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ira [324]
3 years ago
14

Confusing, fill the venn diagram?

Physics
1 answer:
Irina18 [472]3 years ago
4 0

Answer:

electromagnetic waves= LIGHT, TRAVELS IN A VACUUM, VIBRATIONS IN MAGNETIC FIELD

matter/mechanical= REQUIRES A MEDIUM, SOUND, COMPRESSIONS,

both= TRANSFERS ENERGY, AMPLITUDE, FREQUENCY, WAVELENGTH,

Explanation:

electromagnetic waves don't need particles/atoms to travel through whilst matter/mechanical waves do.

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which of the following cannot be increased by using a machine of some kind? work, force, speed, torque
Lemur [1.5K]

Explanation:

Work cannot be increased by using a machine of some kind.

8 0
2 years ago
Choose the correct statement concerning units of power or energy. a. A kilojoule (kJ) is a unit of power. b. A gigawatt (GW) is
sergiy2304 [10]

Answer:

A kilowatt (kW) is a unit of power.

Explanation:

The power of an object is given by :

P=\dfrac{E}{t}

Here,

E is the energy required

t is time

The SI unit of power is Watts and the SI unit of energy is Joule. the commercial unit of energy is kilowatt per hour.    

Option (1) :  A kilojoule (kJ) is a unit of power is incorrect.

Option (2) : A gigawatt (GW) is a unit of energy is incorrect.

Option (3) : A watt (W) is a unit of energy is incorrect.

Option (4) : A kilowatt x hour per year (kWh/yr) is a unit of energy is incorrect.

Option (4) : A kilowatt (kW) is a unit of power is correct.

Hence, the correct option is (d).

3 0
3 years ago
When an object oscillating in simple harmonic motion is at its maximum displacement from the equilibrium position, which of the
ziro4ka [17]

Answer:

E. Zero Maximum

Explanation:

At the point of maximum displacement, the speed is zero while the restoring force is maximum. In fact:

- The restoring force is given by F=kx, where k is the spring constant and x is the displacement - at the point of maximum displacement, x is maximum, so F is maximum as well

- the total energy of the system is sum of kinetic energy and elastic potential energy:

E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2

where m is the mass of the system and v is the speed. Since E (the total energy) is constant due to the law of conservation of energy, we have that when K increases, U decreases, and viceversa. As a result, when x increases, v decreases, and viceversa. At the point of maximum displacement, x is maximum, so v will have its minimum value (which is zero, since the system is changing direction of motion).

4 0
3 years ago
Triton is a moon of Neptune. It has a
Mnenie [13.5K]

Answer:

<em>Well, Your answer will be is </em><em>840.96 mi. </em>

<em>Good Luck!</em>

^{Itsbrazts}

6 0
3 years ago
A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma
raketka [301]

  • Let, the maximum height covered by projectile be \sf{H_m}

\purple{ \longrightarrow  \bf{h_m =  \dfrac{ {v}^{2} \: {sin}^{2} \theta  }{2g} }}

  • Projectile is thrown with a velocity = v
  • Angle of projection = θ

  • Velocity of projectile at a height half of the maximum height covered be \sf{v_0}

\qquad______________________________

Then –

\qquad \pink{  \longrightarrow \bf{ \dfrac{h_m}{2}  = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }}

\qquad \longrightarrow \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2g} \times  \dfrac{1}{2}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{4g}  =  \dfrac{ {v_0}^{2} \: {sin}^{2} \theta  }{2g} }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2}  \: {sin}^{2} \theta  }{2}  =   {v_0}^{2} \: {sin}^{2} \theta }

\qquad\longrightarrow  \sf{ \dfrac{ {v}^{2} }{2}  =   {v_0}^{2} }

\qquad\longrightarrow \bf{v_0 =   \sqrt{ \dfrac{ {v}^{2} }{2} } =  \dfrac{v}{ \sqrt{2} }  }

  • Now, the vertical component of velocity of projectile at the height half of \sf{h_m} will be –

\qquad \longrightarrow   \bf{v_{(y)}=v_0 \: sin \theta }

\qquad \longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} }  \: sin \theta =  \dfrac{v \: sin \: \theta}{ \sqrt{2} }  }

Therefore, the vertical component of velocity of projectile at this height will be–

☀️\qquad\pink {\bf{ \dfrac{v \: sin \:  \theta}{ \sqrt{2} }} }

6 0
2 years ago
Read 2 more answers
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