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3 years ago

6

You are stopped a red light and you are in the front car at an intersection. Thinking as a defensive driver, describe what your

actions will be when the light turns green.

1 answer:

Airida [17]3 years ago

8 0

Proceed to go when the light turns green with caution watching out for other drivers to prevent and accident

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8. At what position does the mass have the greatest acceleration?

gulaghasi [49]

Answer:

Option (e)

Explanation:

If a mass attached to a spring is stretched and released, it follows a simple harmonic motion.

In simple harmonic motion, velocity of the mass will be maximum, kinetic energy is maximum and acceleration is 0 at equilibrium position (at 0 position).

At position +A, mass will have the minimum kinetic energy, zero velocity and maximum acceleration.

Therefore, Option (e) will be the answer.

6 0

2 years ago

The wave function of a particle is exp(i(kx-omegat)), where x is distance, t is time, and k and co are positive real numbers. Th

Step2247 [10]

Answer:

Momentum, $p=\hbar k$

Explanation:

The wave function of a particle is given by :

$y=exp[i(kx-\omega t)]$ ...............(1)

Where

x is the distance travelled

t is the time taken

k is the propagation constant

$\omega$ is the angular frequency

The relation between the momentum and wavelength is given by :

$p=\dfrac{h}{\lambda}$ ............(2)

From equation (1),

$k=\dfrac{2\pi}{\lambda}$

$\lambda=\dfrac{2\pi}{k}$

Use above equation in equation (2) as :

$p=\dfrac{h k}{2\pi }$

Since, $\dfrac{h}{2\pi}=\hbar$

$p=\hbar k$

So, the x-component of the momentum of the particle is $\hbar k$ . Hence, this is the required solution.

8 0

3 years ago

The earth has a vertical electric field at the surface,pointing down, that averages 102 N/C. This field is maintained by various

Schach [20]

Answer:

$q = -461532.5 \ C$

Explanation:

From the question we are told that

The electric filed is $E = 102 \ N/C$

Generally according to Gauss law

=> $E A = \frac{q}{\epsilon_o }$

Given that the electric field is pointing downward , the equation become

$- E A = \frac{q}{\epsilon_o }$

Here $q$ is the excess charge on the surface of the earth

$A$ is the surface area of the of the earth which is mathematically represented as

$A = 4\pi r^2$

Where r is the radius of the earth which has a value $r = 6.3781*10^6 m$

substituting values

$A = 4 * 3.142 * (6.3781*10^6 \ m)^2$

$A =5.1128 *10^{14} \ m^2$

So

$q = -E * A * \epsilon _o$

Here $\epsilon_o$ s the permitivity of free space with value

$\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

substituting values

$q = -102 * 5.1128 *10^{14} * 8.85 *10^{-12}$

$q = -461532.5 \ C$

6 0

3 years ago

Why do you think football players are usually large in size?

kiruha [24]

Football players in the NFL are large in size due to their eating habits since as players they have to work out constantly to stay and shape and eat to gain weight and be strong, so its important for them to be strong against their opponents in gameplay. Football players sometimes take steroids which make their muscles stronger and harder, and also gives them a growth spurt but the effects are worse and makes them sicker, high blood pressure, and even more.

8 0

3 years ago

Read 2 more answers

A spider begins to spin a web by first hanging from a ceiling by his fine, silk fiber. He has a mass of 0.025 kg and a charge of

Rasek [7]

Answer:

a) (5.59 × 10³) N/C

b) 0.226 N directed away from the spider.

Explanation:

a) Electric field, E, felt as a result of point charge, Q, at a distance, d away is given by

E = kQ/d²

So, magnitude of the electric field due to the charge on the second spider at the position of the first spider

Q = 4.2 µC = 4.2 × 10⁻⁶ C

k = Coulomb's constant = 8.99 × 10⁹ Nm²/C

d = 2.6 m

E = (8.99 × 10⁹ × 4.2 × 10⁻⁶)/2.6²

E = 5.59 × 10³ N/C

b) Tension in the silk fiber above the spider is the net force due to the weight of spider one and the force of repulsion due the two charges.

Force due to the two charges = Eq

where q now represents the charge of the first spider at the first point, feeling the electric field calculated in (a)

F = 5.59 × 10³ × 3.4 × 10⁻⁶ = 0.01901 N directed upwards. (That is, F = + 0.019 N)

Weight of the spider = mg = 0.025 × 9.8 = 0.245 N directed downwards. (That is, W = -0.245 N)

Net force, T = mg - F = 0.245 - 0.019 = 0.226 N (that is, 0.226 N, directed upwards, away from the spider).

8 0

3 years ago