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3 years ago

Why should it take significantly more energy to move a beam of alpha particles than a beam of beta minus (β–) particles?

Physics

2 answers:

lana [24]3 years ago

5 0

Because the mass of an alpha particle is significantly greater than the mass of a beta particle <span />

a_sh-v [17]3 years ago

4 0

An 'alpha particle' is the same thing as the nucleus of a helium atom ...
a little bundle made of 2 protons and 2 neutrons.

A 'beta' particle is an electron.

The mass of an alpha particle is more than 7,000 times the mass of
an electron, so it certainly takes more energy to get it moving.

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The world will face energy crisis in near future justify this statement

jeka57 [31]

Answer:

Nowadays most of our works are being done through different types of energy which are non-renewable. People are wasting lot of energy (hydel, dolar etc.) due to which in future we can face energy crisis.

5 0

2 years ago

A planet orbits a star, in a year of length 3.37 x 107 s, in a nearly circular orbit of radius 1.04 x 1011 m. With respect to th

PIT_PIT [208]

Answer

Given,

Time period of star,T = 3.37 x 10⁷ s

Radius of circular orbit,R = 1.04 x 10¹¹ m

a) Angular speed of the planet

$\omega = \dfrac{2\pi}{T}=\dfrac{2\pi}{3.37\times 10^{7}}$

$\omega = 1.864\times 10^{-7}\ rad/s$

b) tangential speed

$v = r \omega = 1.04\times 10^{11}\times 1.864 \times 10^{-7}$

v = 1.94 x 10⁴ m/s

c) centripetal acceleration magnitude

$a = \dfrac{v^2}{r}= \dfrac{(1.94\times 10^4)^2}{1.04\times 10^{11}}$

a = 3.62 x 10⁻³ m/s²

8 0

3 years ago

An air-conditioning system is to be filled from a rigid container that initially contains 5 kg of liquid R-134a at 24°C. The val

OverLord2011 [107]

Answer:

Answer

The Final Quality of teh R-134a in the container is 0.5056

The Total Heat transfer is $Q_{in} = 22.62 KJ$

Explanation:

Explanation is in the following attachments

3 0

3 years ago

Two blocks A and B with mA = 2.2 kg and mB = 0.84 kg are connected by a string of negligible mass. They rest on a frictionless h

madreJ [45]

Answer:

(a) a = 1.875 m/s²

(b)T = 1.575 N

(c)T increase

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the blocks on the horizontal surface and the y-axis in the direction perpendicular to it.

Forces acting on the block A

WA: Weight of of the A block : In vertical direction downaward (-y)

NA : Normal force of the A block :In vertical direction upaward (+y)

F= 5.7 N In in the direction parallel to the movement of the blocks (+x)

T : tension of the string: In in the direction (-x)

Forces acting on the block B

WB: Weight of the B block: In vertical direction downaward (-y)

NB : Normal force of the B block :In vertical direction upaward (+y)

T : tension of the string: In in the direction (+x)

Data

mA = 2.2 kg

mB = 0.84 kg

(a) Magnitude of the acceleration of the blocks

Newton's second law to B block:

∑Fx = m*a

T = (0.84)*a Equation (1)

Newton's second law to A block:

∑Fx = m*a

5.7 - T = (2.2)*a We replace T of the Equation (1)

5.7 - (0.84)*a = (2.2)*a

5.7 = (2.2)*a + (0.84)*a Equation (2)

5.7 =(3.04)*a

a =5.7 / (3.04)

a = 1.875 m/s²

(b)Tension (in N) in the string connecting the two blocks

We replace data in the Equation (1)

T = (0.84)*a

T = (0.84)*(1.875)

T = 1.575 N

(c) How will the tension in the string be affected if mA is decreased?

We observed Equation (2)

5.7 = (2.2)*a + (0.84)*a

5.7 = (mA)*a + (0.84)*a

5.7 =a*( mA+ 0.84)

if mA decrease , then, the acceleration increase and T increase

4 0

3 years ago

A person pushes horizontally on a 50-kg crate, causing it to accelerate from rest and slide across the surface. If the push caus

Marina CMI [18]

Answer:

$v_{f} =4.9\frac{m}{s}$ : velocity of the crate after the person has pushed the crate a distance of 6 meters

Explanation:

Crate kinetics

Crate moves with uniformly accelerated movement

v f²=v₀²+2a*d (formula 1)

d:displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Known data

v₀=0 The speed of the crate is equal to zero because part of the rest.

a= 2m/s²

d= 6m

Distance calculating

We replace data in the Formula (1)

v f²=0+2*2*d

v f²=2*2*6

v f²=24

$v_{f} =\sqrt{24}$

$v_{f} =4.9\frac{m}{s}$

7 0

3 years ago