Which types of a magnet will have a bigger magnetic feild area?

Luis and Aisha conducted an experiment. They exerted different forces on four objects. Their results are shown in the table.

lesya692 [45]

Answer:

Object 3 has greatest acceleration.

Explanation:

Objects Mass Force

1 10 kg 4 N

2 100 grams 20 N

3 10 grams 4 N

4 1 kg 20 N

Acceleration of object 1,

$a_1=\dfrac{F_1}{m_1}\\\\a_1=\dfrac{4}{10}\\\\a_1=0.4\ m/s^2$

Acceleration of object 2,

$a_2=\dfrac{F_2}{m_2}\\\\a_2=\dfrac{20}{0.1}\\\\a_2=200\ m/s^2$

Acceleration of object 3,

$a_3=\dfrac{F_3}{m_3}\\\\a_3=\dfrac{4}{0.01}\\\\a_3=400\ m/s^2$

Acceleration of object 4,

$a_4=\dfrac{F_4}{m_4}\\\\a_4=\dfrac{20}{1}\\\\a_3=20\ m/s^2$

It is clear that the acceleration of object 3 is $400\ m/s^2$ and it is greatest of all. So, the correct option is (3).

4 0

2 years ago

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WILL GIVE BRAINLIEST AND 50 POINTS!

7nadin3 [17]

Answer:

1) Current decreases; 2) Inverse proportionally; 3) 1[A]

Explanation:

1)

As we can see as the resistance increases the current decreases, if we take two points as an example, when the resistance is equal to 50 [ohms] the current is equal to 1[amp] and when the resistance is equal to 200 [ohms] the current tends to have a value below 0.5 [amp]. Thus demonstrating the decrease in current.

2)

Inverse proportionally, by definition we know that the law of ohm determines the voltage according to resistance and amperage. This is the voltage will be equal to the product of the voltage by the resistance.

$V=I*R\\V = voltage [volts]\\I = current[amp]\\R = resistance [ohms]$

where:

$R =\frac{V}{I} \\or\\I=\frac{V}{R}$

And whenever we have in a fractional number the denominator the variable we are interested in, we can say that this is inversely proportional to the value we are interested in determining. In this case, we can see from the two previous expressions that both the current and the resistance appear in the denominator, therefore they are inversely proportional to each other.

3)

If we place ourselves on the graph on the resistance axis, we see that at 50 [ohm] will correspond a current value equal to 1 [A].

4 0

2 years ago

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Identify the conditions for an elastic collision in a closed system. Check all that apply.

s344n2d4d5 [400]

Answer:

In an elastic collision:

There is no external net force acting. Thus, Momentum before and after collision is equal. Momentum remains conserved.

Total energy always remains conserved as energy cannot be created nor destroyed. It can change from one form to another.

There is no lost due to friction in elastic collision. So the kinetic energy is also conserved.

Velocities may change after collision. If the masses are equal, the velocities interchange.

When one object is stationary:

Final velocity of object 1:

v₁ = (m₁ - m₂)u₁/(m₁ +m₂)

Final velocity of object 2:

v₂ = (2 m₁ u₁)/(m₁+m₂) =

Objects do not stick together in elastic collision. They stick together in inelastic collision.

One object may be stationary before the elastic collision.

Thus, conditions for an elastic collision:

Energy is conserved.
Velocities may change.
Momentum is conserved.
Kinetic energy is conserved.
One object may be stationary before the elastic collision.

7 0

2 years ago

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Posting on social media as a business can be a free way of generating interest in a childcare center's services.

max2010maxim [7]

We know this already

4 0

2 years ago

A 10-cm-long thin glass rod uniformly charged to 14.0 nC and a 10-cm-long thin plastic rod uniformly charged to - 14.0 nC are pl

Sonbull [250]

Answer:

A. At a point 1cm from the glass rod, E = 1.36 * 10⁶N/C

B. At a point 2cm from the glass rod, E = 5.17 * 10⁵N/C

C. At a point 3cm from the glass rod, E = 6.993 * 10⁵N/C

Explanation:

Parameters given:

Charge of glass rod, Q = 14nC = 14 * 10⁻⁹ C

Charge of plastic rod, q = 14nC = 14 * 10⁻⁹ C

Distance between both rods = 4.5cm = 0.045

A. Electric field strength at a point 1.0cm (0.01m) from the glass rod is the sum of electric field strength due to both rods i.e.

E = E₁ + E₂

Where

E₁ = electric field strength due to glass rod

E₂ = electric field strength due to plastic rod

E₁ = kQ/0.01²

E₂ = kq/(0.045 - 0.01)² = kq/(0.035)²

E = kQ/0.01² + kq/(0.035)² = k(Q/0001 + q/0.001225)

E = 9 * 10⁹ [(14 * 10⁻⁹ / 0.001) + (14 * 10⁻⁹)/0.001225]

E = 9 * 10⁹[(14 * 10⁻⁵) + (1.143 * 10⁻⁵)]

E = 9 * 10⁹ * 15.143* 10⁻⁵

E = 1.36 * 10⁶N/C

B. Electric field strength at a point 2.0cm (0.02m) from the glass rod is the sum of electric field strength due to both rods i.e.

E = E₁ + E₂

E₁ = kQ/0.02²

E₂ = kq/(0.045 - 0.02)² = kq/(0.025)²

E = kQ/0.02² + kq/(0.025)² = k(Q/0.0004 + q/0.000625)

E = 9 * 10⁹ [(14 * 10⁻⁹ / 0.0004) + (14 * 10⁻⁹)/0.000625]

E = 9 * 10⁹[(3.5 * 10⁻⁵) + (2.24 * 10⁻⁵)]

E = 9 * 10⁹ * 5.74 * 10⁻⁵

E = 5.17 * 10⁵N/C

C. Electric field strength at a point 3.0cm (0.03m) from the glass rod is the sum of electric field strength due to both rods i.e.

E = E₁ + E₂

E₁ = kQ/0.03²

E₂ = kq/(0.045 - 0.03)² = kq/(0.015)²

E = kQ/0.03² + kq/(0.015)² = k(Q/0009 + q/0.000225)

E = 9 * 10⁹ [(14 * 10⁻⁹ / 0.009) + (14 * 10⁻⁹)/0.000225]

E = 9 * 10⁹[( 1.55 * 10⁻⁵) + (6.22 * 10⁻⁵)]

E = 9 * 10⁹ * 7.77 * 10⁻⁵

E = 6.993 * 10⁵N/C

4 0

3 years ago

Which types of a magnet will have a bigger magnetic feild area?​

Which types of a magnet will have a bigger magnetic feild area?