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elena-14-01-66 [18.8K]
3 years ago
13

PLEASE HELP WILL GIVE BRAINLIEST!!!

Physics
1 answer:
Elan Coil [88]3 years ago
6 0

The spring constant is 181.0 N/m

Explanation:

We can solve the problem by applying the law of conservation of energy. In fact, the elastic potential energy initially stored in the compressed spring is completely converted into gravitational potential energy of the dart when the dart is at its maximum height. Therefore, we can write:

\frac{1}{2}kx^2 = mgh

where the term on the left represents the elastic potential energy of the spring while the term on the right is the gravitational potential energy of the dart at maximum height, and where

k is the spring constant of the spring

x = 2.08 cm = 0.0208 m is the compression of the spring

m = 12.3 g = 0.00123 kg is the mass of the dart

g=9.8 m/s^2 is the acceleration due to gravity

h = 3.25 m is the maximum height of the dart

Solving for k, we find:

k=\frac{2mgh}{x^2}=\frac{2(0.00123)(9.8)(3.25)}{(0.0208)^2}=181.0 N/m

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

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Phantasy [73]

Force equals mass time acceleration. Weight is a force and it can replace force in the equation. The acceleration would be gravity, which is an acceleration.

1.)

Fw (weight) = m (mass) · g (gravity, 9.8 m/s²)

Fw = m * 9.81 m/s²

560N = m · 9.81 m/s²

m ≈ 57.08 kg

2.)

d = 350 meters

t = 65 seconds

velocity = d/t

velocity = 350 meters / 65 seconds

velocity ≈ 5.38 meters/sec

3.)

Force = 35N

Distance = 2 meters

Work = Force · Distance

Work = 35N · 2 meters

Work = 70 J

3 0
2 years ago
A 70.0 kg person climbs stairs, gaining 3.90 meters in height. Find the work done (in J) to accomplish this task.
Veronika [31]

Answer:

Work done, W = 2675.4 J

Given:

mass, m = 70.0 kg

height, H = 3.90 m

Solution:

According to the question, as the person jumps the stairs up, there is an increase in the potential energy of the person which is provided by the work done in climbing the stairs and is given by:

Work done, W = mgH

where

g = acceleration due to gravity = 9.8 m/s^{2}[tex][tex]W = 70.0\times 9.8\times 3.90 = 2675.4 J

6 0
3 years ago
Find the equivalent resistance of this parallel circuit with two strands.
svlad2 [7]
In a parallel circuit, the total resistance calculated from the individual resistances is computed from the formula: 1/Rt = 1/R1 + 1/R2. substituting R1 and R2, then 
1/Rt = 1/7 + 1/49 
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Rt = 49/8 <span>Ω

The total resistance hence is </span>49/8 Ω
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An early method of measuring the speed of light makes use of a rotating slotted wheel. A beam of light passes through one of the
Readme [11.4K]

Answer:

4.7 x 10³ rad / s

Explanation:

During the time light goes and comes back , one slot is replaced by next slot while rotating  before the light source

Time taken by light to travel a distance of 2 x 500 m is

= (2 x 500) / 3 x 10⁸

= 3.333 x 10⁻⁶ s .

In this time period, two consecutive slots come before the source of light one after another by rotation. There are 400 slots so time taken to make one rotation

= 3.333 x 10⁻⁶ x 400

= 13.33 x 10⁻⁴ s

This is the time period so

T = 13.33 X 10⁻⁴

Angular speed

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= \frac{2\times3.14}{13.33\times10^{-4}}

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3 years ago
An object islaunched at velocity of 20 m/s in a direction making an angle of 25 degree upward with the horizontal. what is the m
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Answer:

\displaystyle y_m=3.65m

Explanation:

<u>Motion in The Plane</u>

When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.

The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of v_o and \theta\\ as the initial speed and angle, then we have

\displaystyle v_x=v_o\ cos\theta

\displaystyle v_y=v_o\ sin\theta-gt

\displaystyle x=v_o\ cos\theta\ t

\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}

If we want to know the maximum height reached by the object, we find the value of t when v_y becomes zero, because the object stops going up and starts going down

\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt

Solving for t

\displaystyle t=\frac{v_o\ sin\theta }{g}

Then we replace that value into y, to find the maximum height

\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2

Operating and simplifying

\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}

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\displaystyle v_o=20\ m/s,\ \theta=25^o

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\displaystyle y_m=3.65m

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3 years ago
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