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3 years ago

What are the three properties of a magnet?

2 answers:

8 0

1. Magnets attract iron and other ferromagnetic materials such as neodymium and cobalt.

2. Magnets attract or repel other magnets.

3. In addition one part of a magnet will always point north when allowed to swing freely.

belka [17]3 years ago

5 0

The properties of a magnet is

all magnets have magnetic fields around them

the field emerges from one pole of the magnet conventionally known as the North Pole or returns to the other such South Pole

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Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds.

s344n2d4d5 [400]

This question is incomplete, the complete question is;

A student dropped a textbook from the top floor of his dorm and it fell according to the formula s(t) = -16t² + 8√t, where t is the time in seconds and s(t) is the distance in feet from the top of the building.

(a) Write a formula for the average velocity of the ball for t near 4.

(b) Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds

Answer:

Average velocity, (Vavg) of the ball for t near 4.

Vavg = [s(4) - s(0)] / (4 - 0)

Where s(4) = -16 × 4² + 8 × √4= - 240 m

s(0) = -16 × 0 + 8 * 0 = 0

duration = 1 sec

Vavg = [s(5) - s(4)] / (5 - 4)

s(5) = -16 × 52 + 8 × √5 = - 382 m

s(4) = -16 × 42 + 8 √4 = - 240 m

Vavg = (-382 - (-240)) / (5 - 4)

Vavg = - 142.1 m/s

duration = 0.5 sec

Vavg = [s(4.5) - s(4)] / (4.5 - 4)

s(4.5) = -16 × 4.52 + 8 × √4.5 = - 307 m

s(4) = -16 × 42 + 8 × √4 = - 240 m

Vavg = (-307 - (-240)) / (4.5 - 4)

Vavg= - 134.1 m/s

duration = 0.05 sec

Vavg = [s(4.05) - s(4)] / (4.05 - 4)

s(4.05) = -16 × 4.052 + 8 × √4.05 = - 246 m

s(4) = -16 × 42 + 8 × √4 = - 240 m

Vavg = (-246 - (-240)) / (4.05 - 4)

Vavg= - 126.8 m/s

Instantaneous velocity, v = ds/dt

= - 16 × 2 × t + 8 ×× (0.5 / √t )

= - 32 × t + 4/√t

ds/dt at t = 4 is,

v = - 32 × 4 + 4 / √4

= - 126 m/s

5 0

3 years ago

What does that mean?

Veseljchak [2.6K]

Answer:

acceleration

acceleration is the rate at which velocity change

i think

8 0

3 years ago

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rodikova [14]

Answer:

flip the cards over

Explanation:

7 0

3 years ago

Read 2 more answers

4) A racing car undergoing constant acceleration covers 140 m in 3.6 s. (a) If it’s moving at 53 m/s at the end of this interval

Scorpion4ik [409]

Answer

given,

distance = 140 m

time, t = 3.6 s

moving speed = 53 m/s

a) distance = (average velocity) x time

$D = \dfrac{v_0 + v_1}{2}\times t$

$140 = \dfrac{v_0 + 53}{2}\times 3.6$

v₀ + 53 = 77.78

v₀ = 24.78 m/s or 25 m/s

b) $a = \dfrac{v-u}{t}$

$a = \dfrac{53-25}{3.6}$

a = 7.8 m/s²

using equation of motion

v₀² = v₁² + 2 a s

53² = 0²+ 2 x 7.8 x s

s = 180 m

3 0

3 years ago

A railroad car moving at a speed of 3.46 m/s overtakes, collides and couples with two coupled railroad cars moving in the same d

I am Lyosha [343]

Answer:

$2.09\ \text{m/s}$

$22298.4\ \text{J}$

Explanation:

m = Mass of each the cars = $1.6\times 10^4\ \text{kg}$

$u_1$ = Initial velocity of first car = 3.46 m/s

$u_2$ = Initial velocity of the other two cars = 1.4 m/s

v = Velocity of combined mass

As the momentum is conserved in the system we have

$mu_1+2mu_2=3mv\\\Rightarrow v=\dfrac{u_1+2u_2}{3}\\\Rightarrow v=\dfrac{3.46+2\times 1.4}{3}\\\Rightarrow v=2.09\ \text{m/s}$

Speed of the three coupled cars after the collision is $2.09\ \text{m/s}$ .

As energy in the system is conserved we have

$K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}$

The kinetic energy lost during the collision is $22298.4\ \text{J}$ .

6 0

3 years ago