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3 years ago

Which statement explains why a short eyeball causes farsighted

2 answers:

8 0

I think the correct answer would be the rays of light focus behind the retina. It is a condition called farsightedness or hyperopia. Due to his, objects that are close to the eyes seems to be blurry and as it worsens everything would become blurry to the eye.

Nesterboy [21]3 years ago

3 0

Answer: the rays of light focus beyond the retina

Explanation:

You might be interested in

What are the assumptions on which the cosmological principle is based?.

prisoha [69]

Homogeneity and isotropy, On large enough scales, the Universe looks pretty much the same in all directions. The big bang theory is based on two assumptions: the first is centered around Einstein's general theory of relativity, which accurately describes gravity and the interactions of matter; and the second, also known as the cosmological principle, asserts that the universe is homogeneous and isotropic on a large enough scale.

Hope this helps!
Please give Brainliest!

4 0

3 years ago

You connect a 250-2 resistor, a 1.20-mH inductor, and a 1.80-uF capacitor in series across a 60.0-Hz, 120-V (peak) source. When

Basile [38]

Answer:

Explanation:

reactance of inductor = wL = 2 X 3.14 X 60 X 1.2 X 10⁻³ = .45 ohm.

reactance of capacitor = 1/wC = 1/( 2 X 3.14 X 60 X 1.8 X 10⁻⁶ ) = 1474.4

Impedence of the circuit =[ R² + ( I/wC - wL) ]¹/² = [250² + ( 1474.4-.45 )]¹/²

Impedence = 1495 ohm.

RMS Voltage = 120/ 1.414 = 84.86 V

current = 84.86 / 1495 = 0.0576

Potential over resistance = 0.0576 x 250 = 14.2 V.

4 0

4 years ago

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

6 0

3 years ago

In Europe, a large circular walking track with a

Elina [12.6K]

Answer: 10.67

Explanation:

First, we needed to calculate the radius and this will be:

= Diameter / 2

= 0.900km / 2

= 0.450km

Since s = 3.00miles, we need to convert it to kilometers and this will be:

1mile = 1.6km

3 miles = 3 × 1.6km = 4.8km

Radians = s/r = 4.8/0.45 = 10.67

4 0

3 years ago

A basketball player spins the ball with an angular acceleration of 10 rad/s2. What is the ball’s final angular velocity if the b

Lena [83]

Explanation:

It is given that,

The angular acceleration of the basketball, $\alpha=10\ rad/s^2$

Time taken, t = 3 seconds

We need to find the ball’s final angular velocity if the ball starts from rest. It can be calculated using definition of angular acceleration i.e.

$\alpha=\dfrac{\omega_f-\omega_i}{t}$

$\omega_i=0\ (rest)$

$\omega_f=\alpha t$

$\omega_f=10\ rad/s^2\times 3\ s$

$\omega=30\ rad/s$

So, the ball's final angular velocity is 30 rad/s. Hence, this is the required solution.

8 0

3 years ago