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2 years ago

While going 60 mph, how many feet will it take you to stop?

1 answer:

6 0

While you are going 60 mph, it will take approximately 180 feet to stop, A is the correct choice. I know some of the question is missing, so I researched it and found the correct answer.

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Which best describes a difference between laser light and regular light?

Mashcka [7]

Answer:

Explanation:

6 0

2 years ago

A container of gas molecules is at a pressure of 2 atm and has amass density of 1.7 grams per liter. All of the molecules in the

Tpy6a [65]

Answer:

The speed of nitrogen molecule is 1.87 m/s.

Explanation:

Given that,

Pressure = 2 atm

Density = 1.7 grams/liter

Atomic weight = 28 grams

We need to calculate the temperature

Using formula of idea gas

$PV=nRT$

$P=\dfrac{WRT}{VM}$

$P=\dfrac{\rho RT}{M}$

$T=\dfrac{PM}{\rho R}$

Put the value into the formula

$T=\dfrac{2\times28}{1.7\times0.0821}$

$T=401.2\ K$

We need to calculate the speed of nitrogen molecule

Using formula of RMS speed

$V_{rms}=\sqrt{\dfrac{3RT}{M}}$

$V_{rms}=\sqrt{\dfrac{3\times0.0821\times401.2}{28}}$

$V_{rms}=1.87\ m/s$

Hence, The speed of nitrogen molecule is 1.87 m/s.

5 0

3 years ago

A positive charge moving up enters a magnetic field pointing out of the screen. What is the direction of the magnetic force on t

erma4kov [3.2K]

According to the right-hand thumb rule, the forefinger gives the velocity of charge, the thumb gives the magnetic force and the center finger gives the direction of magnetic field.

then, as shown in the picture, the <span>direction of the magnetic force on the charge is in the right direction.</span>

6 0

3 years ago

Read 2 more answers

A car is driving at 85 km/h and the driver spots a stop sign ahead. What coefficient of friction is needed to stop the car at th

almond37 [142]

Answer:

μ = 0.0315

Explanation:

Since the car moves on a horizontal surface, if we sum forces equal to zero on the Y-axis, we can determine the value of the normal force exerted by the ground on the vehicle. This force is equal to the weight of the cart (product of its mass by gravity)

N = m*g (1)

The friction force is equal to the product of the normal force by the coefficient of friction.

F = μ*N (2)

This way replacing 1 in 2, we have:

F = μ*m*g (2)

Using the theorem of work and energy, which tells us that the sum of the potential and kinetic energies and the work done on a body is equal to the final kinetic energy of the body. We can determine an equation that relates the frictional force to the initial speed of the carriage, so we will determine the coefficient of friction.

$\frac{1}{2} *m*v_{i}^{2}-(F*d)= \frac{1}{2} *m*v_{f}^{2}$

where:

vf = final velocity = 0

vi = initial velocity = 85 [km/h] = 23.61 [m/s]

d = displacement = 900 [m]

F = friction force [N]

The final velocity is zero since when the vehicle has traveled 900 meters its velocity is zero.

Now replacing:

(1/2)*m*(23.61)^2 = μ*m*g*d

0.5*(23.61)^2 = μ*9,81*900

μ = 0.0315

5 0

2 years ago

13. Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distanc

Gennadij [26K]

This question is incomplete, the complete question is;

Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distance and the direction. We will investigate in detail just two directions. With charges available in the simulation (all the charges are either positive or negative 1 nC increments).

how do you create a dipole with dipole moment 1 x 10-9 Cm with a direction for the dipole moment pointing to the right. Make a table below that shows the amounts of charge and the distance between the charges. There are many correct answers

Answer:

Given the data in question;

Dipole moment P = 1 × 10⁻⁹ C.m

now dipole pointing to the right;

P→

$_{-\theta }$ (-) ---------------->(+) $_{+\theta }$

so let distance between the dipoles be d

∴ P = d $\Theta$

Let $\Theta_{1}$ = 1 nC

P = d $\Theta$

1 × 10⁻⁹ = 1 × 10⁻⁹ × d

d = (1 × 10⁻⁹) / (1 × 10⁻⁹)

d = 1 m

Also Let $\Theta_{2}$ = 2 nC

P = d $\Theta$

1 × 10⁻⁹ = 2 × 10⁻⁹ × d

d = (1 × 10⁻⁹) / (2 × 10⁻⁹)

d = 0.5 m

Also Let $\Theta_{3}$ = 3 nC

P = d $\Theta$

1 × 10⁻⁹ = 3 × 10⁻⁹ × d

d = (1 × 10⁻⁹) / (3 × 10⁻⁹)

d = 0.33 m

such that;

charge distance

1 nC 1.00 m

2 nC 0.50 m

3 nc 0.33 m

4 nC 0.25 m

5 nC 0.20 m

5 0

2 years ago