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3 years ago

14

A 2.0-kg block is on a perfectly smooth (frictionless) ramp that makes an angle of 30^\circ30 ∘ with the horizontal. What is

the force of the ramp on the block?

1 answer:

Nastasia [14]3 years ago

7 0

Answer:

Explanation:

Since the surface is frictionless therefore there will be no friction force on block but there will be weight of block which we can divide in to two components i.e. mgcosθ &mgsinθ which is perpendicular and parallel to the surface respectively.

In response to mgcosθ ramp will apply a normal force to the block which will be of equal magnitude to that of mgcosθ.

Therefore Ramp will apply a Force of mgcosθ on block where m is the mass of block.

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elena55 [62]

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A dielectric (insulator) is a substance that is poorly conducting or not conducting at all. The concentration of free charge carriers in a dielectric does not exceed 108 cm-3. The main property of the dielectric is the ability to polarize in an external electric field. From the point of view of the band theory of a solid body, a dielectric is a substance with a band gap greater than 3 eV.

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A baseball with a mass of 0.8 kg is given an acceleration of 20m/s.how much force was applied to the ball

Delicious77 [7]

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3 years ago

Five factors predisposing a person to stroke

insens350 [35]

Answer:

the answers maybe→↓

Explanation:

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3 years ago

Read 2 more answers

You are at the controls of a particle accelerator, sending a beam of 3.60 x10^7 m/s protons (mass m) at a gas target of an unkno

matrenka [14]

Answer:

a) mass of unknown nucleus = 0.04245 mp, where mp is the proton mass

b) Speed of the unknown nucleus = (7.067 x 10^7) m/s

Explanation:

Considering the initial conditions, the observed collisions are ellastic, i.e, the total kinetic energy are conserved. The proton's mass will refer as $m_{p}$ .

(a)

Total kinetic energy conservation

$\frac{1}{2}m_{p}v_{p_0}^{2}+\frac{1}{2}m_{u}v_{u_o}^{2}=\frac{1}{2}m_{p}v_{p_f}^{2}+\frac{1}{2}m_{u}v_{u_f}^{2}$

where $v_{u_o}$ represents the initial velocity of the unknown element, $m_{u}$ the mass of the unknown element, and $v_{u_f}$ the final velocity of the unknown element

Linear momentum conservation

$m_{p}v_{p_0}+m_{u}v_{u_o}=m_{p}v_{p_f}+m_{u}v_{u_f}$

Using the initial speed of the target nucleus (unknown) is negligible, i.e, its speed is zero. Thereby, using the relation of linear momentum conservation given above, it is possible to find an expression of the final speed of the unknown nucleus in terms of its mass, which can be inserted in the relation of the kinetic energy conservation to obtain the value of the mass of the unknown elements, as follows;

$m_{u}v_{u_f}=m_{p}v_{p_0}-m_{p}v_{p_f}\\\\v_{u_f}=\frac{m_{p}(v_{p_0}-v_{p_f})}{m_{u}}$

Substituting this expression in the relation of total kinetic energy conservation,

$m_{p}(v^{2}_{p_0}-v^{2}_{p_f})={m_{u}}v^{2}_{u}_{f}$

Then,

$m_{p}(v^{2}_{p_0}-v^{2}_{p_f})={m_{u}}\frac{m^{2}_{p}(v_{p_0}-v_{p_f})^{2}}{m^{2}_{u}}\\\\m_{u}= \frac{m_{p}(v_{p_0}-v_{p_f})^{2}}{(v^{2}_{p_0}-v^{2}_{p_f})}$

Replacing the given data

$m_{u}= \frac{m_{p}(3.6x10^{7}-3.3x10^{7})^{2}}{((3.6x10^{7})^{2}-(3.3x10^{7})^{2})}$

Then,

$m_{u}=0.04245m_{p}$

(b) Using the relation of the final speed from linear momentum conservation and the above result, the speed of the unknown nucleus is calculated

$v_{u_f}=\frac{m_{p}(v_{p_0}-v_{p_f})}{m_{u}}\\\\v_{u_f}=\frac{m_{p}(3.6x10^{7}-3.3x10^{7})}{0.04245m_{p}}\\\\v_{u_f}=7.067x10^{7} m/s$

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3 years ago

An alpha particle collides with an oxygen nucleus, initially at rest. The alpha particle is scattered at an angle of 25.0° above

Greeley [361]

Answer:

$v_{i}= 19\times 10^5\ m/s$

Explanation:

given,

scattering angle of alpha particle = 25.0° above its initial direction of motion

oxygen nucleus recoils at = 50.0° below this initial direction.

final speed of the oxygen = 2.08×10⁵ m/s

mass of alpha particle = 4.0 u

mass o oxygen nucleus = 16 u

momentum conservation along x- axis

$m_{a}v_{i} = m_a v_a cos\theta + m_o v_o cos\theta$

$4v_{i} = 4\times v_a cos25^0 + 16\times 2.08 \times 10^5 cos50^0$

$v_{i}= \dfrac{3.625\times v_a+ 21.39\times 10^5}{4}$ ....(1)

Along y-direction

$0 = m_av_a sin \theta - m_ov_o sin\theta$

$0 = 4\times v_a sin 25 - 16\times 2.08 \times 10^5 sin50^0$

$v_a = \dfrac{25.49 \times 10^5}{1.69}$

$v_a = 15.082\times 10^5\ m/s$

putting value in equation (1)

$v_{i}= \dfrac{3.625\times 15.082\times 10^5+ 21.39\times 10^5}{4}$

$v_{i}= 19\times 10^5\ m/s$

5 0

3 years ago