t₂ = v₀´ / g
t₂ = 220 / 9.8
t₂ = 22.45 s
Now let's calculate the time it takes to get from this point (y₀´´ = 2954.39 m) to the floor
y = y₀´´ + v₀´´ t - ½ g t²
0 = y₀´´ - ½ g t²
t = √ (2 y₀´´ / g)
t = √ (2 2954.39 / 9.8)
t = 24.55 s
the total flight time is
t_total = t₁ + t₂ + t₃
t_total = 4 + 22.45 + 24.55
t_total = 51 s
d) the veloicda right now
v = vo + g t
v = 9.8 24.55
v = 240.59 m / s
(a) At an air show a jet flies directly toward the stands at a speed of 1220 km/h, emitting a frequency of 3270 Hz, on a day when the speed of sound is 342 m/s. What frequency (in Hz) is received by the observers?
Answer:
f' = 358442.3 Hz
Explanation:
This is a problem about the Doppler's effect. To find the perceived frequency of the observers you use the following formula:
[tex]f'=f(\frac{v}{v\pm v_s})[/tex]
v: speed of sound = 342 m/s
vs : speed of the source (jet) = 1220 km/h (1h/3600s)*(1000 m/ 1km) = 338.88 m/s (it is convenient to convert the units of vs to m/s)
f: frequency emitted by the source = 3270 Hz
f': perceived frequency
Due to the jet is getting closer to the observers, the sing of the denominator in equation (1) is minus (-). Then, you replace the values of f, vs and v:
[tex]f'=(3270s^{-1})(\frac{342m/s}{342m/s-338.88m/s})=358442.3 Hz[/tex]
Hence, the perceived frequency by the observers is 358442.3 Hz
Answer:
[tex]f_o=359466.42Hz[/tex]
Explanation:
You can solve this problem using doppler effect formula. The Doppler effect is the phenomenon by which the frequency of the waves perceived by an observer varies when the emitting focus or the observer itself moves relative to each other
Doppler effect general case is given by:
[tex]f_o=f\frac{v\pm v_o}{v \mp v_s} \\\\Where:\\\\f=Actual\hspace{3}frequency\\f_o=Observed\hspace{3}frequency\\v=Speed\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves\\v_o=Velocity\hspace{3}of\hspace{3}the\hspace{3}observer\\v_s=Velocity\hspace{3}of\hspace{3}the\hspace{3}source[/tex]
Now, you can use the following facts:
[tex]+v_o[/tex] Is used when the observer moves towards the source
[tex]-v_o[/tex] Is used when the observer moves away from the source
[tex]-v_s[/tex] Is used when the source moves towards the observer
[tex]+v_s[/tex] Is used when the source moves away from the observer
Since the source is alone in motion towards the observer, the formula is given by:
[tex]f_o=f\frac{v}{v-v_s} \\\\Where:\\\\v_s=1220km/h\approx 338.8888889m/s\\v=342m/s\\f=3270Hz[/tex]
Therefore:
[tex]f_o=(3270) \frac{342}{342-338.8888889} = 359466.42Hz[/tex]
A man paddles a canoe in a long, straight section of a river. The canoe moves downstream with constant speed 4 m/s relative to the water. The river has a steady current of 2 m/s relative to the bank. The man's hat falls into the river. Seven minutes later, he notices that his hat is missing and immediately turns the canoe around, paddling upriver with the same constant speed of 4 m/s relative to the water. How long does it take the man to row back upriver to reclaim his hat
Answer:
7 minutes
Explanation:
From the question, the flow of the river applies equally to both the hat and the canoe. The best way to approach this problem is to look at it from a perspective that follows the water, and ignores the bank.
Hence, the water and the hat are stationary, and the man paddles away from the hat and then paddles back.
Now, the hat is at rest and relative to the water and the canoe always travels at a speed of 4 m/s relative to the water.
We are told that it takes 7 minutes for the man to travel away from the hat.
Since he is travelling at a constant speed, it will therefore take same time of 7 minutes for him to return.
So, it will take 7 minutes for the man to row back upriver to reclaim his hat.
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An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the earth with a speed of 0.99537 c relative to the earth. A scientist at rest on the earth's surface measures that the particle is created at an altitude of 40.0 km .
As measured by the scientist, how much time does it take the particle to travel the 40.0 km to the surface of the earth?
Answer:
Explanation:
Velocity relative to earth = .99537 c
= .99537 x 3 x 10⁸ m /s
= 2.98611 x 10⁸ m /s
distance relative to earth = 40 km = 40 x 10³ m .
Time measured by scientist
= 40 x 10³ / 2.98611 x 10⁸
= 13.395 x 10⁻⁵ s
= 134 μs.
What is potential energy
Answer:
Potential Energy is when the energy is kept in place the object isnt moving so it isnt creating Kinetic energy.
Answer:
[tex]\boxed{\mathrm{view \ explanation}}[/tex]
Explanation:
Potential energy is the energy stored in a body due to the body’s positioning or state.
A 12-V battery is connected to an air-filled capacitor that consists of two parallel plates,
each plate has an area of 7.60 cm2
. The separation between the plates of the capacitor is d =
0.30 cm. (Assume the electric field between the plates to be uniform).
a. Draw the situation.
b. Find the magnitude of the electric field between the plates.
Now, a proton (q = 1.6 x10-19 C) is released from rest at the positive plate of the capacitor.
c. Calculate the electric potential energy gained by the proton just before it touches the negative
plate.
A slab of Teflon of dielectric constant k =2.1 is then inserted between the plates of the
capacitor.
d. What is the new capacitance of the capacitor?
e. Calculate the change in the total energy stored in the capacitor before and after inserting the
dielectric slab.
Answer:
E = 4000 V / m
U = 1.92*10^-18 J
C' = 4.71 pF
1.2 times greater with di-electric
Explanation:
Given:-
- The potential difference between plates, V = 12 V
- The area of each plate, A = 7.6 cm^2
- The separation between plates, d = 0.3 cm
- The charge of the proton. q = 1.6*10^-19 C
- The initial velocity of proton, vi = 0 m/s
Solution:-
- The electric field ( E ) between the parallel plates of the air-filled capacitor is determined from the applied potential difference by the battery on the two ends of the plates.
- The separation ( d ) between the two plates allows the charge to be stored and the Electric field between two charged plates would be:
E = V / d
E = 12 / 0.003
E = 4,000 V/m ... Answer
- The amount of electrostatic potential energy stored between the two plates is ( U ) defined by:
U = q*E*d
U = (1.6 x10^-19)*(4000)*(0.003)
U = 1.92*10^-18 J ... Answer
- The electrostatic energy stored between plates is ( U ) when the proton moves from the positively charges plate to negative charged plate the energy is stored within the proton.
- A slab of di-electric material ( Teflon ) is placed between the two plates with thickness equal to the separation ( d ) and Area similar to the area of the plate ( A ).
- The capacitance of the charged plates would be ( C ):
C = k*ε*A / d
Where,
k: the di-electric constant of material = 2.1
ε: permittivity of free space = 8.85 × 10^-12
- The new capacitance ( C' ) is:
C' = 2.1*(8.85 × 10^-12) *( 7.6 / 100^2 ) / 0.003
C' = 4.71 pF
- The new total energy stored in the capacitor is defined as follows:
U' = 0.5*C'*V^2
U' = 0.5*(4.71*10^-12)*(12)^2
U' = 3.391 * 10^-10 J
- The increase in potential energy stored is by the amount of increase in capacitance due to di-electric material ( Teflon ). The di-electric constant "k" causes an increase in the potential energy stored before and after the insertion.
- Hence, the new potential energy ( U' ) is " k = 2.1 " times the potential energy stored in a capacitor without the di-electric.
an object that is 42kg is a free fall, experiencing an acceleration from gravity (g=9.8m/s^2). what is the force of gravity on the object, assuming no air resistance
Answer:
411.6 N
Explanation:
The force is given by ...
F = ma . . . . m represents mass
F = (42 kg)(9.8 m/s^2) = 411.6 N . . . . m represents meters
The force on the object due to gravity is 411.6 newtons.
What is the difference between reflection and refraction
Answer:
Reflection is when light bounces off an object, while refraction is when light bends while passing through an object.
Explanation:
I just learned about this 2 weeks ago actually.
A 2300 kg truck has put its front bumper against the rear bumper of a 2500 kg SUV to give it a push. With the engine at full power and good tires on good pavement, the maximum forward force on the truck is 18,000 N. What is the maximum possible acceleration the truck can give the SUV?
Answer:
Acceleration a = 3.75 m/s^2
Explanation:
Given that
Mass M1 = mass of the truck = 2300kg
Mass M2 = mass of the SUV = 2500kg
Force applied F = 18000N
From Newton second law,
F = ma
Where a = acceleration
We must consider the entire mass involved in the system. Therefore,
F = (M1 + M2)a
Make a the subject of formula
a = F/(M1 + M2)
Acceleration a = 18000/( 2300 + 2500)
Acceleration a = 18000/ 4800
Acceleration a = 3.75 m/s^2
How long does it take a car travelling at 60 m.p.h to cover 5 miles?
Answer:
Time = 5 minutes
Answer:
2minutes per mile
Explanation:
Not sure if it helps
A learner driver travels at a speed of 3km/h for an hour and a half. What distance does it travel?
Answer:
distance covered is 4.5km or 1.25m
Explanation:
GIVEN DATA:
SPEED=v=3km/h
TIME=t=1.5h
TO FIND:
DISTANCE COVERED=d=?
SOLUTION:
As we know that
speed=distance covered /time taken
here we have to find distance
speed×time taken=distance covered
3km/h×1.5h=distance covered
4.5km=distance covered or in international system of unit the answer is 1.25m
An object is dropped from a bridge. A second object is thrown downwards 1.0 s later. They both reach the water 20 m below at the same instant. What was the initial speed of the second object
Answer:
vi = 14.610
Explanation:
initial velocity of the first object (vi) = 0 m/s, because it was dropped
distance (y) = -20 m
initial position (y0) = 0 m
acceleration due to gravity (g) = -9.8 m/s^2
y = 1/2 gt^2 + vi*t + y0
-20 = 1/2(-9.8)t^2 + 0 + 0
-20 = -4.9t^2
4.081 = t^2
+√4.081 = t
t = 2.020
time of second object = 2.020 - 1 = 1.020
Now we can plug in the new time to solve for vi of the second object.
y = 1/2 gt^2 + vi*t + y0
-20 = 1/2(-9.8)1.020^2 + vi*1.020 + 0
-20 = -5.098 + 1.020vi
14.902 = 1.020vi
vi = 14.610 m/s
In April 1974, Steve Prefontaine completed a 10.0 km race in a time of 27 min , 43.6 s . Suppose "Pre" was at the 7.43 km mark at a time of 25.0 min . If he accelerated for 60 s and then maintained his increased speed for the remainder of the race, calculate his acceleration over the 60 s interval. Assume his instantaneous speed at the 7.43 km mark was the same as his overall average speed up to that time.
Answer:
Acceleration, a = 0.101 m/s²
Explanation:
Average speed = total distance / total time.
At the 7.43km mark, total distance = 7.43km or 7430m
Total time = 25 * 60 s = 1500s
Average speed = 7430m/1500s = 4.95m/s
He then covers (10 - 7.43)km = 2.57 km = 2570 m
in t = 27m43.6s - 25min = 2m43.6s = 163.6 s
Then he accelerates for 60 s, and maintains this velocity V, for the remaining (163.6 - 60)s = 103.6 s.
From V = u + at; V = 4.95m/s + a *60s
Distance covered while accelerating is
s = ut + ½at² = 4.95m/s * 60s + ½ a *(60s)² = 297m + a*1800s²
Distance covered while at constant velocity, v after accelerating is
D = velocity * time
Where v = 4.95m/s + a*60s
D = (4.95m/s + a*60s) * 103.6s = 512.82m + a*6216s²
Total distance covered after initial 7.43 km, S + D = 2570 m, so
2570 m = 297m + a*1800s² + 512.82m + a*6216s²
2570 = 809.82 + a*8016
a = 809.82m / 8016s² = 0.101 m/s²
That the electric field is uniform between the electrodes means that the electric field lines: A.are more closely spaced at the positive electrode than at the negative one. B.intersect halfway between the electrodes. C.are more closely spaced at the negative electrode than at the positive one. D.are equally spaced at both electrodes and between them.
Answer: Option D.
Are equally spaced at both electrodes and between them.
Explanation:
An electric field is the region that surrond electrically charged particle which may cause attraction or repelling of the particle due to the force that is exerted on it.
The electric field between two electrodes due to electrostatic forces exist in the air and there is an increase of external voltage is added to it. Electric field are uniform between electrodes which signify that the electric field lines are evenly spaced at both electrodes and between them.
Where does the energy released in nuclear reactions come from
Answer:
Nuclear energy comes from tiny mass changes in nuclei as radioactive processes occur. In fission, large nuclei break apart and release energy; in fusion, small nuclei merge together and release energy.
I hoep that was helpful! Let me know:)
Explanation:
A 1kw electric heater is switched on for ten minutes. how much heat does it produce
H= P × t
1kW= 1000w
10 min = 600s
H= 1000×600=600,000J
=> 143.40kcal
PLZZZ HELP
Question 2 Multiple Choice Worth 2 points)
(04.05 LC)
Which of the following scenarios describe an object being acted on by a non-conservative force?
O Apuck hovers and glides across an air track at constant velocity.
A dishwasher was pulled up to a window by a rope using a pulley system.
O A feather falls from one end of a tube to the other inside a vacuum.
O A charged particle was pushed away by another charged particle.
Answer:
C
Explanation:
A feather falls from one end of a tube to the other inside a vacuum.
In which situation would you see a partial lunar eclipse?
Answer:
C
Explanation:
It happens when the a portion of the moon passes through the earth's shadow
A canon is tilled back 30.0 degrees and shoots a cannon ball at 155 m/s. What is the
highest point that the cannon ball reaches?
Answer:
[tex]= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m[/tex]
Therefore, highest point that the cannon ball reaches is 168.7m
Explanation:
the cannon is fired at an angle 30 o to the horizonatal with a speed of 155 m/s
highest point that the cannon ball reaches?
[tex]H_{max}=\frac{V^2\sin ^2 \theta}{2g}[/tex]
g = 9.8m/s2
[tex]= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m[/tex]
Therefore, highest point that the cannon ball reaches is 168.7m
A 30-kg skater moving at 3 m/s overtakes a 35-kg skater moving at 1 m/s in the same direction and collides with her. The two skaters stick together. It’s final speed is 1.9 m/s. How much kinetic energy is lost? Round all your answers to one decimal place.
Answer:
The lost in kinetic energy is [tex]KE_l = 125.5 \ J[/tex]
Explanation:
From the question we are told that
The mass of the first skater is [tex]m_1 = 30 \ kg[/tex]
The speed of the first skater is [tex]v_1 = 3 \ m/s[/tex]
The mass of the second skater is [tex]m_2 = 35 \ kg[/tex]
The speed of the second skater is [tex]v_2 = 1 \ m/ s[/tex]
The final speed of both skater are [tex]v_f = 1.9 m/s[/tex]
The initial kinetic energy of both skaters is mathematically represented as
[tex]KE_i = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2[/tex]
substituting values
[tex]KE_i = \frac{1}{2} * 30 * 3^2 + \frac{1}{2} * 35 * 1^2[/tex]
[tex]KE_i = 242.5 \ J[/tex]
The final kinetic energy of both skaters is mathematically represented as
[tex]KE_f = \frac{1}{2} * (m_1 + m_2 ) v_f^2[/tex]
substituting values
[tex]KE_f = \frac{1}{2} * (30 + 35 ) * 1.9^2[/tex]
[tex]KE_f = 117 \ J[/tex]
The lost in kinetic energy is
[tex]KE_l = 242.5 -117[/tex]
[tex]KE_l = 125.5 \ J[/tex]
ASAP pls answer right I will mark brainiest . All I know is 4. Is A
Answer:
Q1: B.2 Q2: B.Waxing crescent Q3: A.Waxing Gibbous
Explanation:
The starship Enterprise approaches the planet Risa at a speed of 0.8c relative to the planet. On the way, it overtakes the intergalactic freighter Astra. The relative speed of the two ships as measured by the navigator on the Enterprise is 0.5c. At what speed is Astra approaching the planet?
Answer:
0.3c
Explanation:
The speed of Enterprise relative to Risa is 0.8c
Relative speed of both ships as measured from Enterprise is 0.5c
therefore, relative speed of Astra to Enterprise is 0.8c - 0.5c = 0.3c
this is also the relative speed with which Astra approaches the planet Risa since Enterprise's speed was calculated relative to Risa.
A spherical shell contains three charged objects. The first and second objects have a charge of −18.0 nC−18.0 nC and 38.0 nC38.0 nC , respectively. The total electric flux through the shell is −218 N⋅m2/C−218 N⋅m2/C . What is the charge on the third object?
Answer:
q3 = 21.9 nC
Explanation:
By the Gauss theorem you have that the electric flux in a Gaussian surface is given by:
[tex]\Phi_E=\frac{Q}{\epsilon_o}[/tex] (1)
ФE: electric flux = -218Nm^2/C
Q: net charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/(Nm^2)
You can consider the spherical shell as a Gaussian surface. Then, the net charge inside the surface is:
[tex]Q=-18.0nC+38.0nC+q_3[/tex] (2)
where charge q3 is unknown charge of the third object:
You replace the equation (2) into the equation (1), and you solve for q3:
[tex]\epsilon_0 \Phi_E=-18.0*10^{-9}C+38.0*10^{-9}C+q_3\\\\\epsilon_0 \Phi_E=20*10^{-9}C+q_3\\\\q_3=(8.85*10^{-12}C^2/(Nm^2))(-218Nm^2/C)-20*10^{-9}C\\\\q_3=2.19*10^{-9}C=21.9nC[/tex]
hence, the charge of the third object is 21.9 nC
