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3 years ago

A car has a kinetic energy of 4.32 x 10^5 J when traveling at a speed of 23 m/s.

Physics

1 answer:

Kisachek [45]3 years ago

4 0

Answer:

I think the most interesting 3rd century 3g durga was 5f5

Explanation:

I 50degree and I have not done a good morning mam sushant since I am not pregnant and have no

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Do electromagnetic waves require matter to travel?

VMariaS [17]

Answer:

No.

Explanation:

Electromagnetic waves do not require a medium of matter to move through, electromagnetic waves are used in things like your cell phone and telecommunications.

6 0

2 years ago

A horizontal spring is attached to a wall at one end and a mass at the other. The mass rests on a frictionless surface. You pull

Kaylis [27]

Answer:

Δt'/ T% = 90.3%

Explanation:

Simple harmonic movement is described by the expression

x = A cos (wt)

we find the time for the two points of motion

x = - 0.3 A

-0.3 A = A cos (w t₁)

w t₁ = cos -1 (-0.3)

remember that angles are in radians

w t₁ = 1.875 rad

x = 0.3 A

0.3 A = A cos w t₂

w t₂ = cos -1 (0.3)

w t₂ = 1,266 rad

Now let's calculate the time of a complete period

x= -A

w t₃ = cos⁻¹ (-1)

w t₃ = π rad

this angle for the forward movement and the same time for the return movement in the oscillation to the same point, which is the definition of period

T = 2 t₃

T = 2π / w s

now we can calculate the fraction of time in the given time interval

Δt / T = (t₁ -t₂) / T

Δt / T = (1,875 - 1,266) / 2pi

Δt / T = 0.0969

This is the fraction for when the mass is from 0 to 0.3, for regions of oscillation of greater amplitude the fraction is

Δt'/ T = 1 - 0.0969

Δt '/ T = 0.903

Δt'/ T% = 90.3%

4 0

3 years ago

A sled of mass 50 kg is pulled along a snow-covered, flat ground. The static friction coefficient is 0.3 and the kinetic frictio

Diano4ka-milaya [45]

Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

b) The weight of the sled is 490.35 newtons.

c) A force of 147.105 newtons is needed to start the sled moving.

d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

$F$ - External force exerted on the sled, measured in newtons.

$f$ - Friction force, measured in newtons.

$N$ - Normal force from the ground on the mass, measured in newtons.

$W$ - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

$W = m\cdot g$ (1)

Where:

$m$ - Mass, measured in kilograms.

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

If we know that $m = 50\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the weight of the sled is:

$W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$W = 490.35\,N$

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

$F_{min,s} = \mu_{s}\cdot W$ (2)

Where:

$\mu_{s}$ - Static coefficient of friction, dimensionless.

$W$ - Weight of the sled, measured in newtons.

If we know that $\mu_{s} = 0.3$ and $W = 490.35\,N$ , then the force needed to start the sled moving is:

$F_{min,s} = 0.3\cdot (490.35\,N)$

$F_{min,s} = 147.105\,N$

A force of 147.105 newtons is needed to start the sled moving.

d) The minimum force needed to keep the sled moving at constant velocity is:

$F_{min,k} = \mu_{k}\cdot W$ (3)

Where $\mu_{k}$ is the kinetic coefficient of friction, dimensionless.

If we know that $\mu_{k} = 0.1$ and $W = 490.35\,N$ , then the force needed to keep the sled moving at a constant velocity is:

$F_{min,k} = 0.1\cdot (490.35\,N)$

$F_{min,k} = 49.035\,N$

A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

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3 years ago

If the sound level at a point is 98.0 dB, what is the intensity of sound at that point (Unit=W/m^2)? please help will give brain

Firlakuza [10]

Answer: The decibel scale is a logarithmic scale where each bel or 10 decibels correspondents to a factor of ten. A power intensity of 10^(-12) watts per square meter is the standard reference for a SPL of 0 db. So an SPL of 98 db corresponds to a power intensity of 10^(9.8)*10^(-12) or 10^(9.8–12) w/m^2.

0.006309573 w/m^2.

You can also readily find the value for any given SPL using the online calculator at: http://www.sengpielaudio.com/calculator-soundlevel.htm

Explanation:

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2 years ago

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Free 25 points cause why the hell not

hodyreva [135]

Answer:

I love nice people. Doing anything interesting for Thanksgiving?

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3 years ago

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