Answer:
density d = 1.59 g/cm^3
The density of the rock is 1.59 g/cm^3
Explanation:
The density of an object can be derived by measuring its mass and then measuring its volume by submerging it in a graduated cylinder.
Density = mass/volume of water displaced
d = m/v ........1
Given;
mass m = 344 g
Volume of water displaced v = 216 cm^3
from equation 1, we can calculate the value of the density;
Substituting the given values;
d = 344/216 g/cm^3
d = 1.592592592592 g/cm^3
d = 1.59 g/cm^3
The density of the rock is 1.59 g/cm^3
In a high school swim competition, a student takes 1.6 s to complete 1.5 somersaults. Determine the average angular speed of the diver, in rad/s, during this time interval.
Answer:
The angular speed is [tex]w = 5.89 \ rad/s[/tex]
Explanation:
From the question we are told that
The time taken is [tex]t = 1.6 s[/tex]
The number of somersaults is n = 1.5
The total angular displacement during the somersault is mathematically represented as
[tex]\theta = n * 2 * \pi[/tex]
substituting values
[tex]\theta = 1.5 * 2 * 3.142[/tex]
[tex]\theta = 9.426 \ rad[/tex]
The angular speed is mathematically represented as
[tex]w = \frac{\theta }{t}[/tex]
substituting values
[tex]w = \frac{9.426}{1.6}[/tex]
[tex]w = 5.89 \ rad/s[/tex]
Two children sit on different sides of a seesaw. The first child of mass 27 kg sits 1.5 m from the center. How far must the second child of mass 23 kg sit from the center on the opposite side in order to balance the seesaw
The distance the second child needs to sit from the center in order to balance the seasaw is 1.76 m.
What is distance?The can be defined as the total length between two points.
To calculate the distance the second child needs to sit from the center, we use the formula below.
Note: For the seesaw to be balanced, sum of clockwise moment is equal to sum of anti clockwise moment acting on it.
Formula:
mgd = MgD......... Equation 1Make D the subject of the equation
D = md/M ............ Equation 2Where:
D = Distance of the second child from the centerm = Mass of the first childd = Distance of the first child from the centerM = Mass of the second childFrom the question,
Given:
m = 27 kgd = 1.5 mM = 23 kgSubstitute these values into equation 1
D = (27×1.5)/23D = 1.76 mHence, the distance of the second child from the center is 1.76 m.
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what happens to the sound waves when the source of the sound is moving toward you
Answer:
The perceived frequency is higher than the actual emitted sound frequency. that means that the received sound waves are of shorter wavelength.
Explanation:
When the source of a sound wave is moving toward the observer, the perceived frequency of the wave changes in relation to the observer producing a change in pitch. The effect is called Doppler effect in honor of the physicist who formulated the physical explanation.
In the case of the sound source approaching the observer, the perceived frequency is higher than that actually emitted by the sound source.
A proton moving in the positive x direction with a speed of 9.9 105 m/s experiences zero magnetic force. When it moves in the positive y direction it experiences a force of 1.6 10-13 N that points in the positive z direction. Determine the magnitude and direction of the magnetic field.
Answer:
The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction
Explanation:
In order to calculate the magnitude and direction of the magnetic field, you take into account the following equation for the magnetic force on the proton:
[tex]\vec{F_B}=q\vec{v}\ X\ \vec{B}[/tex] (1)
v: speed of the proton = 9.9*10^5 m/s
q: charge of the proton = 1.6*10^-19C
B: magnetic field = ?
FB: magnetic force on the proton = 1.6*10^-13N
When the proton travels in the positive y direction (^j), you have that the proton experiences a force in the positive z direction (+^k). To obtain this direction of the magnetic force on the proton, it is necessary that the magnetic field points in the negative x direction, in fact, you have:
^j X (-^i) = -(-^k)=^k
To obtain the magnitude of the magnetic field you use:
[tex]F_B=qvBsin90\°=qvB\\\\B=\frac{F_B}{qv}=\frac{1.6*10^{-13}N}{(1.6*10^{-19}C)(9.9*10^5m/s)}\\\\B=1.01T[/tex]
The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction
Match each term to the best description. ::
1. Coherent
2. Diffraction
3. Grating
4. Interference
5. Specular dot
a. Composed of numerous narrowly spaced parallel slits or grooves
b. Having the same wavelength, frequency, and in-phase
c. Interaction of waves where they meet in space
d. The bending of waves near a boundary or as a wave passes through an opening
e. The zeroth order direct reflection fringe
A 95 N force exerted at the end of a 0.35 m long torque wrench gives rise to a torque of 15 N · m. What is the angle (assumed to be less than 90°) between the wrench handle and the direction of the applied force?
Answer:
The angle between the wrench handle and the direction of the applied force is 26.8°
Explanation:
Given;
applied force, F = 95 N
length of the wrench, r = 0.35 m
torque on the wrench due to the applied force, τ = 15 N.m
Torque is calculated as;
τ = rFsinθ
where;
r is the length of the wrench
F is the applied force
θ is the angle between the applied force and the wrench handle
Make Sin θ the subject of the formula;
Sinθ = τ / rF
Sinθ = 15 / (0.35 x 95)
Sinθ = 0.4511
θ = Sin⁻¹(0.4511)
θ = 26.8°
Therefore, the angle between the wrench handle and the direction of the applied force is 26.8°
A 90.0-kg ice hockey player hits a 0.150-kg puck, giving the puck a velocity of 45.0 m/s. If both are initially at rest and if the ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal 15.0 m away
Answer:
0.0241 m
Explanation:
mass of the hockey player m1 = 90 kg
mass of puck m2 = 0.150 kg
puck velocity v1= 45 m/s
distance traveled by puck to reach the goal =15.0 m.
now accoding to momentum conservation law
90×45+0.15×v2 = 0 [ since, If both are initially at rest and if the ice is frictionless,]
therefore, v2= -0.0725 m/s.
Now time taken by the puck to reach the goal
t= 15/45 = 1/3 sec.
therefore, how far does the player recoil in the time
=0.0725×1/3= 0.0241 m.
the distance travelled by the player( recoil ) in the time the puck reach the goal is 0.025m.
Given the data in the question
Mass of the player; [tex]m_1 = 90.0kg[/tex]Mass of puck; [tex]m = 0.150kg[/tex]Since they were both at rest initially
Initial velocity of player; [tex]u_1 = 0[/tex]Initial velocity of puck; [tex]u = 0[/tex]Velocity of player after the hit; [tex]v_1 = \ ?[/tex]Velocity of puck after the hit; [tex]v = 45.0m/s[/tex]Distance to the goal; [tex]s = 15.0m[/tex]Using conservation of liner momentum:
[tex]mu + m_1u_1 = mv+ m_1v_1[/tex]
Now, Since they were both at rest initially
[tex]0 = mv+ m_1v_1[/tex]
We substitute in our values to find the velocity of the player after the hit ( recoil velocity )
[tex]0 =[ 0.150kg * 45.0m/s ] + [ 90.0kg * v_1 ]\\\\0 = 6.75kg.m/s + [ 90.0kg * v_1 ]\\\\90.0kg * v_1 = -6.75kg.m/s \\\\v_1 = -\frac{6.75kg.m/s}{90.0kg} \\\\v_1 =- 0.075m/s[/tex]
{ The negative sign shows that the velocity of both the player and the puck are in opposite direction }
Hence, recoil velocity of the player is 0.075m/s
Now, we determine the time taken for the puck to trach the goal using the relation between distance, velocity and time .
Time = Distance / Velocity
We substitute our values into the expression
[tex]t = \frac{s}{v} \\\\t = \frac{15.0m}{45m/s} \\\\t = 0.3333s[/tex]
Hence, the time taken for the puck to reach the goal is 0.3333 seconds.
Next, we determine the distance travelled by the player( recoil ) in the time the puck reach the goal using the relation between distance, velocity and time .
Time = Distance / Velocity
We substitute in our values
[tex]t = \frac{s}{v}\\\\0.3333s = \frac{s}{0.075m/s} \\\\s = 0.3333s * 0.075m/s\\\\s = 0.025m[/tex]
Therefore, the distance travelled by the player( recoil ) in the time the puck reach the goal is 0.025m.
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g Skin cancers from direct sunlight are caused by what region of the electromagnetic spectrum? a. Visible Light b. Ultraviolet Light c. Cosmic d. Gamma
Answer: b. Ultraviolet Light
Explanation:
Electromagnetic wave is defined as the wave which is associated with both electrical and magnetic component associated with them. They can travel in vacuum as well and travel with the speed of light.
The electromagnetic radiations consist of radio waves, microwaves, infrared ,Visible , ultraviolet, X rays and gamma rays arranged in order of increasing frequency and decreasing wavelengths.Thus ultraviolet light has more energy than visible light as energy and frequency are directly proportional.
Ultraviolet Light is more energetic than visible light, and have potential to damage skin cells and lead to skin cancer.
Angular velocity in the z direction of a flywheel is w(t)=A + Bt2 The numerical values of the constants are A=2.75 and B=1.50. What is the angular acceleration α(t) when t=0s and t=5.00s?
Answer:
α(0) = 0 rad/s²
α(5) = 15 rad/s²
Explanation:
The angular velocity of the flywheel is given as follows:
w(t) = A + B t²
where, A and B are constants.
Now, for the angular acceleration, we must take derivative of angular velocity with respect to time:
Angular Acceleration = α (t) = dw/dt
α(t) = (d/dt)(A + B t²)
α(t) = 2 B t
where,
B = 1.5
AT t = 0 s
α(0) = 2(1.5)(0)
α(0) = 0 rad/s²
AT t = 5 s
α(5) = 2(1.5)(5)
α(5) = 15 rad/s²
A vector quantity has direction, a scalar quantity does not.
Explanation:
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What is the final temperature of 0.3kg of water initially at 20 Celsius after 35 KJ Of heat is added?
Answer:
Option A. 48°C
Explanation:
The following data were obtained from the question:
Mass (m) = 0.3 Kg
Initial temperature (T1) = 20°C
Heat (Q) added = 35 KJ
Specific heat capacity (C) = 4.18 KJ/Kg°C
Final temperature (T2) =..?
The final temperature of water can be obtained as follow:
Q = MC(T2 – T1)
35 = 0.3 x 4.18 (T2 – 20)
35 = 1.254 (T2 – 20)
Clear the bracket
35 = 1.254T2 – 25.08
Collect like terms
1254T2 = 35 + 25.08
1.254T2 = 60.08
Divide both side by the coefficient of T2 i.e 1.254
T2 = 60.08/1.254
T2 = 47.9 ≈ 48°C
Therefore, the final temperature of the water is 48°C.
A standard 1 kilogram weight is a cylinder 48.5 mm in height and 49.0 mm in diameter. What is the density of the material? kg/m3
Answer:
Density = 10,933.93 kg/m^3
the density of the material is 10,933.93 kg/m^3
Explanation:
Density is the mass per unit volume
Density = mass/volume = m/V
Volume of a cylinder V = πr^2 h
Given;
Height h = 48.5mm = 0.0485 m
Radius r = diameter/2 = 49mm÷2 = 24.5mm = 0.0245m
Substituting the values;
Volume V = π×(0.0245^2)×0.0485
V = 0.000091458438030 m^3
V = 0.000091458 m^3
The mass is given as;
Mass = 1 kg
So, the density can be calculated as;
Density = 1/0.000091458
Density = 10933.92825785 kg/m^3
Density = 10,933.93 kg/m^3
the density of the material is 10,933.93 kg/m^3
A projectile is fired at time t = 0.0 s from point o at the edge of a cliff, with initial velocity components of Vox = 30 m/s and Voy = 100 m/s. The projectile rises, and then falls into the sea
at point P. The time of flight of the projectile is 25 s. Assume air resistance is negligible.
t
What is the height of the cliff?
560 m
450 m
780 m
400 m
640 m
Answer:
It would be 450 or 640. My final answer would be about 450
Explanation: Because it would't be to high if it was shot Voy = 100
btw i think i know what i know what i am talking about.
The answer would be about 450 m.
What peak is considered a cliff?The top isn't the standard for a cliff to be reckoned as a cliff as such. Any steep rock face particularly at the edge of the sea can be specified as a cliff.
A 'clifftop' just refers to any pinnacle of a cliff. A 'plateau' is any flat extended geologic floor. An 'overhang' is a part of a structure or formation that protrudes from the primary frame and rests such that it is 'overhanging' the ground (striking above it).
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I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block and it oscillates with a period of 0.13 s. What is the amplitude of oscillation
Answer:
The amplitude of the oscillation is 2.82 cm
Explanation:
Given;
mass of attached block, m = 4.1 kg
energy of the stretched spring, E = 3.8 J
period of oscillation, T = 0.13 s
First, determine the spring constant, k;
[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]
where;
T is the period oscillation
m is mass of the spring
k is the spring constant
[tex]T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\[/tex]
Now, determine the amplitude of oscillation, A;
[tex]E = \frac{1}{2} kA^2[/tex]
where;
E is the energy of the spring
k is the spring constant
A is the amplitude of the oscillation
[tex]E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm[/tex]
Therefore, the amplitude of the oscillation is 2.82 cm
Which phrases accurately describe an elliptical galaxy? Check all that apply.
may be egg-shaped
may be spiral-shaped
has no recognizable shape
has no new stars being formed
has almost no gas or dust between stars
Answer:
May be egg shaped
Has no new stars being formed.
Has almost no gas or dust between stars.
Explanation:
Elliptical galaxy is the collection of many stars which are bounded together gravitationally, which is smooth and ellipsoidal and shape and the appearance is featureless.
Elliptical galaxy is ovoid or spherical masses of stars.
It is found in galaxy clusters and compact galaxies.
It has no gas or dust between stars which result in low rates of star formation.
It is formed When two spirals collide, they lose their familiar shape, morphing into the less-structured elliptical galaxies.
Elliptical galaxy is made of old stars and have no gas and dust.
An example is elliptical galaxy m60 which shines brightly and is egg shaped.
1. A sphere with a mass of 10 kg and radius of 0.5 m moves in free fall at sea level (where the air density is 1.22 kg/m3). If the object has a drag coefficient of 0.8, what is the object’s terminal velocity? What is the terminal velocity at an altitude of 5,000 m, where the air density is 0.736 kg/m3? Show all calculations in your answer.
Answer:
The terminal velocity at sea level is 7.99 m/s
The terminal velocity at an altitude of 5000 m is 10.298 m/s
Explanation:
mass of sphere m = 10 kg
radius of sphere r = 0.5 m
air density at sea level p = 1.22 kg/m^3
drag coefficient Cd = 0.8
terminal velocity = ?
Area of the sphere A = [tex]4\pi r^{2}[/tex] = 4 x 3.142 x [tex]0.5^{2}[/tex] = 3.142 m^2
terminal velocity is gotten from the relationship
[tex]Vt = \sqrt{\frac{2mg}{pACd} }[/tex]
where g = acceleration due to gravity = 9.81 m/s^2
imputing values into the equation
[tex]Vt = \sqrt{\frac{2*10*9.81}{1.22*3.142*0.8} }[/tex] = 7.99 m/s
If at an altitude of 5000 m where air density = 0.736 kg/m^3, then we replace value of air density in the relationship as 0.736 kg/m^3
[tex]Vt = \sqrt{\frac{2mg}{pACd} }[/tex]
[tex]Vt = \sqrt{\frac{2*10*9.81}{0.736*3.142*0.8} }[/tex] = 10.298 m/s
The value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.
Given data:
The mass of sphere is, m = 10 kg.
The radius of sphere is, r = 0.5 m.
The density of air is, [tex]\rho = 1.22 \;\rm kg/m^{3}[/tex].
The drag coefficient of object is, [tex]C_{d}=0.8[/tex].
The altitude is, h = 5000 m.
The density of air at altitude is, [tex]\rho' =0.736 \;\rm kg/m^{3}[/tex].
The mathematical expression for the terminal velocity of an object is,
[tex]v_{t}=\sqrt\dfrac{2mg}{\rho \times A \times C_{d}}[/tex]
here,
g is the gravitational acceleration.
A is the area of sphere.
Solving as,
[tex]v_{t}=\sqrt{\dfrac{2 \times 10 \times 9.8}{1.22 \times (4 \pi r^{2}) \times C_{d}}}\\\\\\v_{t}=\sqrt{\dfrac{2 \times 10 \times 9.8}{1.22 \times (4 \pi \times 0.5^{2}) \times 0.8}}\\\\\\\v_{t}=7.99 \;\rm m/s[/tex]
Now, the terminal velocity at the altitude of 5000 m is given as,
[tex]v_{t}'=\sqrt\dfrac{2mg}{\rho' \times A \times C_{d}}[/tex]
Solving as,
[tex]v_{t}'=\sqrt{\dfrac{2 \times 10 \times 9.8}{0.736 \times (4 \pi \times 0.5^{2}) \times 0.8}}\\\\\\v_{t}'=10.30 \;\rm m/s[/tex]
Thus, we can conclude that the value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.
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A cylindrical shell of radius 7.00 cm and length 2.21 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 15.2 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. (a) Find the net charge on the shell.
Answer:
The net charge on the shell is 30x10^-9C
Explanation:
Pls see attached file
A small cylinder made of a diamagnetic material is brought near a bar magnet. One end of the cylinder is brought near the North pole of the magnet and is repelled. What happens when the other end of the cylinder is brought near the North pole of the magnet? Group of answer choices
Answer:
it attracts
Explanation:
since in a magnetic body there are two poles
(north and south poles)if the first pole was repeled when brought near the North Pole therefore the other end is going to attarct because the first end was also a North Pole while the second end will be a south pole
A small ferryboat is 4.70 m wide and 6.10 m long. When a loaded truck pulls onto it, the boat sinks an additional 5.00 cm into the river. What is the weight of the truck
Answer:
M = 1433.5 kg
Explanation:
This exercise is solved using the Archimedean principle, which states that the hydrostatic thrust is equal to the weight of the desalinated liquid,
B = ρ g V
with the weight of the truck it is in equilibrium with the push, we use Newton's equilibrium condition
Σ F = 0
B-W = 0
B = W
body weight
W = M g
the volume is
V = l to h
rho_liquid g (l to h) = M g
M = rho_liquid l a h
we calculate
M = 1000 4.7 6.10 0.05
M = 1433.5 kg
What is the change in potential energy of a 2.00 nC test charge, Uelectric, b - Uelectric, a, as it is moved from point a at x
The question is incomplete. Here is the complete question.
A uniform electric field of 2kN/C points in the +x-direction.
(a) What is the change in potential energy of a +2.00nC test charge, [tex]U_{electric,b} - U_{electric,a}[/tex] as it is moved from point a at x = -30.0 cm to point b at x = +50.0 cm?
(b) The same test charge is released from rest at point a. What is the kinetic energy when it passes through point b?
(c) If a negative charge instead of a positive charge were used in this problem, qualitatively, how would your answers change?
Answer: (a) ΔU = 3.2×[tex]10^{-6}[/tex] J
(b) KE = 2×[tex]10^{-6}[/tex] J
Explanation: Potential Energy (U) is the amount of work done due to its position or condition and its unit is Joule (J). Kinetic Energy (KE) is the ability to do work by virtue of velocity and the unit is also (J). Mechanical Energy is the sum of Potential and Kinetic Energies of a system.
(a) Related to electricity, Potential Energy can be calculated as:
ΔU = Eqd
where E is the electric field (in N/C);
q is the charge (in C);
d is the distance between plaques (in m);
For a at x = - 30cm and b at x = 50 cm:
E = 2×[tex]10^{3}[/tex] N/C
q = 2×[tex]10^{-9}[/tex] C
d = 50 - (-30) = 80×[tex]10^{-2}[/tex] = 8×[tex]10^{-1}[/tex]m
ΔU = [tex]U_{electric,b} - U_{electric,a}[/tex] = Eqd
[tex]U_{electric,b} - U_{electric,a}[/tex] = 2×[tex]10^{3}[/tex] . 2×[tex]10^{-9}[/tex] . 8×[tex]10^{-1}[/tex]
ΔU = 3.2×[tex]10^{-6}[/tex] J
(b) Mechanical Energy is constant, so:
[tex]KE_{i} + U_{i} = KE_{f} + U_{f}[/tex]
Since the initial position is zero and there is no initial kinetic energy:
[tex]KE_{f} = - U{f}[/tex]
[tex]KE_{f} =[/tex] - (2×[tex]10^{3}[/tex]. 2×[tex]10^{-9}[/tex] . 5×[tex]10^{-1}[/tex])
[tex]KE_{f} = - 2.10^{-6}[/tex] J
(c) If the charge is negative, electric field does positive work, which diminishes the potential energy. The charge flows from the negative side towards the positive side and stays, not doing anything.
If a negative point charge is placed at P without moving the original charges, the net electrical force the charges ±Q will exert on it is
Answer:
The particle P moves directly upwards
Explanation:
Lets designate the negative point charge at point P as particle P
The +Q charge will exert an attractive force on the particle P.
The -Q charge will exert a repulsive force on the particle P
The +Q charge exerts an upwards and leftward force on particle P
The -Q charge exerts an upwards and rightward force on particle P
Since the charges are equidistant from the particle P, and are of equal magnitude, the rightward force and the leftward force will cancel out, leaving just the upward force on the particle P.
The effect of the upward force is that the particle P moves directly upwards
Devise and draw a circuit using a long, straight wire resistor, instead of a decade box, that would allow the study of the variation of the voltage with resistance (IS constant). According to Ohm's law, what would
Answer:
...
Explanation:
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Which compound is composed of oppositely charged ions?
Answer:
Option A. Li2O
Explanation:
To know which of the compound contains oppositely charged ions, let us determine the nature of each compound. This is illustrated below:
Li2O is an ionic compound as it contains a metal (Lithium, Li) and non metal (oxygen, O). Ionic compounds are charactized by the presence of aggregate positive and negative charge ions. This is true because they are formed by the transfer of electron(s) from the metallic atom to the non-metallic atom.
2Li —> 2Li^+ + 2e
O2 + 2e —> O^2-
2Li + O2 + 2e —> 2Li^+ + O^2- + 2e
2Li + O2 —> 2Li^+ O^2- —> Li2O
OF2 is a covalent compound as it contains non metals only (i.e oxygen, O and fluorine, F). Covalent compounds are characterised by the presence of molecules. This is true because they are formed from the sharing of electron(s) between the atoms involved.
PH3 is a covalent compound as it contains non metals only (i.e phosphorus, P and hydrogen, H).
SCl2 is a covalent compound as it contains non metals only (i.e sulphur, S and chlorine, Cl).
From the above information, we can see that only Li2O contains oppositely charged ions.
Answer:
A
Explanation:
Just took the test
ir temperature in a desert can reach 58.0°C (about 136°F). What is the speed of sound (in m/s) in air at that temperature?
Answer:
363m.s-1
Explanation:
Which kind of energy is found in an atom's nucleus? (A) Nuclear (B) Elastic (C) Thermal (D) Electromagnetic
Answer:
The answer is option A.
NuclearHope this helps you
Answer:
Answer is A Nuclear
Explanation:
Just answered this question on my test
A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnitude 0.7 G directed at 70 below the horizontal north direction. Find the current in the loop.
Answer:
The current is [tex]I = 6.68 \ A[/tex]
Explanation:
From the question we are told that
The radius of the loop is [tex]r = 6 \ cm = 0.06 \ m[/tex]
The earth's magnetic field is [tex]B_e = 0.7G= 0.7 G * \frac{1*10^{-4} T}{1 G} = 0.7 *10^{-4} T[/tex]
The number of turns is [tex]N =1[/tex]
Generally the magnetic field generated by the current in the loop is mathematically represented as
[tex]B = \frac{\mu_o * N * I}{2 r }[/tex]
Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth
[tex]B = B_e[/tex]
=> [tex]B_e = \frac{\mu_o * N * I }{ 2 * r}[/tex]
Where [tex]\mu[/tex] is the permeability of free space with value [tex]\mu _o = 4\pi * 10^{-7} N/A^2[/tex]
[tex]0.7 *10^{-4}= \frac{ 4\pi * 10^{-7} * 1 * I}{2 * 0.06}[/tex]
=> [tex]I = \frac{2 * 0.06 * 0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}[/tex]
[tex]I = 6.68 \ A[/tex]
The current in the loop will be "6.68 A".
Magnetic fieldAccording to the question,
Radius of loop, r = 6 cm or,
= 0.06 m
Earth's magnetic field, [tex]B_e[/tex] = 0.7 G or,
= 0.7 × [tex]\frac{1\times 10^{-4}}{1 G}[/tex]
= 0.7 × 10⁻⁴ T
Number of turns, N = 1
We know the relation,
→ B = [tex]\frac{\mu_0\times N\times I}{2r}[/tex]
or,
B = [tex]B_e[/tex]
then,
→ [tex]B_e[/tex] = [tex]\frac{\mu_0\times N\times I}{2r}[/tex]
By substituting the values,
0.7 × 10⁻⁴ = [tex]\frac{4 \pi\times 10^{-7}\times 1\times I}{2\times 0.06}[/tex]
hence,
The current will be:
I = [tex]\frac{2\times 0.06\times 0.7\times 10^{-4}}{4 \pi\times 10^{-7}\times 1}[/tex]
= 6.68 A
Thus the above approach is correct.
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Can you name some people who have shown perseverance?
Answer:
Milton Hershey
Steve Jobs
Simon Cowell
Thomas Edison
Explanation:
Faraday's Law states that the negative of the time rate of change of the flux of the magnetic field through a surface is equal to which of the following quantities?
a. The flux of the magnetic field through a surface which has the loop as its boundary.
b. The negative of the time rate of change of the flux of the magnetic field through a surface which has the loop as its boundary.
c. The line integral of the magnetic field around the closed loop.
d. The flux of the electric field through a surface which has the loop as its boundary.
Answer:
(C). The line integral of the magnetic field around a closed loop
Explanation:
Faraday's law states that induced emf is directly proportional to the time rate of change of magnetic flux.
This can be written mathematically as;
[tex]EMF = -\frac{\delta \phi _B}{\delta t}[/tex]
[tex](\frac{\delta \phi _B}{\delta t} )[/tex] is the rate of change of the magnetic flux through a surface bounded by the loop.
ΔФ = BA
where;
ΔФ is change in flux
B is the magnetic field
A is the area of the loop
Thus, according to Faraday's law of electric generators
∫BdL = [tex]\frac{\delta \phi _B}{\delta t}[/tex] = EMF
Therefore, the line integral of the magnetic field around a closed loop is equal to the negative of the rate of change of the magnetic flux through the area enclosed by the loop.
The correct option is "C"
(C). The line integral of the magnetic field around a closed loop
Faraday's Law states that the negative of the time rate of change of the flux of the magnetic field through a surface is equal to: D. The flux of the electric field through a surface which has the loop as its boundary.
In Physics, the surface integral with respect to the normal component of a magnetic field over a surface is the magnetic flux through that surface and it is typically denoted by the symbol [tex]\phi[/tex].
Faraday's Law states that the negative of the time rate of change ([tex]\Delta t)[/tex] of the flux of the magnetic field ([tex]\phi[/tex]) through a surface is directly proportional to the flux ([tex]\phi[/tex]) of the electric field through a surface which has the loop as its boundary.
Mathematically, Faraday's Law is given by the formula:
[tex]E.m.f = -N\frac{\Delta \phi}{\Delta t}[/tex]
Where:
N is the number of turns.Read more: https://brainly.com/question/15121836
Suppose the ring rotates once every 4.30 s . If a rider's mass is 53.0 kg , with how much force does the ring push on her at the top of the ride?
The complete question is;
In an amusement park ride called The Roundup, passengers stand inside a 16-m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane.
Suppose the ring rotates once every 4.30 s . If a rider's mass is 53.0 kg , with how much force does the ring push on her at the top of the ride?
Answer:
F_top = 385.36 N
Explanation:
We are given;
mass;m = 52 kg
Time;t = 4.3 s
Diameter;d = 16m
So,Radius;r = 16/2 = 8m
The formula for the centrifugal force is given as;
F_c = mω²R
Where;
R = radius
Angular velocity;ω = 2πf
f = frequency = 1/t = 1/4.3 Hz
F_c = 53 × (2π × 1/4.3)² × 8 = 905.29 N.
The force at top would be;
F_top = F_c - mg
F_top = 905.29 - (9.81 × 53) N
F_top = 385.36 N
The force at the top of ride will be "385.36 N".
Force and mass:According to the question,
Rider's mass, m = 52 kg
Time, t = 4.3 s
Diameter, d = 16 m
Radius, r = [tex]\frac{16}{2}[/tex] = 8 m
Frequency, f = [tex]\frac{1}{t}[/tex] = [tex]\frac{1}{4.3}[/tex] Hz
We know the formula,
Centrifugal force, [tex]F_c[/tex] = mω²R
or,
Angular velocity, ω = 2πf
By substituting the values in the above formula,
[tex]F_c = 53(2\pi \times (\frac{1}{4.3})^2\times 8 )[/tex]
[tex]= 905.29[/tex] N
hence,
The top force will be:
→ [tex]F_{top} = F_c[/tex] - mg
By substituting the values,
[tex]= 905.29-(9.81\times 53)[/tex]
[tex]= 385.36[/tex] N
Thus the above response is correct.
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Find the displacement. Will give brainliest!
Answer:
1000 m upwards
Explanation:
Displacement Formula: Average Velocity = Displacement/Total Time
Simply plug in our known variables and solve:
100 m/s = x m/10 seconds
100 m/s(10 s) = x m
m = 1000