Answer:
SO THAT
EACH APPLIANCE CAN GET SUFFICIENT POTENTIAL DIFF. TO RUNWhat is the highest point at which weather will generally occur?
Answer:
At thestratosphere: it 20- 25km
I need help physics
A surveyor measures the distance across a river that flows straight north by the following method. Starting directly across from a tree on the opposite bank, the surveyor walks distance, D = 130 m along the river to establish a baseline. She then sights across to the tree and reads that the angle from the baseline to the tree is an angle θ = 25°. How wide is the river?
Answer:
The width of the river is [tex]z = 60.62 \ m[/tex]
Explanation:
From the question we are told that
The distance of the base line is D = 130 m
The angle is [tex]\theta = 25^o[/tex]
A diagram illustration the question is shown on the first uploaded image
Applying Trigonometric Rules for Right-angled Triangle,
[tex]tan 25 = \frac{z}{130}[/tex]
Now making z the subject
[tex]z = 130 * tan (25)[/tex]
[tex]z = 60.62 \ m[/tex]
assume that the initial speed is 25 m/s and the angle of projection is 53 degree above the hroizontal. the cannon ball leaves the uzzel of the cannon at a highet of 200 m.( the cannon is at the edge of the cliff) A: find the horizontal distance the cannon travles. B: when does the cannon ball reach the ground? C: find the maximum highet the cannon ball reaches.
Answer:
A. xmax = 131.49 m
B. t = 8.74 s
C. ymax = 220.33 m
Explanation:
A. In order to find the horizontal distance which cannon travels you first calculate the flight time. The flight time can be calculated by using the following formula:
[tex]y=y_o+v_osin\theta-\frac{1}{2}gt^2[/tex] (1)
yo: height from the projectile is fired = 200m
vo: initial velocity of the projectile = 25m/s
g: gravitational acceleration = 9.8 m/s^2
θ: angle between the direction of the initial motion of the ball and the horizontal = 53°
t: time
You need the value of t when the projectile hits the ground. Then, in th equation (1) you make y = 0m.
When you replace the values of all parameters in the equation (1), you obtain the following quadratic formula:
[tex]0=200+(25)sin53\°t-\frac{1}{2}(9.8)t^2\\\\0=200+19.96t-4.9t^2[/tex] (2)
You use the quadratic formula to obtain the value of t:
[tex]t_{1,2}=\frac{-19.96\pm\sqrt{(19.96)^2-4(-4.9)(200)}}{2(-4.9)}\\\\t_{1,2}=\frac{-19.96\pm65.71}{-9.8}\\\\t_1=8.74s\\\\t_2=-4.66s[/tex]
You use the positive value because it has physical meaning.
Now, you can calculate the horizontal range of the projectile by using the following formula:
[tex]x_{max}=v_ocos\theta t[/tex]
[tex]x_{max}=(25m/s)(cos53\°)(8.74s)=131.49m[/tex]
The cannon ball travels a horizontal distance of 131.49 m
B. The cannon ball reaches the canon for t = 8.74s
C. The maximum height is obtained by using the following formula:
[tex]y_{max}=y_o+\frac{v_o^2sin^2\theta}{2g}[/tex] (3)
By replacing in the equation (3) the values of all parameters you obtain:
[tex]y_{max}=200m+\frac{(25m/s)^2(sin53\°)^2}{2(9.8m/s^2)}\\\\y_{mac}=200m+20.33m=220.33m[/tex]
The maximum height reached by the cannon ball is 220.33m
A factory worker pushes a 30.0 kg crate a distance of 3.7 m along a level floor at constant velocity by pushing downward at an angle of 30∘ below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.25.
Required:
a. What magnitude of force must the worker apply?
b. How much work is done on the crate by this force?
c. How much work is done on the crate by friction?
d. How much work is done on the crate by the normal force? By gravity?
e. What is the total work done on the crate?
Answer:
a) [tex]F = 210.803\,N[/tex], b) [tex]W_{F} = 779.971\,J[/tex], c) [tex]W_{f} = 235.683\,J[/tex], d) [tex]W_{N} = 0\,J[/tex]; [tex]W_{g} = 544.289\,J[/tex], e) [tex]W_{net} = 0\,J[/tex]
Explanation:
a) The net force exerted on the crate is:
[tex]\Sigma F = F - m\cdot g \cdot \sin \theta - \mu_{k}\cdot m\cdot g \cdot \cos \theta = 0[/tex]
The magnitud of the force that the work must apply to the crate is:
[tex]F = m\cdot g \cdot \sin \theta + \mu_{k}\cdot m\cdot g \cdot \cos \theta[/tex]
[tex]F = (30\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \sin 30^{\circ} + 0.25 \cdot (30\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 30^{\circ}[/tex]
[tex]F = 210.803\,N[/tex]
b) The work done on the crate due to the external force is:
[tex]W_{F} = (210.803\,N)\cdot (3.7\,m)[/tex]
[tex]W_{F} = 779.971\,J[/tex]
c) The work done on the crate due to the external force is:
[tex]W_{f} = (63.698\,N)\cdot (3.7\,m)[/tex]
[tex]W_{f} = 235.683\,J[/tex]
d) The work done on the crate due the normal force is zero, since such force is perpendicular to the motion direction.
[tex]W_{N} = 0\,J[/tex]
And, the work done by gravity is:
[tex]W_{g} = (147.105\,N)\cdot (3.7\,m)[/tex]
[tex]W_{g} = 544.289\,J[/tex]
e) Lastly, the total work done is:
[tex]W_{net} = W_{F} - W_{f} - W_{g} - W_{N}[/tex]
[tex]W_{net} = 779.971\,J - 235.683\,J - 0\,N - 544.289\,J[/tex]
[tex]W_{net} = 0\,J[/tex]
A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 2.90 m/s2 . At 20.0s after blastoff, the engines suddenly fail, which means that the force they produce instantly stops.
(A) How high above the launch pad will the rocket eventually go?
(B) Find the rocket's velocity at its highest point.
(C) Find the magnitude of the rocket's acceleration at its highest point.
(D) Find the direction of the rocket's acceleration at its highest point.
(E) How long after it was launched will the rocket fall back to the launch pad?
(F) How fast will it be moving when it does so?
Answer:
A) 580m
B) 0 m/s
C) 9.8m/s^2
D) downward
E) 10.87s
F) 106.62 m/s
Explanation:
A) The distance traveled by the rocket is calculated by using the following expression:
[tex]y=\frac{1}{2}at^2[/tex]
a: acceleration of the rocket = 2.90 m/s^2
t: time of the flight = 20.0 s
[tex]y=\frac{1}{2}(2.90\frac{m}{s^2})(20.0s)^2=580m[/tex]
B) In the highest point the rocket has a velocity with magnitude zero v = 0m/s because there the rocket stops.
C) The engines of the rocket suddenly fails in the highest point. There, the acceleration of the rocket is due to the gravitational force, that is 9.8 m/s^2
D) The acceleration points downward
E) The time the rocket takes to return to the ground is given by:
[tex]t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2(580m)}{9.8m/s^2}}=10.87s[/tex]
10.87 seconds
F) The velocity just before the rocket arrives to the ground is:
[tex]v=\sqrt{2gy}=\sqrt{2(9.8m/s ^2)(580m)}=106.62\frac{m}{s}[/tex]
You are on a train traveling east at speed of 19 m/s with respect to the ground. 1)If you walk east toward the front of the train, with a speed of 1.5 m/s with respect to the train, what is your velocity with respect to the ground
Answer:
Vbg = 20.5 m/s
your velocity with respect to the ground Vbg = 20.5 m/s
Explanation:
Relative velocity with respect to the ground is;
Vbg = velocity of train with respect to the ground + your velocity with respect to the train
Vbg = Vtg + Vbt ......1
Given;
velocity of train with respect to the ground;
Vtg = 19 m/s
your velocity with respect to the train;
Vbt = 1.5 m/s
Substituting the given values into the equation 1;
Vbg = 19 m/s + 1.5 m/s
Vbg = 20.5 m/s
your velocity with respect to the ground Vbg = 20.5 m/s
You have just landed on Planet X. You take out a ball of mass 100 gg , release it from rest from a height of 16.0 mm and measure that it takes a time of 2.90 ss to reach the ground. You can ignore any force on the ball from the atmosphere of the planet. How much does the 100-g ball weigh on the surface of Planet X?
Answer:
0.173 N.
Explanation:
We will calculate the mass and then use the following calculations on the surface of planet X that is :
[tex]W=mg[/tex]
We would use the following equation to get the value of g for planet X that is :
[tex]y_f-y_i=v_{yi}t+\frac{1}{2}gt^2[/tex]
Then, put the values in the above equation.
[tex]16=0+\frac{1}{2}\times g\times(2.90)^2[/tex]
[tex]\bf\mathit{g=3.80\;m/s^2}[/tex]
Now, we will measure the ball weight on planet X's surface:
[tex]m=\frac{100}{1000} \;\;\;\;\;\;\;\;\;\;[1kg=1000g][/tex]
Then, we have to put the value in the above equation.
[tex]W=0.1\times 1.73=0.173\:N[/tex]
What do you call a group of sea turtles?
Answer:
a bale
Explanation:
a bale is a group of turtles
Answer:
A bale or nest
Explanation:
2. If rain is falling vertically downward, and you are running for shelter, should you hold your umbrella
vertically, tilted forward, or tilted backward to keep the driest? Please explain.
Answer:
Tilted forward to keep the driest.
Explanation:
The rain is falling vertically so there is no wind. In these circumstances the umbrella should be tilted vertically forward.
The situation is the same as if you would stand still and the rain would come under an angle from the front.
You are watching an object that is moving in SHM. When the object is displaced 0.560 m to the right of its equilibrium position, it has a velocity of 2.45 m/s to the right and an acceleration of 8.60 m/s2 to the left. Part A How much farther from this point will the object move before it stops momentarily and then starts to move back to the left
Answer:
2.95m
Explanation:
The farthest distance the object can move is the radius of the circle of which the Simple harmonic motion is assumed to be a part
But V = w× r; where V is velocity,
w is angular velocity and r is radius.
Also,
a= w2r; where a is linear acceleration
but a = v× r ; by comparing both equations
Hence r = a/v =8.6/2.45 =3.51m
But the horizontal distance of the motion is given by:
X = rcosx ; where x is the angle
X is the distance covered.
We know that the maximum value of cos x is 1 which is 0°
When the object moves in a fashion directly parallel to an horizontal distance, maximum distance would be reached and hence:
X = r=3.51m
Meaning the object needs to travel 3.51-0.56=2.95m further.
Note: the acceleration of the motion is constant whether it is swinging towards the left or right.
When the object is displaced 0.560 m to the right of its equilibrium position, it has a velocity of 2.45 m/s to the right and an acceleration of 8.60 m/s2 to the left and the amplitude of motion A = 0.732 m.
What is Amplitude of motion?
The distance between the central and extreme points for a moving particle is known as the amplitude of motion.
The given data to find the amplitude of motion,
Object displaced = 0.560 m
Velocity = 2.45 m/s
Acceleration = 8.60 m/s²
Starting with sine:
x(t) = Asin(ωt)
so that t = 0, x = 0
x(t) = 0.56 m = Asin(ωt)
v(t) = x(t)'= 2.45 m/s = Aωcos(ωt)
a(t) = v(t)'= -8.60 m/s² = -Aω²sin(ωt)
x(t) / a(t) = Asin(ωt) / -Aω²sin(ωt)
0.56m / -8.60 m/s² = -1 / ω²
ω² = 15.3571 rad^2/s^2
ω = 3.91881 rad/s
x(t) / v(t) = Asin(ωt) / Aωcos(ωt)
0.560m / 2.45m/s = tan(3.91t) / 3.91rad/s
0.8937= tan(3.91t)
t = 0.176 s
x(0.176) = Asin(3.59×0.176)
0.65 m= Asin(0.631)
A = 0.732 m is the amplitude of motion.
To know more about Amplitude of motion,
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If an instalment plan quotes a monthly interest rate of 4%, the effective annual/yearly interest rate would be _____________. 4% Between 4% and 48% 48% More than 48%
Answer:
More than 48%
Explanation:
If the interest is computed monthly on the outstanding balance, it has an effective annual rate of ...
(1 +4%)^12 -1 = 60.1% . . . . more than 48%
The effective annual or yearly interest rate would be=30.56% which is Between 4% and 48%
Calculation of Annual Interest rateThe formula used to calculate annual Interest rate =
[tex](1+ \frac{i}{n} ) {}^{n} - 1[/tex]
where i= nominal interest rate = 4%
n= number of periods= 12 months
Annual Interest rate=
[tex](1 + \frac{4\%}{12} ) {}^{12} - 1[/tex]
= (1+0.333)^12 -1
= (1.333)^12-1
= 31.56 - 1
= 30.56%
Therefore, the effective annual or yearly interest rate would be= 30.56%
Learn more about interest rate here:
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A standing wave on a string that is fixed at both ends has frequency 80.0 Hz. The distance between adjacent antinodes of the standing wave is 12.0 cm. What is the speed of the waves on the string, in m/s
Answer:
v = 19.2 m/s
Explanation:
In order to find the speed of the string you use the following formula:
[tex]f=\frac{v}{2L}[/tex] (1)
f: frequency of the string = 80.0Hz
v: speed of the wave = ?
L: length of the string = 12.0cm = 0.12m
The length of the string coincides with the wavelength of the wave for the fundamental mode.
Then, you solve for v in the equation (1), and replace the values of the other parameters:
[tex]v=2Lf=2(0.12m)(80.0Hz)=19.2\frac{m}{s}[/tex]
The speed of the wave is 19.2 m/s
A 1,269 kg rocket is traveling at 413 m/s with 2,660 kg of fuel on board. If the rocket fuel travels at 1,614 m/s relative to the rocket, what is the rockets final velocity after it uses half of its fuel?
Answer:
About 2104m/s
Explanation:
[tex]F=ma \\\\F=\dfrac{2660kg}{2}\cdot 1614m/s=2,146,620N \\\\2,146,620N=1,269kg\cdot a \\\\a\approx 1691m/s \\\\v_f=v_o+at=413m/s+1691m/s=2104m/s[/tex]
Hope this helps!
The International Space Station is about 90 meters across and about 380 kilometers away. One night t appears to be the same angular size as Jupiter. Jupiter is 143,000 km in size. Use serxa to figure out how far away Jupiter is in AU Note: 1 AU= 1.5 x 10-km
a) 6.0 x 10 Au
b) 4.0 AU
c) 9.1 x 1010 AU
d) 4.0 x 10 AU
Complete Question
The complete question is shown on the first uploaded image
Answer:
The distance is [tex]r_2 = 4 \ AU[/tex]
Explanation:
From the question we are told that
The size of Jupiter is [tex]s_2 = 143,000 \ km[/tex]
The length of the International Space Station is [tex]r_1 = 380\ km[/tex]
The size of the International Space Station is [tex]s_1 = 90 \ m =0.09 \ km[/tex]
The angular size where the same one night and this angular size is mathematically represented as
[tex]\theta = \frac{s}{r}[/tex]
Since [tex]\theta[/tex] is constant
[tex]\frac{s_1}{r_1} = \frac{s_2}{r_2}[/tex]
substituting values
[tex]\frac{0.09}{380} = \frac{143000}{r_2}[/tex]
=> [tex]r_2 = 6.04 * 10^{9} \ km[/tex]
Now we are told to convert to AU and 1 AU [tex]= 1.5 * 10^8 \ km[/tex]
So
[tex]r_2 = \frac{6.04 * 10^8}{1.5*10^{8}}[/tex]
[tex]r_2 = 4 \ AU[/tex]
b. A locomotive of a train exerts a constant force of 280KN on a train while pulling
it at 50 km/h along a level track. What is:
[4 marks)
i. Workdone in quarter an hour and
[4 marks]
Answer:
Work-done in quarter an hour = 3.5 × 10⁶ J
Explanation:
Given:
Force (F) = 280 KN = 280,000 N
Velocity (V) = 50 km / h
Time (t) = 1 / 4 = 0.25 hour
Find:
Work-done in quarter an hour
Computation:
⇒ Displacement = Velocity (V) × Time
⇒ Displacement = 50 × 0.25
⇒ Displacement = 12.5 km
⇒ Work-done = Force (F) × Displacement
⇒ Work-done in quarter an hour = 280,000 × 12.5
⇒ Work-done in quarter an hour = 3,500,000
Work-done in quarter an hour = 3.5 × 10⁶ J
A student drives 105.0 mi with an average speed of 61.0 mi/h for exactly 1 hour and 30
minutes for the first part of the trip. What is the distance in miles traveled during this
time?
Answer:
91.5 miles
Explanation:
61 miles per hour so 61(x amount of hours)
so 61 x 1.5 hours is 91.5 miles
An object, with mass 70 kg and speed 21 m/s relative to an observer, explodes into two pieces, one 4 times as massive as the other; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observer's reference frame
Answer:
K = 3.9 kJ
Explanation:
The kinetic energy ([tex]K_{T}[/tex]) added is given by the difference between the final kinetic energy and the initial kinetic energy:
[tex] K_{T} = K_{f} - K_{i} [/tex]
The initial kinetic energy is:
[tex] K_{i} = \frac{1}{2}m_{1}v_{1}^{2} [/tex]
Where m₁ is the mass of the object before the explosion and v₁ is its velocity
[tex] K_{i} = \frac{1}{2}m_{1}v_{1}^{2} = \frac{1}{2}70 kg*(21 m/s)^{2} = 1.54 \cdot 10^{4} J [/tex]
Now, the final kinetic energy is:
[tex] K_{f} = \frac{1}{2}m_{2}v_{2}^{2} + \frac{1}{2}m_{3}v_{3}^{2} [/tex]
Where m₂ and m₃ are the masses of the 2 pieces produced by the explosion and v₁ and v₂ are the speeds of these pieces
Since m₂ is 4 times as massive as m₃ and v₃ = 0, we have:
[tex] K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}v_{2}^{2} + \frac{1}{2}*\frac{1}{5}m_{1}*0 [/tex] (1)
By conservation of momentum we have:
[tex] p_{i} = p_{f} [/tex]
[tex] m_{1}v_{1} = m_{2}v_{2} + m_{3}v_{3} [/tex]
[tex] m_{1}v_{1} = \frac{4}{5}m_{1}v_{2} + \frac{1}{5}m_{1}*0 [/tex]
[tex] v_{2} = \frac{5}{4}v_{1} [/tex] (2)
By entering (2) into (1) we have:
[tex] K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}(\frac{5}{4}v_{1})^{2} = \frac{1}{2}*\frac{4}{5}70 kg(\frac{5}{4}*21 m/s)^{2} = 1.93 \cdot 10^{4} J [/tex]
Hence, the kinetic energy added is:
[tex] K_{T} = K_{f} - K_{i} = 1.93 \cdot 10^{4} J - 1.54 \cdot 10^{4} J = 3.9 \cdot 10^{3} J [/tex]
Therefore, the kinetic energy added to the system during the explosion is 3.9 kJ.
I hope it helps you!
g A top-fuel dragster starts from rest and has a constant acceleration of 44.0 m/s2. What are (a) the final velocity of the dragster at the end of 2.1 s, (b) the final velocity of the dragster at the end of of twice this time, or 4.2 s, (c) the displacement of the dragster at the end of 2.1 s, and (d) the displacement of the dragster at the end of twice this time, or 4.2 s?
The dragster's velocity v at time t with constant acceleration a is
[tex]v=at[/tex]
since it starts at rest.
After 2.1 s, it will attain a velocity of
[tex]v=\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)[/tex]
or 92.4 m/s.
Doubling the time would double the final velocity,
[tex]v=a(2t)=2at[/tex]
so the velocity would be twice the previous one, 184.8 m/s.
The dragster undergoes a displacement x after time t with acceleration a of
[tex]x=\dfrac12at^2[/tex]
if we take the starting line to be the origin.
After 2.1 s, it will have moved
[tex]x=\dfrac12\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)^2[/tex]
or 88 m.
Doubling the time has the effect of quadrupling the displacement, since
[tex]x=\dfrac12a(2t)^2=4\left(\dfrac12at^2\right)[/tex]
so after 4.2 s it will have moved 352 m.
Plaskett's binary system consists of two stars that revolve In a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal . Assume the orbital speed of each star is |v | = 240 km/s and the orbital period of each is 12.5 days. Find the mass M of each star. (For comparison, the mass of our Sun is 1.99 times 1030 kg Your answer cannot be understood or graded.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The mass is [tex]M =1.43 *10^{32} \ kg[/tex]
Explanation:
From the question we are told that
The mass of the stars are [tex]m_1 = m_2 =M[/tex]
The orbital speed of each star is [tex]v_s = 240 \ km/s =240000 \ m/s[/tex]
The orbital period is [tex]T = 12.5 \ days = 12.5 * 2 4 * 60 *60 = 1080000\ s[/tex]
The centripetal force acting on these stars is mathematically represented as
[tex]F_c = \frac{Mv^2}{r}[/tex]
The gravitational force acting on these stars is mathematically represented as
[tex]F_g = \frac{GM^2 }{d^2}[/tex]
So [tex]F_c = F_g[/tex]
=> [tex]\frac{mv^2}{r} = \frac{Gm_1 * m_2 }{d^2}[/tex]
=> [tex]\frac{v^2}{r} = \frac{GM}{(2r)^2}[/tex]
=> [tex]\frac{v^2}{r} = \frac{GM}{4r^2}[/tex]
=> [tex]M = \frac{v^2*4r}{G}[/tex]
The distance traveled by each sun in one cycle is mathematically represented as
[tex]D = v * T[/tex]
[tex]D = 240000 * 1080000[/tex]
[tex]D = 2.592*10^{11} \ m[/tex]
Now this can also be represented as
[tex]D = 2 \pi r[/tex]
Therefore
[tex]2 \pi r= 2.592*10^{11} \ m[/tex]
=> [tex]r= \frac{2.592*10^{11}}{2 \pi }[/tex]
=> [tex]r= 4.124 *10^{10} \ m[/tex]
So
[tex]M = \frac{v^2*4r}{G}[/tex]
=> [tex]M = \frac{(240000)^2*4*(4.124*10^{10})}{6.67*10^{-11}}[/tex]
=> [tex]M =1.43 *10^{32} \ kg[/tex]
A 148 g ball is dropped from a tree 11.0 m above the ground. With what speed would it hit the ground
Answer:
14.68m/s
Explanation:
As per the question, the data provided is as follows
Mass = M = 0.148 kg
Height = h = 11 m
Initial velocity = U = 0 m/s
Final velocity = V
Gravitational force = F
Mass = M
Based on the above information, the speed that hit to the ground is
As we know that
Work to be done = Change in kinetic energy
[tex]F ( S) = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]
[tex]M g h = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]
[tex]g h = (\frac{1}{2} ) ( V^2 - U^2 )[/tex]
[tex]V^2 - U^2 = 2gh[/tex]
[tex]V^2 - 0 = 2gh[/tex]
[tex]V = \sqrt{2 g h}[/tex]
[tex]= \sqrt{2\times9.8\times11}[/tex]
= 14.68m/s
I need someone that knows physics. I have a test in 10 hrs and Im not good at it. Can anyone help me?
Answer:
I can help! What level of physics is it and what are your main topics?
A CD is spinning on a CD player. In 220 radians, the cd has reached an angular speed of 92 r a d s by accelerating with a constant acceleration of 14 r a d s 2 . What was the initial angular speed of the CD
Answer:
The initial angular speed is [tex]w_i = 48 \ rad/s[/tex]
Explanation:
From the question we are told that
The angular displacement is [tex]\theta = 220 \ rad[/tex]
The angular speed is [tex]w_f = 92 \ rad/s[/tex]
The acceleration is [tex]\alpha = 14 \ rad/s^2[/tex]
Generally the initial angular speed can be evaluated as
[tex]w_f ^2 = w_i ^2 + 2 * \alpha * \theta[/tex]
=> [tex]w_i ^2 = w_f ^2 - 2 * \alpha * \theta[/tex]
substituting values
=> [tex]w_i ^2 = 92 ^2 - 2 * 14 * 220[/tex]
=> [tex]w_i ^2 = 2304[/tex]
=> [tex]w_i = 48 \ rad/s[/tex]
A Ferris wheel has radius 5.0 m and makes one revolution every 8.0 s with uniform rotation. A person who normally weighs 670 N is sitting on one of the benches attached at the rim of the wheel. What is the apparent weight (the normal force exerted on her by the bench) of the person as she passes through the highest point of her motion? ( type in your answer with no units in form xx0)
Answer:
The apparent weight of the person as she pass the highest point is [tex]N = 458.8 \ N[/tex]
Explanation:
From the question we are told that
The radius of the Ferris wheel is [tex]r = 5.0 \ m[/tex]
The period of revolution is [tex]T = 8.0 \ s[/tex]
The weight of the person is [tex]W = 670 \ N[/tex]
Generally the speed of the wheel is mathematically represented as
[tex]v = \frac{2 \pi r}{T }[/tex]
substituting values
[tex]v = \frac{2 * 3.142 * 5}{8 }[/tex]
[tex]v = 3.9 3 \ m/s[/tex]
The apparent weight (the normal force exerted on her by the bench) at the highest point is mathematically evaluated as
[tex]N = mg - \frac{mv^2}{r}[/tex]
Where m is the mass of the person which is mathematically evaluated as
[tex]m = \frac{W}{g}[/tex]
substituting values
[tex]m = \frac{670}{9.8}[/tex]
[tex]m = 68.37 \ kg[/tex]
So
[tex]N = 68.37 * 9.8 - \frac{68.37 * {3.93}^2}{5}[/tex]
[tex]N = 458.8 \ N[/tex]
In a 2 dimensional Cartesian system, the x-component of a vector is known, and the angle between vector and x-axis is known. Which operation is used to calculate the magnitude of the vector? (taken with respect to the x-component)
a. dividing by cosine
b. dividing by sine
c. multiplying by cosine
d. multiplying by sine
Answer:
The correct answer is a
Explanation:
The cosine function is
cos θ = ca / H
done ca is the adjacent leg (x-axis) and H is the hypotenuse (vector module)
we clear
H = ca / cos θ
therefore, to find the magnitude of the vector, the cathete is divided into the cosine.
The correct answer is a
Suppose your hair grows at the rate of 1/55 inch per day. Find the rate at which it grows in nanometers per second. Because the distance between atoms in a molecule is on the order of 0.1 nm, your answer suggests how rapidly atoms are assembled in this protein synthesis.
Answer:5.35nm
Explanation:
Consider that 1 inch is = 0.0254m
we have,
1m= 1x10^9 nm
While:
0.0254m = 2.54x10^7nm
1/55 (2.54x10^7) = 4.6181 x 10^5nm
1 day= 24 hrs
= (24x60) when calculating in min
= (24x60x60) calculating in seconds we have:
= 8.64x10⁴sec
In 8.64x10^4 seconds, the hair grows by 4.6181 x 10^5nm
Therefore, the amount by which the hair grows in 1 second will be;
= (4.6181 x 10^5)/(8.64x10^4)
= 5.35nm
The rate of growth will be 5.35nm
Michelson and Morley's experiment is widely considered to have been:______
a. a success because it detected a shift in the interference pattern.
b. a failure because it detected a shift in the interference pattern.
c. a success because it did not detect a shift in the interference pattern.
d. a failure because it did not detect a shift in the interference pattern.
e. lacking the necessary precision to determine a shift in the interference pattern.
Answer:
The correct answer is option (c) a success because it did not detect a shift in the interference pattern.
Explanation:
In Michelson and Morley experiment it was considered to be successful.
They both found out that the experiment that was carried out was not a failure since it did not detect any shift in the interference pattern.
With this findings it was widely regarded as correct and precise.
A 54.0 kg ice skater is gliding along the ice, heading due north at 4.10 m/s . The ice has a small coefficient of static friction, to prevent the skater from slipping sideways, but Uk = 0. Suddenly, a wind from the northeast exerts a force of 3.70 N on the skater.a) Use work and energy to find the skater's speed after gliding 100 m in this wind.b) What is the minimum value of Ug that allows her to continue moving straight north?
Answer:
a. 2.668 m/s
b. 0.00494
Explanation:
The computation is shown below:
a. As we know that
[tex]W = F\times d[/tex]
[tex]KE = 0.5\times m\times v^2[/tex]
As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.
F = 3.70 cos 45 = 2.62 N
[tex]W = F \times d = 2.62 N \times 100 m[/tex]
[tex]W = 261.6 N\times m[/tex]
We know that
KE1 = Initial kinetic energy
KE2 = kinetic energy following 100 m
The energy following 100 meters equivalent to the initial kinetic energy less the energy lost to the work performed by the wind on the skater.
So, the equation is
KE2 = KE1 - W
[tex]0.5 m\times v2^2 = 0.5 m\ v1^2 - W[/tex]
Now solve for v2
[tex]v2 = \sqrt{v1^2 - {\frac{2W}{M}}}[/tex]
[tex]= \sqrt{4.1 m/s)^2 - \frac{2 \times 261.6 N\times m}{54.0 kg}}[/tex]
= 2.668 m/s
b. Now the minimum value of Ug is
As we know that
Ff = force of friction
Us = coefficient of static friction
N = Normal force = weight of skater
So,
[tex]Ff = Us\times N[/tex]
Now solve for Us
[tex]= \frac{Ff}{N}[/tex]
[tex]= \frac{3.70 N \times cos 45 }{54.0 kg \times 9.81 m/s^2}[/tex]
= 0.00494
an aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. the hole is 30mm in diameter and is 30mm and is 100mm long. if modulus of elasticity for the aluminium is 85GN/m2, calculate the total contraction on the bar due to a compressive load of 180KN
Answer:
ΔL = 1.011 mm
Explanation:
Let's begin by listing out the given information:
Length (L) = 600 mm = 0.6 m,
Diameter (D) = 40 mm = 0.04 m ⇒ Radius (r) = 20 mm = 0.2 m,
Area (cross sectional) = πr² = 3.14 x .02² = 0.001256 m²,
Modulus of Elasticity (E) = 85 GN/m²,
Compressive load (F) = 180 KN
Using the formula, Stress = Load ÷ Area
Mathematically,
σ = F ÷ A = 180 x 10³ ÷ 0.001256
σ = 143312.1 KN/m²
Modulus of elasticity = stress ÷ strain
E = σ ÷ ε
ε = ΔL/L
85 x 10⁹ = 143312.1 x 10³ ÷ (ΔL/L)
ΔL = 143312.1 x 10³ ÷ 85 X 10⁹ = 1686.02 * 10⁻⁶
ΔL = L x 1686.02 * 10⁻⁶
ΔL = 0.6 * 1686.02 * 10⁻⁶ = 1011.61 x 10⁻⁶
ΔL = 1.011 x 10⁻³ m
ΔL = 1.011 mm
∴The bar contracts by 1.011 mm
An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference of 1.2 x 106 V and then enters a uniform magnetic field whose strength is 2.2 T. The alpha particle moves perpendicular to the field. Calculate (a) the speed of the alpha particle, (b) the magnitude of the magnetic force exerted on it, and (c) the radius of its circular path.
Answer:
a) v = 1.075*10^7 m/s
b) FB = 7.57*10^-12 N
c) r = 10.1 cm
Explanation:
(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:
[tex]K=qV[/tex] (1)
q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C
V: potential difference = 1.2*10^6 V
You replace the values of the parameters in the equation (1):
[tex]K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J[/tex]
The kinetic energy of the particle is also:
[tex]K=\frac{1}{2}mv^2[/tex] (2)
m: mass of the particle = 6.64*10^⁻27 kg
You solve the last equation for v:
[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}[/tex]
the sped of the alpha particle is 1.075*10^6 m/s
b) The magnetic force on the particle is given by:
[tex]|F_B|=qvBsin(\theta)[/tex]
B: magnitude of the magnetic field = 2.2 T
The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1
[tex]|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N[/tex]
the force exerted by the magnetic field on the particle is 7.57*10^-12 N
c) The particle describes a circumference with a radius given by:
[tex]r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm[/tex]
the radius of the trajectory of the electron is 10.1 cm
The speed, magnetic force and radius are respectively; 10.75 * 10⁶ m/s; 7.57 * 10⁻¹² N; 0.101 m
What is the Magnetic force?
A) We know that the formula for kinetic energy can be expressed as;
K = qV
where;
q is charge of the particle = 2e = 2(1.6 × 10⁻¹⁹ C) = 3.2 × 10⁻¹⁹ C
V is potential difference = 1.2 × 10⁶ V
K = 3.2 × 10⁻¹⁹ * 1.2 × 10⁶
K = 3.84 × 10⁻¹³ J
Also, formula for kinetic energy is;
K = ¹/₂mv²
where v is speed
Thus;
v = √(2K/m)
v = √(2 * 3.84 × 10⁻¹³)/(6.64 * 10⁻²⁷)
v = 10.75 * 10⁶ m/s
B) The magnetic force is given by the formula;
F_b = qvB
F_b = (3.2 × 10⁻¹⁹ * 10.75 * 10⁶ * 2.2)
F_b = 7.57 * 10⁻¹² N
C) The formula to find the radius is;
r = mv/qB
r = (6.64 * 10⁻²⁷ * 10.75 * 10⁶)/(1.6 × 10⁻¹⁹ * 2.2)
r = 0.101 m
Read more about magnetic field at; https://brainly.com/question/7802337
