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Phantasy [73]
3 years ago
6

A person looks across a stadium at the scoreboard, which appears blurry. The person looks down at the program, which is complete

ly in focus. What vision condition is described?
a) color blindness

b) astigmatism

c) farsightedness

d) nearsightedness
Physics
2 answers:
Alecsey [184]3 years ago
5 0

Answer:

d) nearsightedness

Explanation:

Nearsightedness (myopia) is a common vision condition in which you can see objects near to you clearly, but objects farther away are blurry. It occurs when the shape of your eye causes light rays to bend (refract) incorrectly, focusing images in front of your retina instead of on your retina

Pepsi [2]3 years ago
3 0
Answer is D. Nearsightedness is when a person can see near, but not far. Everything appears blurry from far away, but as you get close to it, it becomes more focused
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Please help me fill out the chart.??
LekaFEV [45]

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5 0
4 years ago
Un movil de masa 12Kg sobre el cual estan actuando varias fuerzas F_1=48N, F_2=60N y F_3=30N Calcular la aceleracion con la cual
Nikitich [7]

Answer:

Lamentablemente el problema está incompleto, pues no sabemos la dirección en la que se aplican las fuerzas. Por ello, voy a resolver el problema asumiendo dos casos. (abajo se puede ver una imagen donde se describe cada caso)

1) Todas las fuerzas están en la misma dirección.

Entonces la fuerza neta será la suma de las 3 fuerzas, entonces:

F = 48N + 60N + 30N = 138N

Y por la segunda ley de Newton sabemos que:

F = m*a

fuerza igual a masa por aceleración.

Entonces la aceleración está dada por:

a = F/m = 138N/12kg = 11.5 m/s^2

2) Segundo caso, suponemos que F1 es opuesta a F2 y F3

En este caso, la fuerza neta será:

F = F2 + F3 - F1 = 60N + 30N - 48N = 42N

En este caso, la aceleración será:

a = 42N/12kg = 3.5 m/s^2

7 0
3 years ago
An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance
jeka94

Answer:

1.8\times 105 N/C

Explanation:

We are given that

u=2\times 10^7 m/s

v=4\times 10^7 m/s

d=1.9 cm=\frac{1.9}{100}=0.019 m

Using 1m=100 cm

We have to find the electric field strength.

v^2-u^2=2as

Using the formula

(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)

16\times 10^{14}-4\times 10^{14}=0.038a

0.038a=12\times 10^{14}

a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2

q=1.6\times 10^{-19} C

Mass of electron,m=9.1\times 10^{-31} kg

E=\frac{ma}{q}

Substitute the values

E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}

E=1.8\times 105 N/C

7 0
4 years ago
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