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Luden [163]
3 years ago
11

you can produce a sound by plucking a string or by blowing in a pipe. how are these two ways different?​

Physics
1 answer:
Shkiper50 [21]3 years ago
4 0

Answer:

a differenceis that the string needs force to be used. without something pulling on it then it wont make the noise. meanwhile the pipe only needs air to be blow Wich can be caused naturally or intentionally

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A 2.4-kg ball falling vertically hits the floor with a speed of 2.5 m/s and rebounds with a speed of 1.5 m/s. What is the magnit
gayaneshka [121]

Answer:

9.6 Ns

Explanation:

Note: From newton's second law of motion,

Impulse = change in momentum

I = m(v-u).................. Equation 1

Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.

Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)

Substitute into equation 1

I = 2.4[2.5-(-1.5)]

I = 2.4(2.5+1.5)

I = 2.4(4)

I = 9.6 Ns

4 0
3 years ago
The magnetic field at the equator points north. If you throw a positively charged object (for example, a baseball with some elec
PilotLPTM [1.2K]

Answer:

The magnetic force points in the positive z-direction, which corresponds to the upward direction.

Option 2 is correct, the force points in the upwards direction.

Explanation:

The magnetic force on any charge is given as the cross product of qv and B

F = qv × B

where q = charge on the ball thrown = +q (Since it is positively charged)

v = velocity of the charged ball = (+vî) (velocity is in the eastern direction)

B = Magnetic field = (+Bj) (Magnetic field is in the northern direction; pointing forward)

F = qv × B = (+qvî) × (Bj)

F =

| î j k |

| qv 0 0|

| 0 B 0

F = i(0 - 0) - j(0 - 0) + k(qvB - 0)

F = (qvB)k N

The force is in the z-direction.

We could also use the right hand rule; if we point the index finger east (direction of the velocity), the middle finger northwards (direction of the magnetic field), the thumb points in the upward direction (direction of the magnetic force). Hence, the magnetic force is acting upwards, in the positive z-direction too.

Hope this Helps!!!

5 0
3 years ago
A baseball thrown by a pitcher is hit by a batter. At the moment when the ball hits the bat, the force exerted on the bat by the
Slav-nsk [51]

Newton's Laws of motion describe the motion of an object based on the applied force

The correct option that gives the relationships between the forces is the the option;

\underline{F_{ball}}<u> on bat</u> = \underline{F_{bat}}<u> on ball</u>

<em>(Either option A or B without the minus symbol before </em>F_{bat}<em>, likely typographical error)</em>

<em />

Reason:

Force exerted on the bat by the ball = F_{ball}

Force exerted on the ball by the bat = F_{bat}

Given that the batter hits the ball with the bat, the force exerted by the bat on the ball, F_{bat}, is reacted to by the force of the ball acting on the bat, F_{ball}

According to Newton's third law of motion, action and reaction are equal and opposite. Therefore, at the moment when the ball hits the bat, we have;

\underline{F_{ball}}<u> on bat</u> = \underline{F_{bat}}<u> on ball</u>

<em>Where the force of the bat is high, the ball is accelerated to travel at high speed</em>

Learn more Newton's Laws of motion here:

brainly.com/question/24522313

8 0
3 years ago
A piano wire has a tension of 650 N and a mass per unit length of 0.060 g/cm. What is the speed of waves on this wire
Trava [24]

Answer:

The speed of waves on this wire is 329.14 m/s

Explanation:

Given;

tension of the wire, T = 650 N

mass per unit length, μ = 0.06 g /cm  = 0.006 kg/m

(convert the unit of mass per length in g/cm to kg/m by dividing by 10 = 0.06 / 10 = 0.006 kg/m)

The speed of waves on this wire is given as;

v = \sqrt{\frac{T}{\mu} } \\\\v = \sqrt{\frac{650}{0.006} }\\\\v = 329.14 \ m/s

Therefore, the speed of waves on this wire is 329.14 m/s

4 0
3 years ago
A scuba diver at 70 m below the surface of a lake, where the temperature is 4 degrees C, releases an air bubble with a volume of
posledela

Answer:

121.3 cm^3

Explanation:

P1 = Po + 70 m water pressure (at a depth)

P2 = Po (at the surface)

T1 = 4°C = 273 + 4 = 277 K

V1 = 14 cm^3

T2 = 23 °C = 273 + 23 = 300 K

Let the volume of bubble at the surface of the lake is V2.

Density of water, d = 1000 kg/m^3

Po = atmospheric pressure = 10^5 N/m^2

P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2

Use the ideal gas equation

\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

By substituting the values, we get

\frac{8\times 10^5\times 14}{277}=\frac{10^{5} \timesV_{2}}{300}

V2 = 121.3 cm^3

Thus, the volume of bubble at the surface of lake is 121.3 cm^3.

6 0
3 years ago
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