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3 years ago

You have given a power supply, copper wire, and an iron nail. What should you do to decrease the strength of the electro magnet?

1 answer:

8 0

If you wrap some of the wire around the nail in one direction and some of the wire in the other direction, the magnetic fields from the different sections fight each other and cancel out, reducing the strength of your magnet.

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Here's a basketball problem: A 87.2 kg basketball player is running in the positive direction at 7.0 m/s. She is met head-on by

Ray Of Light [21]

Answer:

2.47 m/s backwards

Explanation:

From the law of conservation of momentum,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂...................... Equation 1

Where m₁ and m₂ = mass of the first basketball player and second basket ball player respectively, u₁ and u₂ = initial velocity of the first basket player and the second basketball player respectively, v₁ and v₂ = The final velocity of the first basket ball player and second basket ball player respectively.

Making v₁ the subject of the equation,

v₁ = (m₁u₁ + m₂u₂ - m₂v₂)/m₁.......................... Equation 2.

Given: m₁ = 87.2 kg, m₂ = 102.0 kg, u₁ = 7.0 m/s, u₂ = -5.2 m/s, v₂ = 2.9 m/s

Note: u₂ is negative because it moves towards the first basket ball player.

Substitute into equation 2

v₁ = [87.2(7.0)+102(-5.2) - (102×2.9)]/87.2

v₁ = (610.4-530-295.8)/87.2

v₁ = -215.4/87.2

v₁ = -2.47 m/s.

Thus the velocity of the 87.2 kg player = 2.47 m/s backwards.

7 0

3 years ago

A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i

Vera_Pavlovna [14]

Answer:

Explanation:

1. We use the conservation of momentum for before the raining and after. And also we take into account that in 0.5h the accumulated water is

100kg/h*0.5h = 50kg

$p_{b}=p_{a}\\M_{b}v_{b}=(M_{b}+m_{r})v_{a}\\v_{a}=\frac{M_{b}v_{b}}{M_{b}+m_{r}}=\frac{250kg*1\frac{m}{s}}{250kg+50kg}=0.83\frac{m}{s}$

2. the momentum does not conserve because the drag force of water makes that the boat loses velocity

3. If we assume that the force of the boat before the raining is

$F=ma=m\frac{v-v_{0}}{t-t_{0}}=250kg\frac{1m}{s^{2}}=250N$

where we have assumed that the acceleration of the boat is 1m/s{2} just before the rain starts

And if we take the net force as

$F_{net}=M_{b}a_{net}=F-F_{d}=250N-bv^{2}\\F_{net}=250N-0.5N\frac{s^{2}}{m^{2}}(1\frac{m}{s})^{2}=249.5N\\a_{net}=\frac{249.5N}{M_{b}}=\frac{249.5N}{250kg}=0.99\frac{m}{s^{2}}$

where we take v=1m/s because we are taking into account tha velocity just after the rain stars.

I hope this is useful for you

regards

5 0

3 years ago

The Goodyear blimps, which frequently fly over sporting events, hold approximately 1.65_105ft3 of helium.

Sonbull [250]

Applying the ideal gas equation:
PV = nRT
PV = (mass/Mr)RT
mass = PVMr/RT
mass = (101325 x 4672.2 x 4) / (8.314 x 297)
= 766887.3 kg
= 7.7 x 10⁵ kg

7 0

3 years ago

How to calculate final speed when the mass is 40,000kg height id 2.5km and 500,000N of force

Tju [1.3M]

The final speed when the mass is 40,000kg height is 2.5km and 500,000N of force is 176.8 m / s

According to Newton's second law of motion,

F = m a

F = Force

m = Mass

a = Acceleration

m = 40000 kg

F = 500000 N

a = F / m

a = 500000 / 40000

a = 12.5 m / s²

a = v / t

v = d / t

v = Velocity

t = Time

d = Distance

d = 2.5 km = 2500 m

a = d / t²

12.5 = 2500 / t²

t² = 200

t = 14.14 s

v = 2500 / 14.14

v = 176.8 m / s

Therefore, the final speed is 176.8 m / s

To know more about Newton's second law of motion

brainly.com/question/13447525

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8 0

11 months ago

At the presentation ceremony, a championship bowler is presented a 1.60-kg trophy which he holds at arm's length, a distance of

nikdorinn [45]

Answer:

a)10.28 Nm

b)9.93 Nm

Explanation:

Let g = 9.81m/s2. First we can calculate the weight of the trophy

W = mg = 1.6 * 9.81 = 15.696 N

(a) The torque is product of force and its moment arm

T = WL = 15.696 * 0.655 = 10.28 Nm

(b) Suppose his arm makes an angle of 15 degree with respect to the horizontal line. We can still calculate the arm length, or the horizontal distance from the trophy to the champion:

$L_2 = Lcos(15^0) = 0.655cos(15^0) = 0.655*0.966 = 0.633 m$

Again, torque is product of force and its moment arm

$T_2 = WL_2 = 15.696 * 0.633 = 9.93 Nm$

8 0

3 years ago