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2 years ago

During the motion of the slinky in a transverse wave, what do the particles of the slinky coil do?

Physics

1 answer:

Mazyrski [523]2 years ago

5 0

Answer:

C.) The slinky particles move up and down

Explanation:

Transverse Wave-

A wave that has a disturbance perpendicular to the wave motion

Hello! This is the correct answer! Have a blessed day! :)

If you are in K12, please review the lesson! :) It will give you some very helpful definitions! I hope this helped!

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Susan, driving north at 53 mphmph , and Shawn, driving east at 63 mphmph , are approaching an intersection. Part A What is Shawn

mafiozo [28]

Answer:

Shawn's speed relative to Susan's speed = 10 mph

Resultant velocity = 82.32 mph

Explanation:

The given data :-

i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.

ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.

iii) The speed of both Susan and Shawn is relative to earth.

iv) The angle between Susan in north and Shawn in east is 90°.

We have to find Shawn's speed relative to Susan's speed.

v₂₁ = v₂ - v₁ = 63 - 53 = 10 mph

Resultant velocity,

$v = \sqrt{v_{2} ^{2}+ v_{1} ^{2} } =\sqrt{63^{2} +53^{2} }$

v = 82.32 mph

5 0

3 years ago

if a ball with an original velocity of 0 is dropped from a tall structure and takes 7 Seconds to hit the ground what velocity do

krok68 [10]

$a_y=\dfrac{v_y-v_{0y}}t\implies-9.81\,\dfrac{\mathrm m}{\mathrm s^2}=\dfrac{v_y-0}{7\,\mathrm s}\implies v_y=-68.7\,\dfrac{\mathrm m}{\mathrm s}$

5 0

3 years ago

A charge of q= +15 uC moves in a Northeast direction with a speed 5 m/s, 25 degrees East of a magnetic field, pointing North, wi

grin007 [14]

Explanation:

It is given that,

Magnitude of charge, $q=15\ \mu C=15\times 10^{-6}\ C$

It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.

Magnetic field, $B=0.08\ j$

Velocity, $v=(5\ cos25)i+(5\ sin25)j$

$v=[(4.53)i+(2.11)j]\ m/s$

We need to find the magnitude of force on the charge. Magnetic force is given by :

$F=q(v\times B)$

$F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]$

Since, $i\times j=k\ and\ j\times j=0$

$F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]$

$F=0.00000543\ kN$

$F=5.43\times 10^{-6}\ kN$

So, the force acting on the charge is $5.43\times 10^{-6}\ kN$ and is moving in positive z axis. Hence, this is the required solution.

6 0

3 years ago

A tissue sample at 275 K is submerged in 2 kg of liquid nitrogen at 70 K for cryopreservation. The final temperature of the nitr

kirill115 [55]

Answer:

The heat capacity of a sample is 37.7 J/K.

Explanation:

Given that,

Submerged temperature of tissue sample = 275 K

Mass of liquid nitrogen= 2 kg

Temperature = 70 K

Final temperature = 75 K

We need to calculate the heat

Using formula of heat

$Q=mc(T_{f}-T_{i})$

Put the value into the formula

$Q=2\times1.039\times10^{3}\times(75-70)$

$Q=10390\ J$

We need to calculate the heat capacity of a sample

Using formula of heat capacity

$\Delta S=\dfrac{Q}{T}$

Put the value into the formula

$\Delta S=\dfrac{10390}{275}$

$\Delta S=37.7\ J/K$

Hence, The heat capacity of a sample is 37.7 J/K.

7 0

3 years ago

Read 2 more answers

Fast need answers Find the amount of force if the body has pressure of 178 Pa on 18 mm²

guajiro [1.7K]

Answer:

Force = 3.204Newton

Explanation:

Given the following data;

Pressure = 178

Area = 18 mm² to meter = 18/1000 = 0.018 m²

To find the force;

Force = pressure * area

Force = 178 * 0.018

Force = 3.204 Newton.

4 0

3 years ago