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BabaBlast [244]

3 years ago

11

Brian has a medical condition called cone dystrophy, which is a disorder of the cone cells in the eye. What symptoms would you e

xpect Brian to have?

1 answer:

icang [17]3 years ago

4 0

Answer:

Poor Color Vision
Sensitivity to Bright Light
Vision Loss

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Apollo astronauts took a "nine iron" to the Moon and hit a golf ball about 180 m.

Studentka2010 [4]

So in calculating this one its is really hard to explain how i get it on solve it but you must consider this factors that i give in getting the answer. First is the distance cover by the ball when it is hit by the club, Second is you must estimate both of those data when it is in the moon and in the earth whre the gravity of the earth is 9.8m/s^2 so by calculating the Gravity of the moon or gMoon is equal 1.74m/s^2

7 0

3 years ago

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Two power lines run parallel for a distance of 220 m and are separated by a distance of 40.0 cm. If the current in each of the t

daser333 [38]

Answer:

The magnitude of force is 1.86 N and the direction of force is towards the other wire.

Explanation:

Given:

Current flowing through each power line, I = 130 A

Distance between the two power lines, d = 40 cm = 0.4 m

Length of power lines, L = 220 m

The force exerted by the power lines on each other is given by the relation:

$F = \frac{\mu_{0}LII }{2\pi d}$

Substitute the suitable values in the above equation.

$F = \frac{4\pi\times10^{-7}\times220\times130\times130 }{2\pi\times0.4}$

F = 1.86 N

Since the direction of current flowing through the power lines are opposite to each other, so the force is attractive in nature. Hence, the direction of force experienced by the power lines on each other is towards the each other.

5 0

3 years ago

A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for

nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block ( $W$ ), measured in newtons, is:

$W = m\cdot g$ (1)

Where:

$m$ - Mass of the block of ice, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 14\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the magnitudes of the weight and normal force of the block of ice are, respectively:

$N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$N = W = 137.298\,N$

And the pull force is:

$F_{pull} = 98\,N$

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

$m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}$ (2)

Where:

$v_{o}$ , $v_{f}$ - Initial and final speeds of the block, measured in meters per second.

$\Sigma F$ - Horizontal net force, measured in newtons.

$\Delta t$ - Impact time, measured in seconds.

Now we clear the final speed in (2):

$v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $m = 14\,kg$ , $\Sigma F = 98\,N$ and $\Delta t = 1.40\,s$ , then final speed of the ice block is:

$v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}$

$v_{f} = 9.8\,\frac{m}{s}$

The final speed of the block of ice is 9.8 meters per second.

6 0

2 years ago

A 45.00 kg person in a 43.00 kg cart is coasting with a speed of 19 m/s before it goes up a hill. there is no friction, what is

HACTEHA [7]

Answer:

the maximum vertical height the person in the cart can reach is 18.42 m

Explanation:

Given;

mass of the person in cart, m₁ = 45 kg

mass of the cart, m₂ = 43 kg

acceleration due to gravity, g = 9.8 m/s²

final speed of the cart before it goes up the hill, v = 19 m/s

Apply the principle of conservation of energy;

$mgh_{max} = \frac{1}{2}mv^2_{max}\\\\ gh_{max} = \frac{1}{2}v^2_{max}\\\\h_{max} = \frac{v^2_{max}}{2g} \\\\h_{max} =\frac{(19)^2}{2\times 9.8} \\\\h_{max} = 18.42 \ m$

Therefore, the maximum vertical height the person in the cart can reach is 18.42 m

5 0

2 years ago

How does a generator produce electrical energy?

Daniel [21]

The correct answer is B. it converts mechanical energy to electrical energy.

3 0

2 years ago

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