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lora16 [44]
3 years ago
14

A spring balance used to weigh candy is built with a spring. The spring stretches 3.00 cm when a 15 N weight is placed in the pa

n. When the weight is replaced with 3kg of candy what distance will the spring stretch?
7.0cm
4.5cm
3.8cm
5.9cm
Physics
1 answer:
OleMash [197]3 years ago
6 0

Answer:

5.9 cm

Explanation:

We'll begin by calculating the spring constant (K) of the spring. This can be obtained as follow:

Force (F) = 15 N

Extention (e) = 3 cm

Spring constant (K) =.?

F = Ke

15 = K × 3

Divide both side by 3

K = 15 / 3

K = 5 N/cm

Therefore, the spring constant (K) is 5 N/cm.

Now, we shall determine how far the spring will stretch when a 3 kg candy is placed on the spring balance.

This can be obtained as follow:

First, we shall determine the weight of the object.

Mass of object = 3 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Weight (W) =?

W = mg

W = 3 × 9.8

W = 29.4 N

Finally, we shall determine how far the spring will stretch as follow:

Weight (W) = Force (F) = 29.4 N

Spring constant (K) = 5 N/cm

Extention (e) =?

F = ke

29.4 = 5 × e

Divide both side by 5

e = 29.4 / 5

e = 5.88 ≈ 5.9 cm

Therefore, the spring will stretch 5.9 cm when a 3 kg candy is placed on the spring balance.

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First we need to find the speed of the dolphin sound wave in the water. We can use the following relationship between frequency and wavelength of a wave:
v=\lambda f
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Using \lambda = 2 cm = 0.02 m and f=22 kHz = 22000 Hz, we get
v=(0.02 m)(22000 Hz)=440 m/s

We know that the dolphin sound wave takes t=0.42 s to travel to the tuna and back to the dolphin. If we call L the distance between the tuna and the dolphin, the sound wave covers a distance of S=2 L in a time t=0.42 s, so we can write the basic relationship between space, time and velocity for a uniform motion as:
v= \frac{S}{t}= \frac{2L}{t}
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A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimumcoefficient of static friction
zhenek [66]

Answer:

c. 0.816

Explanation:

Let the mass of car be 'm' and coefficient of static friction be 'μ'.

Given:

Speed of the car (v) = 40.0 m/s

Radius of the curve (R) = 200 m

As the car is making a circular turn, the force acting on it is centripetal force which is given as:

Centripetal force is, F_c=\frac{mv^2}{R}

The frictional force is given as:

Friction = Normal force × Coefficient of static friction

f=\mu N

As there is no vertical motion, therefore, N=mg. So,

f=\mu mg

Now, the centripetal force is provided by the frictional force. Therefore,

Frictional force = Centripetal force

f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu=\frac{v^2}{Rg}

Plug in the given values and solve for 'μ'. This gives,

\mu = \frac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}\\\\\mu=\frac{1600\ m^2/s^2}{1960\ m^2/s^2}\\\\\mu=0.816

Therefore, option (c) is correct.

7 0
3 years ago
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