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Sonbull [250]
3 years ago
8

A fish of 108 N is attached to the end of a dangling spring, stretching the spring by 14.0 cm. What is the mass of a fish that w

ould stretch the spring by 23.0 cm?
Physics
1 answer:
siniylev [52]3 years ago
4 0

Answer:

The mass of the fish is 18.1 kg.

Explanation:

Given;

weight of the fish, F = 108 N

extension of the spring by the given weight, x = 14 cm = 0.14 m

First, determine the elastic constant of the spring by applying Hook's law;

F = kx

where;

k is the spring constant

k = F/x

k = 108 / 0.14

k = 771.43 N/m

When the spring is stretched to 23cm, the mass of the fish is calculated as follows;

F = mg = Kx\\\\m = \frac{Kx}{g} \\\\m = \frac{771.43\times 0.23}{9.8} \\\\m = 18.1 \ kg

Therefore, the mass of the fish is 18.1 kg.

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A piece of copper metal is initially at 100.0°C. It is dropped into a coffee cup calorimeter containing 50.0 g of water at a tem
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A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0o above the horizontal towar
Artist 52 [7]

Answer:

a)   v₀ = 32.64 m / s , b)  x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

         x = v₀ₓ t

         y = v_{oy} t - ½ g t²

We look for the components of speed with trigonometry

         sin 43 = v_{oy} / v₀

         cos 43 = v₀ₓ / v₀

         v_{oy} = v₀ sin 43

         v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

         t = x / v₀ cos 43

         y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

          y = x tan 43 - ½ g x² / v₀² cos² 43

          1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

           v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

          v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

          v₀ = √ (35280 / 33.11)

          v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

         y = v_{oy} t - ½ g t²

         .y = v₀ sin43 t - ½ g t²

        25 = 32.64 sin 43 t - ½ 9.8 t²

        4.9 t² - 22.26 t + 25 = 0

         t² - 4.54 t + 5.10 = 0

We solve the second degree equation

         t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

         t = (4.54 ± 0.46) / 2

         t₁ = 2.50 s

         t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

         x = v₀ₓ t

         x = v₀ cos 43 t

         x = 32.64 cos 43  2.50

         x = 59.68 m

8 0
3 years ago
object P1 and P2 are in a straight line with the normal to a plane mirror.If P1 and P2 are 18m and 21m away from the mirror. Cal
VARVARA [1.3K]

The distance between object P1 and its image formed is determined as 36 m.

<h3>Distance of the image</h3>

The distance of the image formed by object P1 is calculated as follows;

In a plane mirror; object distance = image distance

image distance of P1 = 18 m

distance between object and image = 18m + 18 m = 36 m

Thus, the distance between object P1 and its image formed is determined as 36 m.

Learn more about plane mirrors here: brainly.com/question/1126858

#SPJ1

8 0
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