Answer:
W_planet=(W_earth*a)/g
Explanation:
The mass of the object does not change. It is the same at Earth and at the other planet.
W_earth=Mg weight at Earth
W_planet=Ma weight at other planet
If we divide the last equations:
W_planet/W_earth=a/g
W_planet=(W_earth*a)/g
A busy chipmunk runs back and forth along a straight line of acorns that has been set out between its burrow and a nearby tree. At some instant, it moves with a velocity of −1.29 m/s−1.29 m/s . Then, 2.91 s2.91 s later, it moves with a velocity of 1.77 m/s1.77 m/s . What is the chipmunk's average acceleration during the 2.91 s2.91 s time interval?
Answer:
1.05 ms⁻²
Explanation:
Acceleration = change in velocity / Time
Change in velocity = Final velocity - initial velocity
= 1.77 - (-1.29)
= 1.77 + 1.29
= 3.06 m/s
Time = 2.91
Acceleration = 3.06 / 2.91
= 1.05 ms⁻² .
What is the acceleration of a 20 kg cart if the net force on it is 40 N?
Answer:
Acceleration of the cart will be [tex]a=2m/sec^2[/tex]
Explanation:
We have given force F = 40 N
Mass of the cart = 20 kg
From newton's second law we know that force, mass and acceleration are related to each other
From second law of motion force on any object moving with acceleration a is given by
F = ma, here m is mass and a is acceleration
So [tex]40=20\times a[/tex]
[tex]a=2m/sec^2[/tex]
Alex climbs to the top of a tall tree while his friend Gary waits on the ground below. Alex throws down a ball at 8 m/s from 50 m above the ground at the same time Gary throws a ball up. At what speed must Gary throw a ball up in order for the two balls to cross paths 25 m above the ground? The starting height of the ball thrown upward is 1.5 m above the ground. Ignore the effects of air resistance. whats the answer in m/s?
Answer:22.62 m/s
Explanation:
Given
two balls are separated by a distance of 50 m
Alex throws the ball from a height of 50 m with a velocity of 8 m/s and Gary launches a ball with some velocity exactly at the same time.
ball from ground travels a distance of 25 m in t sec
For Person on tree
[tex]25=ut+\frac{1}{2}gt^2[/tex]
[tex]25=8t+\frac{1}{2}\times 9.81\times t^2--------1[/tex]
For person at ground
[tex]23.5=ut-\frac{1}{2}gt^2---------2[/tex]
Solve equation (1)
[tex]50=16t+9.81t^2[/tex]
[tex]9.81t^2+16t-50=0[/tex]
[tex]t=\frac{-16\pm\sqrt{256+4\times 50\times 9.81}}{2\times 9.81}=\frac{47.1-16}{19.62}=1.58 s[/tex]
put the value of t in equation 2
[tex]23.5=u\times 1.58-\frac{9.81\times 1.58^2}{2}[/tex]
[tex]u=\frac{35.744}{1.58}=22.62 m/s[/tex]
A package is dropped from a helicopter moving upward at 15m/s. If it takes 10 s before the package strikes the ground, how high above the ground was the ground was the package when it was released if air resistance is negligible?
(A) 408 m
(B) 272 m
(C) 204 m
(D) 340 m
Answer:
The right answer is (D) 340m
Explanation:
If the package is dropped from a helicopter moving upward and the air resistance is negligible, the only acting force in the package is the gravity force. Therefore you can consider this as a Vertical Projectile Motion problem.
The initial speed of the package V₀ is 15m/s.
The acceleration of the package G is 9.8m/s².
The initial hight is Y₀. Therefore:
Y(t)=Y₀+V₀·t-0.5·G·t²
But we know than T(10s)=0m
0m=Y₀+150m-490m
Y₀=340m
The density of a rock will be measured by placing it into a graduated cylinder partially filled with water, and then measuring the volume of water displaced. The density D is given by D = m/(V1 − V0), where m is the mass of the rock, V0 is the initial volume of water, and V1 is the volume of water plus rock. Assume the mass of the rock is 750 g, with negligible uncertainty, and that V0 = 500.0 ± 0.1 mL and V1 = 813.2 ± 0.1 mL. Estimate the density of the rock, and find the uncertainty in the estimate.
Answer:
[tex]\rho = 2.39 g/mL[/tex]
[tex]\Delta \rho = 1.53 \times 10^{-3} mL[/tex]
Explanation:
As we know that density is the ratio of mass and volume of the object
here we know that
mass of the rock is
[tex]m = 750 g[/tex]
volume of the rock is given as
[tex]V = V_1 - V_o[/tex]
here we know that
[tex]V_1 = 813.2 \pm 0.1 mL[/tex]
[tex]V_2 = 500.0 \pm 0.1 mL[/tex]
now we have
[tex]V = 313.2 \pm 0.2 mL[/tex]
now density is given as
[tex]\rho = \frac{750}{313.2}[/tex]
[tex]\rho = 2.39 g/mL[/tex]
now uncertainty of density is given as
[tex]\Delta \rho = \frac{\Delta V}{V} \rho[/tex]
[tex]\Delta \rho = \frac{0.2}{313.2}(2.39)[/tex]
[tex]\Delta \rho = 1.53 \times 10^{-3} mL[/tex]
The density of the rock, given its mass is 750 g and the volume of water displaced is 313.2 mL, is 2.394 g/mL. The uncertainty in this measurement is ± 0.0015 g/mL, considering an uncertainty of ± 0.1 mL for both the initial and final volume measurements.
Given that the mass (m) of the rock is 750 g, the initial volume of water (V0) is 500.0 mL, and the volume of water plus the rock (V1) is 813.2 mL, we can determine the density (D) and the uncertainty in the density.
Using the formula:
D = m / (V1 - V0)
Therefore, D = 750 g / (813.2 mL - 500.0 mL)
= 750 g / 313.2 mL
= 2.394 g/mL.
Using the uncertainties in V0 and V1, which are both ± 0.1 mL. Since we subtract these volumes, the total volume uncertainty is ± (0.1 mL + 0.1 mL) = ± 0.2 mL. Thus, the uncertainty in the density (ΔD) can be approximated by the formula:
ΔD = D × (ΔV / (V1 - V0))
where ΔV is the total volume uncertainty. Substituting the values, we get ΔD = 2.394 g/mL × (0.2 mL / 313.2 mL) = ± 0.00153 g/mL (rounded to four significant figures).
Therefore, the estimated density of the rock is 2.394 ± 0.0015 g/mL.
The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about 100 cm . a) What is their initial "launch" speed off the ground?
b)How long are they in the air?
Answer:
a) 4.45 m/s
b) 0.9 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
[tex]v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 1}\\\Rightarrow u=4.45\ m/s[/tex]
a) The vertical speed when the player leaves the ground is 4.45 m/s
[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-4.45}{-9.81}\\\Rightarrow t=0.45\ s[/tex]
Time taken to reach the maximum height is 0.45 seconds
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1\times 2}{9.81}}\\\Rightarrow t=0.45\ s[/tex]
Time taken to reach the ground from the maximum height is 0.45 seconds
b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds
Consider steady heat transfer between two large parallel plates at constant temperatures of T1 = 210 K and T2 = 150 K that are L = 2 cm apart. Assume that the surfaces are black (emissivity ε = 1). Determine the rate of heat transfer between the plates per unit surface area assuming the gap between the plates is filled with atmospheric air.
Answer:
[tex]Q=81.56\ W/m^2[/tex]
Explanation:
Given that
[tex]T_1= 210 K[/tex]
[tex]T_2= 150 K[/tex]
Emissivity of surfaces(∈) = 1
We know that heat transfer between two surfaces due to radiation ,when both surfaces are black bodies
[tex]Q=\sigma (T_1^4-T_2^4)\ W/m^2[/tex]
So now by putting the values
[tex]Q=\sigma (T_1^4-T_2^4)\ W/m^2[/tex]
[tex]Q=5.67\times 10^{-8}(210^4-150^4)\ W/m^2[/tex]
[tex]Q=81.56\ W/m^2[/tex]
So rate of heat transfer per unit area
[tex]Q=81.56\ W/m^2[/tex]
A man pushes a lawn mower on a level lawn with a force of 195 N. If 37% of this force is directed downward, how much work is done by the man in pushing the mower 5.7 m?
Final answer:
The work done by the man in pushing the lawn mower is 699.245 J, calculated by determining the horizontal force component and multiplying by the distance pushed.
Explanation:
To calculate how much work is done by the man in pushing the lawn mower, we need to consider only the component of the force that acts in the direction of the movement. Since 37% of the 195 N force is directed downward, only the remaining 63% is contributing to the horizontal movement. Therefore, the horizontal component of the force is 0.63 × 195 N = 122.85 N.
The formula to calculate work (W) is W = force (F) × distance (d) × cosine(θ), where θ is the angle between the force and the direction of movement. In this case, the force and movement are in the same direction, so θ = 0 and cosine(θ) = 1. Thus, the work done is:
W = 122.85 N × 5.7 m × 1 = 699.245 J
In terms of energy expended while pushing a lawn mower, this work is a relatively small amount when compared to a person's daily intake of food energy.
Starting from rest, a runner reaches a speed of 2.8 m/s in 2.1 s. In the same time 2.1 s time, a motorcycles increases speed from 37. 0 to 43.0 m/s. In both cases, assume the acceleration is constant: (a). What is the acceleration (magnitude only) of the runner?(b). What is the acceleration (magnitude only) of the motorcycle?(c). Does the motorcycle travel farther than the runner during the 2.1 s? (yes or no)If som how much father? (if not, enter zero)?
Answer:
a) Acceleration of runner is 1.33 m/s²
b) Acceleration of motorcycle is 2.85 m/s²
c) The motorcycle moves 84.21-2.94 = 81.06 m farther than the runner.
Explanation:
t = Time taken
u = Initial velocity = 0
v = Final velocity
s = Displacement
a = Acceleration
Equation of motion
[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{2.8-0}{2.1}\\\Rightarrow a=1.33\ m/s^2[/tex]
Acceleration of runner is 1.33 m/s²
[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{43-37}{2.1}\\\Rightarrow a=2.85\ m/s^2[/tex]
Acceleration of motorcycle is 2.85 m/s²
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{2.8^2-0^2}{2\times 1.33}\\\Rightarrow s=2.94\ m[/tex]
The runner moves 2.94 m
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{43^2-37^2}{2\times 2.85}\\\Rightarrow s=84.21\ m[/tex]
The motorcycle moves 84.21 m
The motorcycle moves 84.21-2.94 = 81.06 m farther than the runner.
A car is driven east for a distance of 47 km, then north for 23 km, and then in a direction 32° east of north for 27 km. Determine (a) the magnitude of the car's total displacement from its starting point and (b) the angle (from east) of the car's total displacement measured from its starting direction.
Answer:
(a). The car's total displacement from its starting point is 76.58 m.
(b). The angle of the car's total displacement measured from its starting direction is 36.81°.
Explanation:
Given that,
Distance = 47 km in east
Distance = 23 km in north
Angle = 32° east of north
Distance = 27 km
According to figure,
Angle = 90-32 = 58°
(a). We need to calculate the magnitude of the car's total displacement from its starting point
Using Pythagorean theorem
[tex]AC=\sqrt{AB^2+BC^2}[/tex]
[tex]AC=\sqrt{(47+27\cos58)^2+(23+27\sin58)^2}[/tex]
[tex]AC=76.58\ m[/tex]
The magnitude of the car's total displacement from its starting point is 76.58 m.
(b). We need to calculate the angle (from east) of the car's total displacement measured from its starting direction
Using formula of angle
[tex]\tan\theta=\dfrac{y}{x}[/tex]
put the value into the formula
[tex]\theta=tan^{-1}\dfrac{23+27\sin58}{47+27\cos58}[/tex]
[tex]\theta=tan^{-1}0.7486[/tex]
[tex]\theta=36.81^{\circ}[/tex]
Hence, (a). The car's total displacement from its starting point is 76.58 m.
(b). The angle of the car's total displacement measured from its starting direction is 36.81°.
Two objects are dropped at the same time from heights of 0.49 m and 1.27 m. How long after the first object hits the ground does the second object hit the ground?
Answer:
After 0.19288 sec second object hit the ground after hitting first object
Explanation:
We have given two objects are dropped from at the same time from height 0.49 m and 1.27 m
As the object is dropped so its initial velocity will be zero u = 0 m/sec
For object 1 height h = 0.49 m
From second equation of motion
[tex]h=ut+\frac{1}{2}gt^2[/tex]
[tex]0.49=0\times t+\frac{1}{2}\times 9.8\times t^2[/tex]
[tex]t^2=0.1[/tex]
t = 0.31622 sec
For second object
[tex]h=ut+\frac{1}{2}gt^2[/tex]
[tex]1.27=0\times t+\frac{1}{2}\times 9.8\times t^2[/tex]
[tex]t^2=0.1[/tex]
t = 0.5091 sec
So difference in time = 0.5091-0.31622 = 0.19288 sec
So after 0.19288 sec second object hit the ground after hitting first object
A football is kicked from ground level at an angle of 38 degrees. It reaches a maximum height of 9.7 meters before returning to the ground. How long will the football spend in the air, in seconds?
Answer:
Football will be in air for 2.8139 sec
Explanation:
We have given maximum height h = 9.7 meters
Angle of projection [tex]\Theta =38^{\circ}[/tex]
We know that maximum height is given by [tex]h=\frac{u^2sin^2\Theta }{2g}[/tex]
So [tex]9.7=\frac{u^2sin^{2}38^{\circ} }{2\times 9.8}[/tex]
[tex]u^2=501.5842[/tex]
u = 22.396 m/sec
Time of flight is given by
[tex]T=\frac{2usin\Theta }{g}=\frac{2\times 22.396\times \times sin38^{\circ}}{9.8}=2.8139sec[/tex]
38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobaric heating to a temperature of 935.9°C. (a) What is the final volume of the gas?(in cm^3 ) (b) It is then isothermally compressed to a volume 29.3cm^3; what is its final pressure?(in Pa )
Answer:
Final volumen first process [tex]V_{2} = 98,44 cm^{3}[/tex]
Final Pressure second process [tex]P_{3} = 1,317 * 10^{10} Pa[/tex]
Explanation:
Using the Ideal Gases Law yoy have for pressure:
[tex]P_{1} = \frac{n_{1} R T_{1} }{V_{1} }[/tex]
where:
P is the pressure, in Pa
n is the nuber of moles of gas
R is the universal gas constant: 8,314 J/mol K
T is the temperature in Kelvin
V is the volumen in cubic meters
Given that the amount of material is constant in the process:
[tex]n_{1} = n_{2} = n [/tex]
In an isobaric process the pressure is constant so:
[tex]P_{1} = P_{2} [/tex]
[tex]\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }[/tex]
[tex]\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }[/tex]
[tex]V_{2} = \frac{T_{2} V_{1} }{T_{1} }[/tex]
Replacing : [tex]T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}[/tex]
[tex]V_{2} = 98,44 cm^{3}[/tex]
Replacing on the ideal gases formula the pressure at this piont is:
[tex]P_{2} = 3,92 * 10^{9} Pa[/tex]
For Temperature the ideal gases formula is:
[tex]T = \frac{P V }{n R }[/tex]
For the second process you have that [tex]T_{2} = T_{3} [/tex] So:
[tex]\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }[/tex]
[tex]P_{2} V_{2} = P_{3} V_{3} [/tex]
[tex]P_{3} = \frac{P_{2} V_{2}}{V_{3}} [/tex]
[tex]P_{3} = 1,317 * 10^{10} Pa[/tex]
The brakes on your automobile are capable of creating a deceleration of 4.6 m/s^2. If you are going 114 km/h and suddenly see a state trooper, what is the minimum time in which you can get your car under the 79 km/h speed limit? (The answer reveals the futility of braking to keep your high speed from being detected with a radar or laser gun.)
Answer:
The minimum time to get the car under max. speed limit of 79 km/h is 2.11 seconds.
Explanation:
[tex]a=\frac{V_f-V_0}{t}[/tex]
isolating "t" from this equation:
[tex]t=\frac{V_f-V_0}{a}[/tex]
Where:
a=[tex]-4.6m/s^2[/tex] (negative because is decelerating)
[tex]V_f= 79 km/h[/tex]
[tex]V_0= 114 km/h[/tex]
First we must convert velocity from km/h to m/s to be consistent with units.
[tex]79\frac{km}{h}*\frac{1000m}{1 km}*\frac{1h}{3600s}=\frac{79*1000}{3600}=21.94 m/s[/tex]
[tex]114\frac{km}{h}*\frac{1000m}{1 km}*\frac{1h}{3600s}=\frac{114*1000}{3600}=31.67 m/s[/tex]
So;
[tex]t=\frac{V_f-V_0}{a}=\frac{21.94 m/s-31.66m/s}{-4.6 m/s^2}=2.11 s[/tex]
Tom Sawyer runs 10 m/s down the dock and leaps on to his floating raft already moving 1.5 m/s away from shore. If Tom weighs 70 kg and the raft weighs 130 kg, what speed will they both be moving? A. 895 kg m/sB. 11.5 m/sC. 6.4 m/sD. 5.75 m/s
Answer:
Not the right answer in the options, speed is 4.47 m/s, and the procedure is coherent with option A
Explanation:
Answer A uses mass and velocity units, which are momentum units. By using the conservation of momentum:
.[tex]p_{initial} =p_{final} \\m_{Tom}*v_{Tom}+m_{raft}*v_{raft}=(m_{Tom}+m_{raft})*v_{both} \\70*10+130*1.5 kg*m/s=895kg*m/s\\v_{both}=\frac{895 kg*m/s}{200 kg} =4.47 m/s[/tex]
Since Tom stays in the raft, then both are moving with the same speed. From the options, the momentum is in agreement with option A, however, the question asks for speed.
A vacuum gage attached to a power plant condenser gives a reading of 27.86 in. of mercury. The surrounding atmospheric pressure is 14.66 lbf/in. Determine the absolute pressure inside the condenser, in lbf/in. The density of mercury is 848 lb/ft and the acceleration of gravity is g = 32.0 ft/s?.
Answer:
absolute pressure = 1.07 lbft/in^2
Explanation:
given data:
vaccum gauge reading h = 27.86 inch = 2.32 ft
we know that
gauge pressure p is given as
[tex]p = \rho gh[/tex]
[tex]p = 848 \ lb/ft^3 * 32 \ ft/s^2 *2.32 ft = 62955.52 \ lbft/s^2 * 1/ft^2[/tex]
we know that [tex]1\ lb ft\s^2 = \frac{1}{32.174}\ lbft[/tex]
1 ft = 12 inch
therefore [tex]p = 62955.52 * \frac{1}{32.174} * \frac{1}{12^2}\ lbf/in^2[/tex]
[tex]p = 13.59\ lbft/in^2[/tex]
[tex]P_{atm} = 14.66\ lbf/in^2[/tex]
so absolute pressure = [tex]P_{atm} - p[/tex]
= 14.66 - 13.59 = 1.07 lbft/in^2
An object is dropped from rest and falls through height h. It travels 0.5h in the last 1 second of fall. Find the total time & height of the fall. (Hint: use two triangles!)
Answer:
3.41 s
114 m
Explanation:
The object is falling in free fall, accelerated by the surface gravity of Earth. We can use the equation for position under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
We set up a frame of reference with the origin at the point the object was released and the X axis pointing down. Then X0 = 0. Since the problem doesnt mention an initial speed we assume V0 = 0.
It travels 0.5h in the last 1 second of the fall. This means it also traveled in the rest of the time of the fall. t = t1 is the moment when it traveled 0.5*h.
0.5*h = 1/2 * a * t1^2
h = a * t1^2
It travels 0.5*h in 1 second.
h = X(t1 + 1) = 1/2 * a * (t1+1)^2
Equating both equations:
a * t1^2 = 1/2 * a * (t1+1)^2
We simplify a and expand the square
t1^2 = 1/2 * (t1^2 + 2*t1 + 1)
t1^2 - 1/2 * t1^2 - t1 - 1/2 = 0
1/2 * t1^2 - t1 - 1/2 = 0
Solving electronically:
t1 = 2.41 s
total time = t1 + 1 = 3.41.
Now
h = a * t1^2
h = 9.81 * 3.41^2 = 114 m
A hippo drives 42 km due East. He then turns and drives 28 km at 25° East of South. He turns again and drives 32 km at 40° North of East. a) Sketch a plot of the vector sum of this motion. b) Use vector math to find his total displacement in component form. c) Convert to magnitude and direction form. d) How far is the hippo from his starting point? Note: this is distance, a scalar. What total distance has the hippo traveled?
Answer:
a) Please, see the attched figure
b) Total displacement R = (78.3 km; -4.8 km)
c) R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))
d) The hippo is 78.4 km from his starting point.
The total distance traveled is 102 km
Explanation:
a)Please, see the attached figure.
b) The vector A can be expressed as:
A = (magnitude * cos α; magnitude * sin α)
Where
magnitude = 42 km
α= 0
Then,
A = (42 km ; 0) or 42 km i
In the same way, we can proceed with the other vectors:
B = ( Bx ; By)
where
(apply trigonometry of right triangles: sen α = opposite / hypotenuse and
cos α = adjacent / hypotenuse. See the figure to determine which component of vector B is the opposite and adjacent side to α)
Bx = 28 km * sin 25 = 11.8 km
By = 28 km * cos 25 = -25.4 km (it has to be negative since it is directed towards the negative vertical region according to our reference system)
B = (11.8 km; -25.4 km) or 11.8 km i - 25.4 km j
C = (Cx; Cy)
where
Cx = 32 km * cos 40° = 24.5 km
Cy = 32 km * sin 40 = 20.6 km
C = (24.5 km; 20.6 km)
Then:
R = A+B+C = (42 km + 11.8 km + 24.5 km; 0 - 25.4 km + 20.6 km)
= (78.3 km; -4.8 km) or 78.3 km i -4.8 km j
c) R = (78.3 km; -4.8 km)
The magnitude of R is:
[tex]magnitude = \sqrt{(78.3)^{2 }+ (-4.8)^{2}}= 78.4 km[/tex]
Using trigonometry, we can calculate the angle:
Knowing that
tan α = opposite / adjacent
and that
opposite = Ry = -4.8 km
adjacent = Rx = 78.3 km
Then:
tan α = -4.8 km / 78.4 km
α = -3.5°
We can now write the vector R in magnitude and direction form:
R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))
d) The displacement of the hipo relative to the starting point is the magnitude of vector R calculated in c):
magnitude R = 78. 4 km
The total distance traveled is the sum of the magnitudes of each vector:
Total distance = 42 km + 28 km + 32 km = 102 km
A motorist drives north for 35.0 min at 85.0 km/h and then stops for 15.0 min. He then continues north, traveling 130 km in 2.00 h. (a) what is his total displacement? b) what is his average velocity?
Final answer:
The motorist's total displacement is 179.58 km north, and his average velocity is approximately 63.38 km/h north for the entire trip.
Explanation:
Displacement and Average Velocity Calculation
For the given problem, we can find the motorist's total displacement by adding the distances traveled in each segment of the journey, all in the same direction (north). First, the motorist drives north for 35.0 minutes at 85.0 km/h. The distance for this part of the trip is:
Distance = Speed × Time = 85.0 km/h × (35.0 min / 60 min/h) = 49.58 km (rounded to 2 decimal places)
The next segment involves a stop, so the displacement remains unchanged.
Finally, the motorist drives an additional 130 km north.
Therefore, the total displacement is the sum of these distances: Total Displacement = 49.58 km + 130 km = 179.58 km north.
To find the average velocity, we need to consider the total displacement and the total time, including the stop:
Total time = Driving time + Stopping time = (35.0 min + 2.00 h + 15.0 min) = 2.833 hours (2 hours and 50 minutes).
Average Velocity = Total Displacement / Total Time = 179.58 km / 2.833 h ≈ 63.38 km/h north.
The motorist's total displacement is 179.58 km north, and his average velocity over the entire trip is approximately 63.38 km/h north.
Bragg reflection results in a first-order maximum at 15.0°. In this case, at what angle would the second-order maximum occur?
Answer:
31.174°
Explanation:
Bragg's condition is occur when the wavelength of radiation is comparable with the atomic spacing.
So, Bragg's reflection condition for n order is,
[tex]2dsin\theta=n\lambda[/tex]
Here, n is the order of maxima, [tex]\lambda[/tex] is the wavelength of incident radiation, d is the inter planar spacing.
Now according to the question, first order maxima occur at angle of 15°.
Therefore
[tex]2dsin(15^{\circ})=\lambda\\sin(15^{\circ})=\frac{\lambda}{2d}[/tex]
Now for second order maxima, n=2.
[tex]2dsin\theta=2\times \lambda\\sin\theta=\frac{2\lambda}{2d}[/tex]
Put the values from above conditions
[tex]sin\theta=2\times sin(15^{\circ})\\sin\theta=2\times 0.258819045103\\sin\theta=0.517638090206\\\theta=sin^{-1}0.517638090206\\ \theta=31.174^{\circ}[/tex]
Therefore, the second order maxima occurs at 31.174° angle.
If a lens has a power of -14.50, what is the focal length in mm?
Answer:
Focal length of the lens, f = - 68 mm
Explanation:
Given that,
Power of a lens, P = -14.50 D
We need to find the focal length of the lens. We know that the focal length and the power of lens has inverse relationship. Mathematically, it is given by :
[tex]f=\dfrac{1}{P}[/tex]
f is the focal length of the lens
[tex]f=\dfrac{1}{-14.50}[/tex]
f = -0.068 m
or
f = -68 mm
So, the focal length of the lens is (-68 mm). Hence, this is the required solution.
Considere un campo vectorial U que apunta radialmente hacia afuera de un punto P. La magnitud del campo en un punto Q una distancia r del punto P es a K donde K es una constante positiva. Calcule el flujo de este campo vectorial a través de una superficie esférica de radio R centrada en el punto P
Answer:
El flujo del campo U a través de dicha superficie es [tex]\Phi_U=4\pi r^2K[/tex]
Explanation:
El flujo vectorial de un campo U a través de una superficie S está dado por
[tex]\Phi_U=\int_S \vec{U} \cdot \hat{n} \ dS[/tex]
donde [tex]\hat{n}[/tex] representa el vector unitario normal a la superficie. Teniendo en cuenta que el campo se encuentra en dirección radial, se tiene
[tex]\Phi_U=\int_S K\hat{r} \cdot \hat{r} \ dS = K\int_S dS = KS=K*4\pi r^2[/tex]
Air around the center of surface low pressure systems in the Northern Hemisphere is spinning ______ and _______ the system's center. O counter-clockwise; converging towards O counter-clockwise; diverging away from O clockwise; diverging away from O clockwise; converging towards
Answer:
Clockwise, converging towards
Explanation:
At the center of surface , system with low pressure in Northern hemisphere have air around the center that rotates in anti clockwise direction.
As a result of low pressure, the air is directed slightly inwards thus converges towards the center of the system.
In the high pressure systems, air rotates in clockwise direction and diverges away from the center of the system.
Force is a vector, while mass is a scalar. Why can we use mass as an indicator of the magnitude of the force vector?
Answer:
Explanation:
Force = mass x acceleration
[tex]\overrightarrow{F}=m\overrightarrow{a}[/tex]
Force is always vector and acceleration also vector but the mass is a saclar quanity.
here, the direction of force vector is same as the direction of acceleration vector but the magnitude of force depends on the magnitude of mass of the body.
Is mass is more, force is also more.
Thus, the mass is like an indicator of the magnitude of force.
Which of the following combinations of position (x) and direction of motion would give a velocity in the x-direction that has a negative value? a. Positive x, moving towards the origin.b. Negative x, moving away from the origin.c. Both a and b.d. Positive x, moving away from the origin.e. Negative x, moving towards the origin.
Answer:
c) Both a) and b) are the combinations that have a negative velocity.
Explanation:
The velocity is given by this equation:
v = Δx / Δt
Where
Δx = final position - initial position
Δt = elapsed time
Now let´s evaluate the options. We have to find those combinations in which Δx < 0 since Δt is always positive.
a) If the initial position, x, is positive and you move towards the origin, the final position will be a smaller value than x. Then:
final position < initial position
final position - initial position < 0 The velocity will be negative.
b) If x is negative and you move away from the origin, the final position will be a more negative number than x. Again:
final position < initial position
final position - initial position < 0 The velocity will be negative.
Let´s do an example to show it:
initial position = -5
final position = -10 (since you moved away from the origin)
final position - initial position = -10 -(-5) = -5
d) If x is positive and you move away from the origin, the final position will be a greater value than the initial position. Then:
final position > initial position
final position - initial position > 0 The velocity will be positive.
e) If x is negative and you move towards the origin, the final position will be a greater value than the initial position. Then:
final position > initial position
final position - initial position > 0 The velocity will be positive.
Let´s do an example:
initial position = -10
final position = -5
final postion - initial position = -5 - (-10) = 5
or with final position = 0
final postion - initial position = 0 -(-10) = 10
And so on.
The right answer is c) Both a) and b) are the combinations that have a negative velocity.
A thin metal bar, insulated along its sides, is composed of five different metal connected together. The left end bar is immersed in a heat bath at 100°C and right end in a heat bath at 0°C. Starting at the left end, the pieces and lenghts are steel(2cm), brass(3cm), copper(1cm), aluminum(5cm) and silver(1cm). What is the temperature of the steel/brass interface?
Answer:
T = 61.06 °C
Explanation:
given data:
a thin metal bar consist of 5 different material.
thermal conductivity of ---
K {steel} = 16 Wm^{-1} k^{-1}
K brass = 125 Wm^{-1} k^{-1}
K copper = 401 Wm^{-1} k^{-1}
K aluminium =30Wm^{-1} k^{-1}
K silver = 427 Wm^{-1} k^{-1}
[tex]\frac{d\theta}{dt} = \frac{KA (T_2 -T_1)}{L}[/tex]
WE KNOW THAT
[tex]\frac{l}{KA} = thermal\ resistance[/tex]
total resistance of bar = R steel + R brass + R copper + R aluminium + R silver
[tex]R_{total} =\frac{1}[A} [\frac{0.02}{16} +\frac{0.03}{125} +\frac{0.01}{401} +\frac{0.05}{30} +\frac{0.01}{427}][/tex]
[tex]R_{total} =\frac{1}[A} * 0.00321[/tex]
let T is the temperature at steel/brass interference
[tex]\frac{d\theta}{dt}[/tex] will be constant throughtout the bar
therefore we have
[tex]\frac{100-0}{R_{total}} = \frac{100-T}{R_{steel}}[/tex]
[tex]\frac{100-0}{0.00321} *A = \frac{100-T}{0.00125} *A[/tex]
solving for T we get
T = 61.06 °C
A standard 1 kilogram weight is a cylinder 48.0 mm in height and 55.0 mm in diameter. What is the density of the material?
Answer:8769 Kg/m^3
Explanation: First of all we have to calculate the volume of the cylinder so it is equal to:
Vcylinder= π*r^2*h where r and h are the radius and heigth, respectively.
Vcylinder= π*0.0275^2*0.048=1.14*10^-4 m^3
The density is defined by ρ=mass/volume
the we have
ρ=1Kg/1.14*10^-4 m^3=8769 Kg/m^3
A mechanics shop has a 16,500 N car sitting on a large car lift of 125 cm^2 piston. How much force is needed to be applied to the small piston of area 4.51 cm^2 to lift it?
Answer:595.32 N
Explanation:
Given
Mechanic shop has a car of weight([tex]F_1[/tex]) 16,500 N
Area of lift([tex]A_1[/tex])=[tex]125 cm^2[/tex]
Area of small piston([tex]A_2[/tex])=[tex]4.51 cm^2[/tex]
According to pascal's law pressure transmit will be same in all direction in a closed container
[tex]P_1=P_2[/tex]
[tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]
[tex]F_2=F_1\frac{A_2}{A_1}[/tex]
[tex]F_2=16,500\times \frac{4.51}{125}=595.32 N[/tex]
An apple falls (from rest) from a tree. It hits the ground at a speed of about 4.9 m/s. What is the approximate height (in meters) of the tree above the ground? The magnitude of the gravitational acceleration g = 9.8 m/s2 Enter your answer in meters. Keep 2 decimal places.
Answer:
The inicial height of the apple is 1.22 meters
Explanation:
Using the equation for conservarion of mechanical energy:
[tex]E=V+K=constant[/tex]
[tex]K_i=\frac{1}{2}mv_i^2[/tex] where v is the velocity
[tex]V=mgh[/tex]where h is the height
We equate the initial mechanical energy to the final:
Since [tex]v_0=0\ and h_f=0 [/tex]:
[tex]\frac{1}{2}mv_0^2+mgh_0= \frac{1}{2}mv_f^2+mgh_f\\gh_0= \frac{1}{2}v_f^2[/tex]
Solving for h:
[tex]h_0=\frac{4.9^2}{2g}= 1.22 m[/tex]
Two very small 8.55-g spheres, 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of them.
(a)Disregarding all other forces, how many electrons would you have to add to each sphere so that the two spheres will accelerate at 25.0g when released?
(b)Which way will they accelerate?
Answer:
1.43 x 10¹⁷.
They will accelerate away from each other.
Explanation:
Force on each charged sphere F = mass x acceleration
= 8.55 x 10⁻³ x 25 x 9.8
= 2.095 N
Let Q be the charge on each sphere
F = [tex]\frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}[/tex]
2.095 =[tex]\frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}[/tex]
Q² =[tex]\frac{2.095\times(15)^2\times10^{-4}}{9\times10^9}[/tex]
Q = 2.289 X 10⁻⁶
No of electrons = Charge / charge on a single electron
= [tex]\frac{2.289\times10^{-6}}{1.6\times10^{-19}}[/tex]
=1.43 x 10¹³.
They will accelerate away from each other.
The number of electrons would have to add to each sphere to accelerate at 25.0g is 1.43×10¹³ and accelerate away from each other.
What is electric force?Electric force is the force of attraction of repulsion between two bodies.
According to the Coulombs law, the force of attraction of repulsion between charged two bodies is directly proportional to the product of charges of them and inversely proportional to the square of distance between them. It can be given as,
[tex]F=\dfrac{KQ_1Q_2}{r^2}[/tex]
Here, (k) is the coulombs constant, (q1 and q2) is the charges of two bodies and (r) is the distance between the two charges.
Two very small 8.55-g spheres, 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of them.
(a) The Number of electrons would have to add to each sphere-Two spheres accelerate at 25.0g when released, and the mass of each sphere is 8.55 g. Thus, the force on the sphere can be given as,
[tex]F=8.55\times{10^{-3}}\times25\times9.8\\F=2.095\rm \;N[/tex]
Two very small spheres are 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of them.As the charge on both the sphere is same (say Q) Put these values in the above formula as,
[tex]2.095=\dfrac{(9\times10^{9})QQ}{(15\times10^{-2})^2}\\Q=2.289\times10^{-6}\rm \;C[/tex]
It is known that the charge on one electron is 1.6×10⁻¹⁹ C. Thus the number of electron in the above charge is,
[tex]n=\dfrac{2.289\times10^{-6}}{1.6\times10^{-19}}\\n=1.43\times10^{13}[/tex]
(b)Direction of acceleration-The direction of acceleration of both the sphere will be opposite to each other. Thus, they accelerate away from each other.
Hence, the number of electrons would have to add to each sphere to accelerate at 25.0g is 1.43×10¹³ and accelerate away from each other.
Learn more about the electric force here;
https://brainly.com/question/14372859