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Zielflug [23.3K]
3 years ago
15

The ability to clearly see objects at a distance but not close up is properly called ________. The ability to clearly see object

s at a distance but not close up is properly called ________.
Physics
1 answer:
melamori03 [73]3 years ago
5 0

Answer:

Hyperopia

Explanation:

     In hyperopia ,people face difficulties  to see close up object , but can see object easily which are at a distance.

The main reason of hyperopia is our eyeball.When our eyeball become too  short , then light focus behind the retina. Sowe will face problem to see near object but we can see distance object easily. Hyperopia is the opposite of nearsightedness. Hyperopia can be corrected by using contact lenses.

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assume that the initial speed is 25 m/s and the angle of projection is 53 degree above the hroizontal. the cannon ball leaves th
finlep [7]

Answer:

A.  xmax = 131.49 m

B.  t = 8.74 s

C.  ymax = 220.33 m

Explanation:

A. In order to find the horizontal distance which cannon travels you first calculate the flight time. The flight time can be calculated by using the following formula:

y=y_o+v_osin\theta-\frac{1}{2}gt^2      (1)

yo: height from the projectile is fired = 200m

vo: initial velocity of the projectile = 25m/s

g: gravitational acceleration = 9.8 m/s^2

θ: angle between the direction of the initial motion of the ball and the horizontal = 53°

t: time

You need the value of t when the projectile hits the ground. Then, in th equation (1) you make y = 0m.

When you replace the values of all parameters in the equation (1), you obtain the following quadratic formula:

0=200+(25)sin53\°t-\frac{1}{2}(9.8)t^2\\\\0=200+19.96t-4.9t^2 (2)

You use the quadratic formula to obtain the value of t:

t_{1,2}=\frac{-19.96\pm\sqrt{(19.96)^2-4(-4.9)(200)}}{2(-4.9)}\\\\t_{1,2}=\frac{-19.96\pm65.71}{-9.8}\\\\t_1=8.74s\\\\t_2=-4.66s

You use the positive value because it has physical meaning.

Now, you can calculate the horizontal range of the projectile by using the following formula:

x_{max}=v_ocos\theta t      

x_{max}=(25m/s)(cos53\°)(8.74s)=131.49m

The cannon ball travels a horizontal distance of 131.49 m

B. The cannon ball reaches the canon for t = 8.74s

C. The maximum height is obtained by using the following formula:

y_{max}=y_o+\frac{v_o^2sin^2\theta}{2g}     (3)

By replacing in the equation (3) the values of all parameters you obtain:

y_{max}=200m+\frac{(25m/s)^2(sin53\°)^2}{2(9.8m/s^2)}\\\\y_{mac}=200m+20.33m=220.33m

The maximum height reached by the cannon ball is 220.33m

3 0
3 years ago
Two particles, one with charge − 3.77 μC −3.77 μC and one with charge 4.39 μC, 4.39 μC, are 4.34 cm 4.34 cm apart. What is the m
Fudgin [204]

Answer:

the magnitude of the force that one particle exerts on the other is 79.08 N

Explanation:

given information:

q₁ = 3.77 μC = -3.77 x 10⁻⁶ C

q₂ = 4.39 μC = 4.39 x 10⁻⁶ C

r = 4.34 cm = 4.34 x 10⁻² m

What is the magnitude of the force that one particle exerts on the other?

lFl = kq₁q₂/r²

   = (9 x 10⁹) (3.77 x 10⁻⁶) (4.39 x 10⁻⁶)/(4.34 x 10⁻²)²

   = 79.08 N

8 0
3 years ago
What is the acceleration of a Porsche that can go from 15 mi/hr to 75 mi/hr in 4 seconds?
Elodia [21]

Hi there!

Acceleration = change in velocity / change in time = Δv/Δt

Thus:

a = (75 - 15)/4 = 60/4 = 15 mi/hr²

8 0
2 years ago
Read 2 more answers
A small aircraft accelerated down a runway at 4.0 m/s²
Radda [10]

Given data in the problem :-

  • Acceleration (a) = 4.0 m/s^2
  • Initial velocity (u) = 0 m/s
  • Final velocity (v) = 34 m/s
  • Distance travelled by aircraft (S) =  ?

From Newton's Laws of Motion we know that ,

v = u + at  [t = Time taken by aircraft to cover the distance]

⇒ 34 = 0 + 4t

⇒ t = 34/4 s

∴  t = 8.5 s

From Newton's Laws of Motion we also know that ,

S = u.t + 1/2a.t^2

⇒ S = 0×8.5 + 1/2 × 4 × (8.5)^2 m

∴  S = 144.50 m

Thus the distance travelled by the aircraft while accelerating is 144.50 meter .

4 0
2 years ago
What is relative density​
Alja [10]

Answer:

Relative density, or specific gravity, is the ratio of the density (mass of a unit volume) of a substance to the density of a given reference material.

Explanation:

4 0
3 years ago
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