Answer:
a) 1 & 2The pressure at the surface of water and just outside the hole are both atmospheric pressure. { mean value of 101,325 pascals (roughly 14.6959 pounds per square inch).}
3. P(bottom) = 2987.04 Pa
b) v = 1.0844 m/s
c) v₁ = 0.11 m/s
Explanation:
a)
1) The pressure at the surface of water is same as the atmospheric pressure { mean value of 101,325 pascals (roughly 14.6959 pounds per square inch).}
2) The pressure at just outside the hole is also same as the atmospheric pressure. { mean value of 101,325 pascals (roughly 14.6959 pounds per square inch).}
3) At the bottom of the container, the gauge pressure is given by:
P(bottom) = pgh
where h = 12in = 0.3084m
P(bottom) = 1000 kg/m³ × 9.8 m/s² × 0.3084m
P(bottom) = 2987.04 Pa
b)
The velocity at which the water leaves the hole can be obtained from Bernouilli's equation. Point 1 is at the surface of water, point 2 is just outside the hole;
p₁ + 1/2pv₁² + pgh₁ = p² + 1/2pv₂² + pgh₂
Pressures on both sides are the same and the velocity of the water at the surface is approximately zero (since the hole's cross sectional area is much smaller than the container's). The difference in depths h1 - h2 is just the height of the water level
so
pgh = 1/2pv₂²
v = √(2gh)
we substitute
v = √( 2 × 9.8 m/s² × 0.06 m
v = 1.0844 m/s
c)
Now the velocity of the water level can't be neglected, we can use the continuity equation:
v₂ = (A₁/A²)V₁
so Back to Bernouilli's:
1/2pv₁² + pgh₁ = 1/2 (10v₁)² + pgh₂
phg = 99/2pv₁²
v₁ = √(2/99gh)
v₁ = √(2/99 × 9.8 m/s² × 0.06 m)
v₁ = 0.11 m/s
Exercise 2.4.5: Suppose we add possible friction to Exercise 2.4.4. Further, suppose you do not know the spring constant, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz, for the 2 kg weight you measured 0.8 Hz. a) Find k (spring constant) and c (damping constant). Find a formula for the mass in terms of the frequency in Hz. Note that there may be more than one possible mass for a given frequency. b) For an unknown object you measured 0.2 Hz, what is the mass of the object
Answer:
a). C = b/2 and C = b/4
b). [tex]$ \therefore T = 2 \pi \sqrt{\frac{m_1 +m_2}{k (m_1 + m_2)}} = 2 \pi \sqrt{ \mu/k}$[/tex]
c). m = 63.4 kg (approx.)
Explanation:
Ex. 2.4.4
The total force acting on mass m is [tex]$ F = F_{spring }= -kx $[/tex] , where x is the displacement from the equilibrium position.
The equation of motion is [tex]$ m {\overset{..}x} + kx = 0 $[/tex]
or [tex]$ {\overset{..}x}+ \frac{k}{m}x=0 $[/tex] or [tex]$ {\overset{..}x} + w_0^2 x = 0 $[/tex] , where [tex]$ w_0 = \sqrt{\frac{k}{m}} $[/tex]
The solution is [tex]$ x = A \cos (w_0t + \phi) $[/tex] , where A and Ф are constants.
A is amplitude of motion
[tex]$ w_0$[/tex] is the angular frequency of motion
Ф is the phase angle.
Now, [tex]$ w_0 = 2 \pi f_0 = \sqrt{k/m} $[/tex]
or [tex]$ m = \frac{k}{4\pi f_0^2} $[/tex]
Given [tex]$ f_0 = 0.8 Hz , k = 4 N/m $[/tex]
a). [tex]$ m = \frac{4}{4(3.14)^2(0.8)^2} = 0.158\ kg$[/tex]
b). [tex]$ w_0^2 = k/m $[/tex]
or [tex]$ m = k/ w_0^2 = k / (2\pi f_0)^2 = k / 4 \pi^2 f_0^2 $[/tex]
Ex. 2.4.5
a). Total force acting on the mass m is [tex]$F = F_{spring}+f $[/tex]
[tex]$ = -kx-bv $[/tex]
The equation of motion is [tex]$ m {\overset{..}x}= -kx-b{\overset{.}x} $[/tex]
or [tex]$ w_0 = \sqrt{\frac{k}{m}} $[/tex] , angular frequency of the undamped oscillation.
γ = b/2m is called the damping coefficient (γ=C)
[tex]$ k = m w_0^2 = 4 \pi^2 m f_0^2 $[/tex]
for 1 kg weight (= 9.8 N), [tex]$ f_0$[/tex] = 1.1 Hz
k = 4 x (3.14)² x (9.8) x 1.1² = 4.6 x 10² N/m
For 2 kg weight (= 19.6 N), [tex]$ f_0$[/tex] = 0.8 Hz
k = 4 x 9.8596 x 2 x 9.8 x 0.8² = 5 x [tex]$ 10^7$[/tex] N/m
[tex]$ \gamma = \frac{b}{2m_1} = \frac{b}{2m_2} $[/tex]
or [tex]$ \gamma = \frac{b}{2 \times 1} = \frac{b}{2 \times 2} $[/tex]
γ = b/2 (for 1 kg) and γ = b/4 (for 2 kg)
C = b/2 and C = b/4
b). [tex]$ w_0^2 = \frac{k}{m} \Rightarrow \frac{k}{w_0^2} = \frac{k}{(2 \pi f_0)^2} = \frac{k}{4 \pi^2 f_0^2} $[/tex]
For two particle problem,
[tex]$ w'_0^2 = \sqrt{\frac{k(m_1+m_2)}{m_1 +m_2}} $[/tex]
[tex]$ \therefore T = 2 \pi \sqrt{\frac{m_1 +m_2}{k (m_1 + m_2)}} = 2 \pi \sqrt{ \mu/k}$[/tex]
where, μ is the reduced mass.
This time period is same for both the particles.
c). [tex]$ m =\frac{k}{4 \pi^2 f_0^2}$[/tex]
[tex]$ = \frac{5 \times 10^2}{4 \times 9.14^2 \times 0.2} = 63.4\ kg $[/tex] ( approx.)
A rectangular object was found to have a mass of 1.278 kg and density of 4.98 g/cm3. Suppose that you knew that the length was 47 mm and the width was 61 mm. Using this information, compute the height of the rectangle in cm.
Answer:
89.6 cm
Explanation:
From the question,
Volume of the rectangular object = Mass/Density.
V = m/D.................. Equation 1
Given: m = 1.278 kg, D = 4.98 g/cm³ = 4980 kg/m³
Substitute into equation 1
V = 1.278/4980
V = 2.57×10⁻⁴ m³.
But,
V = lwh............... Equation 2
Where l = length of the rectangular object, w = width of the rectangular object, h = height of the rectangular object.
make h the subject of the equation
h = V/lw........... Equation 3
Given: V = 2.57×10⁻⁴ m³, l = 0.047 m, w = 0.061 m.
Substitute into equation 3
h = 2.57×10⁻⁴/(0.047×0.061)
h = 0.896 m
h = 89.6 cm
a torque of 6Nm is required to accelerate a wheel from rest to 7.5rev/s In a time of 5.0s.find the moment of inertia of the wheel
Your question has been heard loud and clear
Torque = 6Nm
Angular acceleration= 7.5 revolutions / second
Moment of inertia = ?
Torque = Moment of inertia * Angular acceleration
Therefore , moment of intertia = Torque / Angular acceleration
Moment of inertia = 6 / 7.5 = 0.8 Kgm^2
Moment of inertia of wheel = 0.8 Kgm^2
Thankyou
An object floats in water with 58 of its volume submerged. The ratio of the density of the object to that of water is
Complete Question
An object floats in water with 5/8 of its volume submerged. The ratio of the density of the object to that of water is:
(a) 8/5
(b) 1/2
(c) 3/8
(d) 5/8
(e) 2/1
Answer:
The correct option is d
Explanation:
From the question we are told that
The ratio of the volume of the object submerged to the total volume of the object is [tex]\frac{V_w}{V_o} = \frac{5}{8}[/tex]
Generally the buoyancy force acting on the object is equal to the weight of the water displaced and this is mathematically represented as
[tex]F_b = W[/tex]
Now the mass of the water displaced is mathematically represented as
[tex]m_w = \rho_w * V_w[/tex]
While the mass of the object is mathematically represented as
[tex]m_o = \rho_o * V_o[/tex]
So
[tex]F_b = W \ \equiv \ \rho * V_o * g = \rho * V_w * g[/tex]
=> [tex]\frac{V_w}{V_o} = \frac{\rho_o}{\rho_w}[/tex]
From the question that it volume of the water displace (equivalent to the volume of the object in water ) to the volume of the total object is
[tex]\frac{V_w}{V_o} = \frac{5}{8}[/tex]
So
[tex]\frac{\rho_o}{\rho_w} = \frac{5}{8}[/tex]
A point on the string of a violin moves up and down in simple harmonic motion with an amplitude of 1.24 mm and frequency of 875 Hz. a) what is the max speed of that point in SI units? b) what is the max acceleration of the point in SI units?
Using
V = Amplitude x angular frequency(omega)
But omega= 2πf
= 2πx875
=5498.5rad/s
So v= 1.25mm x 5498.5
= 6.82m/s
B. .Acceleration is omega² x radius= 104ms²
Answer:
a
[tex]v _{max } = 6.82 \ m/s[/tex]
b
[tex]a_{max} = 37489.5 \ m/s^2[/tex]
Explanation:
From the question we are told that
The amplitude is [tex]A = 1.24 \ mm = 1.24 * 10^{-3} \ m[/tex]
The frequency is [tex]f = 875 \ Hz[/tex]
Generally the maximum speed is mathematically represented as
[tex]v _{max } = A * 2 * \pi * f[/tex]
=> [tex]v _{max } = 1.24*10^{-3} * 2 * 3.142 * 875[/tex]
=> [tex]v _{max } = 6.82 \ m/s[/tex]
Generally the maximum acceleration is mathematically represented as
[tex]a_{max} = A * (2 * \pi * f)[/tex]
=> [tex]a_{max} = 1.24*10^{-3} * (2 * 3.142 * 875 )^2[/tex]
=> [tex]a_{max} = 37489.5 \ m/s^2[/tex]
an electric field of magnitude 200 N/C in the positive x- direction. calculate the acceleration in (m/s^2) of a charged particle of mass 1g and charge 1mC that is released from rest in this field?
1/200 that should get your
answer
¿Qué resistencia debe ser conectada en paralelo con una de 20 Ω para hacer una
resistencia combinada de 15 Ω?
Answer:
La resistencia que debe ser conectada en paralelo con una de 20 Ω para hacer una resistencia combinada de 15 Ω tiene un valor de 60 Ω
Explanation:
Las resistencias son aquellos dispositivos en los circuitos eléctricos que suelen emplearse para oponerse al paso de la corriente eléctrica.
Se denomina resistencia resultante o equivalente al valor de la resistencia que se obtiene al considerar un conjunto de ellas.
Cuando tenes resistencias en paralelo la corriente se divide y circula por varios caminos.
La resistencia equivalente de un circuito de resistencias en paralelo es igual al recíproco de la suma de los inversos de las resistencias individuales:
[tex]R=\frac{1}{\frac{1}{R_{1} } +\frac{1}{R_{2} } +...+\frac{1}{R_{N} }}[/tex]
Esto también puede ser expresado como:
[tex]\frac{1}{R} =\frac{1}{R_{1} } +\frac{1}{R_{2} } +...+\frac{1}{R_{N} }[/tex]
Entonces, en este caso sabes:
R= 15 ΩR1= 20 ΩR2=?Reemplazando:
[tex]\frac{1}{15} =\frac{1}{20}+\frac{1}{R2}[/tex]
y resolviendo:
[tex]\frac{1}{R2} =\frac{1}{15} -\frac{1}{20}[/tex]
[tex]\frac{1}{R2} =\frac{1}{60}[/tex]
se obtiene:
R2=60 Ω
La resistencia que debe ser conectada en paralelo con una de 20 Ω para hacer una resistencia combinada de 15 Ω tiene un valor de 60 Ω
The non reflective coating on a camera lens with an index of refraction of 1.29 is designed to minimize the reflection of 636-nm light. If the lens glass has an index of refraction of 1.56, what is the minimum thickness of the coating that will accomplish this task
Answer:
The minimum thickness is [tex]t = 1.0192 *10^{-7} \ m[/tex]
Explanation:
From the question we are told that
The refractive index is [tex]n = 1.29[/tex]
The wavelength of the light is [tex]\lambda = 636 \ nm = 636 *10^{-9} \ m[/tex]
The refractive index of the glass lens is [tex]n_g = 1.56[/tex]
Generally the condition for destructive interference of a films is
[tex]2t = [ m + \frac{1}{2} ] \frac{\lambda}{n}[/tex]
for minimum m = 0
[tex]2t = [ 0 + \frac{1}{2} ] \frac{\lambda}{n}[/tex]
=> [tex]2t = [ 0 + \frac{1}{2} ] \frac{636 *10^{-9}}{1.56}[/tex]
=> [tex]t = 1.0192 *10^{-7} \ m[/tex]
A 18-kg hammer strikes a nail at a velocity of 7.6 m/s and comes to rest in a time interval of 8.3 ms .
Required:
a. What is the impulse given to the nail?
b. What is the average force acting on the nail?
Answer:
(a) -136.8 Ns.
(b) -1.135 N
Explanation:
(a)
Impulse: This can be defined as the change in momentum.
From the question,
I = mv-mu.................. Equation 1
Where I = impulse, m = mass of the hammer, v = final velocity, u = initial velocity.
Given: m = 18 kg, u = 7.6 m/s, v = 0 m/s (to rest)
Substitute these values into equation 1
I = 18(0)-18(7.6)
I = -136.8 Ns.
(b)
Average force = It.............. Equation 2
Where t = time.
Given: t = 8.3 ms = 0.0083 s.
Average force = -136.8(0.0083)
Average force = -1.135 N
The force of gravity will make it easier to stop your car if you are going uphill, and more difficult to stop your car if you are going downhill.
A race-car drives around a circular track of radius RRR. The race-car speeds around its first lap at linear speed v_iv i v, start subscript, i, end subscript. Later, its speed increases to 4v_i4v i 4, v, start subscript, i, end subscript. How does the magnitude of the car's centripetal acceleration change after the linear speed increases
Answer and Explanation: Centripetal Acceleration is the change in velocity caused by a circular motion. It is calculated as:
[tex]a_{c}=\frac{v^{2}}{r}[/tex]
v is linear speed
r is radius of the curve the object in traveling along
For its first lap:
[tex]a_{c}_{1}=\frac{v_{i}^{2}}{R}[/tex]
After a while:
[tex]a_{c}_{2}=\frac{(4v_{i})^{2}}{R}[/tex]
[tex]a_{c}_{2}=\frac{16v_{i}^{2}}{R}[/tex]
Comparing accelerations:
[tex]\frac{a_{c}_{2}}{a_{c}_{1}}=\frac{16.v_{i}^{2}}{R}.\frac{R}{v_{i}^{2}}[/tex]
[tex]\frac{a_{c}_{2}}{a_{c}_{1}}=\frac{16.v_{i}^{2}}{R}.\frac{R}{v_{i}^{2}}[/tex]
[tex]\frac{a_{c}_{2}}{a_{c}_{1}}=16[/tex]
[tex]a_{c}_{2}=16a_{c}_{1}[/tex]
With linear speed 4 times faster, centripetal acceleration is 16 times greater.
how many atoms of oxygen are there in one molecule of cardon dioxide , if the chemical formula is CO2
Answer:
2 oxygen 1 carbon
Explanation:
Answer:
2 oxygen, 1 carbon
Explanation:
A 60.5-kg hiker starts at an elevation of 1280 m and climbs to the top of a peak 2570 m high.
(a) What is the hiker's change in potential energy?
(b) What is the minimum work required of the hiker?
(c) Can the actual work done be greater than this? Explain.
Answer:
A) Change in potential energy is approximately 766 KJ
B) The minimum work required by the hiker is 765 KJ
C). Yes, the actual work can be greater than this
Explanation:
A) The hiker's potential energy can be calculated at the two different elevations, and then subtracted.
P.E at 1280 m = m X g X h = 60.5kg X 9.81 m/s2 X 1280 = 759, 686 J
P.E at 2570 m = m X g X h = 60.5kg X 9.81 m/s2 X 2570 = 1, 525, 307.85 J
Change in P.E = 1, 525, 307.85 J -759, 686 J = 765, 621.85 J = 766 KJ
B) Work which was done = Force X distance moved.
The force can be got from the effect of gravity on the hiker's mass = 60.5 kg X 9.81 = 593.5 Newton.
The distance moved can be obtained by subtracting the two elevations = 2570 -1280 = 1290 m
Work done = 593 X 1290 = 765, 615 J
The minimum work required by the hiker is 765 KJ
C) Yes, the actual work can be greater than this. This is because we can now put into consideration the fact that the hiker is climbing up the elevation against the force of gravity. This will mean that the hiker is actually doing more work than if he covered the distance on flat terrain,
An ideal gas, initially at a pressure of 11.2 atm and a temperature of 299 K, is allowed to expand adiabatically until its volume doubles.
Required:
What is the gas’s final pressure, in atmospheres, if the gas is diatomic?
Answer:
The pressure is [tex]P_2 = 4.25 \ a.t.m[/tex]
Explanation:
From the question we are told that
The initial pressure is [tex]P_1 = 11.2\ a.t.m[/tex]
The temperature is [tex]T_1 = 299 \ K[/tex]
Let the first volume be [tex]V_1[/tex] Then the final volume will be [tex]2 V_1[/tex]
Generally for a diatomic gas
[tex]P_1 V_1 ^r = P_2 V_2 ^r[/tex]
Here r is the radius of the molecules which is mathematically represented as
[tex]r = \frac{C_p}{C_v}[/tex]
Where [tex]C_p \ and\ C_v[/tex] are the molar specific heat of a gas at constant pressure and the molar specific heat of a gas at constant volume with values
[tex]C_p=7 \ and\ C_v=5[/tex]
=> [tex]r = \frac{7}{5}[/tex]
=> [tex]11.2*( V_1 ^{\frac{7}{5} } ) = P_2 * (2 V_1 ^{\frac{7}{5} } )[/tex]
=> [tex]P_2 = [\frac{1}{2} ]^{\frac{7}{5} } * 11.2[/tex]
=> [tex]P_2 = 4.25 \ a.t.m[/tex]
The final pressure of the gas will be 4.244 atm.
Given information:
The initial pressure of the gas is [tex]P_1=11.2\rm\;atm[/tex].
The initial temperature of the gas is [tex]T_1=299\rm\; K[/tex].
Let the initial volume of the gas be V. So, the final volume will be double or 2V.
The given diatomic gas is expanded adiabatically. So, the equation of the adiabatic process will be,
[tex]P_1V_1^{\gamma}=P_2V_2^{\gamma}[/tex]
where [tex]\gamma[/tex] is the ratio of specific heats of the gas which is equal to 1.4 for a diatomic gas.
So, the final pressure [tex]P_2[/tex] can be calculated as,
[tex]P_1V_1^{\gamma}=P_2V_2^{\gamma}\\11.2\times V^{1.4}=P_2\times (2V)^{1.4}\\P_2=4.244\rm\;atm[/tex]
Therefore, the final pressure of the gas will be 4.244 atm.
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a father and his son want to play on a seesaw. where on the seesaw should each of them sit to balance the torque
A father and his son want to play on a seesaw. Due to his larger size than that of the son, the father should outweigh the boy on the opposing sides.
What is seesaw?A seesaw is a long, narrow board with a single pivot point, which is often situated in the middle of both ends. As one end rises, the other falls.
What is outweigh?Rugby continues to have far more health advantages than hazards.
A father and his son want to play on a seesaw. Due to his larger size than that of the son, the father should outweigh the boy on the opposing sides.
To know more about seesaw and outweigh
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(A) Calculate the one temperature at which Fahrenheit and Celsius thermometers agree with each other.
(B) Calculate the one temperature at which Fahrenheit and Kelvin thermometers agree with each other.
Answer:
A) -40° C and -40° F
B) 574.25° K and 574.25° F
Explanation:
see attachment for calculation and explanation
A hot piece of iron is thrown into the ocean and its temperature eventually stabilizes. Which of the following statements concerning this process is correct? (There may be more than one correct choice.)
A. The entropy gained by the iron is equal to the entropy lost by the ocean.
B. The ocean gains less entropy than the iron loses.
C. The change in the entropy of the iron-ocean system is zero.
D. The entropy lost by the iron is equal to the entropy gained by the ocean.
E. The ocean gains more entropy than the iron loses.
Answer:
E. The ocean gains more entropy than the iron loses.
Explanation:
When there is a spontaneous process , entropy of the system increases . Here hot iron is losing entropy and ocean is gaining entropy . Net effect will be gain of entropy . That means entropy gained by ocean is more than entropy lost by iron .
Hence option E is correct .
If an electron in an atom moves from an energy level of 5 to an energy level of 10:____.
a. a photon of energy 5 is absorbed.
b. a photon of energy 15 is absorbed.
c. a photon of energy 5 is emitted.
d. a photon of energy 15 is emitted.
Answer
Answer:
a. a photon of energy 5 is absorbed
Explanation:
Because when an electron in a lower energy state absorbs energy, in form of photons it moves to higher energy stage in this case 5 photons because it moved from 5 to 10
The howler monkey is the loudest land animal and can be heard up to a distance of 2.5 km. Assume the acoustic output of a howler to be uniform in all directions. At 2.5 km away from the monkey, what would be the intensity of the sound
Answer:
10⁻¹² W / m²
Explanation:
The feeble sound that a man can hear is of the vale which measures 0 on decibel scale . The intensity of sound in terms of J / m² .s is 10⁻¹² W / m² .
So the intensity of sound of monkey at 2.5 km must be 10⁻¹² W / m² .
Why is area a scalar quantity?
Answer:
Area is a scalar quantity because there is no need of direction to define and also follow the algebraic summation. When we talk about vector there exists a frame of reference with a certain origin. It actually depends on the fact of the physical area, but if that factor changes to a non-directional object such as a rug spread on the floor, you can consider the area or a region as a scalar. (*Scalars are quantities that are fully described by a magnitude (or numerical value) alone.)
Hope this helps!
A beak ball is thrown in an arc and in 2.0s its shadow on the ground travels 180 m in a straight line. What is the average speed of the shadow
Answer: 90 m/s
Explanation:
The formula for speed is distance/time. The distance is 180 m and the time it took for the ball to travel is 2 s.
180 m/2 s = 90 m/s
20 copper wires of length l and diameter d are connected in parallel to form a single composite conductor of resistance R. What is the ratio of the diameter of a single copper wire to d, if its length is also l and it has the same resistance?
Please show work and type out answer.
Answer:
D = 4.47d
Explanation:
given that
1/R(eq) = 20/R
R(eq) = R/20
also, we know that the formula for resistance is given by the relation
R = pl/A, where A is πd²/4
If we substitute the value of A, we have
R = pl/(πd²/4)
R = 4pl/πd²
Now, we substitute this in the earlier derived equation
R(eq) = (4pl/πd²) / 20
R(eq) = pl/5πd²
To find the resistance of a single wire made of the same material, the resistance is
R(D) = 4pl / πD²
R(eq) = R(D), and thus
pl/5πd² = 4pl/πD²
1/5d² = 4/D²
D² = 20d²
D = √20d²
D = 4.47 d
A turntable of radius R1 is turned by a circular rubberroller of radius R2 in contact with it at their outeredges. What is the ratio of their angular velocities,ω1 / ω2 ?
Answer:
The ratio is [tex]\frac{w_1}{w_2} = \frac{R_2}{R_1}[/tex]
Explanation:
From the question we are told that
The first radius is [tex]R_1[/tex]
The second radius is [tex]R_2[/tex]
Generally the angular speed of the turntable is mathematically represented as
[tex]w_1 = \frac{ v_k }{R_1 }[/tex]
Generally the angular speed of the rubber roller is mathematically represented as
[tex]w_2 = \frac{ v_k }{R_2 }[/tex]
Where [tex]v_k[/tex] is the velocity of both turntable and rubber roller
So
[tex]\frac{w_1}{w_2} = \frac{\frac{v_k}{R_1} }{\frac{v_k}{R_2} }[/tex]
[tex]\frac{w_1}{w_2} = \frac{R_2}{R_1}[/tex]
In a region of space where the magnetic field of the earth has a magnitude of 80 μT and is directed 30° below the horizontal, a 50-cm length of wire oriented horizontally along an east-west direction is moved horizontally to the south with a speed of 20 m/s. What is the magnitude of the induced potential difference between the ends of this wire?
Answer:
V_ind = 4 × 10^(-4) V
Explanation:
We are given;
Magnetic field; B = 80 μT = 8 × 10^(-5) T
Angle;θ = 30°
Lenght;L = 50 cm = 0.5 m
Speed; v = 20 m/s
Now, formula for the induced potential difference is known as;
V_ind = NBLVsin θ
Where;
V_Ind = Induced potential difference/voltage
N = Number of turns
B = Magnetic field
V = velocity
L = length
Number of turns in this case is 1 since it's just a wire between both ends.
Thus, plugging in the relevant values, we have;
V_ind = 1 × 8 × 10^(-5) × 20 × 0.5 × sin 30
V_ind = 4 × 10^(-4) V
A vertical scale on a spring balance reads from 0 to 220 N. The scale has a length of 15.0 cm from the 0 to 220 N reading. A fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.70 Hz. Ignoring the mass of the spring, what is the mass m of the fish?
Mass of fish is 5.09Kg
Explanation:
First to find the spring constant K using
k = F/s
= 220/0.15 = 1466.7 N/m
So using the formula
T = 2π√(m/k)
f = 1/T = 1/2πx √(k/m)
f² x 4π²= k/m
So
m = k/(f² x π²)
m = 1466.7/(2.7² x 4π²)
m = 5.09 kg
Which of the following statements is true regarding electromagnetic waves traveling through a vacuum?
a. All waves have the same wavelength.
b. All waves have the same frequency.
c. All waves have the same speed.
d. The speed of the waves depends on their wavelength.
e. The speed of the waves depends on their frequency.
Answer:
C. All waves have the same speed.
Explanation:
Wave equation is given as;
V = fλ
where;
V is the speed of the wave
f is the frequency of the wave
λ is the wavelength
The speed of the wave depends on both wavelength and frequency
The speed of the electromagnetic waves in a vacuum is 3 x 10⁸ m/s, this also the speed of light which is constant for all electromagnetic waves.
Therefore, the correct option is "C"
C. All waves have the same speed.
Which planet orbits in a different plane than all of the others?
Although they're all 'close', none of the planets orbits in the same plane as any other planet. They're all in slightly different planes.
The farthest out compared to all the others is Pluto, with an orbit inclined about 17 degrees compared to the ecliptic plane (Earth's orbit). But Pluto is officially not a planet, so I don't think it's a good answer.
The next greatest inclination compared to Earth's orbit is Mercury. That one is about 7 degrees.
The other six planets are all in different orbital planes inclined less than 7 degrees compared to Earth's orbit.
(4)
The electric field inside an uncharged metal sphere is initially zero. If the sphere is
then charged positively, the field at the center of the sphere will be :
A) zero
B) finite and directed radially inward
C) nearly infinite
D) finite and directed radially outward
Answer:
Option (A) : Zero
Explanation:
Electric field Intensity inside a metallic body is always ZERO.
Suppose the width of your fist is 4.1 inches and the length of your arm is 35.4 inches. Based on these measurements, what will be the angular width (in degrees) of your fist held at arm’s length?
Answer:
7 degree
Explanation:
given data
width = 4.1 inches
length = 35.4 inches
solution
we consider as per fig
O is mid point of BC
so OB = 2.05 inches
and
AB = [tex]\sqrt{OB^2 + OA^2}[/tex]
AB = [tex]\sqrt{2.05^2 + 35.4^2}[/tex]
AB = 35.078 inches
so
[tex]sin \frac{\alpha }{2} = \frac{OB}{AB}[/tex]
[tex]sin \frac{\alpha }{2} = \frac{2.05}{35.078}[/tex]
[tex]\alpha = 7 degree[/tex]
Solve for A if F=MA and F=100 and M=25?
Answer:
[tex] \boxed{ \boxed{ \bold{A = 4}}}[/tex]Explanation:
Given,
F = 100 , M = 25
Now, let's find the value of A
[tex] \sf{F = MA}[/tex]
plug the values
⇒[tex] \sf{100 = 25 A}[/tex]
Swap the sides of the equation
⇒[tex] \sf{25A = 100}[/tex]
Divide both sides of the equation by 25
⇒[tex] \sf{ \frac{25A}{25} = \frac{100}{25} }[/tex]
Calculate
⇒[tex] \sf{A = 4}[/tex]
Hope I helped!
Best regards!!
Given:-
Force,F = 100 NMass,m = 25 kgTo find out:-
Calculate the acceleration, a ?
Formula applied:-
F = m × a
Solution:-
We know,
[tex] \sf{F = m × a}[/tex]
Substituting the values of mass and acceleration,we get
[tex] \sf\implies \: 100 = 25 \times a[/tex]
[tex] \sf \implies a = \cancel \dfrac{100}{25} [/tex]
[tex] \sf \implies a = 4 \: ms {}^{ - 1} [/tex]