Please help me out !!!!!!

Your friend tells you that a shiny element has to be a metal. Do you agreed? Explain your reasoning

Svetradugi [14.3K]

no i dont agree, because a glass bowl is shiny but its not a metal. just because some metals are shiny doesnt make all metals shiny

3 0

3 years ago

Read 2 more answers

the other end of the pipe. With the aide of the pipe, does the applied force produce a smaller torque, a greater torque, or the

pav-90 [236]

Answer:

With the addition of the pipe we have a greater torque.

Explanation:

We need to complete the description of the problem, searchin in internet we have:

"Sometimes, even with a wrench, one cannot loosen a nut that is frozen tightly to a bolt. It is often possible to loosen the nut by slipping one end of a long pipe over the wrench handle and pushing at the other end of the pipe. With the aid of the pipe, does the applied force produce a smaller torque, a greater torque, or the same torque on the nut?"

With the addition of the pipe we have a greater torque, as it increases the distance or radius of torque.

We know that torque is defined, as the product of force by distance, in this way we have:

T = F * d

where:

T = torque [N*m]

F = force [N]

d = distance [m]

We can see in the above equation, that increasing the distance increases torque proportionally.

3 0

3 years ago

Suppose an object is launched from a point 320 feet above the earth with an initial velocity of 128 ft/sec upward, and the only

Ne4ueva [31]

Answer:

(a)Therefore the highest altitude attained by the object is =576 ft .

(b)Therefore the object takes 6 sec to fall to the ground.

Explanation:

Initial velocity: Initial velocity is a velocity from which an object starts to move.

u is usually used for notation of initial notation.

Final velocity: Final velocity is a velocity of an object after certain second from starting.

The final velocity is denoted by v.

Acceleration: The difference of final velocity and initial velocity per unit time

The S.I unit of acceleration is m/s².

(a)

Given that u= 128 ft\sec and g = 32 ft/sec².

At highest point the velocity of the object is 0 i.e v=0

Since the displacement is opposite to the gravity.

Therefore acceleration( a)= -g = -32 ft/sec².

To find the time this to happen we use the following formula

$v=u+at$

Here v=0

⇒0=128+(-32) t

⇒32t=128

⇒t = 4 sec

To determine the height we use the following formula

$s=ut+\frac{1}{2} at^2$

$\Rightarrow s= (128\times4)+\frac{1}{2}\times (-32) \times4^2$

⇒s= 256 ft

Therefore the highest altitude attained by the object is =(320+256)ft=576 ft .

(b)

At the highest point the velocity of the object is 0.

so u=0. a=g= 32 ft/sec² [ since the direction of gravity and the displacement are same] s= 576 ft

To determine the time to fall we use the following formula

$s=ut+\frac{1}{2} at^2$

$\Rightarrow 576 = (0\times t)+\frac{1}{2} \times 32 \times t^2$

$\Rightarrow 16\times t^2=576$

$\Rightarrow t^2=\frac{576}{16}$

$\Rightarrow t^2=36$

⇒t=6 sec

Therefore the object takes 6 sec to fall to the ground.

8 0

3 years ago

A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

$f'=(\frac{v+v_r}{v+v_s})f$

where

f is the original frequency

f is the apparent frequency

$v$ is the velocity of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

$v_s = v_A = -28.7 m/s$ (surface A is the source, which is moving towards the receiver)

$v_r = +62.2 m/s$ (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

$f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz$

(b) 0.232 m

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

$\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m$

(c) 1481.2 Hz

Again, we can use the same formula

$f'=(\frac{v+v_r}{v+v_s})f$

In the source frame (= on surface A), we have

$v_s = v_B = -62.2 m/s$ (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

$v_r = +28.7 m/s$ (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

$f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz$

(d) 0.225 m

The wavelength of the wave is given by

$\lambda=\frac{v}{f}$

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

8 0

3 years ago

A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne

Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

$F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz$

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0

3 years ago