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avanturin [10]
2 years ago
10

The diagram shows an image of an object being

Physics
1 answer:
mihalych1998 [28]2 years ago
8 0

Answer:

i think the answer is x

Explanation:

have a great day<3

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Rather than ascribing the increased kinetic energy of the stone to the work of gravity, we now (when using potential energy rath
Sophie [7]

Answer:

Change/ Potential

Explanation:

Work is the amount of energy required to perform an action that is for a force to cause a displacement.

From work-energy theorem, work done by body is equal to change in its kinetic energy.

Work of gravity is basically the potential energy stored in the body due to gravity. From the law of conservation of mechanical energy, increased kinetic energy comes from the change of the potential energy of the stone.

8 0
3 years ago
A vehicle that goes from 5m/s to 45m/s in 8s. what is its acceleration?
GaryK [48]

Answer: 5m/s^2

Explanation:

V= 45m/s

U = 5m/s

t = 8s

a =?

V = u + at

45 = 5 + 8a

8a = 45 — 5

8a = 40

a = 40 / 8

a = 5m/s^2

3 0
4 years ago
Why won't a very bright beam of red light impart more energy to an ejected electron than a feeble beam of violet light?
bearhunter [10]
This is related to the energy carried by photons of light the energy of each photon is proportional to the frequency of the light since red light has a lower frequency then violet light and photons of red light carry less energy than the photons of violet light as a result the red protons eject electrons that have less energy than the ejected electrons by Violet photons
3 0
3 years ago
3 In a television tube, an electron starting from rest experiences a force of 4.0 × 10−15 N over a distance of 50 cm. The final
MAXImum [283]

Answer:

The final speed of the electron = 2.095×10⁸ m/s

Explanation:

From newton's fundamental equation of dynamics,

F = ma ........................Equation 1

Where F = force, m = mass of the electron, a = acceleration of the electron.

making a the subject of the equation,

a = F/m.................... Equation 2

Given: F = 4.0×10⁻¹⁵ N,

Constant: m =  9.109×10⁻³¹ kg.

Substituting into equation 2

a = 4.0×10⁻¹⁵/9.109×10⁻³¹

a = 4.39×10¹⁶ m/s².

Using newton's equation of motion,

v² = u²+2as .......................... Equation 3

Where v = final velocity of the electron, u = initial velocity of the electron, a = acceleration of the electron, s = distance covered by the electron.

Given: u = 0 m/s(at rest), s = 50 cm = 0.5 m, a = 4.39×10¹⁶ m/s²

Substituting into equation 3

v² = 0² + 2(0.5)(4.39×10¹⁶)

v = √(4.39×10¹⁶)

v = 2.095×10⁸ m/s

Thus the final speed of the electron = 2.095×10⁸ m/s

7 0
3 years ago
An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1
motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment M_{l}=2.90\times10^{-28}\ kg

Mass of the heavier fragment M_{h}=1.62\times10^{-27}\ Kg

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}

\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}

\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}

\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}

\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c

\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2

\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}

\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}

\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}

\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})

\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}

v_{2}^2=\dfrac{c^2}{8.93}

v_{2}=0.335c

Hence, The speed of the heavier fragment is 0.335c.

7 0
3 years ago
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