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astraxan [27]
3 years ago
15

Pwease help me with thesse three

Physics
1 answer:
DiKsa [7]3 years ago
7 0

Answer:

1. Spring

2. 23.5

3. Summer

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Q|C A string with linear density 0.500 g/m is held under tension 20.0 N. As a transverse sinusoidal wave propagates on the strin
julia-pushkina [17]

A string with linear density 0.500 g/m.

Tension 20.0 N.

The maximum speed  v_{y, max}

The energy contained in a section of string 3.00 m long as a function of v_{y, max}.

We are given following data for string with linear density held under tension :

μ = 0.5 \frac{g}{m}

  = 0.5 x 10⁻³ \frac{kg}{m}

T = 20 N

If string is L = 3m long, total energy as a function of v_{y, max} is given by:

E = 1/2 x μ x L x ω² x A²

  = 1/2 x μ x L x v^{2} _{y, max}

  = 7.5 x 10⁻⁴ v^{2} _{y, max}

So, The total energy as a function of  v^{2} _{y, max} = 7.5 x 10⁻⁴ v^{2} _{y, max}

Learn more about linear density problem here:

brainly.com/question/17190616

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3 0
2 years ago
Why is it that when riding in a car, you don't feel like you're moving?
UkoKoshka [18]

This may shock you:

We NEVER feel speed, velocity, or motion, as long as it's constant.

We only feel CHANGES in speed, velocity, or motion.

That means speeding up, slowing down, or changing direction.

As long as we're moving in a straight line at a constant speed, we don't feel anything.

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3 years ago
What do wind and water do to rocks at the surface?
RUDIKE [14]
Igneous rock can<span> form underground, where the magma cools slowly. Or, igneous </span>rock can<span> form above ground, where the magma cools quickly. When it pours out on Earth's </span>surface, magma is called lava. ... On Earth's Surface<span>, </span>wind and water can<span> break </span>rock<span> into pieces.</span>
6 0
4 years ago
Read 2 more answers
Which of the following statements best support the claim that human actions decreased the whooping crane population? (1
Anna35 [415]

Answer:

D

Explanation:

A is incorrect because it would actually increase the population...

B is incorrect because other birds do not matter, it is only whooping cranes...

C is incorrect because it would also increase the population.

8 0
3 years ago
Figure 8-56 shows a solid, uniform cylinder of mass 7.00 kg and radius 0.450 m with a light string wrapped around it. A 3.00-N t
AVprozaik [17]

Answer:

a) The cylinder has an angular acceleration of 3.810 radians per square second, b) The frictional force has a magnitude of 9 newtons and has the same direction of tension force.

Explanation:

The external force exerted on string creates a tension force that tries to move the cylinder in translation, but it is opposed by the friction force between cylinder and ground that generates rolling on cylinder. The Free Body Motion on cylinder-string system is presented below as attachment. Given that cylinder is a rigid body in planar motion, two equations of equilibrium for translation and an equation of equilibrium for rotation are needed to represent the system, which are now described:

\Sigma F_{x} = T + f = M\cdot R\cdot \alpha

\Sigma F_{y} = N - M\cdot g = 0

\Sigma M_{G} = (T-f)\cdot R = I_{G}\cdot \alpha

Where:

T - Tension, measured in newtons.

f - Friction force, measured in newtons.

M - Mass of the cylinder, measured in kilograms.

R - Radius of the cylinder, measured in meters.

\alpha - Angular acceleration, measured in radians per square second.

N - Normal force from ground exerted on cylinder, measured in newtons.

g - Gravitational acceleration, measured in meters per square second.

I_{G} - Moment of inertia of the cylinder with respect to its center of mass, measured in kilogram-square meters.

The moment of inertia of the cylinder is:

I_{G} = \frac{1}{2}\cdot M\cdot R^{2}

a) The angular acceleration is determined by solving on first and third equation after eliminating  friction force:

f = M\cdot R \cdot \alpha - T

(T-M\cdot R\cdot \alpha+T) \cdot R = I_{G}\cdot \alpha

2\cdot T\cdot R = (I_{G} + M\cdot R^{2})\cdot \alpha

\alpha = \frac{2\cdot T\cdot R}{I_{G}+M\cdot R^{2}}

\alpha = \frac{2\cdot T \cdot R}{\frac{1}{2}\cdot M\cdot R^{2}+M\cdot R^{2} }

\alpha = \frac{4\cdot T}{3\cdot M\cdot R}

If T = 3\,N, M = 7\,kg and R = 0.45\,m, then:

\alpha = \frac{4\cdot (3\,N)}{(7\,kg)\cdot (0.45\,m)}

\alpha = 3.810\,\frac{rad}{s^{2}}

The cylinder has an angular acceleration of 3.810 radians per square second.

b) The magnitude of the frictional force can be determined with the help of the following expression:

f = M\cdot R \cdot \alpha - T

Given that T = 3\,N, M = 7\,kg, R = 0.45\,m and \alpha = 3.810\,\frac{rad}{s^{2}}, the magnitude of the friction force is:

f = (7\,kg)\cdot (0.45\,m)\cdot \left(3.810\,\frac{rad}{s^{2}} \right)-3\,N

f = 9\,N

The frictional force has a magnitude of 9 newtons and has the same direction of tension force.

5 0
4 years ago
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