Vf=V0+at solve for a

The density of a material is calculated by:

anyanavicka [17]

Answer:

The density of the material is defined as the mass of the material to its volume.

ρ = M / V

Explanation:

The density of the material is a scalar quantity.

If the mass and the volume of the material is known the density of the material can be calculated using the formula,

ρ = M / V

The mass of the material can be found using a physical balance.

The volume of a geometrically symmetrical material can be found using the geometrical measurement. For a brick,

V = l x b x h

For non-geometrically shaped material the volume can be found from the displacement method using water.

5 0

3 years ago

A small steel wire of diameter 1.0mm is connected to an oscillator and is under a tension of 5.7N . The frequency of the oscilla

Fiesta28 [93]

Answer:

a) $P = 2 \pi^2 (57 Hz)^2 (0.0054 m)^2 \sqrt{\pi(0.0005m)^2 (7800 \frac{kg}{m^3}) (5.7N)}= 0.349 W$

b) For this case we se that $P \propto A^2 f^2$

Since the power is constant but the frequency is doubled we will see that $A^2 \propto \frac{P}{f^2}$

So the original amplitude is $A_i \propto \sqrt{\frac{P}{f^2}}$

And if the frequency is doubled we have that:

$A^2_f \propto \frac{P}{(2f)^2}= \frac{P}{4f^2}$

$A_f \propto \sqrt{\frac{P}{4f^2}}= \frac{1}{2} \sqrt{\frac{P}{f^2}}$

So then we will see that the amplitude would be reduced the half and for this case:

$A_f = \frac{0.54cm}{2}= 0.27 cm = 0.0027 m$

Explanation:

For this case we have the following data given:

$D= 1mm = 0.001 m$ represent the diameter

$r = D/2= 0.0005m$ represent the radius

$T= 5.7 N$ represent the tension

$f = 57 Hz$ represent the frequency of the oscillator

$A= 0.54 cm = 0.0054 m$ represent the amplitude of the wave

Part a

For this case we can assume that the power transmitted to the wave is the same power of the oscillator. and we have the following formula for the power:

$P= 2 \pi^2 \rho S v f^3 A^2$

This expression can be written in different ways:

$P= 2 \pi^2 \rho S \sqrt{\frac{T}{\mu}} f^2 A^2$

$P= 2\pi^2 \rho S \sqrt{\frac{T}{\rho S}} f^2 A^2$

$P= 2 \pi^2 f^2 A^2 \sqrt{S \rho T}$

Where f is the frequency , T the tension $rho= 7800 \frac{kg}{m^3}$ the density of the steel, A the amplitude and $S= \pi r^2$ the area, so then we have everuthing in order to replace and we got:

$P = 2 \pi^2 (57 Hz)^2 (0.0054 m)^2 \sqrt{\pi(0.0005m)^2 (7800 \frac{kg}{m^3}) (5.7N)}= 0.349 W$

Part b

For this case we se that $P \propto A^2 f^2$

Since the power is constant but the frequency is doubled we will see that $A^2 \propto \frac{P}{f^2}$

So the original amplitude is $A_i \propto \sqrt{\frac{P}{f^2}}$

And if the frequency is doubled we have that:

$A^2_f \propto \frac{P}{(2f)^2}= \frac{P}{4f^2}$

$A_f \propto \sqrt{\frac{P}{4f^2}}= \frac{1}{2} \sqrt{\frac{P}{f^2}}$

So then we will see that the amplitude would be reduced the half and for this case:

$A_f = \frac{0.54cm}{2}= 0.27 cm = 0.0027 m$

8 0

4 years ago

At 20°c, the resistance of a sample of nickel is 525 ohms. what is the resistance when the sample is heated to 70°C?

Ipatiy [6.2K]

The resistance of the sample is $682.5\Omega$

Explanation:

The relationship between resistance of a material and temperature is given by the equation

$R(T)=R_0(1+\alpha (T-T_0))$

where

$R_0$ is the resistance at the temperature $T_0$

$\alpha$ is the temperature coefficient of resistance

For the sample of nickel in this problem, we have:

$R_0 = 525 \Omega$ when the temperature is $T_0 = 20^{\circ}C$

While the temperature coefficient of resistance of nickel is

$\alpha = 0.006/^{\circ}C$

Therefore, the resistance of the sample when its temperature is

$T=70^{\circ}C$

is

$R=(525)(1+0.006(70-20))=682.5 \Omega$

Learn more about resistance:

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brainly.com/question/10597501

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#LearnwithBrainly

3 0

3 years ago

What is the x component of this vector: 8 N [North 50 degrees East]?

maksim [4K]

Answer:

Explanation:

Given

Force = 8N

Direction = N50°E

Required

x component of the Force

Fx = F cos theta

Fx = 8cos 50°

Fx = 8(0.6428)

Fx = 5.1423N

Hence the x component of the force is 5.1423N

4 0

3 years ago

What is the acceleration of a wagon of mass 20kg if a horizontal force of 64 is applied to it

zlopas [31]

Here we have to calculate the acceleration of the wagon.

as per the question the mass of the wagon is given as 20 kg

the horizontal force that is applied on the body is 64 newton.

as the perfect unit of the fore is not mentioned it may be 64N or 64 dyne or 64 kgf.

as per Newton's second law of motion we know that the rate of change of momentum is the applied force.from the quantitative definition of Newton's second law we know that applied force is equal to the product of mass and acceleration.

hence $F=mass *acceleration$

$Acceleration=\frac{force\force}{mass}$

let F= 64 N,then $a=\frac{64N}{20kg}$

a=3.2 m/s^2

if F=64kgf=64×9.8 N,then $a=\frac{64*9.8N}{20 kg}$

=31.36 m/s^2

if F =64 dyne ,the acceleration is found to be 0.0032 cm/s^2 [ans]

6 0

4 years ago