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andrey2020 [161]
3 years ago
14

What will most likely happen if a sound wave moves from the air through a solid?

Physics
2 answers:
stich3 [128]3 years ago
6 0

Answer:

It will increase in speed

Explanation:

kobusy [5.1K]3 years ago
5 0

Answer:

C. it will increase in speed

Explanation:

Edge 2021

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How much charge must pass by a point in a wire in 1.5 s for the current inb the wire to be 2.0 A?
skelet666 [1.2K]

Answer:

3 Coulombs

Explanation:

Q = Current x time

Q = 2.0 x 1.5

Q = 3 Coulombs

3 0
4 years ago
Plzzzz urgent <br><br><br>solve this​
Mariana [72]

Answer:

\large{ \tt{☄ \: EXPLANATION}} :

  • Before solving , You'll have to know - When an object starts from the state of rest , in this case , initial velocity ( u ) = 0

  • Notice that we're provided the time ( t ) in minutes. So , first thing we have to do is convert the minutes into seconds. It would be - Time ( t ) = 5 minutes = 5 × 60 sec = 300 sec [ 1 min = 60 sec ]

  • Here , We're provided - Initial velocity ( u ) = 0 , Final velocity ( v ) = 60 m / s , Time taken ( t ) = 300 seconds & We're asked to find out the acceleration ( a ) & distance covered by the jeep ( s ) .

\large{ \tt{♨ \:LET'S \: START}} :

  • Acceleration is defined as the rate of change of velocity. We know :

\large{ \boxed{ \tt{❁ \: ACCELERATION \: (a) =  \frac{FINAL \: VELOCITY(v) - INITIAL \: VELOCITY(u)}{TIME \: TAKEN \: ( \: t \: )}}}}

- Plug the values & then simplify !

\large{ \bf{↬a =  \frac{60 - 0}{300}  =  \frac{60}{300} =  \boxed{ \bold{ \bf{0.2 \: m {s}^{ - 2} }}} }}

  • The acceleration of the jeep is 0.2 m/s²

\large{ \tt{۵ \: AGAIN, \: USING\: SECOND \: EQUATION \: OF \: MOTION}} :

\boxed{ \large{ \bf{✾ \: s =  \frac{u + v}{2}  \times t}}}

- Plug the values & then simplify !

\large{ \bf{↦s =  \frac{0 + 60}{2}  \times 300 =  \boxed{ \bold{ \bf{9000 \: m}}}}}

  • The distance covered by the jeep is 9000 m .

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7 0
3 years ago
What's an electron transitions involves a photon emitting the least amount of energy?
FrozenT [24]
Give me some answer choices and i will be happy to help
4 0
3 years ago
Consider a uniform sphere, which has a mass of 4.80 kg and a radius of 22.0 cm. A tangential force of 11.2 N is applied to the o
Tcecarenko [31]

Answer:

The moment of inertia of this sphere is 0.0929\ kg-m^2.                  

Explanation:

It is given that,

Mass of the sphere, m = 4.8 kg

Radius of the sphere, r = 22 cm = 0.22 m

Tangential force, F = 11.2 N

The moment of inertia of the uniform sphere is given by :

I=\dfrac{2}{5}mr^2

I=\dfrac{2}{5}\times 4.8\ kg\times (0.22\ m)^2

I=0.0929\ kg-m^2

So, the moment of inertia of this sphere is 0.0929\ kg-m^2. Hence, this is the required solution.              

8 0
3 years ago
A proton moves through an electric potential created by a number of source charges. Its speed is 2.5 x 105 m/s at a point where
Kaylis [27]

Answer:

Proton’s speed, a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s

Explanation:

Given;

initial speed of proton, u = 2.5 x 10⁵ m/s

initial potential, V = 1500 V

mass of proton = 1.67 x 10⁻²⁷ kg

Work done, W = eV= ΔK.E = ¹/₂mu²

eV = ¹/₂mu² (J)

where;

e is the charge of the proton in coulombs

V is the electric potential in volts

m is the mass of the proton in kg

u is the speed of the proton in m/s

m =\frac{2eV_1}{u_1{^2}} = \frac{2eV_2}{u_2{^2}} = \frac{V_1}{u_1{^2}}} =\frac{V_2}{u_2{^2}}

\frac{V_1}{u_1{^2}}} =\frac{V_2}{u_2{^2}} = \frac{1500}{(2.5*10^5)^2} = \frac{500}{u_2{^2}} \\\\u_2{^2} =\frac{500*(2.5*10^5)^2}{1500} = 0.333*6.25*10^{10}\\\\u_2 = \sqrt{0.333*6.25*10^{10}} =1.4 *10^5 \ m/s

Therefore, proton’s speed a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s

4 0
4 years ago
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