Answer:
The original volume of the first bar is half of the original volume of the second bar.
Explanation:
The coefficient of cubic expansivity of substances is given by;
γ = ΔV ÷ ([tex]V_{1}[/tex]Δθ)
Given: two metal bars with equal change in volume, equal change in temperature.
Let the volume of the first metal bar be represented by [tex]V_{1}[/tex], and that of the second by [tex]V_{2}[/tex].
Since they have equal change in volume,
Δ[tex]V_{1}[/tex] = Δ[tex]V_{2}[/tex] = ΔV
For the first metal bar,
2γ = ΔV ÷ ([tex]V_{1}[/tex]Δθ)
⇒ Δθ = ΔV ÷ (2γ[tex]V_{1}[/tex])
For the second metal bar,
γ = ΔV ÷ ([tex]V_{2}[/tex]Δθ)
⇒ Δθ = ΔV ÷ ([tex]V_{2}[/tex]γ)
Since they have equal change in temperature,
Δθ of first bar = Δθ of the second bar
ΔV ÷ (2γ[tex]V_{1}[/tex]) = ΔV ÷ ([tex]V_{2}[/tex]γ)
So that;
(1 ÷ 2[tex]V_{1}[/tex]) = (1 ÷ [tex]V_{2}[/tex])
2[tex]V_{1}[/tex] = [tex]V_{2}[/tex]
[tex]V_{1}[/tex] = [tex]\frac{V_{2} }{2}[/tex]
Thus, original volume of the first bar is half of the original volume of the second bar.
what is the most likely elevation of point x?
A. 150 ft
B. 200 ft
C. 125 ft
D. 250 ft
A skydiver stepped out of an airplane at an altitude of 1000m fell freely for 5.00s opened her parachute and slowed to 7.00m/s in a negligible time what was the total elapsed time from leaving the airplane to landing on the ground
Answer:
t = 17.68s
Explanation:
In order to calculate the total elapsed time that skydiver takes to reache the ground, you first calculate the distance traveled by the skydiver in the first 5.00s. You use the following formula:
[tex]y=y_o-v_ot-\frac{1}{2}gt^2[/tex] (1)
y: height for a time t
yo: initial height = 1000m
vo: initial velocity = 0m/s
g: gravitational acceleration = 9.8m/s^2
t: time = 5.00 s
You replace the values of the parameters to get the values of the new height of the skydiver:
[tex]y=1000m-\frac{1}{2}(9.8m/s^2)(5.00s)^2\\\\y=877.5m[/tex]
Next, you take this value of 877.5m as the initial height of the second part of the trajectory of the skydiver. Furthermore, use the value of 7.00m/s as the initial velocity.
You use the same equation (1) with the values of the initial velocity and new height. We are interested in the time for which the skydiver arrives to the ground, then y = 0
[tex]0=877.5-7.00t-4.9t^2[/tex] (2)
The equation (2) is a quadratic equation, you solve it for t with the quadratic formula:
[tex]t_{1,2}=\frac{-(-7.00)\pm \sqrt{(-7.00)^2-4(-4.9)(877.5)}}{2(-4.9)}\\\\t_{1,2}=\frac{7.00\pm 131.33}{-9.8}\\\\t_1=12.68s\\\\t_2=-14.11s[/tex]
You use the positive value of t1 because it has physical meaning.
Finally, you sum the times of both parts of the trajectory:
total time = 5.00s + 12.68s = 17.68s
The total elapsed time taken by the skydiver to arrive to the ground from the airplane is 17.68s
A toy rocket, launched from the ground, rises vertically with an acceleration of 20 m/s2 for 6.0 s until its motor stops. Disregarding any air resistance, what maximum height above the ground will the rocket achieve?
Answer:
h = 1094.69m
The maximum height above the ground the rocket will achieve is 1094.69m.
Explanation:
The maximum height h is;
h = height covered during acceleration plus height covered when the motor stops.
h = h1 + h2 .......1
height covered during acceleration h1 can be derived using the equation of motion;
h1 = ut + 0.5at^2
Initial speed u = 0
h1 = 0.5at^2
acceleration a = 20 m/s^2
Time t = 6.0 s
h1 = 0.5×(20 × 6^2)
h1 = 0.5(20×36)
h1 = 360 m
height covered when the motor stops h2 can be derived using equation of motion;
h2 = ut + 0.5at^2 .......2
Where;
a = g = acceleration due to gravity = -9.8 m/s^2
The speed when the motor stops u;
u = at = 20 m/s^2 × 6.0 s = 120 m/s
Time t2 can be derived from;
v = u - gt
v = 0 (at maximum height velocity is zero)
u = gt
t = u/g
t = 120m/s / 9.8m/s^2
t = 12.24 seconds.
Substituting the values into equation 2;
h2 = 120(12.24) - 0.5(9.8×12.24^2)
h2 = 734.69376 m
h2 = 734.69 m
From equation 1;
h = h1 + h2 . substituting the values;
h = 360m + 734.69m
h = 1094.69m
The maximum height above the ground the rocket will achieve is 1094.69m.
Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m . If the track is completely flat and the race car is traveling at a constant 26.5 m/s (about 59 mph ) around the turn, what is the race car's centripetal (radial) acceleration
Answer:
The centripetal acceleration of the car will be 12.32 m/s² .
Explanation:
Given that
radius ,R= 57 m
Velocity , V=26.5 m/s
We know that centripetal acceleration given as follows
[tex]a_c=\dfrac{V^2}{R}[/tex]
Now by putting the values in the above equation we get
[tex]a_c=\dfrac{26.5^2}{57}=12.32\ m/s^2[/tex]
Therefore the centripetal acceleration of the car will be 12.32 m/s² .
Giving quadrilateral a(2,-1 ) b ( 1,3) c(6,5) d(7,1) you want to prove that it is a parallelogram by showing opposite sides are congruent . what formula would you use ? show that sb is congruent to cd
Answer:
AB = CD = √17
Explanation:
The distance formula is used to find the length of a line segment between two points. Here, we want to show the distance AB is the same as the distance CD.
d = √((x1 -x1)² +(y2 -y1)²)
__
AB: d = √((1 -2)² +(3 -(-1))²) = √((-1)² +4²) = √17
CD: d = √((7-6)² +(1-5)²) = √(1² +(-4)²) = √17 . . . . same as AB
Segment AB is congruent to segment CD.
Tether ball is a game children play in which a ball hangs from a rope attached to the top of a tall pole. The children hit the ball, causing it to swing around the pole. What is the total initial acceleration of a tether ball on a 2.0 m rope whose angular velocity changes from 13 rad/s to 7.0 rad/s in 15 s
Answer:
a_total = 14.022 m/s²
Explanation:
The total acceleration of a uniform circular motion is given by the following formula:
[tex]a=\sqrt{a_c^2+a_T^2}[/tex] (1)
ac: centripetal acceleration
aT: tangential acceleration
Then, you first calculate the centripetal acceleration by using the following formula:
[tex]a_c=r\omega^2[/tex]
r: radius of the circular trajectory = 2.0m
w: final angular velocity of the ball = 7.0 rad/s
[tex]a_c=(2.0m)(7.0rad/s)^2=14.0\frac{m}{s^2}[/tex]
Next, you calculate the tangential acceleration. aT is calculate by using:
[tex]a_T=r\alpha[/tex] (2)
α: angular acceleration
The angular acceleration is:
[tex]\alpha=\frac{\omega_o-\omega}{t}[/tex]
wo: initial angular velocity = 13 rad/s
t: time = 15 s
Then, you use the expression for the angular acceleration in the equation (1) and solve for aT:
[tex]a_T=r(\frac{\omega_o-\omega}{t})=(2.0m)(\frac{7.0rad/s-13.0rad/s}{15s})=-0.8\frac{m}{s^2}[/tex]
Finally, you replace the values of aT and ac in the equation (1), in order to calculate the total acceleration:
[tex]a=\sqrt{(14.0m/s^2)^2+(-0.8m/^2)^2}=14.022\frac{m}{s^2}[/tex]
The total acceleration of the ball is 14.022 m/s²
A 2-kg block is released from rest at the top of a 20-mlong frictionless ramp that is 4 m high. At the same time, an identical block is released next to the ramp so that it drops straight down the same 4 m. What are the values for each of the following for the blocks just before they reach ground level.
Required:
a. Gravitational potential energy Block a_____ J Block b _____ J
b. Kinetic energy Block a _____ J Block b _____
c. Speed Block a _____ J Block b _____ J
d. Momentum Block a _____ J Block b _____ J
Answer:
A.) 78.4 J for both
B.) 78.4 J for both
C.) 8.85 m/s for both
D.) 17.7 kgm/s
Explanation:
Given information:
Mass m = 2 kg
Distance d = 20 m
High h = 4 m
A.) Gravitational potential energy can be calculated by using the formula
P.E = mgh
P.E = 2 × 9.8 × 4
P.E = 78.4 J
Since the two objects are identical, the gravitational potential energy of the block for both a and b will be 78.4 J
B.) According to conservative energy,
Maximum P.E = Maximum K.E.
Therefore, the kinetic energy of the two blocks will be 78.4 J
C.) Since K.E = 1/2mv^2 = mgh
V = √(2gh)
Solve for velocity V by substituting g and h into the formula
V = √(2 × 9.8 × 4)
V = √78.4
V = 8.85 m/s
The velocities of both block will be 8.85 m/s
D.) Momentum is the product of mass and velocity. That is,
Momentum = MV
Substitute for m and V into the formula
Momentum = 2 × 8.85 = 17.7 kgm/s
Both block will have the same value since the ramp Is frictionless.
A large crate of mass m is place on the flatbed of a truck but not tied down. As the truck accelerates forward with acceleration a, the crate remains at rest relative to the truck. What force causes the crate to accelerate?
Answer:
Friction
Explanation:
There are tiny bumps and grooves on every object, which make them rough and more difficult to rub against each other. Even though the crate remains at rest at first, the frictional force causes it to stay in place and accelerate with the truck. Hope this helps!
A butcher grinds 5 and 3/4 lb of meat then sells it for 2 and 2/3 pounds to the customer what is the maximum amount me that the butcher can sell to the next customer
Answer:
The maximum amount of meat that the butcher can sell is [tex]3\frac{1}{12}\:lb[/tex]
Explanation:
The maximum amount can be found by taking the difference of mixed numbers.
[tex]5\frac{3}{4}-2\frac{2}{3}\\\\\mathrm{Subtract\:the\:numbers:}\:5-2=3\\\\\mathrm{Combine\:fractions:\:}\frac{3}{4}-\frac{2}{3}=\frac{1}{12}\\\\=3\frac{1}{12}\\[/tex]
Best Regards!
. A very fine sample is placed 0.15 cm from the objective of a microscope. The focal length of the objective is 0.14 cm and of the eyepiece is 1.0 cm. The near-point distance of the person using the microscope is 25.0 cm. What is the final magnification of the microscope?
Answer:
Final magnification = -375
Explanation:
The total magnification of a compound microscope is expressed mathematically as the product of the magnifying power of each of the lenses that are combined in the compound microscope.
Final magnification = (magnifying power of the objective lens) × (magnifying power of the eyepiece lens) = m₁ × m₂
Magnifying power of a lens is defined as the ratio of the least distance of distinct vision to the focal length of the lens.
For the eyepiece, the least distance of distinct vision is the distance from the object to the near point = 25 + 1 + 0.15 = 26.15 cm
Focal length = 1 cm
Magnifying power of the eyepiece length = (26.15/1) = 26.15
For the objective lens,
The focal length = 0.14 cm
Least distance of distinct vision = -(1 + 1.00 + 0.15 - 0.14) = -2.01 cm (the negative sign means the image seen is upside down)
Magnifying power of the objective lens = (-2.01/0.14) = -14.357
Final magnification = -14.357 × 26.15 = -375
Hope this Helps!!!
A ray in glass (n = 1.51) reaches a boundary with air at 49.2 deg. Does it reflect internally or refract into the air? Enter 0 for reflect, and 1 for refract.
Answer:
0 - Then, the ray is totally reflected
Explanation:
The ray reaches the boundary between the two mediums at 49.2°.
If the ray is totally reflected it is necessary that the crictical angle is lower that the incidet angle.
You use the following to calculate the critical angle:
[tex]\theta_c=sin^{-1}(\frac{n_2}{n_1})[/tex] (1)
n2: index of refraction of the second medium (air) = 1.00
n1: index of refraction of the first medium (glass) = 1.51
You replace the values of the parameters in the equation (1):
[tex]\theta_c=sin^{-1}(\frac{1.00}{1.51})=41.47\°[/tex]
The critical angle is 41.47°, which is lower than the incident angle 49.2°.
Then, the ray is totally reflected.
0
At least how many Calories does a mountain climber need in order to climb from sea level to the top of a 5.42 km tall peak assuming the muscles of the climber can convert chemical energy to mechanical energy with an efficiency of 16.0 percent. The total mass of the climber and the equipment is 78.4 kg. (Enter your answer as a number without units.)
Answer:
Ec = 6220.56 kcal
Explanation:
In order to calculate the amount of Calories needed by the climber, you first have to calculate the work done by the climber against the gravitational force.
You use the following formula:
[tex]W_c=Mgh[/tex] (1)
Wc: work done by the climber
g: gravitational constant = 9.8 m/s^2
M: mass of the climber = 78.4 kg
h: height reached by the climber = 5.42km = 5420 m
You replace in the equation (1):
[tex]W_c=(78.4kg)(9.8m/s^2)(5420m)=4,164,294.4\ J[/tex] (2)
Next, you use the fact that only 16.0% of the chemical energy is convert to mechanical energy. The energy calculated in the equation (2) is equivalent to the mechanical energy of the climber. Then, you have the following relation for the Calories needed:
[tex]0.16(E_c)=4,164,294.4J[/tex]
Ec: Calories
You solve for Ec and convert the result to Cal:
[tex]E_c=\frac{4,164,294.4}{016}=26,026,840J*\frac{1kcal}{4184J}\\\\E_c=6220.56\ kcal[/tex]
The amount of Calories needed by the climber was 6220.56 kcal
Water flowing through a cylindrical pipe suddenly comes to a section of pipe where the diameter decreases to 86% of its previous value. If the speed of the water in the larger section of the pipe was what is its speed in this smaller section if the water behaves like an ideal incompressible fluid
Answer:
Explanation:
The speed of the water in the large section of the pipe is not stated
so i will assume 36m/s
(if its not the said speed, input the figure of your speed and you get it right)
Continuity equation is applicable for ideal, incompressible liquids
Q the flux of water that is Av with A the cross section area and v the velocity,
so,
[tex]A_1V_1=A_2V_2[/tex]
[tex]A_{1}=\frac{\pi}{4}d_{1}^{2} \\\\ A_{2}=\frac{\pi}{4}d_{2}^{2}[/tex]
the diameter decreases 86% so
[tex]d_2 = 0.86d_1[/tex]
[tex]v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}\\\\=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\\\\\approx1.35v_{1} \\\\v_{2}\approx(1.35)(38)\\\\\approx48.6\,\frac{m}{s}[/tex]
Thus, speed in smaller section is 48.6 m/s
A woman is standing in the ocean, and she notices that after a wave crest passes by, five more crests pass in a time of 50.2 s. The distance between two successive crests is 30.2 m. What is the wave's (a) period, (b) frequency, (c) wavelength, and (d) speed
Explanation:
(a) The period of a wave is the time required for one complete cycle. In this case, we have the time of five cycles. So:
[tex]T=\frac{t}{n}\\\\T=\frac{50.2s}{5}\\T=10.04s[/tex]
(b) The frequency of a wave is inversely proportional to its period:
[tex]f=\frac{1}{T}\\f=\frac{1}{10.04s}\\f=0.01Hz[/tex]
(c) The wavelength is the distance between two successive crests, so:
[tex]\lambda=30.2m[/tex]
(d) The speed of a wave is defined as:
[tex]v=f\lambda\\v=(0.1Hz)(30.2m)\\v=3.02\frac{m}{s}[/tex]
A hollow spherical iron shell floats almost completely submerged in water. The outer diameter is 60.0 cm, and the density of iron is 7.87 g∕c m cubed . Find the inner diameter in cm. Express to 3 sig figs.
Answer:
The inner diameter is 57.3 cm
Explanation:
The inner diameter of the hollow spherical iron shell can be found using the weight of the sphere ([tex]W_{s}[/tex]) and the weight of the water displaced ([tex]W_{w}[/tex]):
[tex] W_{s} = W_{w} [/tex]
[tex] m_{s}*g = m_{w}*g [/tex]
[tex] D_{s}*V_{s} = D_{w}*V_{w} [/tex]
Where D is the density and V is the volume
[tex] D_{s}*\frac{4}{3}\pi*(\frac{d_{o}^{3} - d_{i}^{3}}{2^{3}}) = \frac{4}{3}\pi*(\frac{d_{o}}{2})^{3} [/tex]
Where [tex]d_{o}[/tex] is the outer diameter and [tex]d_{i}[/tex] is the inner diameter
[tex] D_{s}*(d_{o}^{3} - d_{i}^{3}) = d_{o}^{3} [/tex]
[tex] D_{s}*d_{i}^{3} = d_{o}^{3}(D_{s} - 1) [/tex]
[tex] 7.87*d_{i}^{3} = 60.0^{3}(7.87 - 1) [/tex]
[tex] d_{i} = 57.3 cm [/tex]
Therefore, the inner diameter is 57.3 cm.
I hope it helps you!
After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands
Complete question is;
After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands? Assume the leg can be treated as a uniform rod x = 0.98 m long that pivots freely about the hip.
Answer:
Tangential speed of foot just before it lands is; v = 5.37m/s
Explanation:
Let U (potential energy) be zero on the ground.
So, initially, U = mgh
where, h = 0.98/2 = 0.49m (midpoint of the leg)
Now just before the leg hits the floor it would have kinetic energy as;
K = ½Iω²
where ω = v/r and I = ⅓mr²
So, K = ½(⅓mr²)(v/r)²
K = (1/6) × (mr²)/(v²/r²)
K = (1/6) × mv²
From principle of conservation of energy, we have;
Potential energy = Kinetic energy
Thus;
mgh = (1/6) × mv²
m will cancel out to give;
gh = (1/6)v²
Making v the subject, we have;
v = √6gh
v = √(6 × 9.81 × 0.49)
v = √28.8414
v = 5.37m/s
Consider a uniform horizontal wooden board that acts as a pedestrian bridge. The bridge has a mass of 300 kg and a length of 10 m. The bridge is supported by two vertical stone pillars, one 2.0 m from the left end of the bridge and the other 2.0 m from the right end of the bridge. If a 200 kg knight stands on the bridge 4.0 m from the left end, what force is applied by the left support
Answer:
F = 2123.33N
Explanation:
In order to calculate the torque applied by the left support, you take into account that the system is at equilibrium. Then, the resultant of the implied torques are zero.
[tex]\Sigma \tau=0[/tex]
Next, you calculate the resultant of the torques around the right support, by taking into account that the torques are generated by the center of mass of the wooden, the person and the left support. Furthermore, you take into account that torques in a clockwise direction are negative and in counterclockwise are positive.
Then, you obtain the following formula:
[tex]-\tau_l+\tau_p+\tau_{cm}=0[/tex] (1)
τl: torque produced by the left support
τp: torque produced by the person
τcm: torque produced by the center of mass of the wooden
The torque is given by:
[tex]\tau=Fd[/tex] (2)
F: force applied
d: distance to the pivot of the torque, in this case, distance to the right support.
You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:
[tex]-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N[/tex]
Where d1, d2 and d3 are distance to the right support.
You solve the equation for F and replace the values of the other parameters:
[tex]F=\frac{W_pd_2+W_d_3}{d_1}=\frac{(1960N)(2.0m)+(2940N)(3.0m)}{6.0m}\\\\F=2123.33N[/tex]
The force applied by the left support is 2123.33 N
a cannon is fired with an initial horizontal velocity of 20m/s and an initial velocity of 25m/s. After 3s in the air, the cannon hits its target. How far away(in meters) was the cannon from its target
Answer:
60 m
Explanation:
After 3 seconds of travel at 20 m/s, the projectile is 3·20 = 60 meters horizontally from the cannon.
__
The vertical height after 3 seconds is 0.9 m, so the straight-line distance from cannon to target is √(60^2 +0.9^2) ≈ 60.007 meters.
If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance. What is its vertical displacement (in m) after 10 s? (g = 10 m/s2)
Answer:
500mExplanation:
Using the equation of motion S = ut + 1/2 gt²
S = the vertical displacement (in m)
u = initial velocity of the object (in m/s)
g = acceleration due to gravity (in m/s²)
t = time taken (in secs)
Given u = 0m/s, g = 10m/s² and t = 10s, substituting this value into the equation to get the vertical displacement w have;
S = 0+1/2 (10)(10)²
S = 1000/2
S = 500m
The vertical displacement after 10seconds is 500m
A projectile is launched in the horizontal direction. It travels 2.050 m horizontally while it falls 0.450 m vertically, and it then strikes the floor. How long is the projectile in the air
Answer:
0.303s
Explanation:
horizontal distance travel = 2.050 m, vertical distance travel = 0.45 m
Using equation of linear motion
Sy = Uy t + 1/2 gt² Uy is the inital vertical component of the velocity, t is the time taken for the vertical motion in seconds, and S is the vertical distance traveled, taken downward vertical motion as negative
-0.45 = 0 - 0.5 × 9.81×t²
0.45 / (0.5 × 9.81) = t²
t = √0.0917 = 0.303 s
A cart of mass 350 g is placed on a frictionless horizontal air track. A spring having a spring constant of 7.5 N/m is attached between the cart and the left end of the track. The cart is displaced 3.8 cm from its equilibrium position. (a) Find the period at which it oscillates. s (b) Find its maximum speed. m/s (c) Find its speed when it is located 2.0 cm from its equilibrium position.
Answer:
(a) T = 1.35 s
(b) vmax = 0.17 m/s
(c) v = 0.056 m/s
Explanation:
(a) In order to calculate the period of oscillation you use the following formula for the period in a simple harmonic motion:
[tex]T=2\pi\sqrt{\frac{m}{k}}[/tex] (1)
m: mass of the cart = 350 g = 0.350kg
k: spring constant = 7.5 N/m
[tex]T=2\pi \sqrt{\frac{0.350kg}{7.5N/m}}=1.35s[/tex]
The period of oscillation of the car is 1.35s
(b) The maximum speed of the car is given by the following formula:
[tex]v_{max}=\omega A[/tex] (2)
w: angular frequency
A: amplitude of the motion = 3.8 cm = 0.038m
You calculate the angular frequency:
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{1.35s}=4.65\frac{rad}{s}[/tex]
Then, you use the result of w in the equation (2):
[tex]v_{max}=(4.65rad/s)(0.038m)=0.17\frac{m}{s}[/tex]
The maximum speed if 0.17m/s
(c) To find the speed when the car is at x=2.0cm you first calculate the time t by using the following formula:
[tex]x=Acos(\omega t)\\\\t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\t=\frac{1}{4.65rad/s}cos^{-1}(\frac{0.02}{0.038})=0.069s[/tex]
The speed is the value of the following function for t = 0.069s
[tex]|v|=|\omega A sin(\omega t)|\\\\|v|=(4.65rad/s)(0.038m)sin(4.65rad/s (0.069s))=0.056\frac{m}{s}[/tex]
The speed of the car is 0.056m/s
A particle covers equal distance in equal intervals of time. It is said to be?
1.at rest
2.moving with constant acceleration
3.moving with constant velocity
4.moving with constant speed
4.Moving with constant speed
The planet should move around the elliptical orbit, and two segments of the orbit should become shaded in green. What aspect(s) of the orbit and shaded segments are the same?
Answer: not sure
Explanation:
The temperature coefficient of resistivity for the metal gold is 0.0034 (C )1, and for tungsten it is 0.0045 (C )1. The resistance of a gold wire increases by 7.0% due to an increase in temperature. For the same increase in temperature, what is the percentage increase in the resistance of a tungsten wire
Answer:
% increase in resistance of tungsten = 9.27%
Explanation:
We are given:
Co-efficient of resistivity for the metal gold; α_g = 0.0034 /°C
Co-efficient of resistivity for tungsten;α_t = 0.0045 /°C
% Resistance change of gold wire due to temperature change = 7%
Now, let R1 and R2 be the resistance before and after the temperature change respectively.
Thus;
(R2 - R1)/R1) x 100 = 7
So,
(R2 - R1) = 0.07R1
R2 = R1 + 0.07R1
R2 = 1.07R1
The equation to get the change in temperature is given as;
R2 = R1(1 + αΔt)
So, for gold,
1.07R1 = R1(1 + 0.0034*Δt)
R1 will cancel out to give;
1.07 = 1 + 0.0034Δt
(1.07 - 1)/0.0034 = Δt
Δt = 20.59°C
For this same temperature for tungsten, let Rt1 and Rt2 be the resistance before and after the temperature change respectively and we have;
Rt2 = Rt1(1 + α_t*Δt)
So, Rt2/Rt1 = 1 + 0.0045*20.59
Rt2/Rt1 = 1.0927
From earlier, we saw that;
(R2 - R1)/R1) x 100 = change in resistance
Similarly,
(Rt2 - Rt1)/Rt1) x 100 = change in resistance
Simplifying it, we have;
[(Rt2/Rt1) - 1] × 100 = %change in resistance
Plugging in the value of 1.0927 for Rt2/Rt1, we have;
(1.0927 - 1) × 100 = %change in resistance
%change in resistance = 9.27%
4. A rock is thrown from the edge of the top of a 100 m tall building at some unknown angle above the horizontal. The rock strikes the ground at a horizontal distance of 160 m from the base of the building 5.0 s after being thrown. Determine the speed with which the rock was thrown.
Answer:
Explanation:
Let the velocity of projectile be v and angle of throw be θ.
The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m
considering its vertical displacement
h = - ut +1/2 g t²
100 = - vsinθ x 5 + .5 x 9.8 x 5²
5vsinθ = 222.5
vsinθ = 44.5
It covers 160 horizontally in 5 s
vcosθ x 5 = 160
v cosθ = 32
squaring and adding
v²sin²θ +v² cos²θ = 44.4² + 32²
v² = 1971.36 + 1024
v = 54.73 m /s
Answer:
55.42 m/s
Explanation:
Along the horizontal direction, the rock travels at constant speed: this means that its horizontal velocity is constant, and it is given by
u_x = d/t
Where
d = 160 m is the distance covered
t = 5.0 s is the time taken
Substituting, we get
u_x =160/5 = 32 m/s.
Along the vertical direction, the rock is in free-fall - so its motion is a uniform accelerated motion with constant acceleration g = -9.8 m/s^2 (downward). Therefore, the vertical distance covered is given by the
[tex]S=u_yt+\frac{1}{2}at^2[/tex]
where
S = -100 m is the vertical displacement
u_y is the initial vertical velocity
Replacing t = 5.0 s and solving the equation for u_y, we find
-100 = u_y(5) + (-9.81)(5)^2/2
u_y = 45.25 m/s
Therefore, the speed with which the rock was thrown u
[tex]u= \sqrt{u_x^2+u_y^2} \\=\sqrt{32^2+45.25^2}\\ = 55.42 m/s[/tex]
For the instant represented, car A has an acceleration in the direction of its motion, and car B has a speed of 45 mi/hr which is increasing. If the acceleration of B as observed from A is zero for this instant, determine the magnitude of the acceleration of A and the rate at which the speed of B is changing.
Answer:
[tex]\mathbf{a_A = 10.267 \ ft/s^2}[/tex]
[tex]\mathbf{V_B = (a_t)_B =-7.26 \ ft/s^2}[/tex]
Explanation:
Firstly, there is supposed to be a diagram attached in order to complete this question;
I have attached the diagram below in order to solve this question.
From the data given;
The radius of the car R = 600 ft
Velocity of the car B, [tex]V_B = 45 mi / hr[/tex]
We are to determine the magnitude of the acceleration of A and the rate at which the speed of B is changing.
To start with the magnitude of acceleration A;
We all know that
1 mile = 5280 ft and an hour = 3600 seconds
Thus for ; 1 mile/hr ; we have :
5280 ft/ 3600 seconds
= 22/15 ft/sec
However;
for the velocity of the car B = 45 mi/hr; to ft/sec, we have:
= (45 × 22/15) ft/sec
= 66 ft/sec
A free body diagram is attached in the second diagram showing how we resolve the vector form
Now; to determine the magnitude of the acceleration of A; we have:
[tex]^ \to {a_A} = a_A sin 45^0 ^{\to} + a_A cos 45^0 \ j ^{\to} \\ \\ ^\to {a_B} = -(a_t)_B \ i ^ \to + (a_c )_B cos 45 ^0 \ j ^{\to}[/tex]
Where;
[tex](a_c)_B[/tex] = radial acceleration of B
[tex](a_t)_B[/tex] = tangential acceleration of B
From observation in the diagram; The acceleration of B is 0 from A
So;
[tex]a_B ^\to - a_A ^\to = a_{B/A} ^ \to[/tex]
[tex](-(a_t)_B - a_A sin 45^0 ) ^\to i+ ((a_t)_B-a_A \ cos \ 45^0) ^ \to j = 0[/tex]
[tex](a_c)_B = \dfrac{V_B^2}{R}[/tex]
[tex](a_c)_B = \dfrac{(66)^2}{600}[/tex]
[tex](a_c)_B = 7.26 ft/s^2[/tex]
Equating the coefficient of i and j now; we have :
[tex](a_t)_B = -a_A \ sin 45^0 --- (1)\\ \\ (a_c)_B = a_A cos \ 45^0 --- (2)\\ \\[/tex]
From equation (2)
replace [tex](a_c)_B[/tex] with 7.26 ft/s^2; we have
[tex]7.26 \ ft/s^2 = a_A cos \ 45^0 \\ \\ a_A = \dfrac{7.26 \ ft/s^2}{co s \ 45^0}[/tex]
[tex]\mathbf{a_A = 10.267 \ ft/s^2}[/tex]
Similarly;
From equation (1)
[tex](a_t)_B = -a_A \ sin 45^0[/tex]
replace [tex]a_A[/tex] with 10.267 ft/s^2
[tex](a_t)_B = -10.267 \ ft/s^2 * \ sin 45^0[/tex]
[tex]\mathbf{V_B = (a_t)_B =-7.26 \ ft/s^2}[/tex]
Imagine you are in a small boat on a small pond that has no inflow or outflow. If you take an anchor that was sitting on the floor of the boat and lower it over the side until it sits on the ground at the bottom of the pond, will the water level rise slightly, stay the same, or lower slightly?Two students, Ian and Owen, are discussing this. Ian says that the anchor will still displace just as much water when it is sitting on the bottom of the pond as it does when it is in the boat. After all, adding the anchor to the boat causes the water level in the lake to rise, and so would immersing the anchor in the pond. So Ian reasons that both displacements would be equal, and the lake level remains unchanged.
Answer;
The pond's water level will fall.
Explanation;
Archimedes principle explains that a floating body will displace the amount of water that weighs the same as it, whereas a body resting on the bottom of the water displaces the amount of water that is equal to the body's volume.
When the anchor is in the boat it is in the category of floating body and when it is on the bottom of the pond it is in the second category.
Since anchors are naturally heavy and denser than water, the amount of water displaced when the anchor is in the boat is greater than the amount of water displaced when the anchor is on the bottom of the pond since the way anchors are doesn't make for them to have considerable volume.
When the anchor is dropped to the bottom of the pond, the water level will therefore fall. If the anchor doesn't reach the bottom it is still in the floating object category and there will be no difference to the water level, but once it touches the bottom of the pond, the water level of the pond drops.
Hope this Helps!!!
Buoyancy is an upward force exerted by a fluid on a body partially or completely immersed in it
The pond water level will lower slightly
According to Archimedes principle, the up thrust on the boat by the water is given by the volume of the water displaced
When a boat floats, the weight of the boat and all its contents and passengers is equal to the displaced water, so that larger boats with more wider opening can displace more water and therefore, carry more loadWith regards to lowering the anchor from the boat into the pond, the weight of the anchor is no longer carried by the boat but by the bottom of the pond, therefore, the weight of the boat reduces, and the boat rises, while the volume initially occupied by the boat is taken up by the water available, therefore, the water level lowers slightly
Learn more here;
https://brainly.com/question/24529607
What is the main difference between work power and energy
Answer:
Work is the energy required to move an object from one point to another. while power is the energy transferred per unit time.
Energy can also be defined as the ability to do work.
what happen to the volume of liquid displaced
when the density of liquid is changed
explain ?
Answer:
Density depends on the temperature and the gap between particles of the liquid. In most of cases temperature is inversely proportional to density means if the temperature increases then the density decreases and the space between particles of that liquid is also inversely proportional to the density means if the intraparticle space increases then the density decreases.
A cheetah goes from 0m/s to 25m/s in 2.5 s. What is the cheetah's rate of acceleration?
Answer:
10 m/s²
Explanation:
Acceleration: This the rate of change of velocity. The unit of acceleration is m/s²
From the question,
a = (v-u)/t.................... Equation 1
Where a = acceleration of the cheetah, v = final velocity of the cheetah, u = initial velocity of the cheetah, t = time.
Given: u = 0 m/s, v = 25 m/s, t = 2.5 s.
Substitute these values into equation 1
a = (25-0)/2.5
a = 25/2.5
a = 10 m/s²
Hence the acceleration of the cheetah = 10 m/s²