Answer:
Option d)
Solution:
As per the question:
Work done by farm hand, [tex]W_{FH} = 972J[/tex]
Force exerted, F' = 310 N
Angle, [tex]\theta = 23^{\circ}[/tex]
Now,
The component of force acting horizontally is F'cos[tex]\theta[/tex]
Also, we know that the work done is the dot or scalar product of force and the displacement in the direction of the force acting on an object.
Thus
[tex]W_{FH} = \vec{F'}.\vec{d}[/tex]
[tex]972 = 310\times dcos23^{\circ}[/tex]
d = 3.406 m = 3.4 m
The distance the wagon travels is 3.4 m. So the correct option is d.
Explanation:To find the distance the wagon travels, we need to use the work-energy principle. Since the force is applied at an angle, we can find the horizontal component of the force by multiplying the force by the cosine of the angle. The work done is equal to the force multiplied by the distance, so we can rearrange the equation to solve for distance. Therefore, the distance the wagon travels is equal to the work done divided by the force component, or 972 J / (310 N * cos(23°)) = 3.4 m.
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If the force on the tympanic membrane (eardrum) increases by about 1.50 above the force from atmospheric pressure, the membrane can be damaged.When you go scuba diving in the ocean, below what depth could damage to your eardrum start to occur? The eardrum is typically 8.20 in diameter. Take the density of seawater to be equal1.03*10^3 kg/m^3.
Damage to the eardrum might start to occur at a depth of approximately [tex]\[ 2.25 \times 10^6 \, \text{m} \][/tex] underwater when scuba diving in the ocean
To calculate the depth at which damage to the eardrum might occur while scuba diving, we need to consider the increase in pressure as you descend underwater. The pressure increases with depth due to the weight of the water above you. This pressure increase is given by the hydrostatic pressure formula:
[tex]\[ P = \rho \cdot g \cdot h \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure,
- [tex]\( \rho \)[/tex] is the density of the fluid (seawater in this case),
- [tex]\( g \)[/tex] is the acceleration due to gravity, and
- [tex]\( h \)[/tex]is the depth.
The force exerted on the eardrum is equal to the pressure difference between the external pressure and the pressure inside the ear canal, multiplied by the area of the eardrum. This can be represented as:
[tex]\[ F = A \cdot (P_{\text{internal}} - P_{\text{external}}) \][/tex]
Given that the force on the eardrum should not exceed 1.50 times the force from atmospheric pressure, we have:
[tex]\[ F_{\text{limit}} = 1.5 \cdot P_{\text{atm}} \cdot A \][/tex]
We can solve for the depth [tex](\( h \))[/tex] when this pressure difference exceeds the threshold.
First, let's find [tex]\( P_{\text{atm}} \)[/tex], the atmospheric pressure at sea level, which is approximately [tex]\( 101.3 \, \text{kPa} \)[/tex]. Then, we can calculate the pressure at depth and find the depth where the pressure difference reaches the limit.
[tex]\[ P_{\text{atm}} = 101.3 \, \text{kPa} \][/tex]
Now, let's calculate the depth at which damage to the eardrum might occur:
[tex]\[ F_{\text{limit}} = 1.5 \cdot P_{\text{atm}} \cdot A \]\[ P_{\text{internal}} = P_{\text{atm}} + \rho \cdot g \cdot h \]\[ F_{\text{limit}} = A \cdot (P_{\text{atm}} + \rho \cdot g \cdot h - P_{\text{external}}) \][/tex]
Given that [tex]\( P_{\text{external}} \)[/tex] is atmospheric pressure, we can rewrite this as:
[tex]\[ F_{\text{limit}} = A \cdot (\rho \cdot g \cdot h) \][/tex]
Now, we can rearrange to solve for [tex]\( h \):[/tex]
[tex]\[ h = \frac{F_{\text{limit}}}{A \cdot \rho \cdot g} \][/tex]
Substitute the values:
[tex]\[ h = \frac{1.5 \cdot P_{\text{atm}}}{A \cdot \rho \cdot g} \]\[ h = \frac{1.5 \cdot 101.3 \, \text{kPa}}{\pi \cdot (8.20 \, \text{mm})^2 \cdot (1.03 \times 10^3 \, \text{kg/m}^3) \cdot (9.81 \, \text{m/s}^2)} \]\[ h \approx \frac{1.5 \times 10^5 \, \text{Pa}}{\pi \times (8.20 \times 10^{-3} \, \text{m})^2 \times (1.03 \times 10^3 \, \text{kg/m}^3) \times (9.81 \, \text{m/s}^2)} \]\[ h \approx \frac{1.5 \times 10^5}{\pi \times 6.69 \times 10^{-5} \times 1.03 \times 10^3 \times 9.81} \][/tex]
[tex]\[ h \approx \frac{1.5 \times 10^5}{6.66 \times 10^{-2}} \]\[ h \approx 2.25 \times 10^6 \, \text{m} \][/tex]
So, damage to the eardrum might start to occur at a depth of approximately [tex]\[ 2.25 \times 10^6 \, \text{m} \][/tex] underwater when scuba diving in the ocean.
Two tiny conducting spheres are identical and carry charges of -19.8μC and +40.7μC. They are separated by a distance of 3.59 cm. (a) What is the magnitude of the force that each sphere experiences? (b) The spheres are brought into contact and then separated to a distance of 3.59 cm. Determine the magnitude of the force that each sphere now experiences.
Answer:
(a): [tex]\rm -5.627\times 10^3\ N.[/tex]
(b): [tex]\rm 7.626\times 10^2\ N.[/tex]
Explanation:
Given:
Charge on one sphere, [tex]\rm q_1 = -19.8\ \mu C = -19.8\times 10^{-6}\ C.[/tex]Charge on second sphere, [tex]\rm q_2 = +40.7\ \mu C = +40.7\times 10^{-6}\ C.[/tex]Separation between the spheres, [tex]\rm r=3.59\ cm = 3.59\times 10^{-2}\ m.[/tex]Part (a):
According to Coulomb's law, the magnitude of the electrostatic force of interaction between two static point charges is given by
[tex]\rm F=k\cdot\dfrac{q_1q_2}{r^2}[/tex]
where,
k is called the Coulomb's constant, whose value is [tex]\rm 9\times 10^9\ Nm^2/C^2.[/tex]
From Newton's third law of motion, both the spheres experience same force.
Therefore, the magnitude of the force that each sphere experiences is given by
[tex]\rm F=k\cdot\dfrac{q_1q_2}{r^2}\\=9\times 10^9\times \dfrac{(-19.8\times 10^{-6})\times (+40.7\times 10^{-6})}{(3.59\times 10^{-2})^2}\\=-5.627\times 10^3\ N.[/tex]
The negative sign shows that the force is attractive in nature.
Part (b):
The spheres are identical in size. When the spheres are brought in contact with each other then the charge on both the spheres redistributes in such a way that the net charge on both the spheres distributed equally on both.
Total charge on both the spheres, [tex]\rm Q=q_1+q_2=-19.8\ \mu C+40.7\ \mu C = 20.9\ \mu C.[/tex]
The new charges on both the spheres are equal and given by
[tex]\rm q_1'=q_2'=\dfrac Q2 = \dfrac{20.9}{2}\ \mu C=10.45\ \mu C = 10.45\times 10^{-6}\ C.[/tex]
The magnitude of the force that each sphere now experiences is given by
[tex]\rm F'=k\cdot \dfrac{q_1'q_2'}{r^2}'\\=9\times 10^9\times \dfrac{10.45\times 10^{-6}\times 10.45\times 10^{-6}}{(3.59\times 10^{-2})^2}\\=7.626\times 10^2\ N.[/tex]
Comment on why the acceleration due to gravity is less for the plastic ball. Why do the other two balls (steel ball and golf ball) not have such a low value for the acceleration?
Answer and Explanation:
The gravitational acceleration 'g' depend directly on the mass of the object or body and inversely on the distance or radius squared:
[tex]g = \frac{Gm_{o}}{R^{2}}[/tex]
where
[tex]m_{o}[/tex] = mass of the object
G = Gravitational constt
Thus the plastic ball is lighter and have low mass as compared to the steel and golf balls.
This is the reason that a plastic ball have a low value of acceleration as compared to that of steel and golf balls with higher values of acceleration.
Final answer:
The acceleration due to gravity is a constant (g) for all objects in the same gravitational field, and any observed difference in falling rates is likely due to air resistance, not gravitational pull. Experiments have confirmed that objects fall at the same rate regardless of their mass or composition, assuming no air resistance.
Explanation:
The acceleration due to gravity should not vary with the material of the object if we ignore effects such as air resistance. This is because, according to Newton's second law, the force acting on an object is the product of its mass and the acceleration (F = ma). Here, the force due to gravity is mg, where m is the mass and g is the acceleration due to gravity. Since a = F/m, the mass cancels out, leaving a = g, which is constant for all objects in the same gravitational field.
The question assumes that the plastic ball has a lower acceleration due to gravity, which contradicts known physics principles. All objects, regardless of their mass or composition, feel the same acceleration due to gravity near the Earth's surface, assuming no other forces, like air resistance, play a significant role. Historical experiments by scientists such as Galileo and Eötvös have confirmed the equality of gravitational acceleration (g) for different substances within exceptionally high precision.
If you observe differing acceleration rates, this can be often attributed to air resistance, not a difference in gravitational pull. Objects with a larger surface area or less aerodynamic shape, like the plastic ball, may experience greater air resistance and thus appear to fall slower, even though their acceleration due to gravity is the same as denser objects like the steel or golf ball.
Assume everyone in the United States consumes one soft drink in an aluminum can every two days. if there are 280 million americans, how many tons of aluminum need to be recycles each year if each can weight 1/15 pound and once ton=2000 pounds?
Answer:
1.708*10^6 tons.
Explanation:
Number of Aluminum cans used by 1 person in 1 year = 365/2=182.5 say it as 183 cans per year.
Total number of people in US= 280,000,000
Total number of cans used by americans.
[tex] = 5.12×10^10 cans[/tex]
Weight of 1 can =1/15 pounds
Weight of all cans used in 1 year
[tex]= \frac{5.12*10^10}{15} =3.41*10^9pounds.[/tex]
we know that
1ton=2000pounds.
[tex]\frac{3.41*10^9 pounds}{2000} = 1.708*10^6 tons.[/tex]
Lightning can be studied with a Van de Graaff generator, which consists of a spherical dome on which charge is continuously deposited by a moving belt. Charge can be added until the electric field at the surface of the dome becomes equal to the dielectric strength of air. Any more charge leaks off in sparks. Assume the dome has a diameter of 43.0 cm and is surrounded by dry air with a "breakdown" electric field of 3.00 10^6 V/m. (a) What is the maximum potential of the dome?
(b) What is the maximum charge on the dome?
The maximum potential is [tex]6.45 \times 10^5\ V[/tex] and the maximum charge is [tex]71 \mu C[/tex].
The electric field is the gradient of the potential. So, the electric field in terms of the potential is given as:
[tex]V=E \times r[/tex]
Here, V is the potential, E is the electric field and r is the distance.
Given:
Diameter of the dome, [tex]d=43\ cm[/tex]
Maximum electric field, [tex]E=3.00 \times10^6\ V/m[/tex]
(a) The maximum potential is computed as:
[tex]V_{max}={E}{d/2}\\V_{max}= {3.00 \times 10^6} \times \frac{0.43}{2}\\V_{max}= 6.45 \times 10^5\ V[/tex]
(b) The maximum charge is computed as:
[tex]Q= \frac{Er}{k}\\Q= \frac{3.00 \times 10^6 \times 0.215}{9 \times 10^9}\\Q=71 \mu C[/tex]
Therefore, the maximum potential is [tex]6.45 \times 10^5\ V[/tex] and the maximum charge is [tex]71 \mu C[/tex].
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The maximum potential of the Van de Graaff generator with a diameter of 43.0 cm is 645 kV, and the maximum charge it can hold, before sparks occur due to air breakdown, is 5.11 μC.
Explanation:The maximum potential of a Van de Graaff generator can be calculated using the formula V = (kQ)/r, where V is the potential, k is Coulomb's constant, Q is the charge, and r is the radius of the sphere. Given that the Van de Graaff generator has a diameter of 43.0 cm (which means the radius r is 21.5 cm or 0.215 m) and it is surrounded by air with a breakdown electric field of 3.00 x 106 V/m, the maximum surface potential (V) equals the breakdown electric field (E) times the radius (r), which gives V = E × r = 3.00 x 106 V/m × 0.215 m = 645,000 V or 645 kV.
The maximum charge on the dome can be calculated using the relationship between the electric field and charge on a sphere, which is E = (kQ)/(r2). Re-arranging this for Q gives Q = Er2/k. Inserting the known values, we get Q = (3.00 x 106 V/m) × (0.215 m)2 / (8.988 x 109 N·m2/C2) = 5.11 x 10−5 C or 5.11 μC, where k is Coulomb's constant.
If a marathon runner averages 9.39 mi/h, how long does it take him or her to run a 26.22-mi marathon? Express your answers in h, min and s.
Answer:
[tex]t=2.8h[/tex]
[tex]t=10080s[/tex]
[tex]t=168 min[/tex]
Explanation:
From this exercise we have velocity and distance. Using the following formula, we can calculate time:
[tex]v=\frac{d}{t}[/tex]
Solving for t
[tex]t=\frac{d}{v}=\frac{26.22mi}{9.39mi/h} =2.8h[/tex]
[tex]t=2.8h*\frac{3600s}{1h} =10080s[/tex]
[tex]t=2.8h*\frac{60min}{1h} =168min[/tex]
A golfer hits a shot to a green that is elevated 3.20 m above the point where the ball is struck. The ball leaves the club at a speed of 18.1 m/s at an angle of 49.0° above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.
Answer:
16.17 m/s
Explanation:
h = 3.2 m
u = 18.1 m/s
Angle of projection, θ = 49°
Let H be the maximum height reached by the ball.
The formula for the maximum height is given by
[tex]H=\frac{u^{2}Sin^{2}\theta }{2g}[/tex]
[tex]H=\frac{18.1^{2}\times Sin^{2}49 }{2\times 9.8}=9.52 m[/tex]
The vertical distance fall down by the ball, h' H - h = 9.52 - 3.2 = 6.32 m
Let v be the velocity of ball with which it strikes the ground.
Use third equation of motion for vertical direction
[tex]v_{y}^{2}=u_{y}^{2}+2gh'[/tex]
here, uy = 0
So,
[tex]v_{y}^{2}=2\times 9.8 \times 6.32[/tex]
vy = 11.13 m/s
vx = u Cos 49 = 18.1 x 0.656 = 11.87 m/s
The resultant velocity is given by
[tex]v=\sqrt{v_{x}^{2}+v_{y}^{2}}[/tex]
[tex]v=\sqrt{11.87^{2}+11.13^{2}}[/tex]
v = 16.27 m/s
Suppose you're on a hot air balloon ride, carrying a buzzer that emits a sound of frequency f. If you accidentally drop the buzzer over the side while the balloon is rising at constant speed, what can you conclude about the sound you hear as the buzzer falls toward the ground?
(A) The frequency and intensity increase
(B) The frequency decrease and intensity increase
(C) The frequency decrease and intensity decrease
(D) The frequency remains the same, but the intensity decreases.
Answer:
(C) The frequency decrease and intensity decrease
Explanation:
The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source, or the wave source is moving relative to the observer, or both.
if the observer and the source move away from each other as is the case for this problem, the wavelength heard by the observer is bigger.
The frequency is the inverse from the wavelength, so the frequency heard will increase.
The sound intensity depends inversely on the area in which the sound propagates. When the buzzer is close, the area is from a small sphere, but as the buzzer moves further away, the wave area will be from a larger sphere and therefore the intensity will decrease.
If the car’s speed decreases at a constant rate from 64 mi/h to 30 mi/h in 3.0 s, what is the magnitude of its acceleration, assuming that it continues to move in a straight line? What distance does the car travel during the braking period?
Answer:[tex]3.874 m/s^2[/tex]
Explanation:
Given
Car speed decreases at a constant rate from 64 mi/h to 30 mi/h
in 3 sec
[tex]60mi/h \approx 26.8224m/s[/tex]
[tex]34mi/h \approx 15.1994 m/s[/tex]
we know acceleration is given by [tex]=\frac{velocity}{Time}[/tex]
[tex]a=\frac{15.1994-26.8224}{3}[/tex]
[tex]a=-3.874 m/s^2[/tex]
negative indicates that it is stopping the car
Distance traveled
[tex]v^2-u^2=2as[/tex]
[tex]\left ( 15.1994\right )^2-\left ( 26.8224\right )^2=2\left ( -3.874\right )s[/tex]
[tex]s=\frac{488.419}{2\times 3.874}[/tex]
s=63.038 m
A stone is thrown vertically upward from ground level at t = 0. At t=2.50 s, it passes the top of a tall building, and 1.50 s later, it reaches its maximum height. What is the height of the tall building? We assume an answer in meters.
Answer:67.45 m
Explanation:
Given
at t=2.5 s it passes the top of a tall building and after 1.5 s it reaches maximum height
let u is the initial velocity of stone
v=u+at
0=u-gt
[tex]u=9.81\times 4=39.24 m/s[/tex]
Let us take h be the height of building
[tex]h=ut+\frac{-1}{2}gt^2[/tex]
[tex]h=39.24\times 2.5-\frac{1}{2}\times 9.81\times 2.5^2[/tex]
h=67.45 m
Consider the following situations:
A car is moving along a straight road at a constant speed.
A car is moving along a straight road while slowing down.
A car is moving along a straight road while speeding up.
A hockey puck slides along a horizontal icy (frictionless) surface.
A hockey puck slides along a rough concrete surface.
A cockroach is speeding up from rest.
A rock is thrown horizontally; air resistance is negligible.
A rock is thrown horizontally; air resistance is substantial.
A rock is dropped vertically; air resistance is negligible.
A rock is dropped vertically; air resistance is substantial.
Part H
Which of these situations describe the motion shown in the motion diagram at point A?
Type the letters corresponding to all the right answers in alphabetical order. Do not use commas. For instance, if you think that only situations C and D are correct, type CD.
Answer:
AD
Explanation:
A motion diagram is used to represent the motion of an object. Each point represents the location of the object after equal intervals of time, and the velocity of the object is represented using arrows.
The motion diagram for this problem has been attached here. The question is asking us to analyze the motion at point A. We can make the following observations for the motion diagram at point A:
- The direction of the arrows does not change --> this means that the object is moving on a straight line
- The length of the arrows remains constant at A --> this means that the object is moving at constant speed
Of all the situations listed in the question, only two of them are compatible with these observations:
A) A car is moving along a straight road at a constant speed.
D) A hockey puck slides along a horizontal icy (frictionless) surface. (Frictionless surface means there is no friction acting on the puck, so the acceleration is zero and therefore the speed is constant).
All the other options involves a non-zero acceleration, so that the velocity is not constant. Therefore, only A and D are the correct options.
A rocket sled for testing equipment under large accelerations starts at rest and accelerates according to the expression a = (3.6 m/s 3 )t + (5.6 m/s 2 ). How far does the sled move in the time interval t = 0 to t = 1.6 s ? Answer in units of m.
Answer:
9.6 m
Explanation:
This is a case of motion under variable acceleration . So no law of motion formula will be applicable here. We shall have to integrate the given equation .
a = 3.6 t + 5.6
d²x / dt² = 3.6 t + 5.6
Integrating on both sides
dx /dt = 3.6 t² / 2 + 5.6 t + c
where c is a constant.
dx /dt = 1.8 t² + 5.6 t + c
when t = 0 , velocity dx /dt is zero
Putting these values in the equation above
0 = 0 +0 + c
c = 0
dx /dt = 1.8 t² + 5.6 t
Again integrating on both sides
x = 1.8 t³ / 3 + 5.6 x t² /2 + c₁
x = 0.6 t³ + 2.8 t² + c₁
when t =0, x = 0
c₁ = 0
x = 0.6 t³ + 2.8 t²
when t = 1.6
x = .6 x 1.6³ + 2.8 x 1.6²
= 2.4576 + 7.168
= 9.6256
9.6 m
The sled moves approximately 16.896 meters from t=0 s to t=1.6 s when it accelerates according to a = (3.6 m/s 3 )t + (5.6 m/s 2 )
Explanation:The calculation for the distance moved by the sled involves using the equations of motion. For a sled that starts from rest and accelerates according to the function a = (3.6 m/s³) t + (5.6 m/s²), we first find the velocity function by integrating the acceleration function. That is, v(t) = ∫a dt = ∫[(3.6 m/s³)t + (5.6 m/s²)] dt = (1.8 m/s²) t² + (5.6 m/s) t + C. Given that the sled starts from rest, the constant C is zero. Therefore the velocity function is v(t) = (1.8 m/s²) t² + (5.6 m/s) t.
Next, we find the displacement function by integrating the velocity function. That is, x(t) = ∫v dt = ∫[(1.8 m/s²) t² + (5.6 m/s) t] dt = (0.6 m/s) t³ + (2.8 m/s) t² + D. Given the initial condition that the sled starts from rest, the constant D is also zero. So the displacement function is x(t) = (0.6 m/s) t³ + (2.8 m/s) t².
Finally, substituting t = 1.6 s into the displacement function will yield the distance moved by the sled in the time interval from t = 0 to t = 1.6 s, which comes out to be x(1.6 s) = (0.6 m/s) (1.6 s)³ + (2.8 m/s) (1.6 s)² = 9.728 m + 7.168 m = 16.896 m. So, the sled moves approximately 16.896 m.
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Two objects attract each other with a gravitational force
ofmagnitude 1.00 x 10-8 N when separated by 20.0 cm.
Ifthe total mass of the two objects is 5.00 kg, what is the mass
ofeach?
Answer:
mass of bigger object M= 3 kg and mass of smaller object m= 2 kg
Explanation:
force between two objects F= 1.00×10^-8 N
distance between them R= 20 cm = 0.2 m
total mass M+m= 5 kg
We know that
[tex]F= G\frac{Mm}{R^2}[/tex]
putting the values we get
[tex]10^{-8}= 6.67\times10^{-11}\frac{M\times m}{0.2^2}[/tex]
M×m= 6
and M+m= 5
therefore we can calculate that M= 3 kg and m= 2 kg
mass of bigger object M= 3 kg and mass of smaller object m= 2 kg
A person drives a car around a circular cloverleaf with a radius of 63 m at a uniform speed of 10 m/s. (a) What is the acceleration of the car? (b) Compare this answer with the acceleration due to gravity as a percentage?
Answer:
acceleration is 1.59 m/s²
required % is 16.22 %
Explanation:
given data
radius = 63 m
speed = 10 m/s
to find out
acceleration of the car and acceleration due to gravity
solution
we know car is moving in circular path
so acceleration will be, a = [tex]\frac{v^2}{r}[/tex] .........1
put here value
a = [tex]\frac{10^2}{63}[/tex]
a = 1.59 m/s²
so acceleration is 1.59 m/s²
so acceleration due to gravity as a percentage is 9.8 m/s²
required % = [tex]\frac{1.59}{9.8}*100[/tex]
required % = 16.22 %
Recent findings on the topic of brain based research indicate all of the following except
Recent findings on the topic of brain-based research indicate all of the following except
A. early environments matter.
B. all children are born ready to learn.
C. society isn't addressing the needs of young children. ?
D. the brain stops growing at around age two
A charge Q= 2 C is distributed uniformly through out a bar of length L=2.5 m. The bar is placed horizontally in free space. A second charge q = 10−9C is placed along the line of the bar a distance d= 2m away in space, measured from the right end of the bar. What force is exerted on charge q by the charged bar?
Answer:
F = 2 N
Explanation:
Let the rod is made up of large number of point charges
so here force due to one small part which is at distance "x" from the end of the rod on the given charge is
[tex]dF = \frac{kdq q}{(d + x)^2}[/tex]
here we know that
[tex]dq = \frac{Q}{L} dx[/tex]
so we have
[tex]F = \int \frac{k Q q dx}{L(d + x)^2}[/tex]
[tex]F = \int_0^L \frac{kQq dx}{L(d + x)^2}[/tex]
[tex]F = \frac{kQq}{L} ( \frac{1}{d} - \frac{1}{d + L})[/tex]
now plug in all data
[tex]F = \frac{(9\times 10^9)(2C)(10^{-9})}{2.5} (\frac{1}{2} - \frac{1}{(2 + 2.5)})[/tex]
[tex]F = 2N[/tex]
A spelunker is surveying a cave. She follows a passage 180 m straight west, then 230 m in a direction 45° east of south, and then 280 m at 30° east of north. After a fourth unmeasured displacement, she finds herself back where she started. A: Use a scale drawing to determine the magnitude of the fourth displacement. Express your answer using two significant figures.
B: Determine the direction of the fourth displacement. Express your answer using two significant figures.
The problem requires vector operations in two dimensions. Displacement is broken down into x and y-components which follow the east-west and north-south directions respectively. Total displacement being zero means the sum of the x and y components of displacement will also be zero. The fourth displacement is determined by negating the total x and y components of the first three displacements. The magnitude and direction are then obtained using Pythagorean theorem and arctan function respectively.
Explanation:To solve this problem, we need to deal with the changes in displacement in terms of vector operations. Given the different directions, we need to break down the vectors into their x and y components, where x represents east-west direction, and y north-south direction. Since our spelunker starts and ends in the same place, the sum of the displacements in each dimension will also be zero.
For displacement 1, moving 180m west, the x component would be -180, and y component would be 0. For displacement 2, moving 230m in a direction 45° east of south, the x would be -230×sin(45) and y would be -230×cos(45). For displacement 3, moving 280m at 30° east of north, the x would be 280×cos(30) and y would be 280×sin(30).
To determine the fourth displacement, we sum up the x and y components for displacement 1,2 and 3 and then negate them to get the x and y component of the fourth displacement. We then use the Pythagorean theorem to calculate the magnitude of the 4th displacement which is square root of (sum x² + sum y²). The direction can be obtained by calculating the arctan of the total y component / total x component.
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A student aims a cork gun at a small dot on the wall. The wall is 10m from the students, and 8m above the level of the gun. The gun can launch corks with a speed of 12m/s. How high must the gun be aimed to hit the spot?
Answer:[tex]\theta =57.99\approx 58 ^{\circ}[/tex]
Explanation:
Given
wall is 10 m from student
8 m high
Therefore student should launch at an angle \theta such that its maximum height is 8 m and its range is 20 m
For maximum height(h)[tex]=\frac{u^2sin^2\theta }{2g}[/tex]
Range (R)[tex]=\frac{u^2sin2\theta }{g}[/tex]
[tex]8=\frac{12^2sin^2\theta }{2g}-----1[/tex]
[tex]20=\frac{12^2sin2\theta }{g}----2[/tex]
divide 1 & 2
[tex]\frac{20}{8}=\frac{2sin2\theta }{sin^2\theta }[/tex]
[tex]\frac{5}{2}=\frac{2\times 2\times sin\theat \cdot cos\theta }{sin^2\theta }[/tex]
[tex]tan\theta =\frac{8}{5}[/tex]
[tex]\theta =57.99\approx 58 ^{\circ}[/tex]
A student witnesses a flash of lightning and then t=2.5s later the student hears assiciated clap of thunder. (please show work) (Part A) Sound travels at 343 m?s in the air, What distance from the student is the lightning strikes, in meters? (Part B) Light travels at 3.0x108 m/s in the air. How long, t1, in seconds did it take the light to reach the student's eyes after the flash?
Answer:
857.5 m
2.8583×10⁻⁶ seconds
Explanation:
Time taken by the sound of the thunder to reach the student = 2.5 s
Speed of sound in air is 343 m/s
Speed of light is 3×10⁸ m/s
Distance travelled by the sound = Time taken by the sound × Speed of sound in air
⇒Distance travelled by the sound = 2.5×343 = 857.5 m
⇒Distance travelled by the sound = 857.5 m
Time taken by light = Distance the light travelled / Speed of light
[tex]\text{Time taken by light}=\frac{857.5}{3\times 10^8}\\\Rightarrow \text{Time taken by light}=2.8583\times 10^{-6}[/tex]
Time taken by light = 2.8583×10⁻⁶ seconds
We want to find the distance between the student and the lighting given that we know the time that passes between the moment he sees the lighting and the moment he hears it.
The solutions are:
a) D = 857.501 mb) t₁ = 2.858*10^(-6) sNow, notice that there is an intrinsic problem with this question.
The student sees the "flash" when the light impacts on his eyes. Thus, the chain of events is:
Lighting falls.Light travels and impacts the student's eyes, at this moment he sees the flash of light, this is what we consider the t = 0.2.5 seconds after, sound arrives.So there is a little time between when the lighting happens and the student sees it, that will also have an impact on the distance to the lighting.
then we have:
(3*10^8 m/s)*t₁ = D
(343m/s)*t₂ = D
Where t₁ is the time that the light needs to reach the student, t₂ is the time the sound needs to reach the student, and D is the distance between the student and the ligthing.
We also know that:
t₂ - t₁ = 2.5s
Then we can write:
t₂ = t₁ + 2.5s and replace it on the second equation:
(343m/s)*(t₁ + 2.5s) = D
Then we have a system of equations:
(3*10^8 m/s)*t₁ = D
(343m/s)*(t₁ + 2.5s) = D
We can see that both of these left sides are equal to the same thing, then we must have:
(3*10^8 m/s)*t₁ = (343m/s)*(t₁ + 2.5s)
Then we solve it for t₁
t₁*(3*10^8 m/s - 343m/s) = (343m/s)*2.5s = 857.5m
t₁ = (857.5m)/(3*10^8 m/s - 343m/s) = 2.858*10^(-6) s
Now that we know the value of t₁, we can find the value of D.
(3*10^8 m/s)*t₁ = D
(3*10^8 m/s)* 2.858*10^(-6) s = D = 857.501 m
Then the solutions are:
a) D = 857.501 mb) t₁ = 2.858*10^(-6) sIf you want to learn more, you can read:
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The alternator in a car consists of a rectangular coil, with 250 turns of wire and area 0.01 m^2, rotating in a 0.1 T magnetic field. (The field is produced by a direct-current electromagnet.) If the rotation rate is 10^3 rpm, what is the peak output voltage?
Answer:
26.17 V
Explanation:
In a alternator , an ac voltage is produced whose magnitude of voltage varies sinusoidally.
Maximum voltage induced = nBAω
where n is no of turns , B is magnetic field , A is area of the coil and ω is angular velocity .
Rate of rotation n = 1000rpm = 1000/60 rps = 16.67 rps
angular velocity ω = 2π n = 2π x 16.67 = 104.68 rad / s
Putting all values in the given equation above
Max voltage = 250 x .1 x .01 x 104.68
= 26.17 V
Which of the following is not a unit of torque? O pound-foot 0 kilogram-newton Newton-meter O pound-inch
Answer: Kilogram- newton is wrong unit for Torque
Idem pound-inch is also wrong unit for Torque
Explanation: As it is well known the torque is defined as :
Τorque= F x R
so its UNITS are Newton*meter (SI)
or in the Imperial System is often use Pound-foot
A galvanometer has an internal resistance of 100 Ω and deflects full-scale at 2.00 mA. What size resistor should be added to the galvanometer to convert it to a milliammeter capable of reading up to 4.00 mA, and how should this resistor be connected to the galvanometer?
Answer
The resistor has to be 100
Explanation:
We will have to use the Current Divider Rule, that rule states:
[tex]Ig=\frac{Req}{Rg}*(It)\\[/tex]
where:
Ig= Galvanometer current
It= Total current
Rg= Galvanometer Resistor
Req= Equivalent circuit resistor
For the case of two resistor in parallel:
[tex]Req=\frac{R1*Rg}{R1+Rg}[/tex]
now:
[tex]Req=\frac{2mA}{4mA}*100\\[/tex]
Req=50Ω
having the Equivalent resistor we can calculate R1 reformulating the Req formula:
[tex]R1=\frac{(Rg*Req)}{Rg-Req}\\[/tex]
R1=100 Ω
So now when a 4mA current flows into the new circuit, 2mA will go through the Galvanometer deflecting the full scale.
To convert the galvanometer to a milliammeter capable of reading up to 4.00 mA, add a 100 Ω shunt resistor in parallel.
To convert a galvanometer with an internal resistance of 100 Ω and a full-scale deflection of 2.00 mA into a milliammeter capable of reading up to 4.00 mA, you need to add a shunt resistor parallel to the galvanometer.
The shunt resistor will divert the excess current, allowing the galvanometer to properly read the increased current range.
Given values:Calculate the voltage corresponding to the galvanometer's full-scale deflection:
V = I * RV = 2.00 mA * 100 Ω V = 0.2 V.Determine the current that will flow through the shunt resistor:
4.00 mA - 2.00 mA = 2.00 mA.Calculate the resistance of the shunt resistor (Rsh) using Ohm's law:
Rsh = V / I = 0.2 V / 2.00 mA = 100 Ω.Therefore, a resistor of 100 Ω should be connected in parallel to the galvanometer to enable it to read up to 4.00 mA.
What is the electric flux through one side of a cube that has a single point charge of-2.80μC placed at its center? Hint: You do not need to integrate any equations to get the answer
Answer:
The flux through one face of a cube is [tex]- 5.2730\times 10^{4} Wb[/tex]
Given:
Charge, Q = [tex]- 2.80\micro C = -2.80\times 10^{- 6} C[/tex]
Solution:
The electric flux, [tex]\phi_{E} = \frac{Q}{\epsilon_{o}}[/tex]
Since, there are 6 faces in a cube, the flux through one face:
[tex]\phi_{E} = \frac{Q}{6\epsilon_{o}}[/tex]
[tex]\phi_{E} = \frac{-2.80\times 10^{- 6}}{6\epsilon_{o}}[/tex]
[tex]\phi_{E} = - 5.2730\times 10^{4} Wb[/tex]
A quantity of 14.1 cm^3 of water at 8.4°C is placed in a freezer compartment and allowed to freeze to solid ice at -7.2°C. How many joules of energy must be withdrawn from the water by the refrigerator?
Answer:920.31 J
Explanation:
Given
Volume of water (V)[tex]=14.1 cm^3 [/tex]
mass(m)[tex]=\rho \times V=1000\times 14.1\times 10^{-6}=14.1 gm[/tex]
Temperature [tex]=8.4^{\circ} C[/tex]
Final Temperature [tex]=-7.2 ^{\circ}C[/tex]
specific heat of water(c)[tex]=4.184 J/g-^{\circ}C[/tex]
Therefore heat required to removed is
[tex]Q=mc(\Delta T)[/tex]
[tex]Q=14.1\times 4.184\times (8.4-(-7.2))[/tex]
[tex]Q=920.31 J[/tex]
A certain elevator cab has a total run of 218 m and a maximum speed is 319 m/min, and it accelerates from rest and then back to rest at 1.20 m/s^2. (a) How far does the cab move while accelerating to full speed from rest? (b) How long does it take to make the nonstop 218 m run, starting and ending at rest?
Answer:
a)11.6m
b)45.55s
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)\\\\
{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\
X=Xo+ VoT+0.5at^{2} (3)\\
X=(Vf+Vo)T/2 (4)
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
a)
for this problem
Vo=0
Vf=319m/min=5.3m/s
a=1.2m/s^2
we can use the ecuation number 1 to calculate the time
t=(Vf-Vo)/a
t=(5.3-0)/1.2=4.4s
then we use the ecuation number 3 to calculate the distance
X=0.5at^2
X=0.5x1.2x4.4^2=11.6m
b)second part
We know that when the elevator starts to accelerate and decelerate, it takes a distance of 11.6m and a time of 4.4s, which means that if the distance is subtracted 2 times this distance (once for acceleration and once for deceleration)
we will have the distance traveled in with constant speed.
With this information we will find the time, and then we will add it with the time it takes for the elevator to accelerate and decelerate
X=218-11.6x2=194.8m
X=VT
T=X/v
t=194.8/5.3=36.75s
Total time=36.75+2x4.4=45.55s
A sphere of radius r = 5cm carries a uniform volume charge density rho = 400 nC/m^3. Q. What is the total charge Q of the sphere?
Answer:
The total charge Q of the sphere is [tex]2.094\times10^{-10}\ C[/tex].
Explanation:
Given that,
Radius = 5 cm
Charge density [tex]J= 400\ nC/m^3[/tex]
We need to calculate the total charge Q of the sphere
Using formula of charge
[tex]q=\rho V[/tex]
Where, [tex]\rho[/tex] = charge density
V = volume
Put the value into the formula
[tex]q=\rho\times(\dfrac{4}{3}\pi r^3)[/tex]
Put the value into the formula
[tex]q=\dfrac{4}{3}\times\pi\times400\times10^{-9}\times(5\times10^{-2})^3[/tex]
[tex]q=2.094\times10^{-10}\ C[/tex]
Hence, The total charge Q of the sphere is [tex]2.094\times10^{-10}\ C[/tex].
The atomic radii of a divalent cation and a monovalent anion are 0.19 nm and 0.126 nm, respectively. (a) Calculate the force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another). Enter your answer for part (a) in accordance to the question statement N (b) What is the force of repulsion at this same separation distance
Final answer:
The force of attraction between the divalent cation and monovalent anion can be calculated using Coulomb's law. Plugging the values into the formula gives a force of -1.742 N.
Explanation:
The force of attraction between two ions can be calculated using Coulomb's law. The formula is given by F = k * (Q1 * Q2) / r^2, where F is the force, k is the Coulomb constant (9 x 10^9 Nm^2/C^2), Q1 and Q2 are the charges of the ions, and r is the distance between their centers. In this case, since the anion is monovalent and the cation is divalent, the charges would be -1 and +2 respectively.
To calculate the force of attraction, we need to find the equilibrium interionic separation. Given that the atomic radii of the divalent cation and monovalent anion are 0.19 nm and 0.126 nm respectively, their total separation would be the sum of their radii, which is 0.316 nm.
Plugging these values into the formula, we get:
F = (9 x 10^9 Nm^2/C^2) * (-1 C) * (+2 C) / (0.316 x 10^-9)^2 = -1.742 N
What must the charge (sign and magnitude) of a particle of mass 1.47 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 680 N/C ? Use 9.81 m/ s 2 for the magnitude of the acceleration due to gravity. View Available Hint(s) nothing C C Submit Part B What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight? Use 1.67×10−27 kg for the mass of a proton, 1.60×10−19 C for the magnitude of the charge on an electron, and 9.81 m/ s 2 for the magnitude of the acceleration due to gravity. View Available Hint(s) nothing N/C N / C Submit Provide Feedback Next
Answer:
Part a)
[tex]q = -21.2 \times 10^{-6} C[/tex]
Part b)
[tex]E = 1.02 \times 10^{-7} N/C[/tex]
Explanation:
Part a)
As we know that charge and its sign that remains in equilibrium is under gravity must be such that it will balance the gravitational force by electric force
so here we have
[tex]mg = qE[/tex]
[tex]1.47 \times 10^{-3} (9.81) = q(680)[/tex]
[tex]q = 21.2 \times 10^{-6} C[/tex]
and its sign must be negative so that it will have upward electric force
so it is
[tex]q = -21.2 \times 10^{-6} C[/tex]
Part b)
as we know that weight of proton is balanced by electric force
so we will have
[tex]qE = mg[/tex]
[tex](1.6 \times 10^{-19})E = (1.67 \times 10^{-27})(9.81)[/tex]
[tex]E = 1.02 \times 10^{-7} N/C[/tex]
A. The charge (sign and magnitude) of the particle of mass 1.47 g is –2.12×10¯⁵ C
B. The magnitude of the electric field in which the electric force on the proton is equal in magnitude to its weight is 1.02×10¯⁷ N/C
What is electric field strength?This is simply defined as the force per unit charge. Mathematically, it is expressed as:
E = F / Q
Where
E is the electric field strength F is the force Q is the charge A. How to determine the charge Mass (m) = 1.47 g = 1.47 / 1000 = 1.47×10¯³ KgElectric field strength (E) = 680 N/CAcceleration due to gravity (g) = 9.81 m/s²Charge (Q) =?E = F/Q
E = mg / Q
680 = (1.47×10¯³ × 9.81) / Q
Cross multiply
680 × Q = (1.47×10¯³ × 9.81)
Divide both side by 680
Q = (1.47×10¯³ × 9.81) / 680
Q = 2.12×10¯⁵ C
Q = –2.12×10¯⁵ C (since it is stationary)
B. How to determine the electric field strength Mass (m) = 1.67×10¯²⁷ KgAcceleration due to gravity (g) = 9.81 m/s²Charge (Q) = 1.60×10¯¹⁹ CElectric field strength (E) =?E = mg / Q
E = (1.67×10¯²⁷ × 9.81) / 1.60×10¯¹⁹
E = 1.02×10¯⁷ N/C
Learn more about Coulomb's law:
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A mass m = 550 g is hung from a spring with spring constant k = 2.8 N/m and set into oscillation at time t = 0. A second, identical mass and spring next to the first set is also set into motion. At what time t should the second system be set into motion so that the phase difference in oscillations between the two systems is pi/2?
Answer:
The second system must be set in motion [tex]t=0.70s[/tex] seconds later
Explanation:
The oscillation time, T, for a mass, m, attached to spring with Hooke's constant, k, is:
[tex]T=2\pi\sqrt(\frac{m}{k} )[/tex]
One oscillation takes T secondes, and that is equivalent to a 2π phase. Then, a difference phase of π/2=2π/4, is equivalent to a time t=T/4.
If the phase difference π/2 of the second system relative to the first oscillator. The second system must be set in motion [tex]t=\frac{\pi}{2}\sqrt(\frac{m}{k})=\frac{\pi}{2}\sqrt(\frac{0.55}{2.8}= 0.70s)[/tex] seconds later
An elevator ascends from the ground with uniform speed. A time Tį later, a boy drops a marble through a hole in the floor. A time T2 after that (i.e. Ti +T2 after start) the marble hits the ground. Find an expression for the height of the elevator at time Ti. (Local gravity is g.) What checks can you make?
Answer:
The height of the elevator at [tex]T_{i}[/tex] is [tex]\frac{gT_{2}^{2}}{2(1 + \frac{T_{2}}{T_{i}})}[/tex]
Solution:
As per the question:
Let us assume:
The velocity with which the elevator ascends be u'
The height attained by the elevator at time, [tex]T_{i}[/tex] be h
Thus
[tex]u' = \frac{h}{T_{i}}[/tex] (1)_
Now, with the help of eqn (2) of motion, we can write:
[tex]h = - u'T_{2} + \frac{1}{2}gT_{2}^{2}[/tex]
Using eqn (1):
[tex]h = - \frac{h}{T_{i}}T_{2} + \frac{1}{2}gT_{2}^{2}[/tex]
[tex]h + \frac{h}{T_{i}}T_{2} = \frac{1}{2}gT_{2}^{2}[/tex]
[tex]h(1 + \frac{h}{T_{i}}T_{2}) = \frac{1}{2}gT_{2}^{2}[/tex]
[tex]h = \frac{gT_{2}^{2}}{2(1 + \frac{T_{2}}{T_{1}})}[/tex]