Answer:
critical value is known as threshold
Explanation:
solution
when we apply electrical stimulus to the cell of muscle
then magnitude of stimulus will be strong enough for reaching the critical value
and this critical value is threshold because when the size of an stimulus change in requirement for the detection, this is called the threshold
but if change is done by without stimulation then there threshold stimulus is absolute threshold
so correct answer is threshold
How is the magnetic force on a particle moving in a magnetic field different from gravitational and electric forces.
Answer:
The magnetic force on a free moving charge depends on the velocity of the charge and the magnetic field, direction of the force is given by the right hand rule. While gravitational depends on the mass and distance of the moving particle and electric forces depends on the magnitude of the charge and distance of separation.
Explanation:
The magnetic force on a free moving charge depends on the velocity of the charge and the magnetic field and direction of the force is given by the right hand rule. While gravitational depends on the mass and distance of the moving particle and electric forces depends on the magnitude of the charge and distance of separation.
The magnetic force is given by the charge times the vector product of velocity and magnetic field. While gravitational force is given by the square of the particle mass divided by the square its distance of separation. Also electric forces is given by the square of the charge magnitude divided by the square its distance separation.
An arbitrarily shaped piece of conductor is given a net negative charge and is alone in space. What can we say about the electric potential within the conductor? Assume that the electric potential is zero at points that are very far away from the conductor.
Answer:
The electrostatic potential within the conductor will be negative and constant throughout the conductor.
Explanation:
Whenever a conductor is charged the total charge is spread throughout the entire surface uniformly. Inside the conductor the net electric field is always zero. Also we know if [tex]\overrightarrow{E}[/tex] e the electric field and V is the potential, then
[tex]\overrightarrow{E}= - \overrightarrow{\nabla}V[/tex]
Now as in this case electric field within the conductor is zero, so potential (V) has to be constant.
Final answer:
The electric potential within a charged conductor is constant and uniform, though not necessarily zero, while outside the conductor, the potential is nonzero. This is because the electric field inside a conductor in electrostatic equilibrium is zero, leading to a uniform potential throughout the conductor's inside and surface.
Explanation:
When considering an arbitrarily shaped piece of conductor with a net negative charge in space, we can assert that the electric potential within the conductor is constant throughout its volume. By the principle that in electrostatic equilibrium, the electric field inside a conductor is zero, the potential has the same value in all points inside the conductor. However, this potential value is not necessarily zero; it is referenced to a point far away from the conductor, where the potential is assumed to be zero. Outside of the conductor, the potential is nonzero and governed by the distribution of charge on the conductor's surface and the distance from it.
Considering a perfect conductor, when a test charge is moved from one point to another within or on the surface of the conductor, no work is done, because the electric field is zero. Hence, the electric potential difference between any two points is zero, and all points in and on the conductor share the same potential value with respect to infinity or ground.
Despite the conductor’s potential not being zero in general, its uniformity throughout highlights a fundamental concept in electrostatics: that upon reaching equilibrium, charge distributes on the conductor's surface to ensure the electric field within the conductor is nonexistent, thus establishing a uniform potential.
problem A newly established colony on the Moon launches a capsule vertically with an initial speed of 1.445 km/s. Ignoring the rotation of the Moon, what is the maximum height reached by the capsule
Answer:
Maximum height =1031km
Explanation:
Given:
Velocity,Vo= 1.445km/s= 1445m/s
Mass of moon= 7.35×10^22kg
Radius of moon= 1737km= 1737000m
Using conservation energy
Ui + Ki= Uf + Kf
--'>(GMm/R) + 1/2 (m ^2)
-GMm/(R + h) - 0
Vo^2= 2Gm(1/R - 1/R+h)
1445^2= 2× (6.67x10^-11)×(7.35×10^22)[(1/1737000)- (1/1737000 + h)]
h= 1031km
Answer:
104.4km
Explanation:
Projectile motion occurs when object is launched into air and allowed to fall freely under the influence of gravity.
Maximum height reached by the object is expressed as;
H = u²sin²(theta)/2g where;
u is the initial velocity = 1.445km/s
u = 1445m/s since 1000m is equivalent to 1km
theta is the angle of launch
g is the acceleration due to gravity = 10m/s²
theta = 90° (since object is launched vertically)
Substituting the values in the formula we have;
H = 1445²(sin90°)²/2(10)
H = 1445²/20
H = 104,401.25m
H = 104.4km
The the maximum height reached by the capsule is 104.4km
Two 5.0 mm × 5.0 mm electrodes are held 0.10 mm apart and are attached to 7.5 V battery. Without disconnecting the battery, a 0.10-mm-thick sheet of Mylar is inserted between the electrodes.A) What is the capacitor's potential difference before the Mylar is inserted?
B) What is the capacitor's electric field before the Mylar is inserted?
C) What is the capacitor's charge before the Mylar is inserted?
D) What is the capacitor's potential difference after the Mylar is inserted?
E) What is the capacitor's electric field after the Mylar is inserted?
F) What is the capacitor's charge after the Mylar is inserted?
The potential difference across the capacitor is 7.5 V both before and after the Mylar is inserted. To calculate charge and electric field values, additional information about the system, such as the dielectric constant of Mylar or capacitance, is required. Without this information, only the potential difference can be definitively determined.
The subject concerns Physics, specifically the study of capacitors in the context of their behavior before and after a dielectric material is inserted. We are given the dimensions of the capacitor plates and the potential difference applied by a battery, and we need to deduce several properties of the capacitor both before and after the insertion of the Mylar dielectric sheet. The capacitance of the capacitor is not provided directly, but can be inferred using the provided dimensions and the permittivity of free space or air, as required.
The potential difference across a capacitor is determined by the voltage applied by the battery. Thus, the capacitor's potential difference before the Mylar is inserted is 7.5 V.
To find the electric field in the capacitor before the Mylar is inserted, we can use the formula E = V/d, where V is the potential difference and d is the separation between the plates. Thus, E = 7.5 V / 0.1 mm = 75,000 V/m.
The charge on a capacitor is given by Q = CV, where C is the capacitance and V is the potential difference. However, without the capacitance, we cannot calculate the charge directly. More information, like the dielectric constant of the Mylar or the capacitance of the capacitor with air between the plates, would be needed.
Once the Mylar is inserted, if we assume the dielectric is not discharged, the potential difference across the capacitor remains at 7.5 V, since the battery is still connected.
If the Mylar has a dielectric constant, we can calculate the new electric field using the formula E' = E / K, where K is the dielectric constant of the Mylar. But again, without K, we cannot calculate E' directly.
The charge on the capacitor after the Mylar is inserted will be the same as before, provided the capacitor is still connected to the battery, since potential difference and charge are related by Q = CV, and V remains unchanged at 7.5 V.
A) Potential difference before the Mylar is inserted: [tex]\( 7.5 \, \text{V} \).[/tex] B) Electric field before the Mylar is inserted: [tex]\( 7.5 \times 10^4 \, \text{V/m} \)[/tex]. C) Charge before the Mylar is inserted: [tex]\( 1.66 \times 10^{-11} \, \text{C} \)[/tex]. D) Potential difference after the Mylar is inserted: [tex]\( 7.5 \, \text{V} \)[/tex]. E) Electric field after the Mylar is inserted: [tex]\( 7.5 \times 10^4 \, \text{V/m} \).[/tex] F) Charge after the Mylar is inserted: [tex]\( 5.14 \times 10^{-11} \, \text{C} \).[/tex]
To solve this problem, we need to use the concepts of capacitance, electric field, and the effect of a dielectric on a capacitor.
- Electrode area [tex]\( A = 5.0 \, \text{mm} \times 5.0 \, \text{mm} = 25.0 \, \text{mm}^2 = 25.0 \times 10^{-6} \, \text{m}^2 \)[/tex]
- Separation distance [tex]\( d = 0.10 \, \text{mm} = 0.10 \times 10^{-3} \, \text{m} \)[/tex]
- Voltage [tex]\( V = 7.5 \, \text{V} \)[/tex]
- Dielectric constant of Mylar [tex]\( \kappa \approx 3.1 \)[/tex]
A) Potential difference before the Mylar is inserted
The potential difference across the capacitor before the Mylar is inserted is simply the voltage of the battery since the battery is connected directly to the capacitor.
[tex]\[ V_{\text{before}} = 7.5 \, \text{V} \][/tex]
B) Electric field before the Mylar is inserted
The electric field [tex]\( E \)[/tex] in a parallel-plate capacitor is given by:
[tex]\[ E = \frac{V}{d} \][/tex]
Substituting the given values:
[tex]\[ E_{\text{before}} = \frac{7.5 \, \text{V}}{0.10 \times 10^{-3} \, \text{m}} \][/tex]
[tex]\[ E_{\text{before}} = 7.5 \times 10^{4} \, \text{V/m} \][/tex]
C) Charge on the capacitor before the Mylar is inserted
The capacitance [tex]\( C \)[/tex] of a parallel-plate capacitor without a dielectric is given by:
[tex]\[ C = \frac{\epsilon_0 A}{d} \][/tex]
where [tex]\( \epsilon_0 \)[/tex] is the permittivity of free space [tex](\( \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m} \)).[/tex]
[tex]\[ C_{\text{before}} = \frac{(8.85 \times 10^{-12} \, \text{F/m})(25.0 \times 10^{-6} \, \text{m}^2)}{0.10 \times 10^{-3} \, \text{m}} \][/tex]
[tex]\[ C_{\text{before}} \approx 2.21 \times 10^{-12} \, \text{F} \][/tex]
The charge [tex]\( Q \)[/tex] on the capacitor is given by:
[tex]\[ Q = CV \][/tex]
[tex]\[ Q_{\text{before}} = (2.21 \times 10^{-12} \, \text{F})(7.5 \, \text{V}) \][/tex]
[tex]\[ Q_{\text{before}} \approx 1.66 \times 10^{-11} \, \text{C} \][/tex]
D) Potential difference after the Mylar is inserted
When the Mylar is inserted, the potential difference across the capacitor remains the same because the battery is still connected.
[tex]\[ V_{\text{after}} = 7.5 \, \text{V} \][/tex]
E) Electric field after the Mylar is inserted
The electric field [tex]\( E \)[/tex] in a capacitor with a dielectric is given by the same formula:
[tex]\[ E = \frac{V}{d} \][/tex]
Since the voltage and the separation remain the same, the electric field remains the same:
[tex]\[ E_{\text{after}} = \frac{7.5 \, \text{V}}{0.10 \times 10^{-3} \, \text{m}} \][/tex]
[tex]\[ E_{\text{after}} = 7.5 \times 10^{4} \, \text{V/m} \][/tex]
F) Charge on the capacitor after the Mylar is inserted
The capacitance with the dielectric [tex]\( C' \)[/tex] is given by:
[tex]\[ C' = \kappa \cdot C \][/tex]
[tex]\[ C_{\text{after}} = 3.1 \times 2.21 \times 10^{-12} \, \text{F} \][/tex]
[tex]\[ C_{\text{after}} \approx 6.85 \times 10^{-12} \, \text{F} \][/tex]
The charge [tex]\( Q \)[/tex] on the capacitor is given by:
[tex]\[ Q = CV \][/tex]
[tex]\[ Q_{\text{after}} = (6.85 \times 10^{-12} \, \text{F})(7.5 \, \text{V}) \][/tex]
[tex]\[ Q_{\text{after}} \approx 5.14 \times 10^{-11} \, \text{C} \][/tex]
(a) (i) Find the gradient of f. (ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rate is f decreasing? (b) (i) Find the gradient of F. (ii) Find the directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k.
Question:
Problem 14. Let f(x, y) = (x^2)y*(e^(x−1)) + 2xy^2 and F(x, y, z) = x^2 + 3yz + 4xy.
(a) (i) Find the gradient of f.
(ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rate is f decreasing?
(b) (i) Find the gradient of F.
(ii) Find the directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k.
Answer:
The answers to the question are
(a) (i) the gradient of f = ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j
(ii) The direction in which f decreases most rapidly at the point (1, −1), ∇f(x, y) = -1·i -3·j is the y direction.
The rate is f decreasing is -3 .
(b) (i) The gradient of F is (2·x+4·y)i + (3·z+4·x)j + 3·y·k
(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k is ñ∙∇F = 4·x +⅟4 (8-3√3)y+ 9/4·z at (1, 1, −5)
4 +⅟4 (8-3√3)+ 9/4·(-5) = -6.549 .
Explanation:
f(x, y) = x²·y·eˣ⁻¹+2·x·y²
The gradient of f = grad f(x, y) = ∇f(x, y) = ∂f/∂x i+ ∂f/∂y j = = (∂x²·y·eˣ⁻¹+2·x·y²)/∂x i+ (∂x²·y·eˣ⁻¹+2·x·y²)/∂y j
= ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j
(ii) at the point (1, -1) we have
∇f(x, y) = -1·i -3·j that is the direction in which f decreases most rapidly at the point (1, −1) is the y direction.
The rate is f decreasing is -3
(b) F(x, y, z) = x² + 3·y·z + 4·x·y.
The gradient of F is given by grad F(x, y, z) = ∇F(x, y, z) = = ∂f/∂x i+ ∂f/∂y j+∂f/∂z k = (2·x+4·y)i + (3·z+4·x)j + 3·y·k
(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2·i + 3·j −√3·k
The magnitude of the vector 2·i +3·j -√3·k is √(2²+3²+(-√3)² ) = 4, the unit vector is therefore
ñ = ⅟4(2·i +3·j -√3·k)
The directional derivative is given by ñ∙∇F = ⅟4(2·i +3·j -√3·k)∙( (2·x+4·y)i + (3·z+4·x)j + 3·y·k)
= ⅟4 (2((2·x+4·y))+3(3·z+4·x)- √3∙3·y) = 4·x +⅟4 (8-3√3)y+ 9/4·z at point (1, 1, −5) = -6.549
All light waves move through a vacuum with a constant speed. True or False
Answer: True
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.
Electromagnetic spectrum consists of electromagnetic waves called as radio waves, microwaves, infrared, Visible, ultraviolet , X rays and gamma rays in order of increasing frequency and decreasing wavelength.
Thus the statement that All light waves move through a vacuum with a constant speed is true.
The prominent semicircular space above a doorway in a Romanesque church portal is referred to as a ___________ and was often covered in elaborate carvings
Answer: Tympanum
Explanation: A Tympanum could be described as the semicircular space above the doorway or window or entrance of a building. The space is usually designed using elaborate sculptural designs, Rocky costumes and ornaments. The tympanum could also be triangular-shaped and attributed to classical Greece and Roman architecture. It is usually located between the portals archivolt and the lintel. The tympanum is commonly observed in the churches and temples of ancient Rome and Greece.
The semicircular space above a doorway in a Romanesque church portal is called a tympanum and it was often decorated with intricate carvings relating to religious themes.
Explanation:The prominent semicircular space above a doorway in a Romanesque church portal is referred to as a tympanum. These were often covered in elaborate carvings that represented religious imagery or important biblical stories. The Romanesque style of architecture, prevalent in Europe from the 9th to 12th centuries, was known for its massive quality, thick walls, and robust pillars. This style also romanticized elaborate decorative details, among them the beautifully carved tympanum.
Learn more about Romanesque architecture here:https://brainly.com/question/33441054
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fessor drives off with his car (mass 840 kg), but forgot to take his coffee mug (mass 0.41 kg) off the roof. The coefficient of static friction between the mug and the roof is 1.1, and the coefficient of kinetic friction is 0.4. What is the maximum acceleration of the car, so the mug does not slide off
Answer:
Acceleration =4.42m/s^2
Explanation:
As the mug is not sliding on the car roof ,there must be static friction between the car and the mug. The maximum limit of static friction that can act is given by:
Kmg
Where I is coefficient of static friction between car roof and mug
m = mass of mug
g= acceleration due to gravity
Acceleration of the car= Kmg
Acceleration = 1.1 × 0.41 × 9.8 =4.42m/s
Answer:
10.791 m/s^2
Explanation:
Solution:
- As the mug is not sliding on the car roof , there will be static friction between car and mug.
- The maximum limit of static friction that can act is given by = k*m*g
- Where, k = coefficient of static friction between the mug and the roof
m = mass of mug
g = acc. due to gravity
- So the maximum acc. of car ,
(m*a = k*m*g)
a = k*g
- Therefore max. a = 1.1*g = 1.1*9.81 = 10.791 m/s^2
An open container holds ice of mass 0.500kg at a temperature of -16.1?C . The mass of the container can be ignored. Heat is supplied to the container at the constant rate of 850J/minute .
The specific heat of ice to is 2100 J/kg?K and the heat of fusion for ice is 334
Answer: Tmelt = 19.89mins
Explanation: (complete question- 334×10^3j/kg. How much time, Tmelt passes before the ice starts to melt?)
For ice to melt, its temperature must be 0* C.
Let
Q1 = heat required to raise the ice temp from -16.1 °C. to 0* C.
Q2 = heat required to melt the ice
Q1 = MCp(delta T)
Q2 = m(Hf)
where
M = mass of the ice = 0.500 kg (given)
Cp = specific heat of ice = 2100 J/kg K (given)
delta T = temperature change = 0 - -16.1= 16.1 °C.
Hf = heat of fusion of ice = 334 x 10^3 J/jg.
Substituting values,
Q1 = 0.500× 2100× 16.1 = 16905J
Q2 = 0.500 × 334×10^3 = 167,000J
Heat required to melt the ice = Q1 + Q2 = 16905 + 167000 = 183905 J
How much time Tmelts = Q1/ 850 = 16905/850 = 19.89mins.
Two long parallel wires 6.50cm apart carry 18.5A currents in the same direction. Determine the magnetic field strength at a point 12.0cm from one wire and 13.6cm from the other.
Answer:
[tex]5.8\cdot 10^{-5}T[/tex]
Explanation:
The magnetic field produced by a current-carrying wire is:
[tex]B=\frac{\mu_0 I}{2\pi r}[/tex]
where
[tex]\mu_0=4\pi \cdot 10^{-7} H/m[/tex] is the vacuum permeability
I is the current in the wire
r is the distance from the wire
The direction of the field is given by the right-hand rule: the thum is placed in the same direction of the current, and the other fingers "wrapped" around the thumb gives the direction of the field.
First of all, here we are finding the magnetic field strength at a point 12.0cm from one wire and 13.6cm from the other: since the two wires are 6.50 cm apart, this means that the point at which we are calculating the field is located either on the left or on the right of both wires. So, by using the right-hand rule for both, we see that both fields go into the same direction (because the currents in the two wires have same direction).
Therefore, the net magnetic field will be the sum of the two magnetic fields.
For wire 1, we have:
[tex]I_1=18.5 A\\r_1=12.0 cm =0.12 m[/tex]
So the field is
[tex]B_1=\frac{(4\pi \cdot 10^{-7})(18.5)}{2\pi(0.12)}=3.1\cdot 10^{-5} T[/tex]
For wire 2, we have:
[tex]I_2=18.5 A\\r_2=13.6 cm=0.136 m[/tex]
So the field is
[tex]B_2=\frac{(4\pi \cdot 10^{-7})(18.5)}{2\pi(0.136)}=2.7\cdot 10^{-5} T[/tex]
Therefore, the total magnetic field is
[tex]B=B_1+B_2=3.1\cdot 10^{-5}+2.7\cdot 10^{-5}=5.8\cdot 10^{-5}T[/tex]
A 10-newton force is applied to a 2-kg block. The block slides across the floor at a constant speed of 5 m/s. What is a valid conclusion from this situation?
A) The force of friction was 10-newtons.
B) The 10-newton force was an unbalanced force.
C) There was no friction experienced by the block.
D) There is a net a force of 10-newtons on the block.
Answer:
The frictional force will also act on the box. So, we can say that the 10 N force is an unbalanced force.
Explanation:
It is given that, a 10-Newton force is applied to a 2-kg block. The block slides across the floor at a constant speed of 5 m/s.
The block is sliding with a constant speed. This shows the acceleration of the block is zero.
When an object slides that means the force acting on it is unbalanced i.e. the object will move in a particular direction.
A) The force of friction was 10-newtons.
The force of friction was 10-newtons. If the block is moving at constant speed, all the forces on the block must be balanced.
The muzzle velocity of a gun is the velocity of the bullet when it leaves the barrel. The muzzle velocity of one rifle with a short barrel is greater than the muzzle velocity of another rifle that has a longer barrel. In which rifle is the acceleration of the bullet larger
Answer:small barrel gun
Explanation:
Given
Muzzle velocity of bullet is greater in short barrel gun as compared to larger barrel gun
acceleration is given by change in velocity with respect to time
[tex]a=\dfrac{\Delta v}{\Delta t}[/tex]
In case of short barrel bullet time taken by bullet to reach its muzzle velocity is less therefore acceleration of small barrel bullet is more compared to long barrel bullet.
Final answer:
The muzzle velocity of a bullet can be calculated using the formula v = u + at. Using the given acceleration of 6.20 x 10⁵ m/s² and time of 8.10 x 10⁻⁴ s, the muzzle velocity is 502 m/s.
Explanation:
To calculate the muzzle velocity of a bullet, which is the velocity of the bullet as it leaves the barrel of a gun, we can use the formula for uniform acceleration: v = u + at, where v is the final velocity, u is the initial velocity (which is 0 m/s in this case, as the bullet starts from rest), a is the acceleration, and t is the time. Here, we are given an acceleration of 6.20 × 10⁵ m/s² and a time of 8.10 × 10⁻⁴ s. Plugging these values into the formula gives us the muzzle velocity.
Using the formula, we have: v = 0 m/s + (6.20 × 10⁵ m/s²)(8.10 × 10⁻⁴ s), which simplifies to v = 502 m/s. Therefore, the muzzle velocity of the bullet is 502 meters per second. This value represents how fast the bullet is traveling when it exits the gun barrel.
Some Canadian troops are sent (as part of a U.N. peacekeeping force) to a country located on the Earth's equator. At night, when homesickness makes them gaze sleeplessly at the stars, which of the following will be familiar to them (the same at the equator as in Canada):
a. the celestial poles are on the north and south points of the horizon
b. the celestial equator is overhead and passes through the zenith
c. all stars rise and set (none remain in the sky all night long)
d. all-stars are above the horizon exactly half a day
e. none of the above are the same on the equator as in Canada
Answer:
Option E is correct.
none of the listed options are the same on the equator as at the celestial north pole; the region where Canada is.
Explanation:
Firstly, let's explain some terms.
The earth's rotation is not straight forward as a sphere just rotating. The shape of the earth (better described as geoid; spherical, but a bit flattened at the poles), some gravitational forces on the earth and the existence of seasons mean the earth's rotation is tilted at an angle of 23.5° relative to our orbital plane; the plane of the earth's orbit round the sun.
When the tilted earth is considered, the celestial north and south poles are the north and south poles on the tilted earth's rotational axis.
And the celestial equator is when the equator on the tilted earth, equidistant from the celestial north and south poles is projected into space. A portion of the celestial equator passes through the real equator.
Now, the options one by one,
A) the celestial poles are on the north and south points of the horizon
At the celestial north and south poles, the poles are observable at the north and south points of the horizon.
At the celestial north pole, it is usually located by locating the north star, Polaris, and at the celestial South poles, there are about 4 different methods of locating the poles on the horizon.
But these poles aren't noticeable on the horizon at the equator. The northern one can only be noticed at regions around the celestial north pole.
Hence, the statement A isn't correct.
B) the celestial equator is overhead and passes through the zenith.
Zenith describes the the point in the sky directly above one's head.
At some points on the equator, the celestial equator is truly overhead, but the celestial equator doesn't pass overhead at the north pole, So, this isn't something common to the equator and the north pole.
C) all stars rise and set (none remain in the sky all night long)
Stars remaining in the sky all night long only occurs at the poles (north pole especially). Depending on the seasons, there are times and/or places that have 24 hour sunlight or darkness.
But, this is usually not the case at the equator. At the equator, all stars rise and set all year round. Only small noticeable longer days, shorter nights or longet nights, shorter days are experienced at the equator.
Hence, the statement given isn't true for the north pole and the equator.
D) all-stars are above the horizon exactly half a day.
The paths of the stars are vertical and are cut exactly in half by the horizon at the equator. Each star is up half the time and down half the time.
Unlike at the poles, as explained in the previous statement, they enjoy more varied daytime and night times.
Again, the statement described isn't true for the north pole and the equator.
Using a good pair of binoculars, you observe a section of the sky where there are stars of many different apparent brightnesses. You find one star that appears especially dim. This star looks dim because it is:_______.
Answer:
Using a good pair of binoculars, you observe a section of the sky where there are stars of many different apparent brightnesses. You find one star that appears especially dim. This star looks dim because it is farther away or it has a small radius.
Explanation:
Apparent magnitude in astronomy is the apparent brightness of a star that is seen from the Earth, that brightness can variate according to the distance at which the star is from the Earth or due to its radius.
That can be demonstrate with the next equation:
[tex]F = \frac{L}{4\pi r^2}[/tex] (1)
Where F is the radiant flux received from the star, L its intrinsic luminosity and r is the distance.
For example, an observer sees two motorbikes approaching it with its lights on but one of the motorbikes is farther, so the light of this one appears dimmer, even when the two lights emit the same amount of energy per second.
That is because the radiant flux decreases with the square distance, as can be seen in equation 1.
In the other hand, a bigger radius means that the gravity in the surface of the star will be lower, allowing that light can escape more easily:
[tex]g = \frac{GM}{R^2}[/tex] (2)
Where g is the surface gravity in the star, G is the gravitational constant, M is the mass of the star and R is the radius of the star.
Although solid matter is mostly empty space, we don't fall through the floor because atoms are constantly vibrating, even at absolute zero. of electrical forces. of gravitational forces. of nuclear forces. none of the above
We don't fall through the floor because of the electrical forces, specifically the electrostatic repulsion between negatively charged electrons surrounding atoms in both our bodies and the floor.
Although solid matter is mostly empty space, we don't fall through the floor because of electrical forces. Atoms consist mostly of empty space, with the volume of protons, neutrons, and electrons comprising less than 1% of an atom's total volume. The reason we perceive matter as solid and do not pass through the floor is that the electrons surrounding atoms are negatively charged, and the same charges repel each other. This repulsive electrostatic force is stronger at the scale we experience than gravitational force, which is comparatively weak.
Tito knows that one of the reasons people do not return to his electronics store is because of the slow service. How would a SWOT analysis classify the slow service at Tito's electronic store?
It is classified as Weakness in SWOT analysis.
Explanation:
SWOT analysis is performed to understand the characteristics of any start ups. It is the abbreviation of Strength, Weakness, Opportunities and Threats. It helps in self analyzing and to plan strategically for improvements. So in this different loopholes, advantages, benefits, profits and loss attained by any organized is classified in these four options like Strength, Weakness, Opportunities and Threats.
In the present case, the problem of slow service speed by Tito's company lead to reduction in customers. Also many of the customers are not returning back due to its slow speed in servicing any instrument. So this loophole is placed in the box of weakness in SWOT analysis. So it is classified as Weakness in SWOT analysis.
A spring hangs from the ceiling with an unstretched length of x 0 = 0.69 m x0=0.69 m . A m 1 = 7.5 kg m1=7.5 kg block is hung from the spring, causing the spring to stretch to a length x 1 = 0.84 m x1=0.84 m . Find the length x 2 x2 of the spring when a m 2 = 2.1 kg m2=2.1 kg block is hung from the spring. For both cases, all vibrations of the spring are allowed to settle down before any measurements are made.
Answer:
x2=0.732m
Explanation:
We can calculate the spring constant using the equilibrium equation of the block m1. Since the spring is in equilibrium, we can say that the acceleration of the block is equal to zero. So, its equilibrium equation is:
[tex]m_1g-k\Delta x_1=0\\\\\implies k=\frac{m_1g}{\Delta x_1}\\\\k=\frac{(7.5kg)(9.8m/s^{2})}{0.84m-0.69m}=490N/m[/tex]
Then using the equilibrium equation of the block m2, we have:
[tex]m_2g-k\Delta x_2=0\\\\\\implies x_2=x_0+\frac{m_2g}{k} \\x_2=0.69m+\frac{(2.1kg)(9.8m/s^{2})}{490N/m}= 0.732m[/tex]
In words, the lenght x2 of the spring when the m2 block is hung from it, is 0.732m.
A strong base ______ in solution.
A.) dissociates completely
B.) dissociates partially
C.) does not dissociate
D.) always solidifies
Answer:
a) dissociates completely
Explanation: strong base is a base that is completely dissociated in an aqueous solution.
A man stands on the roof of a building of height 14.9 m and throws a rock with a velocity of magnitude 31.9 m/s at an angle of 25.2 ∘ above the horizontal. You can ignore air resistance.Assume that the rock is thrown from the level of the roof.
Calculate the maximum height above the roof reached by the rock.
Calculate the magnitude of the velocity of the rock just before it strikes the ground.
Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.
Answer:
(a) 9.402 meters above the roof
(b) Velocity is 31.193 m/s just before hitting the ground
(c) Horizontal distance traveled is 104.2 meters
Explanation:
Let's first find the horizontal and vertical velocity components of the rock when it is thrown. These are:
[tex]V_x=31.9 * Cos(25.2)[/tex]
[tex]V_y=31.9*Sin(25.2)[/tex]
So we have,
Horizontal velocity = 28.864 m/s
Vertical velocity = 13.582 m/s
Now let's solve the three parts with this information:
(a) To find the maximum height, we need to use the fact that vertical velocity at this point will be zero, as the rock is just about to start falling downward. So we have:
[tex](V_f)^2 - (V_i)^2 = 2*a*s[/tex]
where [tex]V_f[/tex] is the final velocity = 0 m/s
[tex]V_i[/tex] is the initial velocity = 13.582 m/s
and a is the acceleration = -9.81 m/s^2
Solving, we get:
[tex]0^2-13.582^2=2*(-9.81)*s[/tex]
s = 9.402 m (distance above roof)
(b) Using the maximum height from (a), we can solve the following equation:
[tex](V_f)^2 - (V_i)^2 = 2*a*s[/tex]
where [tex]V_f[/tex] is the final velocity we need to find,
[tex]V_i[/tex] is the initial velocity = 0 m/s (from maximum height)
a is the acceleration = 9.81 m/s^2
and s = 14.9 + 9.402 = 24.302 m
Solving, we get:
[tex](V_f)^2 - (0)^2 = 2*(9.81)*(24.302)[/tex]
[tex]V_f = 21.836[/tex] m/s
As this is just the vertical velocity, to find the total velocity we have:
V = [tex]\sqrt{(V_x)^2+(V_y)^2}[/tex]
V = [tex]\sqrt{28.864^2 + 21.836^2}[/tex]
V = 36.193 m/s (Total velocity just before it hits the ground)
(c) To solve for this, we need to know the total time for this projectile motion, we can calculate this as follows:
[tex]s=u*t+\frac{1}{2} (a*t^2)[/tex]
here, s = -14.9 m
u = 13.582 m/s (initial vertical velocity)
a = - 9.81 m/s^2 (acceleration due to gravity)
Solving, we get:
[tex]-14.9 = 13.582t+0.5(-9.81t^2)[/tex]
and get the answers:
t1 = 3.61 s
t2 = -0.84 s
Since t2 isn't possible, our total time is t1 = 3.61 seconds.
Using this and our horizontal velocity, we can find the total distance traveled:
Distance = 28.864 * 3.61
Distance = 104.2 m (horizontal)
A horizontal spring-mass system has low friction, spring stiffness 205 N/m, and mass 0.6 kg. The system is released with an initial compression of the spring of 13 cm and an initial speed of the mass of 3 m/s.
(a) What is the maximum stretch during the motion? m
(b) What is the maximum speed during the motion? m/s
(c) Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system. What is the average power input in watts required to maintain a steady oscillation?
Answer:
a) x_max = 0.20794 m
b) v_max = 3.8436 m/s
c) P = 0.05883 W
Explanation:
Given:
- The stiffness k = 205 N / m
- The mass m = 0.6 kg
- initial compression of the spring xi = 13 cm
- initial speed of the mass vi = 3 m/s
Find:
(a) What is the maximum stretch during the motion? m
(b) What is the maximum speed during the motion? m/s
(c) Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system. What is the average power input in watts required to maintain a steady oscillation?
Solution:
- Conservation of energy principle can be applied that the total energy U of the system remains constant. So the Total energy is:
U = K.E + P.E
U = 0.5*m*v^2 + 0.5*k*x^2
- We will take initial point with given values and maximum compression x_max when v = 0.
0.5*m*vi^2 + 0.5*k*xi^2 = 0.5*k*x_max^2
(m/k)*vi^2 + xi^2 = x_max^2
x_max = sqrt ( (m/k)*vi^2 + xi^2 ) = sqrt ( (.6/205)*3^2 + .13^2
x_max = 0.20794 m
- The angular speed w of the harmonic oscillation is given by:
w = sqrt ( k / m )
w = sqrt ( 205 / 0.6 )
w = 18.48422 rad/s
- The maximum velocity v_max is given by:
v_max = - w*x_max
v_max = - (18.48422)*(0.20794)
v_max = 3.8436 m/s
- The amount of power required to stabilize each oscillation is given by:
P = E_cycle / T
Where, E = Energy per cycle = 0.02 J
T = Time period of oscillation
T = 2π/w
P = E_cycle*w / 2π
P = (0.02*18.48422) / 2π
P = 0.05883 W
The maximum stretch and maximum speed of the spring can be obtained from the conservation of energy. The average power required to maintain a steady oscillation can be calculated using the energy dissipation and the period of oscillation.
Explanation:This problem involves
conservation of energy
(kinetic and potential) which is given by the equation E = K + U where K is the kinetic energy = 0.5*m*v^2, m is the mass and v is the speed. U is the potential energy = 0.5*k*x^2 where k is the spring stiffness and x is the spring displacement. The maximum stretch of the spring occurs when all the kinetic energy has been transferred into potential energy (maximum potential energy). At this maximum stretch, v = 0, thus the total energy E becomes 0.5*k*x_max^2, from which we can calculate x_max. The maximum speed occurs when the spring is at its equilibrium position, at which point all the potential energy has been transferred into kinetic energy (maximum kinetic energy). At this point, x = 0, and the total energy E becomes 0.5*m*v_max^2, from which we can calculate v_max. The average power P required to maintain a steady oscillation with energy dissipation is P = energy dissipation / period of oscillation. The period T of the oscillation of a spring-mass system is given by T = 2*pi*sqrt(m/k).
Learn more about Conservation of Energy here:https://brainly.com/question/35373077
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An electrical cable consists of 200 strands of fine wire, each having 2.26 µΩ resistance. The same potential difference is applied between the ends of all the strands and results in a total current of 0.876 A. (a) What is the current in each strand? (b) What is the applied potential difference? (c) What is the resistance of the cable?
Answer:
4.38 × 10⁻³ A
0.395mΩ
452 µΩ
Explanation:
(a)the current in each strand is = the total current/ number of strand
0.876/ 200 = 4.38 × 10⁻³ A
(b) potential difference = IR
the total resistance = 2.26 µΩ × 0.876
= 0.395mΩ
c) resistance of the cable = 2.26 µΩ × 200
=452 µΩ
A soccer ball kicked on a level field has an initial vertical velocity component of 15.0 mps. assuming the ball lands at the same height from which it was kicked was the total time of the ball in the air?
Answer:
3seconds
Explanation:
Time of flight of an object is the time taken by the object to spend in the air before landing. Mathematically it is represented as;
T = 2u/g
Where g is the acceleration due to gravity = 10m/s²
If the vertical velocity component (u) is given as 15m/s, the time of flight will be;
T = 2(15)/10
T = 30/10
T = 3seconds.
Therefore the total time by the ball in the air is 3seconds
Explanation:
Below is an attachment containing the solution.
what is the best description of an atom with a full valence shell of electrons ?
A. it is a metal
B. it is not very reactive
C. it is very reactive
D. it is likely to form iconic bonds
Answer:
B
Explanation:
Because they have a full valence shell they don't need to make bonds to stabilize because they are already stable. These will be your Noble Gasses (ie Neon)
Which planetary body has the greatest gravitational pull?
Answer:
Jupiter
Explanation:
This is because it has the largest mass. Jupiter is massive and has the highest mass in the solar system.
Gravitational pull is dependent on the mass of body based on the newton's law of gravitation.
I hope this was helpful, please mark as brainliest
Answer:i had this on usatest prep it is jupiter
Explanation:
The blade of a lawn mower is rotating at an angular speed of 128 rad/s. The tangential speed of the outer edge of the blade is 32 m/s. What is the radius of the blade?
Answer:
0.25m
Explanation:
Using the expression that relates the angular velocity(w) and the linear velocity (v).
v= wr where;
w is the angular speed= 128rad/s
v is the linear speed = 32m/s
r is the radius
r = v/w
r = 32/128
r = 0.25m
The radius of the blade is 0.25m
Now suppose the crate is lifted so rapidly that air resistance was significant during the raising. How much work was done by the lifting force as the box was raised 1.5 m?
Answer:
A
Explanation:
- The complete question is:
" A crate is lifted vertically 1.5 m and then held at rest. The crate has weight 100 N (i.e., it is supported by an upward force of 100 N).Now suppose the crate is lifted so rapidly that air resistance was significant during the raising. How much work was done by the lifting force as the box was raised 1.5 m?"
Options:
1. More than 150 J
2. A bit less than 150 J because the air partially supported the crate.
3. No work was done.
4. Still 150 J
5. None of these
Solution:
- As the box is raised the work is done against gravity and air resistance opposes the motion of box. Hence, both the force of gravity ( Weight and air resistance act downward).
- The amount of work done is the sum of both work done against gravity and against air resistance.
- W = W_g + W_r
W > W_g
W > 100*1.5
W > 150 J
- Hence, the work done is greater than 150 J i.e work is also done against ai r resistance.
Consider a car that travels between points A and B. The car's average speed can be greater than the magnitude of its average velocity, but the magnitude of its average velocity can never be greater than its average speed. A) True B) False
Answer: A.True
Explanation: simply put the magnitude of the speed is a function of Distance while the magnitude of velocity is a function of Displacement. Displacement is the average Distance moved in different directions which will be smaller in magnitude compared to the Total Distance used the calculating the magnitude of speed.
A 3 billiard ball (mass = 0.16 kg) moving at 4.0 m/s collides elastically head-on with a cue ball (mass = 0.17 kg) that is initially at rest. Find the final velocities of both the cue ball and the 3 billiard ball after the collision.
Answer: 1.9394m/s
Explanation: m1 (mass of 3 billiard ball) = 0.16kg
U1(initial velocity of the 3 billiard ball) = 4.0m/s
M2 (mass of the cue ball) = 0.17kg
U1 (initial velocity of the cue ball) = 0m/s
Final velocities of both the cue ball and 3 billiard ball after collision = ?
According to the principle of conservation of linear momentum, total momenta before collision = total momenta after collision:
M1U1 + M2U2 = (M1+M2)V
0.16*4.0 + 0.17* 0 = (0.16 + 0.17)V
0.64 + 0 = 0.33V
V = 0.64/0.33 = 1.9394m/s
A bucket of water is being raised from a well using a rope. If the bucket of water has a mass of 6.2 kg, how much force (in N) must the rope exert on the bucket to accelerate it upward at 1 m/s2
Answer:
6.2N force
Explanation:
According to Newton's second law of motion, force is equal to the product of the mass of a body and its acceleration. Mathematically,
Force = mass × acceleration
Given mass of bucket of water = 6.2kg
acceleration of the bucket = 1m/s²
Force exerted on the rope = 6.2 × 1
= 6.2N
In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 2.0 m/s up a 20.0° inclined track. The combined mass of monkey and sled is 24 kg, and the coefficient of kinetic friction between sled and incline is 0.20. How far up the incline do the monkey and sled move? m
Answer:
2.65 m
Explanation:
From work-kinetic energy principle,
workdone by friction + workdone by gravity on sled = kinetic energy change of sled
Let h be the vertical height moved and d the distance moved along the incline. h = dsinθ where θ is the angle of the incline = 20°.
The workdone by gravity on the sled is mghcos180 = -mgh = -mgdsinθ
The frictional force = -μmgcosθ where μ = 0.20 and the work done by friction = -μmgcosθd
The kinetic energy change = 1/2m(v₂² - v₁²) where v₁ = initial speed = 2.0 m/s and v₂ = final speed = 0 m/s (since the sled stops)
So, -mgdsinθ - μmgcosθd = 1/2m(v₂² - v₁²)
-gd(sinθ - μcosθ) = 1/2(v₂² - v₁²)
d = (v₂² - v₁²)/-2g(sinθ - μcosθ)
substituting the values for the variables,
d = (0 - 2²) /[- 2 × 9.8(sin20 - 0.2cos20)]
d = -4 /[- 2 × 9.8(sin20 - 0.2cos20)]
d = -4/ - 1.51 = 2.65 m
It has moved up the incline a distance of 2.65 m.