Answer:
magnitude of the average acceleration is 4.16 m/s²
Explanation:
given data
initial velocity u = 5 m/s
distance s = 3 m
final velocity v = 0
to find out
magnitude of the average acceleration
solution
we will apply here linear motion equation that is
v²-u² = 2×a×s ..............1
put here all these value
v is final velocity and u is initial velocity and s is distance
0²-5² = 2×a×3
a = -4.16
so magnitude of the average acceleration is 4.16 m/s²
Answer: The magnitude of the average acceleration is 4.16m/s^2
An astronaut weights 588 N in Earth surface. How much the astronaut would weight in the surface in the moon in pounds(lbs)?
Answer:
22.02 lb
Explanation:
The weight of astronaut Earth (w) = 588 N
We know that,
[tex]4.45N = 1lb[/tex]
Thus,
[tex]588N = \frac{1}{4.45}\times 588lb[/tex]
588N = 132.13 lb
Acceleration due to gravity on Earth (g) = 9.8 m/s²
Acceleration due to gravity on Moon = g'
[tex]g'=\frac{g}{6}[/tex]
We know that weight of an object on Earth is,
[tex]w = m\times g[/tex]
[tex]m = \frac{w}{g}[/tex]
Similarly, weight on Moon will be
[tex]w' = m\times g'[/tex]
[tex]w' = \frac{w}{g}\times\frac{g}{6}[/tex]
[tex]w' = \frac{132.13}{6}[/tex]
[tex]w' = 22.02[/tex]
Thus the astronaut will weigh 22.02 lb on Moon.
Two moles of He gas are at 25 degrees C and a pressure of 210 kPa. If the gas is heated to 48 degrees C and its pressure reduced to 37 % of its initial value, what is the new volume?
Answer:[tex]0.0686 m^3[/tex]
Explanation:
Given
No of moles=2
[tex]T=25^{\circ}\approx 298 K[/tex]
[tex]P_1=210 KPa[/tex]
[tex]T_2=48 ^{\circ}\approx 321 k[/tex]
[tex]P_2=77.7 KPa[/tex]
PV=nRT
Substitute
[tex]77.7\times V_2=2\times 8.314\times 321[/tex]
[tex]V_2=0.0686 m^3[/tex]
Using local acceleration of gravity data from the Internet, de- termine the weight, in N, of a person whose mass is 80 kg living in: a. Mexico City, Mexico b. Cape Town, South Africa c. Tokyo, Japan d. Chicago, IL e. Copenhagen, Denmark
Answer:
Mexico Citi: 782.32 NCape Town: 783.68 NTokyo: 783.84 NChicago: 784.24 NCopenhagen: 785.2 NExplanation:
The weight w its:
[tex]w \ = \ m \ a[/tex]
where m is the mass and a is the acceleration.
Mexico City, MexicoLocal acceleration of gravity = [tex]9.779 \frac{m}{s^2}[/tex]
weight :
[tex]w = \ 80\ kg \ * \ 9.779 \frac{m}{s^2}[/tex]
[tex]w = \ 782.32 \ N [/tex]
Cape Town, South AfricaLocal acceleration of gravity = [tex]9.796 \frac{m}{s^2}[/tex]
weight :
[tex]w = \ 80\ kg \ * \ 9.796 \frac{m}{s^2}[/tex]
[tex]w = \ 783.68 \ N [/tex]
Tokyo, JapanLocal acceleration of gravity = [tex]9.798 \frac{m}{s^2}[/tex]
weight :
[tex]w = \ 80\ kg \ * \ 9.798 \frac{m}{s^2}[/tex]
[tex]w = \ 783.84 \ N [/tex]
Chicago, ILLocal acceleration of gravity = [tex]9.803 \frac{m}{s^2}[/tex]
weight :
[tex]w = \ 80\ kg \ * \ 9.803 \frac{m}{s^2}[/tex]
[tex]w = \ 784.24 \ N [/tex]
Copenhagen, DenmarkLocal acceleration of gravity = [tex]9.815 \frac{m}{s^2}[/tex]
weight :
[tex]w = \ 80\ kg \ * \ 9.815 \frac{m}{s^2}[/tex]
[tex]w = \ 785.2 \ N [/tex]
Answer
mass of a person is 80 kg
to calculate weight of the person = ?
a) Mexico City, Mexico
acceleration due to gravity in mexico = 9.779 m/s²
weight = 80× 9.779 = 782.32 N
b) Cape Town, South Africa
acceleration due to gravity in cape town = 9.796 m/s²
weight = 80× 9.796 = 783.68 N
c) Tokyo, Japan
acceleration due to gravity in Tokyo = 9.798 m/s²
weight = 80× 9.798 = 783.84 N
d) Chicago, IL
acceleration due to gravity in Chicago = 9.80 m/s²
weight = 80× 9.80 = 784 N
e) Copenhagen, Denmark
acceleration due to gravity in Copenhagen = 9.81 m/s²
weight = 80× 9.81 = 784.8 N
A truck traveling at a constant speed of 24 m/s passes a more slowly moving car. The instant the truck passes the car, the car begins to accelerate at a constant rate of 1.4 m/s^2 and passes the truck 578 m farther down the road. What was the speed of the car (in meters/second) when it was first passed by the truck?
Answer:
7.15 m/s
Explanation:
We use a frame of reference in which the origin is at the point where the trucck passed the car and that moment is t=0. The X axis of the frame of reference is in the direction the vehicles move.
The truck moves at constant speed, we can use the equation for position under constant speed:
Xt = X0 + v*t
The car is accelerating with constant acceleration, we can use this equation
Xc = X0 + V0*t + 1/2*a*t^2
We know that both vehicles will meet again at x = 578
Replacing this in the equation of the truck:
578 = 24 * t
We get the time when the car passes the truck
t = 578 / 24 = 24.08 s
Before replacing the values on the car equation, we rearrange it:
Xc = X0 + V0*t + 1/2*a*t^2
V0*t = Xc - 1/2*a*t^2
V0 = (Xc - 1/2*a*t^2)/t
Now we replace
V0 = (578 - 1/2*1.4*24.08^2) / 24.08 = 7.15 m/s
A rocket fired two engines simultaneously. One produces a thrust 725 N directly forward while the other gives a 513N thrust at 32.4degrees above the forward direction. Find the magnitude and direction ( relative to the forward direction) of the resultant force that these engines exert on the rocket
Answer:
Magnitude of resultant force is 1190.314 N
Direction of force is 13.352°
Explanation:
given data:
thrust force = 725 N
Angle = 32.4 degree
Let x is consider as positive direction
Resultant force in x direction is
Rx = 725 + 513cos32.4 = 1158.14 N
and Resultant force perpendicular to x direction is:
Ry = 513sin32.4 = 274.88 N
Magnitude of resultant force is
[tex]R=\sqrt{R_x^2+R_y^2} = 1190.314N[/tex]
and resultant force direction is
[tex]\theta=tan^{-1}\frac{R_y}{R_x} = 13.352\degree[/tex]
Capacitances of 10uF and 20uF are connected in parallel,
andthis pair is then connected in series with a 30-uF capacitor.
What is the equivalent capacitance of this arrangement?
Answer:
The equivalent capacitance will be [tex]15\mu F[/tex]
Explanation:
We have given two capacitance [tex]C_=10\mu F\ and\ C_2=20\mu F[/tex]
They are connected in parallel
So equivalent capacitance [tex]C=C_1+C_2=10+20=30\mu F[/tex]
This equivalent capacitance is now connected in series with [tex]30\mu F[/tex]
In series combination of capacitors the equivalent capacitance is given by [tex]\frac{1}{C}=\frac{1}{30}+\frac{1}{30}[/tex]
[tex]C=\frac{30}{2}=15\mu F[/tex]
So the equivalent capacitance will be [tex]15\mu F[/tex]
The equivalent capacitance of the arrangement where capacitors of 10 uF and 20 uF are connected in parallel, and in series with a 30 uF capacitor, is 15 uF.
To determine the equivalent capacitance of a combination in which capacitors are connected both in parallel and in series, one must consider the rules for each type of connection. For capacitors connected in parallel, the equivalent capacitance is simply the sum of their individual capacitances. When capacitors are connected in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances. In your case, the 10 uF and 20 uF capacitors are connected in parallel, so their equivalent capacitance is 10 uF + 20 uF = 30 uF. This combined capacitance is then in series with the 30 uF capacitor.
The next step is to calculate the equivalent capacitance for the series connection using the formula:
1/Cseries = 1/C1 + 1/C2
Where C1 is the equivalent parallel capacitance (30 uF) and C2 is the capacitance of the third capacitor (30 uF). Substituting the values:
1/Cseries = 1/30 uF + 1/30 uF = 1/15 uF
So, Cseries = 15 uF. This is the total equivalent capacitance of the combination of capacitors.
Starting from rest, a person runs with a constant acceleration, traveling 40 meters in 10 seconds. What is their final velocity?
To calculate the final velocity of someone running with constant acceleration, we first find the acceleration using the distance and time, and then apply it to determine the final velocity. The final velocity after 10 seconds is 8 m/s.
To find the final velocity of the person running with constant acceleration, we use the kinematic equation v = u + at, where v is final velocity, u is initial velocity, a is acceleration, and t is time. The person starts from rest, so u = 0. We need to find a first, and we can do so using the equation s = ut + (1/2)at², where s is the distance traveled.
Insert the given values: 40m = 0 + (1/2)a(10s)², resulting in a = 0.8 m/s². Now apply the acceleration to the initial equation: v = 0 + 0.8 m/s² * 10s, which gives us v = 8 m/s. Therefore, the person's final velocity after 10 seconds is 8 m/s.
A group of humans traveling in space discover a habitable planet. They settle down there and start populating it. Their population grows at a steady rate and doubles every 10 years. Thousand years after the first group of humans arrived, the planet is filled to half its carrying capacity. How many more years are there to reach the carrying capacity of the planet? Assume there are no resource crisis and they grow at the same rate. A) 500 years
B) 1000 years
C) 5 years
D) 1 year
E) 10 years
Answer:
10 years
Explanation:
As you can understand from the question it is given that the planet is already filled to half of its capacity. Also the population doubles in 10 years. To fill up the planet completely the population needs to double only once. To do that only 10 years are required.
As it is mentioned there are no other factors affecting the growth rate, in 10years the planet will be filled to its carrying capacity.
A red ball is thrown down with an initial speed of 1.1 m/s from a height of 28 meters above the ground. Then, 0.5 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 24.3 m/s, from a height of 0.9 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s^2. What is the height of the blue ball 2 seconds after the red ball is thrown?
How long after the red ball is thrown are the two balls in the air at the same height?
Answer:0.835 s
Explanation:
Given
Red ball initial velocity([tex]u_r[/tex])=1.1 m/s
height of building(h)=28 m
after 0.5 sec blue ball is thrown with a velocity([tex]u_b[/tex])=24.3 m/s
Height of blue after 2 sec red ball is thrown
i.e. height of blue ball at t=1.5 sec after blue ball is thrown upward
[tex]h=24.3\times 1.5-\frac{9.81\times 1.5^2}{2}=25.414 m[/tex]
therefore blue ball is at height of 25.414+0.9=26.314 from ground
moment after the two ball is at same height
for red ball
[tex]14=1.1\times \left ( t+0.5\right )+\frac{9.81\times \left ( t+0.5\right )^2}{2}-----1[/tex]
for blue ball
[tex]13.1=24.3\times t-\frac{9.81\times t^2}{2}-----2[/tex]
add 1 & 2
we get
[tex]27.1=1.1t+0.55+24.3t+\frac{g\left ( t+0.25\right )}{2}[/tex]
27.1=25.4t+0.55+4.905t+1.226
t=0.835 s
Two racing boats set out from the same dock and speed away at the same constant speed of 104 km/h for half an hour (0.500 h), the blue boat headed 24.0° south of west, and the green boat headed 37.7° south of west. During this half-hour (a) how much farther west does the blue boat travel, compared to the green boat, and (b) how much farther south does the green boat travel, compared to the blue boat? Express your answers in km.
Answer:
The blue boat traveled 6.1 km farther west than the green boat
The green boat traveled 10.7 km farther south than the blue boat
Explanation:
The equation for linear uniform speed movement is
X(t) = X0 + v * t
Since we have two coordinates (X, Y) we use
X(t) = X0 + vx * t
Y(t) = Y0 + vy * t
The dock will be the origin of coordinates so X0 and Y0 will be zero. The X axis will be pointing west and the Y axis south.
The blue boat moves with a direction 24° south of west, so it will have speeds:
vxb = 104 * cos(24) = 95 km/h
vyb = 104 * sin(24) = 42.3 km/h
And the green boat:
vxg = 104 * cos(37.7) = 82.3 km/h
vyg = 104 * sin(37.7) = 63.6 km/h
After half hour the boats will have arrived at positions
Xb = 95 * 0.5 = 47.5 km
Yb = 42.3 * 0.5 = 21.1 km
And
Xg = 82.3 * 0.5 = 41.4 km
Yg = 63.6 * 0.5 = 31.8 km
The difference in positions of the boats
47.5 - 41.4 = 6.1 km
31.8 - 21.1 = 10.7 km
Surface waves on water with wavelengths large compared to depth are described by the equatio
d^2h/dx^2 + d^2h/dy^2 = 1/gd . d^2h/dt^2
where g is the acceleration of gravity, d is the equilibrium depth of the water, and h(x,y,t) is the height of the wave above the surface is equilibrium position. What is the speed of traveling waves described by this equation?
Answer with Explanation:
The general wave equation is given by
[tex]\frac{\partial ^2u}{\partial x^2}+\frac{\partial ^2u}{\partial y^2}=\frac{1}{c^2}\frac{\partial ^2u}{\partial t^2}[/tex]
where
'c' is the velocity of the wave
Comparing with the given equation
[tex]\frac{\partial ^2h}{\partial x^2}+\frac{\partial ^2h}{\partial y^2}=\frac{1}{gd}\frac{\partial ^2h}{\partial t^2}[/tex]
We can see that
[tex]c^2=gd\\\\\therefore c=\sqrt{gd}[/tex]
Thus the velocity of wave is given by [tex]v=\sqrt{gd}[/tex]
A stone is thrown vertically upward with a speed of 15.0 m/s from the edge of a cliff 75.0 m high.How much later does it reach the bottom of the cliff?What is its speed just before hitting?What total distance did it travel?
Therefore, the stone takes 3.06 seconds to reach the bottom, has a speed of 29.8 m/s downwards just before hitting, and travels a total distance of 150.0 meters.
Here's how to solve the problems about the falling stone:
1. Time to reach the bottom:
First half of the journey (upward):
Use the equation v = u - gt, where v is the final velocity (0 m/s at the top), u is the initial velocity (15.0 m/s), g is the acceleration due to gravity (9.81 m/s^2), and t is the time.
Solving for t, we get t = u/g = 15.0 m/s / 9.81 m/s^2 = 1.53 seconds.
Second half of the journey (downward):
The stone falls freely again with an initial velocity of 0 m/s. The time taken will be the same as the upward journey, 1.53 seconds.
Total time to reach the bottom:
Add the times for both halves: 1.53 seconds + 1.53 seconds = 3.06 seconds.
2. Speed just before hitting:
Use the same equation v = u - gt, but this time u is 0 m/s (at the top) and t is the total time (3.06 seconds).
Solving for v, we get v = 0 m/s - 9.81 m/s^2 * 3.06 seconds = -29.8 m/s (negative sign indicates downward direction).
Therefore, the stone's speed just before hitting is 29.8 m/s downwards.
3. Total distance traveled:
The stone travels twice the height of the cliff: once going up and once going down.
Total distance = 2 * cliff height = 2 * 75.0 m = 150.0 m.
Therefore, the stone takes 3.06 seconds to reach the bottom, has a speed of 29.8 m/s downwards just before hitting, and travels a total distance of 150.0 meters.
A ball is projected vertically downward at a speed of 4.00 m/s. How far does the bal travel in 1.80 s? What is the velocity of the ball at that time?
Answer:
1) Distance traveled equals 23.1 meters.
2) Final velocity equals 21.658 m/s.
Explanation:
The problem can be solved using second equation of kinematics as
[tex]s=ut+\frac{1}{2}at^{2}[/tex]
where
s is the distance covered
u is the initial speed of the ball
a is the acceleration the ball is under
t is time of travel
Applying the given values in the above equation we get
[tex]s=4.0\times 1.8+\frac{1}{2}\times 9.81\times 1.8^{2}\\\\s=23.1meters[/tex]
Part 2)
The velocity after 't' time can be obtained using first equation of kinematics.
[tex]v=u+at[/tex]
Applying the given values we get
[tex]v=4+9.81\times 1.80\\\\\therefore v=21.658m/s[/tex]
Final answer:
The ball travels 23.07 meters in 1.80 seconds, and the velocity of the ball at that time is 21.66 m/s downward.
Explanation:
The student wants to know how far a ball projected vertically downward at a speed of 4.00 m/s will travel in 1.80 s and the ball's velocity at that time. To solve these questions, we use the kinematic equations that govern linear motion with constant acceleration due to gravity, ignoring air resistance.
The distance traveled by the ball can be found using the equation:
s = ut + [tex]rac{1}{2}at^2[/tex]
Where:
s is the distance traveled,
u is the initial velocity (4.00 m/s),
t is the time (1.80 s),
a is the acceleration due to gravity (9.81 m/s^2).
Plugging in the values, we have:
s = 4.00 m/s imes 1.80 s + 0.5 imes 9.81 [tex]m/s^2[/tex] imes [tex](1.80 s)^2[/tex]
s = 7.20 m + 15.87 m
s = 23.07 m
The ball travels 23.07 meters in 1.80 seconds.
The velocity of the ball after 1.80 seconds can be calculated using the equation:
v = u + at
Where:
v is the final velocity,
u is the initial velocity,
a is the acceleration,
t is the time.
So, v = 4.00 m/s + 9.81 [tex]m/s^2[/tex] imes 1.80 s
v = 4.00 m/s + 17.66 m/s
v = 21.66 m/s
The velocity of the ball after 1.80 seconds is 21.66 m/s downward.
A taxi traveling along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 27.0 m/s. Then the vehicle travels for 41.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s. (a) How long is the taxi in motion (in s)? s (b) What is the average velocity of the taxi for the motion described? (Enter the magnitude in m/s.) m/s
Answer:
Part a)
[tex]T = 59.5 s[/tex]
Part b)
v = 22.8 m/s
Explanation:
Part a)
Time taken by the taxi to reach the maximum speed is given as
[tex]v_f - v_i = at[/tex]
[tex]27 - 0 = 2(t)[/tex]
[tex]t = 13.5 s[/tex]
now it moves with constant speed for next 41 s and then finally comes to rest in next 5 s
so total time for which it will move is given as
[tex]T = 13.5 s + 41 s + 5 s[/tex]
[tex]T = 59.5 s[/tex]
Part b)
Distance covered by the taxi while it accelerate to its maximum speed is given as
[tex]d_1 = \frac{v_f + v_i}{2} t[/tex]
[tex]d_1 = \frac{27 m/s + 0}{2} (13.5)[/tex]
[tex]d_1 = 182.25[/tex]
Now it moves for constant speed for next 41 s so distance moved is given as
[tex]d_2 = (27)(41)[/tex]
[tex]d_2 = 1107 m[/tex]
Finally it comes to rest in next 5 s so distance moved in next 5 s
[tex]d_3 = \frac{v_f + v_i}{2} t[/tex]
[tex]d_3 = \frac{27 + 0}{2}(5)[/tex]
[tex]d_3 = 67.5 m[/tex]
so here we have total distance moved by it is given as
[tex]d = 182.25 + 1107 + 67.5 [/tex]
[tex]d = 1356.75 m[/tex]
average speed is given as
[tex]v = \frac{1356.75}{59.5}[/tex]
[tex]v = 22.8 m/s[/tex]
Final answer:
The taxi is in motion for 59.5 seconds and its average velocity is 19.38 m/s.
Explanation:
(a) To find the total time the taxi is in motion, we need to calculate the time it takes to accelerate to 27.0 m/s, the time it travels at constant speed, and the time it takes to stop. The time to accelerate is given by the equation:
t = (vf - vi) / a = (27.0 m/s - 0 m/s) / 2.00 m/s² = 13.5 s
The time traveling at constant speed is given as 41.0 s, and the time to stop is given as 5.00 s. Therefore, the total time the taxi is in motion is:
Total time = Time to accelerate + Time at constant speed + Time to stop = 13.5 s + 41.0 s + 5.00 s = 59.5 s
(b) The average velocity is given by the total displacement divided by the total time. The displacement during acceleration is given by:
Displacement = vi × t + 0.5 × a × t² = 0 m/s × 13.5 s + 0.5 × 2.00 m/s² × (13.5 s)² = 182.25 m
The displacement during constant speed is equal to the constant speed multiplied by the time:
Displacement = constant speed × time = 27.0 m/s × 41.0 s = 1,107 m
The displacement during deceleration is given by:
Displacement = vf × t + 0.5 × (-a) × t² = 27.0 m/s × 5.00 s + 0.5 × (-2.00 m/s²) × (5.00 s)² = -137.5 m
Therefore, the total displacement is 182.25 m + 1,107 m - 137.5 m = 1,152.75 m.
The average velocity is then:
Average velocity = Total displacement / Total time = 1,152.75 m / 59.5 s = 19.38 m/s
Represent 0.00563 mg as a number between 0.1 and 1000 using an appropriate prefix Express your answer to three significant figures and include the appropriate units. 2 ? "! HAR 0.00563 mg = 5.63 . 10-3 O mg
Answer:
The answer is: 0.00563 mg = 5.63 μg
Explanation:
Significant figures of a number are the figures or digits that carry meaning and contributes to the precision of the given number.
The given quantity 0.00563 mg, has three significant figures as all the leading zeros are not significant.
Now to represent 0.00563 mg between 0.1 and 1000, the given unit milligram (mg) is converted to microgram (μg).
As, 1 mg = 1000 mg
Therefore, 0.00563 mg = 5.63 μg, has three significant figures as non-zeros figures are significant.
Therefore, 5.63 μg lies between between 0.1 and 1000 and has significant figures.
A 0.5 μF and a 11 μF capacitors are connected in series. Then the pair are connected in parallel with a 1.5 μF capacitor. What is the equivalent capacitance? Give answer in terms of mF.
Answer:
[tex]C_{eq}=1.97\ \mu F[/tex]
Explanation:
Given that,
Capacitance 1, [tex]C_1=0.5\ \mu F[/tex]
Capacitance 2, [tex]C_2=11\ \mu F[/tex]
Capacitance 3, [tex]C_3=1.5\ \mu F[/tex]
C₁ and C₂ are connected in series. Their equivalent is given by :
[tex]\dfrac{1}{C'}=\dfrac{1}{C_1}+\dfrac{1}{C_2}[/tex]
[tex]\dfrac{1}{C'}=\dfrac{1}{0.5}+\dfrac{1}{11}[/tex]
[tex]C'=0.47\ \mu F[/tex]
Now C' and C₃ are connected in parallel. So, the final equivalent capacitance is given by :
[tex]C_{eq}=C'+C_3[/tex]
[tex]C_{eq}=0.47+1.5[/tex]
[tex]C_{eq}=1.97\ \mu F[/tex]
So, the equivalent capacitance of the combination is 1.97 micro farad. Hence, this is the required solution.
Dana uses a rope to pull a box that weighs 300 N across alevel
surface with constant velocity. The rope makes an angle of 30above
the horizontal and the tension in the rope is 100 N. What isthe
coefficient of friction?
The coefficient of friction in this scenario is calculated as the ratio of the frictional force to the normal force. After calculating the frictional force to be 50N and the normal force to be 300N, the coefficient of friction is 0.167.
Explanation:Dana is pulling a box on a level surface with a rope that makes an angle of 30 degrees. The box has a weight of 300 N and the tension in the rope is 100 N. We first must understand that the force of friction is balancing the force that Dana is applying to move the box.
The vertical component of the tension (Tsin30) should be equal to the force of friction since they are balancing each other. Therefore, Tsin30 = Friction. Substituting the given values (100N*sin30 = Friction), we can calculate friction as 50 N.
The coefficient of friction can be calculated as the ratio of the frictional force to the normal force. As there is no vertical acceleration, the normal force is equal to the weight of the box which is 300N. Therefore, the coefficient of friction (mu) equals friction/normal force is equal to 50N/300N which is 0.167.
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The coefficient of friction is approximately 0.247.
The coefficient of friction between the box and the surface is given by the ratio of the frictional force to the normal force. The frictional force is the horizontal component of the tension in the rope, and the normal force is the vertical component of the tension in the rope plus the weight of the box acting downward.
First, let's calculate the horizontal and vertical components of the tension in the rope. The horizontal component can be found using the cosine of the angle, and the vertical component can be found using the sine of the angle:
[tex]\[ T_{\text{horizontal}} = T \cdot \cos(\theta) \] \[ T_{\text{vertical}} = T \cdot \sin(\theta) \][/tex]
Given that the tension in the rope (T) is 100 N and the angle [tex](\(\theta\))[/tex] is 30 degrees, we have:
[tex]\[ T_{\text{horizontal}} = 100 \text{ N} \cdot \cos(30^\circ) \] \\T_{\text{vertical}} = 100 \text{ N} \cdot \sin(30^\circ) \][/tex]
Using the values of cosine and sine for 30 degrees:
[tex]\[ T_{\text{horizontal}} = 100 \text{ N} \cdot \frac{\sqrt{3}}{2} \approx 86.6 \text{ N} \]\\T_{\text{vertical}} = 100 \text{ N} \cdot \frac{1}{2} = 50 \text{ N} \][/tex]
The normal force (N) is the sum of the weight of the box (W) and the vertical component of the tension:
[tex]\[ N = W + T_{\text{vertical}} \] \[ N = 300 \text{ N} + 50 \text{ N} \] \[ N = 350 \text{ N} \][/tex]
The frictional force (f) is equal to the horizontal component of the tension since the box is moving at constant velocity (implying that the net force in the horizontal direction is zero):
[tex]\[ f = T_{\text{horizontal}} \] \[ f = 86.6 \text{ N} \][/tex]
Now, the coefficient of friction is the ratio of the frictional force to the normal force:
[tex]\[ \mu = \frac{f}{N} \] \[ \mu = \frac{86.6 \text{ N}}{350 \text{ N}} \] \[ \mu \approx 0.247 \][/tex]
A boat’s speed in still water is 1.60 m/s . The boat is to travel directly across a river whose current has speed 1.05 m/s. Determine the speed of the boat with respect to the shore.
Express your answer using three significant figures and include the appropriate units
The speed of the boat with respect to the shore is 1.91 m/s.
From the information given, we have that;
A boat's speed in still water is 1.60 m/s
The boat is to travel directly across a river whose current has speed 1.05 m/s
We can see that the movement is in both horizontal and vertical directions.
Using the Pythagorean theorem, let use determine the resultant speed of the boat with respect to the shore, we have that;
Resultant speed² = √((boat's speed)² + (current's speed)²)
Substitute the value as given in the information, we have;
= (1.60)² + (1.05 )²)
Find the value of the squares, we get;
= (2.56 + 1.1025 )
Find the square root of both sides, we have;
= √3.6625
Find the square root of the value, we have;
= 1.91 m/s.
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Calculate the work done in compressing adiabatically 3kg of helium (He) to one fifth of its original volume if it is initially at 13°C. Find the change in internal energy of the gas resulting from the compression. (cp/cv for monatomic gases is 1.667; gas constant for helium is 2079 K^-1 kg^-1)
Answer:
Work done,[tex]w=5.12\times10^{6}\ \rm J[/tex]
change in internal Energy ,[tex]\Delta U=5.12\times10^6\ \rm J[/tex]
Explanation:
Given:
Mass of helium gas [tex]m=3\ \rm kg[/tex]initial temperature [tex]T_i=286\ \rm K[/tex]Since It is given that the process is adiabatic process it means that there is no exchange of heat between the system and surroundings
[tex]T_iV_i^{\gamma -1}=T_fV_f^{\gamma -1}\\\\286\times V_i^{\gamma -1}=T_f \left( \dfrac{V_i}{5} \right )^{\gamma -1}\\T_f=840.76\ \rm K[/tex]
Let n be the number of moles of Helium given by
[tex]n=\dfrac{m}{M}\\n=\dfrac{3\times10^3}{4}\\n=0.75\times10^3[/tex]
Work done in Adiabatic process
Let W be the work done
[tex]W=\dfrac{nR(T_1-T_2)}{\gamma-1}\\W=\dfrac{0.75\times10^3\times8.314(286-840.76)}{1.67-1}\\W=-5.12\times 10^6\ \rm J[/tex]
The Internal Energy change in any Process is given by
Let [tex]\Delta U[/tex] be the change in internal Energy
[tex]\Delta U=nC_p\Delta T\\\Delta U=0.75\times10^3\times1.5R\times(840.76-286)\\\Delta U=5.12\times10^6\ \rm J[/tex]
I have a device that can generate sounds with frequencies between 800 Hz and 1600 Hz. I also have an unlabeled tuning fork that I need to work out the fundamental frequency of. I find that it resonates at 920 Hz and 1380 Hz, but no other frequencies in that range. What's the lowest frequency that it will resonate at?
Answer:
460 Hz
Explanation:
the given resonating frequency of the device
f₁ = 920 Hz and f₂ = 1380 Hz
fundamental frequency of the device is
f₀ = n₂ - n₁
= 1380 - 920
= 460 Hz
expression of frequency of organic pipe open at both ends
[tex]f_0=n\dfrac{\nu}{2l}[/tex]
at n = 1
[tex]f_0=\dfrac{\nu}{2l} = 460 Hz[/tex]
the frequency ratios of the closed pipe
[tex]f_0:f_1:f_2: ...... =[1:2:3:.........]f_0[/tex]
=[1:2:3:.........]460 Hz
= 460 Hz : 920 Hz : 1380 Hz
so, the lowest frequency for the pipe open at both end is 460 Hz
An arrow is shot from a height of 1.3 m toward a cliff of height H . It is shot with a velocity of 34 m/s at an angle of 58.1º above the horizontal. It lands on the top edge of the cliff 4.1 s later. What is the height of the cliff?
Answer:
The height of the cliff is 39.655 m
Given:
Height at which the arrow was shot, h = 1.3 m
Velocity of the arrow, u = 34 m/s
Angle, [tex]\theta' = 58.1^{\circ}[/tex]
Time of the fight, t = 4.1 s
Solution:
Let the Height of the cliff be H
Since, the motion of the object is projectile motion and the direction of motion is vertical at some angle
Therefore, we consider the vertical component of velocity, [tex]u_{y} = usin\theta[/tex].
Now,
The Height of the cliff is given by applying the second equation of motion in the projectile:
Thus
[tex]s = u_{y}t - \frac{1}{2}gt^{2}[/tex]
[tex]s = 34sin60^{\circ}\times 4.1 - \frac{1}{2}\times 9.8\times 4.1^{2}[/tex]
s = 38.355 m
Now, the height of the cliff, H:
H = s + h = 38.355 + 1.3 = 39.655 m
A horizontal clothesline is tied between 2 poles, 16 meters apart. When a mass of 3 kilograms is tied to the middle of the clothesline, it sags a distance of 3 meters. What is the magnitude of the tension on the ends of the clothesline? NOTE: Use g=9.8m/s2 for the gravitational acceleration.
The magnitude of the tension on the ends of the clothesline is approximately [tex]\( 41.99 \, \text{N} \).[/tex]
To find the tension on the ends of the clothesline, we can use the principle of equilibrium. When the mass is at rest in the middle of the clothesline, the tension in the clothesline must balance the gravitational force acting on the mass.
Let's denote:
- [tex]\( T \)[/tex] as the tension in the clothesline.
- [tex]\( m \)[/tex] as the mass tied to the clothesline.
- [tex]\( g \)[/tex] as the gravitational acceleration.
The gravitational force acting on the mass is given by [tex]\( F_{\text{gravity}} = m \cdot g \).[/tex]
When the mass is at rest in the middle of the clothesline, the horizontal components of the tensions on either side cancel each other out, leaving only the vertical components to support the weight.
Since the clothesline sags a distance of 3 meters in the middle, each side of the clothesline forms a right triangle with the sagging distance as one leg and half of the distance between the poles as the other leg. Therefore, each side of the clothesline forms a right triangle with legs of 8 meters and 3 meters.
Using the Pythagorean theorem, we can find the length of the clothesline (the hypotenuse of each triangle), which is the same as the length of each segment between the midpoint and the poles:
[tex]\[ L = \sqrt{(8 \, \text{m})^2 + (3 \, \text{m})^2} \]\[ L = \sqrt{64 + 9} \]\[ L = \sqrt{73} \][/tex]
Now, we can set up an equation for the vertical equilibrium:
[tex]\[ 2T \sin(\theta) = F_{\text{gravity}} \][/tex]
Where [tex]\( \theta \)[/tex] is the angle the clothesline makes with the horizontal.
We can find [tex]\( \sin(\theta) \)[/tex] using trigonometry:
[tex]\[ \sin(\theta) = \frac{3}{L} \][/tex]
Now, substitute this expression for [tex]\( \sin(\theta) \)[/tex] into the equation for vertical equilibrium:
[tex]\[ 2T \left(\frac{3}{L}\right) = m \cdot g \]\[ 2T \left(\frac{3}{\sqrt{73}}\right) = 3 \times 9.8 \]\[ T = \frac{3 \times 9.8 \times \sqrt{73}}{2 \times 3} \]\[ T = \frac{9.8 \times \sqrt{73}}{2} \][/tex]
Now, we can calculate the tension:
[tex]\[ T \approx \frac{9.8 \times \sqrt{73}}{2} \]\[ T \approx \frac{9.8 \times 8.544}{2} \]\[ T \approx \frac{83.9712}{2} \]\[ T \approx 41.9856 \][/tex]
So, the magnitude of the tension on the ends of the clothesline is approximately [tex]\( 41.99 \, \text{N} \).[/tex]
The magnitude of the tension at the ends of the clothesline is approximately [tex]\[T = 41.87 \, \text{N}\][/tex]
To solve for the tension in the clothesline, we first consider the forces and the geometry involved.
Given:
Distance between poles, [tex]\( L = 16 \) meters[/tex]
Sagging distance,[tex]\( h = 3 \) meters[/tex]
Mass, [tex]\( m = 3 \) kilograms[/tex]
Gravitational acceleration, [tex]\( g = 9.8 \) m/s\(^2\)[/tex]
Determine the Weight of the Mass
The weight [tex]\( W \)[/tex] of the mass is given by:
[tex]\[W = mg = 3 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 29.4 \, \text{N}\][/tex]
Analyzing the Geometry
[tex]\[\text{Half the distance between poles} = \frac{L}{2} = \frac{16}{2} = 8 \, \text{meters}\][/tex]
[tex]\[\text{Sagging distance} = h = 3 \, \text{meters}\][/tex]
Calculate the Length of the Hypotenuse
[tex]\[\sqrt{(8 \, \text{m})^2 + (3 \, \text{m})^2} = \sqrt{64 + 9} = \sqrt{73} \, \text{meters}\][/tex]
Analyze Forces in Vertical and Horizontal Directions
[tex]\[2T_v = W \quad = \quad T_v = \frac{W}{2} = \frac{29.4 \, \text{N}}{2} = 14.7 \, \text{N}\][/tex]
The vertical component of the tension [tex]\( T \)[/tex] in each half of the clothesline is:
[tex]\[T_v = T \sin \theta\][/tex]
where \( \theta \) is the angle the clothesline makes with the horizontal. From the triangle, we know:
[tex]\[\sin \theta = \frac{h}{\sqrt{(L/2)^2 + h^2}} = \frac{3}{\sqrt{73}}\][/tex]
[tex]\[T \sin \theta = 14.7 \, \text{N}\][/tex]
Solve for the Tension[tex]\( T \)[/tex]
[tex]\[T \sin \theta = 14.7 \, \text{N} \quad = \quad T \cdot \frac{3}{\sqrt{73}} = 14.7 \, \text{N}\][/tex]
[tex]\[T = \frac{14.7 \times \sqrt{73}}{3}\][/tex]
[tex]\[T = 14.7 \times \frac{\sqrt{73}}{3}\][/tex]
[tex]\[T = 14.7 \times \frac{8.544}{3}\][/tex]
[tex]\[T = 14.7 \times 2.848 = 41.87 \, \text{N}\][/tex]
A pilot claims to have seen a UFO moving initially at a speed of about 351 m/s in an easterly direction and then, in a time interval of only 1.0 s, turning 48° south of east and moving at 351 m/s. Compute the UFO's average acceleration during the turn. Find Magnitude and direction
Answer:
Explanation:
This problem can be solved easily if we represent velocity in the form of vector.
The velocity of 351 was towards easterly direction so
V₁ = 351 i
The velocity of 351 was towards south west making - 48° with east or + ve x direction.
V₂ = 351 Cos 48 i - 351 sin 48 j
V₂ = 234.86 i - 260.84 j
Change in velocity
= V₂ - V₁ = 234.86 i - 260.84 j - 351 i
= -116.14 i - 260.84 j
acceleration
= change in velocity / time
(-116.14 i - 260.84 j )/ 1
= -116.14 i - 260.84 j
magnitude = 285.53 ms⁻²
Direction
Tan θ = 260.84 / 116.14 = 2.246
θ = 66 degree south of west .
What is an equipotential line?
An equipotential line is like the contour line of a map that had lines of equal altitude. In this case the "altitude"is the electrical potential or voltage. Equipotential lines are always perpendicular to the electric field. In three dimensions these lines form equipotential surfaces. The movement along an equipotential surface does not perform work, because that movement is always perpendicular to the electric field.
Calculate the The wavelength of the first Balmer series of hydrogen is 6562 following: a) The ionization potential, and b) The first excitation potential of the hydrogen atom.
Answer:
(a) 13.6 eV
(b) 10.2 V
Explanation:
a) Ionization potential energy is defined as the minimum energy required to excite a neutral atom to its ionized state i.e basically the minimum energy required to excite an electron from n=1 to infinity.
Energy of a level, n, in Hydrogen atom is, [tex]E_{n}=-\frac{13.6}{n^{2} }[/tex]
Now ionization potential can be calculated as
[tex]E_{\infty}- E_{0}[/tex]
Substitute all the value of energy and n in above equation.
[tex]=-\frac{13.6}{\infty^{2}}-(-\frac{13.6}{1^{2}})\\=13.6eV[/tex]
Therefore, the ionization potential is 13.6 eV.
b) This is the energy required to excite a atom from ground state to its excited state. When electrons jumps from ground state level(n=1) to 1st excited state(n=2) the corresponding energy is called 1st excitation potential energy and corresponding potential is called 1st excitation potential.
So, 1st excitation energy = E(n 2)- E(n = 1)
[tex]=-\frac{13.6}{2^{2}}-(-\frac{13.6}{1^{2}})\\=-3.4eV - (-13.6eV) \\=10.2eV[/tex]
Now we can find that 1st excitation energy is 10.2 eV which gives,
[tex]eV'=10.2eV\\V'=10.2V[/tex]
Therefore, the 1st excitation potential is 10.2V.
A block slides down a frictionless incline with constant acceleration. After sliding 6.80 m down, it has a speed of 3.80 m/s. What was the speed when the block had slid 3.40 m down? O 1.90 m/s O 2.69 m/s O None of the above
Answer:
speed when the block had slid 3.40 m is 2.68 m/s
Explanation:
given data
distance = 6.80 m
speed = 3.80 m/s
to find out
speed when the block had slid 3.40 m
solution
we will apply here equation of motion that is
v²-u² = 2×a×s ..............1
here s is distance, a is acceleration and v is speed and u is initial speed that is 0
so put here all value in equation 1 to get a
v²-u² = 2×a×s
3.80²-0 = 2×a×6.80
a = 1.06 m/s²
so
speed when distance 3.40 m
from equation 1 put value
v²-u² = 2×a×s
v²-0 = 2×1.06×3.40
v² = 7.208
v = 2.68
so speed when the block had slid 3.40 m is 2.68 m/s
A 50kg trunk is pushed 6.0 m at a constant speed up a 30degree
incline by a constant horizontal force. Thecoefficient of kinetic
energy n\between the trunk and the inclineis 0.20. Calculate the
work done by:
a.) the applied force and
b.) by the weight of the trunk.
c.) How much energy was dissipated by the frictional
forceacting on the trunk?
Answer:
a) 2034 J
b) -1471 J
c) -509 J
Explanation:
The trunk has a mass of 50 kg, so its weight is
f = m * a
f = 50 * 9.81 = 490 N
If the incline is of 30 degrees, the force tangential to the incline is:
ft = f * sin(a)
ft = 490 * sin(30) = 245 N
And the normal force is:
fn = f * cos(a)
fn = 490 * cos(30) = 424 N
The frictional force is:
ff = μ * fn
ff = 0.2 * 424 = 84.8 N
To push the trunk up one must apply a force slightly greater than the opposing forces, the opposing are the tangential component of the weight and the friction force
fp = fn + ff
fp = 245 + 84.8 = 339 N
The work of the applied force is:
L = f * d
Lp = fp * d
Lp = 339 * 6 = 2034 J
The work of the weight is done by the tangential component:
Lw = -245 * 6 = -1471 J (it is negative because the weight was opposed to the direction of movement)
The work of the friction force is
Lf = -84.8 * 6 = -509 J
A car traveling 34 mi/h accelerates uniformly for 4 s, covering 615 ft in this time. What was its acceleration? Round your answer to the nearest 100th place. Answer in units of ft/s^2. What is the final velocity at this time? Answer in units of ft/s.
Answer:
51.94 ft/s²
257.63 ft/s
Explanation:
t = Time taken = 4 s
u = Initial velocity = 34 mi/h
v = Final velocity
s = Displacement = 615 ft
a = Acceleration
Converting velocity to ft/s
[tex]34\ mi/h=\frac{34\times 5280}{3600}=49.87\ ft/s[/tex]
Equation of motion
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow a=2\frac{s-ut}{t^2}\\\Rightarrow a=2\left(\frac{615-49.87\times 4}{4^2}\right)\\\Rightarrow a=51.94\ ft/s^2[/tex]
Acceleration is 51.94 ft/s²
[tex]v=u+at\\\Rightarrow v=49.87+51.94\times 4\\\Rightarrow v=257.63\ ft/s[/tex]
Final velocity at this time is 257.63 ft/s
A rocket sled accelerates at 21.5 m/s^2 for 8.75 s. (a) What's its velocity at the end of that time? (b) How far has it traveled?
Answer:
(a ) vf= 188.12m/s : Final speed at 8.75 s
(b) d= 823.04 m : Distance the rocket sled traveled
Explanation:
Rocket sled kinematics :The rocket sled moves with a uniformly accelerated movement, then we apply the following formulas:
d =vi*t+1/2a*t² Formula (1)
vf= vi+at Formula(2)
Where:
vi: initial speed =0
a: acceleration=21.5 m/s²
t: time=8.75 s
vf: final speed in m/s
d:displacement in meters(m)
Calculation of displacement (d) and final speed (vf)
We replace data in formulas (1) and (2):
d= 0+1/2*21.5*8.75²
d= 823.04 m
vf= 0+21.5*8.75
vf= 188.12m/s
A ball is dropped from a building of height h. Derive an equation for its velocity just before it hits the ground. Assume no air resistance.
Answer:
The final velocity of the ball just hits the ground is [tex]\sqrt{2gh}[/tex].
Explanation:
A ball is dropped from the height h. As the ball is dropped means there is no initial velocity given to the ball.
GIVEN:
The distance traveled by the ball is equal to h.
Initial velocity of the ball is zero.
Concept:
Two type of force comes into play in dropping of ball.
1. Gravitational force
Due to gravitational force ball forced to fall downward towards the surface.
2. Air resistance
Air develops drag force on the ball in opposite direction of the gravitational force.
Assumption:
There is no air resistance in falling of ball. So, there will be no drag force on the ball.
Calculation:
It is given that a ball is dropped from a building of height h.
The Newton’s equations of motion are as follows:
[tex]V^{2}-u^{2}=2as[/tex]
[tex]s=ut+\frac{1}{2}at^{2}[/tex]
[tex]v=u+at[/tex]
STEP 1
Newton’s equation of motion is expressed as follows:
[tex]V^{2}-u^{2}=2as[/tex]
Here, V is final velocity of the ball just before the ball hits the ground, a is acceleration due to gravity, u is the initial velocity of the ball and s is the distance traveled by the ball.
Here, initial velocity is zero. So, u = 0 m/s.
Vertical distance traveled by the ball is h. So, s= h.
Only gravitational force is responsible for motion as well as for acceleration. Thus, the acceleration due to gravity acts on the ball. Acceleration due to gravity is denoted by small g.
Acceleration due to gravity is constant for different planate. The acceleration due to gravity for the earth is 9.8 m/s². Here, no planet is defined. So the acceleration due to gravity is a = g.
Substitute 0 for m/s, g for a and h for s in above equation as follows:
[tex]V^{2}-u^{2}=2as[/tex]
[tex]V^{2}-(0)^{2}=2gh[/tex]
[tex]V^{2}=2gh[/tex]
[tex]v=\sqrt{2gh}[/tex]
Thus, the final velocity of the ball just hits the ground is [tex]\sqrt{2gh}[/tex].