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2 years ago

10

Ryan is experimenting with core materials for an electromagnet. He slides different core materials through a coil of current-car

rying wire. Sort the core materials based on whether they will or won’t increase the strength of Ryan’s electromagnet.

2 answers:

Fudgin [204]2 years ago

7 0

Answer:

We can say that electromagnet is coils of wire which behave like bar magnets with a distinct South and North poles when a current of electricity passes through the coil.

The bigger the strength of Ryan's electromagnet, the better the conductivity of the material.

The material is as follows,

1. Silver and it is the best conductor.

2. Iron.

3. Nickel

4. Steel.

5. Glass which is an insulator, and

6. wood which is an insulator, and we thoroughly know that insulators they don't conduct electricity.

Explanation:

Arte-miy333 [17]2 years ago

5 0

Will increase strength: iron, nickel, steel
won't increase strength: wood, silver, glass

^thats on Plato

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A car traveling at a velocity v can stop in a minimum distance d. What would be the car's minimum stopping distance if it were t

alexira [117]

Answer:

a. 4d.

If the car travels at a velocity of 2v, the minimum stopping distance will be 4d.

Explanation:

Hi there!

The equations of distance and velocity of the car are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the car at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity of the car at time t.

Let´s find the time it takes the car to stop using the equation of velocity. When the car stops, its velocity is zero. Then:

velocity = v0 + a · t v0 = v

0 = v + a · t

Solving for t:

-v/a = t

Since the acceleration is negative because the car is stopping:

v/a = t

Now replacing t = v/a in the equation of position:

x = x0 + v0 · t + 1/2 · a · t² (let´s consider x0 = 0)

x = v · (v/a) + 1/2 · (-a) (v/a)²

x = v²/a - 1/2 · v²/a

x = 1/2 v²/a

At a velocity of v, the stopping distance is 1/2 v²/a = d

Now, let´s do the same calculations with an initial velocity v0 = 2v:

Using the equation of velocity:

velocity = v0 + a · t

0 = 2v - a · t

-2v/-a = t

t = 2v/a

Replacing in the equation of position:

x1 = x0 + v0 · t + 1/2 · a · t²

x1 = 2v · (2v/a) + 1/2 · (-a) · (2v/a)²

x1 = 4v²/a - 2v²/a

x1 = 2v²/a

x1 = 4(1/2 v²/a)

x1 = 4x

x1 = 4d

If the car travels at a velocity 2v, the minimum stopping distance will be 4d.

5 0

1 year ago

To move a suitcase up to the check-in stand at the airport a student pushes with a horizontal force through a distance of 0.95 m

hoa [83]

Answer:

33.68 N

Explanation:

Data

W= 32J

d- 0.95m

F= ?

W=Fd

They are asking for the magnitude which is the force, so you need to solve for force.

F=W/d

= 32J/ 0.95m

= 33.68 N

3 0

1 year ago

Gretchen runs the first 4.0 km of a race at 5.0 m/s. Then a stiff wind comes up, so she runs the last 1.0 km at only 4.0 m/s.

KIM [24]

Answer:

The velocity is $v = 4.76 \ m/s$

Explanation:

From the question we are told that

The first distance is $d_1 = 4.0 \ km = 4000 \ m$

The first speed is $v_1 = 5.0 \ m/s$

The second distance is $d_2 = 1.0 \ km = 1000 \ m$

The second speed is $v_2 = 4.0 \ m/s$

Generally the time taken for first distance is

$t_1 = \frac{d_1 }{v_1 }$

$t_1 = \frac{4000}{5}$

$t_1 = 800 \ s$

The time taken for second distance is

$t_1 = \frac{d_2 }{v_2 }$

$t_1 = \frac{1000}{4}$

$t_1 = 250 \ s$

The total time is mathematically represented as

$t = t_1 + t_2$

=> $t = 800 + 250$

=> $t = 1050 \ s$

Generally the constant velocity that would let her finish at the same time is mathematically represented as

$v = \frac{d_1 + d_2}{t }$

=> $v = \frac{4000 + 1000}{1050 }$

=> $v = 4.76 \ m/s$

7 0

2 years ago

A block moves at 5 m/s in the positive x direction and hits an identical block, initially at rest. A small amount of gunpowder h

Anestetic [448]

Answer:

Speed of 1.83 m/s and 6.83 m/s

Explanation:

From the principle of conservation of momentum

$mv_o=m(v_1 + v_2)$ where m is the mass, $v_o$ is the initial speed before impact, $v_1$ and $v_2$ are velocity of the impacting object after collision and velocity after impact of the originally constant object

$5m=m(v_1 +v_2)$

Therefore $v_1+v_2=5$

After collision, kinetic energy doubles hence

$2m*(0.5mv_o)=0.5m(v_1^{2}+v_2^{2})$

$2v_o^{2}=v_1^{2} + v_2^{2}$

Substituting 5 m/s for $v_o$ then

$2*(5^{2})= v_1^{2} + v_2^{2}$

$50= v_1^{2} + v_2^{2}$

Also, it’s known that $v_1+v_2=5$ hence $v_1=5-v_2$

$50=(5-v_2)^{2}+ v_2^{2}$

$50=25+v_2^{2}-10v_2+v_2^{2}$

$2v_2^{2}-10v_2-25=0$

Solving the equation using quadratic formula where a=2, b=-10 and c=-25 then $v_2=6.83 m/s$

Substituting, $v_1=-1.83 m/s$

Therefore, the blocks move at a speed of 1.83 m/s and 6.83 m/s

6 0

1 year ago

When a car is 100 meters from its starting position traveling at 60.0 m/s., it starts braking and comes to a stop 350 meters fro

NISA [10]

Remember your kinematic equations for constant acceleration. One of the equations is $x_{f} = x_{i} + v_{i}(t) + \frac{1}{2} at^{2}$ , where $x_{f}$ = final position, $x_{i}$ = initial position, $v_{i}$ = initial velocity, t = time, and a = acceleration.

Your initial position is where you initially were before you braked. That means $x_{i}$ = 100m. You final position is where you ended up after t seconds passed, so $x_{f}$ = 350m. The time it took you to go from 100m to 350m was t = 8.3s. You initial velocity at the initial position before you braked was $v_{i}$ = 60.0 m/s. Knowing these values, plug them into the equation and solve for a, your acceleration:
$350\:m = 100\:m + (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\
250\:m = (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\
250\:m = 498\:m +34.445\:s^{2}(a)\\
-248\:m = 34.445\:s^{2}(a)\\
a \approx -7.2 \: m/s^{2}$

Your acceleration is approximately $-7.2 \: m/s^{2}$ .

4 0

2 years ago