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professor190 [17]
2 years ago
6

AYUDA!!!!

Physics
1 answer:
xeze [42]2 years ago
3 0

Answer:

4. B) 69N

5. B) 7.2kg

Explanation:

Dados los siguientes parámetros;

Masa = 7kg

Peso = 71N

La aceleración debida a la gravedad es igual a 9.8m/s².

a. Para determinar el peso de la siguiente masa;

Matemáticamente, el peso viene dada por la fórmula;

Peso = masa * aceleración debida a la gravedad

Peso = 7 * 9.8

Peso = 68.6 ≈ 69 Newton.

b. ¿Cuál es la masa sobre la tierra de un objeto que pesa.

Peso = masa * aceleración debida a la gravedad

71 = masa * 9.8

Masa = 71/9.8

Masa = 7.2 Kilograms.

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Issac and Blaise decide to race. They both start at the same position at the same time. Issac runs at 2m/s but decides to take a
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Let the Blaise runs for time "t" to complete the race

so the total distance he moved is given by

d_1 = 1* t

Now Issac runs for time t = "t - 2*60"

because it took rest for 2 minutes

d_2 = 2*(t - 120)

now it is given that Blaise wins by 10 m distance

d_1 - d_2 = 10

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t - 2t + 240 = 10

t = 230 s

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3 years ago
A student performs a lab measuring the velocities of toy cars of different masses
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The answer is Car 1 and Car 2.
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3 years ago
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Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted
galina1969 [7]

Answer:

1) When d_{o} < d_{i} (hence  d_{o} < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When d_{o} > d_{i} (and d_{o} < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}

Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}

Where:

f = Focal length of the mirror = R/2

d_{i} = Image distance from the mirror

d_{o} = Object distance from the mirror

h_{i} = Image height

h_{o} = Object height

d_{o} is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

d_{i} is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When d_{o} < d_{i} (hence  d_{o} < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When d_{o} > d_{i} (and d_{o} < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

5 0
3 years ago
Two bodies are falling with negligible air resistance, side by side, above a horizontal plane. If one of the bodies is given an
borishaifa [10]

Answer:a

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Given

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since there is no change in vertical acceleration therefore time taken by both bodies is same irrespective of change in horizontal direction

suppose both start from rest from a height of h

time taken

t=\sqrt{\frac{2h}{g}}

The only difference is that body with horizontal acceleration will be some distance away from first                  

4 0
3 years ago
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Water emerges straight down from a faucet with a 2.51-cm diameter at a speed of 3.04 m/s. (because of the construction of the fa
Mkey [24]
This is a question on conservation of energy. That is,
mgh + KE1 = KE2
mgh +1/2mv1^2 = 1/2mv2^2
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Where, h = 0.2 m, v1 =3.04 m/s
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Now, Volumetric flow rate, V/time, t = Surface area, A*velocity, v
Where,
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At 0.2 m below,
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But, A = πr^2
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3 years ago
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