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3 years ago

If the fly experienced a force of 100 N when it hits how big was the force on the vehicle in this collision

Physics

1 answer:

iogann1982 [59]3 years ago

8 0

Answer:

100N

Explanation:

because newton's third law of motion states that if body A exerts a force on the body B, then Body B will exert an equal force but opposite in direction force on body A

so if the fly experienced 100N, then the car will also experience 100N

hope you get it

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A nuetron is the lightest subatomic particle

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3 years ago

In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo

Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of $32^{\circ}$ (when light enters from the air.)

Explanation:

Let $v_{1}$ denote the speed of light in the first medium. Let $v_{\text{air}}$ denote the speed of light in the air. Assume that the light entered the boundary at an angle of $\theta_{1}$ to the normal and exited with an angle of $\theta_{\text{air}}$ . By Snell's Law, the sine of $\theta_{1}\!$ and $\theta_{\text{air}}\!$ would be proportional to the speed of light in the corresponding medium. In other words:

$\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}$ .

When light enters a boundary at the critical angle $\theta_{c}$ , total internal reflection would happen. It would appear as if the angle of refraction is now $90^{\circ}$ . (in this case, $\theta_{\text{air}} = 90^{\circ}$ .)

Substitute this value into the Snell's Law equation:

$\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}$ .

Rearrange to obtain an expression for the speed of light in the first medium:

$v_{1} = v_{\text{air}} \cdot \sin(\theta_{1})$ .

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For $0 < \theta < 90^{\circ}$ , $\sin(\theta)$ is monotonically increasing with respect to $\theta$ . In other words, for $\!\theta$ in that range, the value of $\sin(\theta)\!$ increases as the value of $\theta\!$ increases.

Therefore, compared to the medium in this question with $\theta_{c} = 27^{\circ}$ , the medium with the larger critical angle $\theta_{c} = 32^{\circ}$ would have a larger $\sin(\theta_{c})$ . such that light would travel faster in that medium.

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3 years ago

From Gauss's law, the electric field set up by a uniform line of charge is given by the following expression where is a unit vec

Evgesh-ka [11]

Answer:

$\Delta V=\lambda *ln(r_{2}/r_{1}) /\ (2\pi*\epsilon_{o})$

Explanation:

Using the Gauss Law, we obtain the electric Field for a uniform large line of charge:

$2\pi r L*E=\lambda *L/\epsilon_{o}$

$E=\lambda /\(2 \pi* r *\epsilon_{o})$

We calculate the potential difference from the electric field:

$\Delta V=-\int\limits^{r_{1}}_{r_{2}} E \, dr =-\int\limits^{r_{1}}_{r_{2}} \lambda dr/ (2\pi*r*\epsilon_{o})=\lambda *ln(r_{2}/r_{1}) /\ (2\pi*\epsilon_{o})$

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4 years ago

Calculate the work performed by an ideal Carnot engine as a cold brick warms from 150 K to the temperature of the environment, w

olga nikolaevna [1]

To solve this problem, apply the concepts related to the calculation of the work performed according to the temperature change (in an ideal Carnot cycle), for which you have to:

$W = \int\limit_{T_c}^{T_H} C (1-\frac{T_H}{T})$

Where,

C = Heat capacity of the Brick

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Integrating,

$W = C (T_H-T_C)- T_H C ln (\frac{T_H}{T_C})$

Our values are given as

$T_H= 300K$

$T_C = 150K$

Replacing,

$W = (1) (300-150)-300(1)ln(2)$

$W = 150-300ln2$

$W = -57.94kJ \approx 58kJ$

Therefore the work perfomed by this ideal carnot engine is 58kJ

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4 years ago

g Two long parallel wires are a center-to-center distance of 2.50 cm apart and carry equal anti-parallel currents of 2.70 A. Fin

schepotkina [342]

Answer:

864 mT

Explanation:

The magnetic field due to a long straight wire B = μ₀i/2πR where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current in wire, and R = distance from center of wire to point of magnetic field.

The magnitude of magnetic field due to the first wire carrying current i = 2.70 A at distance R which is mid-point between the wires is B = μ₀i/2πR.

Since the other wire also carries the same current at distance R, the magnitude of the magnetic field is B = μ₀i/2πR.

The resultant magnetic field at B is B' = B + B = 2B = 2(μ₀i/2πR) = μ₀i/πR

Now R = 2.50 cm/2 = 1.25 cm = 1.25 × 10⁻² m and i = 2.70 A.

Substituting these into B' = μ₀i/πR, we have

B' = 4π × 10⁻⁷ H/m × 2.70 A/π(1.25 × 10⁻² m)

B = 10.8/1.25 × 10⁻⁵ T

B = 8.64 × 10⁻⁵ T

B = 864 × 10⁻³ T

B = 864 mT

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3 years ago