If the cornea is reshaped (this can be surgically done or with contact lenses) to correct myopia, should its curvature be made greater or smaller? Explain. Also, explain how hyperopia can be corrected?
Answer:
Myopia curvature of the cornea if it is negative the curvatures are positive,
hypermetry,
Explanation:
Myopia is the visual defect that does not allow to see distant objects, which is why it is corrected with a divergent lens so that the image is formed on the retina, therefore, by reforming the curvature of the cornea if it is negative
therefore the curvature must decrease
To correct hypermetry, the curvatures are positive, consequently the curvature of the lens must increase
When Peter tosses an egg against a sagging sheet, the egg doesn't break due to
A) reduced impulse.
B) reduced momentum.
C) both of these
D) neither of these
It has to do with impulse or force. Just how the sheet has no volume. There is no sufficient impulse to crack the shell.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
The sagging sheet gives the impact with the egg additional time, which prevents the egg from breaking when it is hurled against it. This lessens the force the egg would have applied to the wall had it been flung at it.
It has to do with impulse or force. Just how the sheet has no volume. There is no sufficient impulse to crack the shell.
To learn more about force refer to the link:
brainly.com/question/13191643
#SPJ2
a girl is moving with a uniform velocity of 1.5 m/s then mathematically find her acceleration
Answer:
0
Explanation:
a = dv/dt
if v is constant than the slope of the v graph will be 0, so dv/dt is 0
a= 0
What is the length of the x-component of the vector shown below?
у
6
28°
Answer:
Explanation:
6cos28
=5.3 N
Where is the sun in relation to earths orbit
what will be the gravitational force between two heavenly bodies if the masses of both are tripled keeping the distance between them constant
Answer:
If the mass of one of the objects is tripled, then the force of gravity between them is tripled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces
Define measurements.
Answer:
act or process of measuring
Explanation:
Explanation:
the comparison of an unknown quantity with a known quantity.
A ball has a mass of 4.65kg and approximates a ping pong ball of mass 0.060kg that is at rest by striking it in an elastic collision. The initial velocity of the bowling ball is 5.00m/s, determine the final velocities of both masses after the collision.
Answer:
Look at work
Explanation:
Elastic Collision: Ki=Kf
M1=4.65kg
M2: 0.060kg
v1=5m/s
v2=0m/s
4.65*5+0.060*0=4.65*v1'+0.060*v2'
23.25+0=4.65v1'+0.060v2'
Also since it is an elastic collision we can use
v1+v1'=v2+v2'
4.65+v1'=v2'
4.65+v1'=v2'
Substitute into the earlier equation
23.25=4.65v1'+0.060(4.65+v1')
Expand
23.25=4.65v1'+0.279+0.06v1'
Solve for v1'
22.971=4.71v1'
v1'=4.88m/s
v2'=4.65+4.88=9.53m/s
Please show steps as to how to solve this problem
Thank you!
Answer:
Torques must balance
F1 * X1 = F2 * Y2
or M1 g X1 = M2 g X2
X2 = M1 / M2 * X1 = 130.4 / 62.3 * 10.7
X2 = 22.4 cm
Torque = F1 * X2 =
62.3 gm* 980 cm/sec^2 * 22.4 cm = 137,000 gm cm^2 / sec^2
Normally x cross y will be out of the page
r X F for F1 will be into the page so the torque must be negative
Given a 64.0 V battery and 30.0 Ω and 88.0 Ω resistors, find the current (in A) and power (in W) for each when connected in series. I30.0 Ω = A P30.0 Ω = W I88.0 Ω = A P88.0 Ω = W (b) Repeat when the resistances are in parallel. I30.0 Ω = A P30.0 Ω = W I88.0 Ω = A P88.0 Ω = W
Answer:
a. i. 0.542 A ii. 8.813 W iii. 0.542 A iv. 25.85 W
b. i. 2.13 A ii. 136.53 W iii. 0.727 A iv. 46.55 W
Explanation:
a. Find the current (in A) and power (in W) for each when connected in series.
Since the resistors are connected in series, their combined resistance is R = R₁ + R₂ where R₁ = 30.0 Ω and R₂ = 88.0 Ω.
So, substituting the values of the variables into the equation, we have
R = R₁ + R₂
R = 30.0 Ω + 88.0 Ω
R = 118.0 Ω
Since from Ohm's law, V = IR where V = voltage across circuit = battery voltage = 64.0 V, I = current in circuit and R = total resistance of circuit = 118.0 Ω
So, I = V/R = 64.0V/118.0 Ω = 0.542 A
Since the resistors are in series, the same current flows through them
i. Current in 30.0 Ω
Current in 30.0 Ω is I = 0.542 A since the resistors are in series.
ii Power in the 30.0 Ω
The power in the 30.0 Ω is P₁ = I²R₁ where I = current = 0.542 A and R₁ = resistance = 30.0 Ω
So, P₁ = I²R₁
= (0.542 A)² × 30.0 Ω
= 0.293764 A² × 30.0 Ω
= 8.8129 W
≅ 8.813 W
iii. Current in 88.0 Ω
Current in 88.0 Ω is I = 0.542 A since the resistors are in series.
iv. Power in the 88.0 Ω
The power in the 88.0 Ω is P = I²R₂ where I = current = 0.542 A and R₂ = resistance = 88.0 Ω
So, P₂ = I²R₂
= (0.542 A)² × 88.0 Ω
= 0.293764 A² × 88.0 Ω
= 25.8512 W
≅ 25.85 W
(b) Repeat when the resistances are in parallel.
Since the resistors are connected in parallel, the same voltage is applied across them.
i. Current in 30.0 Ω
Using Ohm's law, V = I₁R₁ where V = voltage = 64.0 V, I₁ = current in 30.0 Ω resistor and R₁ = resistance = 30.0 Ω
So, I₁ = V/R₁ = 64.0 V/30.0 Ω = 2.13 A
ii Power in the 30.0 Ω
The power in the 30.0 Ω resistor is P₁ = V²/R₁ where V = voltage across resistor = 64.0 V and R₁ = resistance = 30.0 Ω
So, P₁ = V²/R₁
P₁ = (64.0 V)²/30.0 Ω
P₁ = 4096 V²/30.0 Ω
P₁ = 136.53 W
iii. Current in 88.0 Ω
Using Ohm's law, V = I₂R₂ where V = voltage = 64.0 V, I₂ = current in 88.0 Ω resistor and R₂ = resistance = 88.0 Ω
So, I₂ = V/R₂ = 64.0 V/88.0 Ω = 0.727 A
iv. Power in the 88.0 Ω
The power in the 30.0 Ω resistor is P₂ = V²/R₂ where V = voltage across resistor = 64.0 V and R₂ = resistance = 88.0 Ω
So, P₂ = V²/R₂
P₂ = (64.0 V)²/88.0 Ω
P₂ = 4096 V²/88.0 Ω
P₂ = 46.55 W
describe the cause of earth's magnetism ?
A 2.90 m segment of wire supplying current to the motor of a submerged submarine carries 1400 A and feels a 2.00 N repulsive force from a parallel wire 4.50 cm away. What is the direction and magnitude (in A) of the current in the other wire? magnitude A direction
Answer:
[tex]I_2=30.9A[/tex]
Explanation:
From the question we are told that:
Wire segment [tex]l_s=2.9m[/tex]
Initial Current [tex]I_1=1400A[/tex]
Force [tex]F=2.00N[/tex]
Distance of Wire [tex]d=4.50cm=>0.0450m[/tex]
Generally the equation for Force is mathematically given by
[tex]F=\frac{\mu_0 * I_1*I_2*l_s}{2 \pi *r}[/tex]
[tex]F=\frac{4 \pi*10^{-7} *1400 I*I_2*2.9}{2 \pi *0.0450}[/tex]
[tex]I_2=\frac{22.5*10^-2}{2*10^{-7}*1400*2.6}[/tex]
[tex]I_2=30.9A[/tex]
Which physical phenomenon is illustrated by the fact that the prism has different refractive indices for different colors
Answer:
The incoming white light is composed of light of different colors,
Since these different colors have different refractive indices they are refracted at different angles from one another.
The output light is then separated by color creating a color spectrum.
Since n is greater for shorter wavelengths (violet colors) these wavelengths are refracted thru the larger angles.
Two spheres are rolling without slipping on a horizontal floor. They are made of different materials, but each has mass 5.00 kg and radius 0.120 m. For each the translational speed of the center of mass is 4.00 m/s. Sphere A is a uniform solid sphere and sphere B is a thin-walled, hollow sphere. Part B How much work, in joules, must be done on the solid sphere to bring it to rest? Express your answer in joules. VO AE4D ? J WA Request Answer Submit Part C How much work, in joules, must be done on the hollow sphere to bring it to rest? Express your answer in joules. Wa Request
Answer:
Explanation:
Moment of inertia of solid sphere = 2/5 m R²
m is mass and R is radius of sphere.
Putting the values
Moment of inertia of solid sphere I₁
Moment of inertia of hollow sphere I₂
Kinetic energy of solid sphere ( both linear and rotational )
= 1/2 ( m v² + I₁ ω²) [ ω is angular velocity of rotation ]
= 1/2 ( m v² + 2/5 m R² ω²)
= 1/2 ( m v² + 2/5 m v²)
=1/2 x 7 / 5 m v²
= 0.7 x 5 x 4² = 56 J .
This will be equal to work to be done to stop it.
Kinetic energy of hollow sphere ( both linear and rotational )
= 1/2 ( m v² + I₂ ω²) [ ω is angular velocity of rotation ]
= 1/2 ( m v² + 2/3 m R² ω²)
= 1/2 ( m v² + 2/3 m v²)
=1/2 x 5 / 3 m v²
= 0.833 x 5 x 4² = 66.64 J .
This will be equal to work to be done to stop it.
A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 N force on the ball.
What is the magnitude of the velocity after it hits the ground?
Answer:
9.25 m/s
Explanation:
Consider a tall building of height 200.0 m. A stone A is dropped from the top (from the cornice of the building). One second later another stone B is thrown vertically up from the point on the ground just below the point from where stone A is dropped.Birthstones meet at half the height of the tower. (a) Find the initial velocity of vertical throw of stone B.(b) Find the velocities of A and B, just before they meet.
Answer:
a) v₀ = 44.27 m / s, b) stone A v = 44.276 m / s, stone B v = 0.006 m / s
Explanation:
a) This is a kinematics exercise, let's start by finding the time it takes for stone A to reach half the height of the building y = 100 m
y = y₀ + v₀ t - ½ gt²
as the stone is released its initial velocity is zero
y- y₀ = 0 - ½ g t²
t = [tex]\sqrt{ -2(y-y_o)/g}[/tex]
t = [tex]\sqrt{ -2(100-200)/9.8}[/tex]
t = 4.518 s
now we can find the initial velocity of stone B to reach this height at the same time
y = y₀ + v₀ t - ½ g t²
stone B leaves the floor so its initial height is zero
100 = 0 + v₀ 4.518 - ½ 9.8 4.518²
100 = 4.518 v₀ - 100.02
v₀ = [tex]\frac{100-100.02}{4.518}[/tex]
v₀ = 44.27 m / s
b) the speed of the two stones at the meeting point
stone A
v = v₀ - gt
v = 0 - 9.8 4.518
v = 44.276 m / s
stone B
v = v₀ -g t
v = 44.27 - 9.8 4.518
v = 0.006 m / s
A 10 kg box is at static equilibrium and the downward pull of gravity acting on the box is 98 Newton’s what is the minimum force that would require to just pick up the box
Explanation:
static equilibrium means its on the floor or something
so slightly greater than 98 newtons in the upward direction
An object is 2.0 cm from a double convex lens with a focal length of 1.5 cm. Calculate the image distance
Answer:
0.857 cm
Explanation:
We are given that:
The focal length for a convex lens to be (f) = 1.5cm
The object distance (u) = - 2.0 cm
We are to determine the image distance (v) = ??? cm
By applying the lens formula:
[tex]\dfrac{1}{f} = \dfrac{1}{u}+\dfrac{1}{v}[/tex]
By rearrangement and making (v) the subject of the above formula:
[tex]v = \dfrac{uf}{u-f}[/tex]
replacing the given values:
[tex]v = \dfrac{(-2.0)(1.5)}{(-2.0 -1.5)}[/tex]
[tex]v = \dfrac{-3.0}{(-3.5)}[/tex]
v = 0.857 cm
How are Newton’s 1 and 2 law related?
In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with zero initial speed and constant acceleration a with arrow of magnitude 0.35 m/s2. What angle between a with arrow and the positive direction of the y axis would result in a collision?
Answer:
59.26°
Explanation:
Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.
Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.
Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration = acosθ.
So, y = ut + 1/2a't²
y = 0 × t + 1/2(acosθ)t²
y = 0 + 1/2(acosθ)t²
y = 1/2(acosθ)t² (1)
Also, both particles must move the same horizontal distance to collide in time, t.
Let x be the horizontal distance,
x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision
Also, using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration = asinθ.
So, x = ut + 1/2a"t²
x = 0 × t + 1/2(ainsθ)t²
x = 0 + 1/2(asinθ)t²
x = 1/2(asinθ)t² (3)
Equating (2) and (3), we have
vt = 1/2(asinθ)t² (4)
From (1) t = √[2y/(acosθ)]
Substituting t into (4), we have
v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²
v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)
v√[2y/(acosθ)] = ytanθ
√[2y/(acosθ)] = ytanθ/v
squaring both sides, we have
(√[2y/(acosθ)])² = (ytanθ/v)²
2y/acosθ = (ytanθ/v)²
2y/acosθ = y²tan²θ/v²
2/acosθ = ytan²θ/v²
1/cosθ = aytan²θ/2v²
Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1
secθ = ay(sec²θ - 1)/2v²
2v²secθ = aysec²θ - ay
aysec²θ - 2v²secθ - ay = 0
Let secθ = p
ayp² - 2v²p - ay = 0
Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have
ayp² - 2v²p - ay = 0
0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0
10.85p² - 15.68p - 10.85 = 0
dividing through by 10.85, we have
p² - 1.445p - 1 = 0
Using the quadratic formula to find p,
[tex]p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125[/tex]
Since p = secθ
secθ = 1.95625 or secθ = -0.51125
cosθ = 1/1.95625 or cosθ = 1/-0.51125
cosθ = 0.5112 or cosθ = -1.9956
Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.
So, cosθ = 0.5112
θ = cos⁻¹(0.5112)
θ = 59.26°
So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.
1. A message signal m(t) has a bandwidth of 5kHz and a peak magnitude of 2V. Estimate the bandwidth of the signal u(t) obtained when m(t) frequency modulates a carrier with a) kf = 10 Hz/V, b) kf = 100 Hz/V, and c) kf = 1000 Hz/V.
Answer:
3v at 5.3 herts
Explanation:
If an object travels on a circular path is an acceleration? What is changing to cause an acceleration?
What is (a) the x component and (b) the y component of the net electric field at the square's center
Answer:
What is (a) the x component and (b) the y component of the net electric field at the square's center
The primary coil in a transformer has 250 turns; the secondary coil has 500. Which is correct?
a. This is a step-down transformer.
b. The voltage in the secondary coil will be higher than in the primary.
c. The power in the secondary coil is greater.
d. The power in the primary coil is greater.
Explanation:
option b is the correct one
Could you show detailed steps in how to solve this problem please
Answer: See attached pic. Hope this helps.
Explanation:
b) Two skaters collide and grab on to each other on a frictionless ice. One of them, of mass 80 kg, is moving to the right at 5.0 m/s, while the other of mass 70 kg is moving to the left at 2.0 m/s. What are the magnitude and direction of the two skaters just after they collide
Answer:
The two skaters move with a speed of 1.73 m/s after the collision in the right direction.
Explanation:
Given that,
The mas of skater 1, m₁ = 80 kg
The speed of skater 1, u₁ = 5 m/s (right)
The mass of skater 2, m₂ = 70 kg
The speed of skater 2, u₂ = -2 m/s (left)
Let v is the magnitude of the two skaters just after they collide. They must have a common speed. So, using the conservation of momentum as follows :
[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\\v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]
Put all the values,
[tex]v=\dfrac{80(5)+70(-2)}{(80+70)}\\\\=1.73m /s[/tex]
So, the two skaters move with a speed of 1.73 m/s after the collision in the right direction.
g a horizontal wheel of radius is rotating about a vertical axis. What is the magnitude of the resultant acceleration of a bug that is hanging tightl on the rim of the wheel
Answer:
a = w² r
Explanation:
In this exercise, indicate that the wheel has angular velocity w, the worm experiences the same angular velocity if it does not move, and has an acceleration towards the center of the circle, according to Newton's second law, called the centripetal acceleration.
a = v² / r
angular and linear variables are related
v = w r
we substitute
a = w² r
where r is the radius of the wheel
How many neutrons are in an isotope of selenium-83
A. 34
B. 83
C. 49
D. 117
Answer:
C.49 is yr ans...
hope it helpsstay safe healthy and happy....At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.58 g
Answer:
w = 1,066 rad / s
Explanation:
For this exercise we use Newton's second law
F = m a
the centripetal acceleration is
a = w² r
indicate that the force is the mass of the body times the acceleration
F = m 0.58g = m 0.58 9.8
F = 5.684 m
we substitute
5.684 m = m w² r
w = [tex]\sqrt{5.684/r}[/tex]
To finish the calculation we must suppose a cylinder radius, suppose it has r = 5 m
w = [tex]\sqrt{ 5.684/5}[/tex]
w = 1,066 rad / s
If there are no other changes, explain what effect reducing the mass of the car will have on its acceleration when starting to move.
Answer:
when the mass of an object is decreased, the acceleration will increase
when mass is increased, acceleration decreases