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3 years ago

12

A 12 kg ball moving at 20 m/s strikes another ball at rest. After a completely inelastic collision, the balls move at 10 m/s. Wh

at was the mass of the second ball?

1 answer:

Alborosie3 years ago

6 0

(12)(20)+(x)(0)=(12+x)(10)
240=120+10x
120=10x
x=12 kg

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If you wish to observe features that are around the size of atoms, say 1 .5 x 100 m, with electromagnetic radiation, the radiati

chubhunter [2.5K]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

If you wish to observe features that around the size of atoms, say 1.5×10⁻¹⁰ m, with electromagnetic radiation, the radiation must have a wavelength about the size of the atom itself.

a) If you had a microscope which was capable of doing this, what would the frequency of electromagnetic radiation be, in hertz that you would have to use?

b) What type of electromagnetic radiation would this be?

Given Information:

Wavelength = λ = 1.5×10⁻¹⁰ m

Required Information:

a) Frequency = f = ?

b) Type of electromagnetic radiation = ?

Answer:

a) Frequency = f = 2×10¹⁸ Hz

b) Type of electromagnetic radiation = X-rays

Explanation:

a) The frequency of the electromagnetic radiation is given by

f = c/ λ

Where λ is the wavelength of the electromagnetic radiation and c is the speed of light and its value is 3×10⁸ m/s

f = 3×10⁸/1.5×10⁻¹⁰

f = 2×10¹⁸ Hz

Therefore, the frequency of the electromagnetic radiation would be 2×10¹⁸ Hz.

b)

The frequency range of X-rays is 3×10¹⁶ Hz to 3×10¹⁹ Hz

The frequency 2×10¹⁸ lies in that range, therefore, the type of electromagnetic radiation is X-rays

5 0

3 years ago

A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium whe

notsponge [240]

Answer:

$Q = \frac{0.068}{E}$

where E = electric field intensity

Explanation:

As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical

So here force due to electrostatic force on the charged ball is in horizontal direction along the direction of electric field

while weight of the ball is vertically downwards

so here we have

$QE = F_x$

$mg = F_y$

since string makes 30 degree angle with the vertical so we will have

$tan\theta = \frac{F_x}{F_y}$

$tan30 = \frac{QE}{mg}$

$Q = \frac{mg}{E}tan30$

$Q = \frac{0.012\times 9.81}{E} tan30$

$Q = \frac{0.068}{E}$

where E = electric field intensity

5 0

3 years ago

A small grinding wheel has a moment of inertia of 4.0*10-5kgm2. What net torque must be applied to the wheel for its angular acc

kvv77 [185]

Hi there!

We can use the rotational equivalent of Newton's Second Law:

$\huge\boxed{\Sigma \tau = I \alpha}$

Στ = Net Torque (Nm)

I = Moment of inertia (kgm²)

α = Angular acceleration (rad/sec²)

We can plug in the given values to solve.

$\Sigma \tau = (4 * 10^{-5})(150) = \boxed{0.006 Nm}$

4 0

2 years ago

Given a wire with a cross section of .45cm^2, a length of 3cm, and an internal resistance of 3 ohms, show your 5 steps to solve

Gre4nikov [31]

Resistance ∞ (proportional) length
resistance ∞ 1/ area

therefore,
(the constant that we take is known as the resistivity)

resistance = (resistivity*length )/ area
resistivity = (resistance * area ) / length
= (3 * 45) / 3 = 135/3 = 45 Ωm

in short your answer is 45 Ωm

4 0

3 years ago

An unbanked (flat) curve of radius 150 m is rated for a maximum speed of 32.5 m/s. At what maximum speed, in m/s, should a flat

Rasek [7]

Answer:

The maximum speed is 21.39 m/s.

Explanation:

Given;

radius of the flat curve, r₁ = 150 m

maximum speed, $v_{max}$ = 32.5 m/s

The maximum acceleration on the unbanked curve is calculated as;

$a_c_{max} = \frac{V_{max}^2}{r} \\\\a_c_{max} = \frac{32.5^2}{150} \\\\a_c_{max} = 7.04 \ m/s^2$

the radius of the second flat curve, r₂ = 65.0 m

the maximum speed this unbanked curve should be rated is calculated as;

$a_c_{max} = \frac{V_{max}^2}{r_2} \\\\V_{max}^2 = a_c_{max} \ \times \ r_2\\\\V_{max} = \sqrt{a_c_{max} \ \times \ r_2} \\\\V_{max} =\sqrt{7.04 \ \times \ 65} \\\\V_{max} = 21.39 \ m/s$

Therefore, the maximum speed is 21.39 m/s.

3 0

3 years ago