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Alexxx [7]
3 years ago
6

On the way home from school, Taylor's car runs out of gas. He has to walk 25m north and 10m west in order to reach the nearest g

as station. Find his distance traveled and his displacement from his car
Physics
1 answer:
spin [16.1K]3 years ago
8 0

Answer:

<em>The distance is 35 m and the magnitude of the displacement is 26.93 m</em>

Explanation:

<u>Displacement  and Distance</u>

These are two related concepts. A moving object constantly travels for some distance at defined periods of time. The total distance is the sum of each individual distance the object traveled. It can be written as:

dtotal=d1+d2+d3+...+dn

This sum is calculated independently of the direction the object moves.

The displacement only takes into consideration the initial and final positions of the object. The displacement, unlike distance, is a vectorial magnitude and can even have magnitude zero if the object starts and ends the movement at the same point.

Taylor walks 25 m north and 10 m west. The total distance is the sum of both numbers:

d = 25 m + 10 m = 35 m

To calculate the displacement, we need to know the final position with respect to the initial position. If we set the coordinates of Taylor's car as the origin (0,0), then his final position is (-10,25), assuming the west direction is negative and the north direction is positive.

The magnitude of the displacement is the distance from (0,0) to (-10,25):

D=\sqrt{(25-0)^2+(-10-0)^2}

D=\sqrt{625+100}=\sqrt{725}

D = 26.93 m

The distance is 35 m and the magnitude of the displacement is 26.93 m

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We have two solenoids: solenoid 2 has twice the diameter, half the length, and twice as many turns as solenoid 1. The current in
leva [86]

Answer:

the field at the center of solenoid 2 is 12 times the field at the center of solenoid 1.

Explanation:

Recall that the field inside a solenoid of length L, N turns, and a circulating current I, is given by the formula:

B=\mu_0\, \frac{N}{L} I

Then, if we assign the subindex "1" to the quantities that define the magnetic field (B_1) inside solenoid 1, we have:

B_1=\mu_0\, \frac{N_1}{L_1} I_1

notice that there is no dependence on the diameter of the solenoid for this formula.

Now, if we write a similar formula for solenoid 2, given that it has :

1) half the length of solenoid 1 . Then L_2=L_1/2

2) twice as many turns as solenoid 1. Then N_2=2\,N_1

3) three times the current of solenoid 1. Then I_2=3\,I_1

we obtain:

B_2=\mu_0\, \frac{N_2}{L_2} I_2\\B_2=\mu_0\, \frac{2\,N_1}{L_1/2} 3\,I_1\\B_2=\mu_0\, 12\,\frac{N_1}{L_1} I_1\\B_2=12\,B_1

5 0
3 years ago
A 8.34 × 103-kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is
Mnenie [13.5K]

Answer:

21870.3156 N

Explanation:

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 1.6 m/s²

Equation of motion

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-18.2^2}{2\times 162}\\\Rightarrow a=-1.02234\ m/s^2

The acceleration of the craft should be 1.02234 m/s²

F=ma\\\Rightarrow F=8.34\times 10^3\times 1.02234\\\Rightarrow F=8526.3156\ N

Weight of the craft

W=mg\\\Rightarrow W=8.34\times 10^3\times 1.6\\\Rightarrow W=13344\ N

Thrust

F_t=F+W\\\Rightarrow F_t=8526.3156+13344\\\Rightarrow F_t=21870.3156\ N

The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N

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3 years ago
How would a parachute fall on the surface of the moon ? Describe. The mass of the Jupiter is
GREYUIT [131]

Explanation:

the answer is 2.46 × 10^12

6 0
2 years ago
A ball rolls along the floor with a constant velocity of 6m/s how far it have to gone after 50 seconds
Alika [10]
50(6)=300.
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8 0
3 years ago
Longitudinal sound waves cannot propagate through
strojnjashka [21]

Answer:

A vacuum

Explanation:

Sound waves are examples of mechanical waves. Mechanical waves are waves which are transmitted through the vibrations of the particles in a medium.

For example, sound waves in air consist of oscillations of the air particles, which vibrate back and forth (longitudinal wave) along the direction of propagation of the wave itself.

Given this definition of mechanical wave, we see that such a wave cannot propagate if there is no medium, because there are no particles that would oscillate. Therefore, among the choices given, the following one:

a vacuum

represent the only situation in which a sound wave cannot propagate through: in fact, there are no particles in a vacuum, so the oscillations cannot occur. In all other cases, instead, sound waves can propagate.

3 0
2 years ago
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