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3 years ago

13

Where is the magnetic field of point P pointed

1 answer:

lisov135 [29]3 years ago

8 0

Field lines point in the direction that leaves north and enters south. So at P they are pointing toward C since it's in the direction of the south pole of the magnet

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a 1.2x10^3 kilogram car is accelerated uniformly from 10. meters per second to 20 meters per second in 5.0 seconds. what is the

irinina [24]

Force , F = ma

F = m(v - u)/t

Where m = mass in kg, v= final velocity in m/s, u = initial velocity in m/s
t = time, Force is in Newton.

m= 1.2*10³ kg, u = 10 m/s, v = 20 m/s, t = 5s

F = 1.2*10³(20 - 10)/5

F = 2.4*10³ N = 2400 N

7 0

3 years ago

While visiting the beach, you enjoy the warm ocean water, but the sand burns your feet. That night you walk along the beach and

garik1379 [7]

The answer would be B..

Since sand can heat up quickly, it will also cool off quickly. But water takes a long time to heat up and cool down.

8 0

3 years ago

What is booklet?<br>how to prepare booklet<br><br>

const2013 [10]

A booklet is a small book—with fewer pages and smaller dimensions than a “real” book, and usually a paper cover.

5 0

3 years ago

A person sitting on a pier observes incoming waves that have a sinusoidal form with a distance of 2.5 m between the crests. Of a

Doss [256]

Answer:

Part(a): The frequency is $\bf{0.2~Hz}$ .

Part(b): The speed of the wave is $\bf{0.5~m/s}$ .

Explanation:

Given:

The distance between the crests of the wave, $d = 2.5~m$ .

The time required for the wave to laps against the pier, $t = 5.0~s$

The distance between any two crests of a wave is known as the wavelength of the wave. So the wavelength of the wave is $\lambda = 2.5~m$ .

Also, the time required for the wave for each laps is the time period of oscillation and it is given by $T = 5.0~s$ .

Part(a):

The relation between the frequency and time period is given by

$\nu = \dfrac{1}{T}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Substituting the value of $T$ in equation (1), we have

$\nu &=& \dfrac{1}{5.0~s}\\~~~&=& 0.2~Hz$

Part(b):

The relation between the velocity of a wave to its frequency is given by

$v = \nu \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Substituting the value of $\nu$ and $\lambda$ in equation (2), we have

$v &=& (0.2~Hz)(2.5~m)\\~~~&=& 0.5~m/s$

5 0

3 years ago

If state law mandates that elevators cannot accelerate more than 4.80 m/s2 or travel faster than 19.8 m/s , what is the minimum

Rudik [331]

Answer:

23.0 s

Explanation:

Given:

v₀ = 0 m/s

v = 19.8 m/s

a = 4.80 m/s²

Find: Δx and t

v² = v₀² + 2aΔx

(19.8 m/s)² = (0 m/s)² + 2 (4.80 m/s²) Δx

Δx = 40.84 m

v = at + v₀

19.8 m/s = (4.80 m/s²) t + 0 m/s

t = 4.125 s

The elevator takes 40.84 m and 4.125 s to accelerate, and therefore also 40.84 m and 4.125 s to decelerate.

That leaves 291.3 m to travel at top speed. The time it takes is:

291.3 m / (19.8 m/s) = 14.71 s

The total time is 4.125 s + 14.71 s + 4.125 s = 23.0 s.

8 0

3 years ago