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3 years ago

What is the force that can cause two pieces of iron to attract each other?

2 answers:

8 0

Is it A ? Yes ! Two pieces of iron DO always attract each other with equal and opposite gravitational forces, although the forces are so small that we never notice them.

Is it B ? Yes, it could be. If at least one of the pieces of iron is magnetized, then they can attract each other with equal and opposite magnetic forces.

Is it C ? Yes, it could be. But the pieces of iron would have to be joined by a stretched rubber band. Then they would attract each other with equal and opposite elastic forces.

Is it D ? Yes, it could be. If one of the pieces of iron carries a net electric charge, then they attract each other with equal and opposite electrostatic forces.

Rudiy273 years ago

6 0

Answer:

B. Magnetic Force

Explanation:

two pieces of irons cannot attract each other unless at least one of them is magnetize. That force is called magnetism.

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A pebble is thrown into the air with a velocity of 19/m at an angle of 36 with respect to the horizontal.

kow [346]

Answer:

The maximum height the pebble reaches is approximately;

A. 6.4 m

Explanation:

The question is with regards to projectile motion of an object

The given parameters are;

The initial velocity of the pebble, u = 19 m/s

The angle the projectile path of the pebble makes with the horizontal, θ = 36°

The maximum height of a projectile, $h_{max}$ , is given by the following equation;

$h_{max} = \dfrac{\left (u \times sin(\theta) \right)^2}{2 \cdot g}$

Therefore, substituting the known values for the pebble, we have;

$h_{max} = \dfrac{\left (19 \times sin(36 ^{\circ}) \right)^2}{2 \times 9.8} = 6.3633894140470403035477570509439$

Therefore, the maximum height of the pebble projectile, $h_{max}$ ≈ 6.4 m.

3 0

3 years ago

What is the escape speed of an electron launched from the surface of a 1.1-cm-diameter glass sphere that has been charged to 8.0

zimovet [89]

At surface,
v = kq/r

And potential energy of an electron is given by,
PE = -ev = -ekq/r

At escape velocity,
PE + KE = 0.
Therefore,
1/2mv^2 - ekq/r =0
1/2mv^2 = ekq/r
v = Sqrt [2ekq/mr], where v = escape velocity, e = 1.6*10^-19 C, k = 8.99*10^9 Nm^2/C^2, m = 9.11*10^-31 kg, r = 1.1*10^-2 m, q = 8*10^-9 C

Substituting;
v = Sqrt [(2*1.6*19^-19*8.99*10^9*8*10^-9)/(9.11*10^-31*1.1*10^-2)] = 47949357.23 m/s ≈ 4.795 *10^7 m/s

5 0

3 years ago

X-rays with frequency 3 ⨉ 10 18 Hz shine on a crystal, producing an interference pattern. If the first bright spot is observed a

FrozenT [24]

Answer:

$5.1645\times 10^{-11}\ m$

Explanation:

n = Order = 1

c = Speed of light = $3\times 10^8\ m/s$

f = Frequency = $3\times 10^{18}\ Hz$

$\theta$ = Angle = $75.5^{\circ}$

Lattice spacing is given by

$d=\dfrac{n\lambda}{2\sin\theta}\\\Rightarrow d=\dfrac{n\times c}{f2\sin\theta}\\\Rightarrow d=\dfrac{1\times 3\times 10^{8}}{3\times 10^{18}\times 2\times \sin75.5^{\circ}}\\\Rightarrow d=5.1645\times 10^{-11}\ m$

The lattice spacing of the crystal is $5.1645\times 10^{-11}\ m$

6 0

3 years ago

One end of a string 4.32 m long is moved up and down with simple harmonic motion at a frequency of 75 Hz . The waves reach the o

max2010maxim [7]

To solve this problem, we will apply the concepts related to the kinematic equations of linear motion, which define speed as the distance traveled per unit of time. Subsequently, the wavelength is defined as the speed of a body at the rate of change of its frequency. Our values are given as,

$\text{Length of the string} = L = 4.32 m$