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almond37 [142]
4 years ago
7

A baseball bat hits a baseball with a force of 100 newtons. What is the force and its direction exerted by the ball on the bat?

Physics
1 answer:
Brut [27]4 years ago
7 0

Answer:

The force exerted by the ball on the bat has a magnitude of 100 N and its direction is exactly opposite to that of the force exerted by the bat on the ball.

Explanation:

Recall that Newton's third law tells us that : "For every action, there is an equal and opposite reaction."

Therefore if the bat acts on the ball with a force of 100 N, the ball acts on the bat with a similar magnitude of force (100 N) but direction opposite to the original force.

F_{ab} = -\,F_{ba}

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For the winter, a duck flies 10.0 m/s due south against a gust of wind with a speed of 2.5 m/s. What is the resultant velocity o
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10.0 m/s + (-2.5 m/s) = 7.5 m/s
6 0
4 years ago
A sound wave of the form s = sm cos(kx - ?t + f) travels at 343 m/s through air in a long horizontal tube. At one instant, air m
Naddik [55]

Answer:

960.24 Hz

Explanation:

Here is the complete question

A sound wave of the form

S=Smcos(kx−ωt+Φ)

travels at 343 m/s through air in a long horizontal tube. At one instant, air molecule A at x = 2.000 m is at its maximum positive displacement of 6.00 nm and air molecule B at x = 2.070 m is at a positive displacement of 2.00 nm. All the molecules between A and B are at intermediate displacements. What is the frequency of the wave?

Solution

Given x₁ = 2.0 m, x₂ = 2.070 m, maximum positive displacement s = 6.00 nm at x₁, positive displacement s = 2.00 nm at x₂, velocity of wave v = 343 m/s, maximum positive displacement s₀ = 6.00 nm

Let t₀ = 0 at x₁ = 2.0 m for maximum displacement.

So, s = s₀cos(kx−ωt+Φ)

     6 = 6cos(2k - 0 + Φ) = 6cos(2k + Φ)⇒ cos(2k + Φ) = 6/6 = 1

cos(2k + Φ) = 1 ⇒ (2k + Φ) = cos⁻¹ (1) = 0 ⇒ 2k + Φ = 0

Let t₀ = 0 at x₂ = 2.070 m for displacement s = 2.00 nm.

So, s = s₀cos(kx−ωt+Φ)

     2 = 6cos(2.070k - 0 + Φ) = 6cos(2.070k + Φ)⇒ cos(2.070k + Φ) = 2/6 = 1/3

cos(2.070k + Φ) = 1/3 ⇒ (2.070k + Φ) = cos⁻¹ (1/3) = 70.53 ⇒ 2.070k + Φ = 70.53.

We now have two simultaneous equations.

2k + Φ = 0   (1)

2.070k + Φ = 70.53.    (2)

Subtracting (2) -(1)

2.070k - 2k = 70.53

0.070k = 70.53

k = 70.53/0.070 = 1007.554π/180 rad/m = 17.59 rad/m

k = 2π/λ ⇒ λ = 2π/k

and frequency, f = v/λ = v/2π/k = kv/2π = 17.59 × 343/2π = 960.24 Hz

4 0
3 years ago
The dielectric strength of rutile is 6.0 × 106 V/m, which corresponds to the maximum electric field that the dielectric can sust
VLD [36.1K]

Answer:

6.48*10⁻⁷ C

Explanation:

  • By definition, the capacitance of a capacitor is expressed as follows:

       C =\frac{Q}{V}

  • where Q is the charge on one of the plates of the capacitor, and V the potential difference between the plates.
  • The maximum electric field, the potential difference, and the distance between plates are related by the following expression:

        V = E_{max} * d

  • Replacing by the givens, we can find V as follows:

        V = 6.0e6 V/m * 0.00108 m= 6480 V

  • Now, we can find the maximum charge Qmax,  as follows:

        Q_{max} = C*V = 1e-10F* 6480 V = 6.48e-7 C

  • The maximum charge that the capacitor can hold is 6.48*10⁻⁷ C.
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