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melamori03 [73]
3 years ago
14

Which of the following selections completes the given radioactive decay? A). Alpha particle B). Gamma radiation. C). Positron. D

). Neutron. E). Beta particle.

Physics
1 answer:
goldenfox [79]3 years ago
8 0

Answer:

gamma radiation

is which I think

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Geologists use the blank test to determine the hardness of a mineral
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Geologists use the Mohs Harness Scale (aka the Scratch Test) to determine the hardness of a mineral.

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8 0
3 years ago
Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an
8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 m/s^2 where as radial the component of acceleration is 8.77 m/s^2

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

r=e^u

So the transverse component of acceleration are given as

a_{\theta}=(ru''+2r'u')\\

Here

r=e^u\\r=e^{\pi/4}\\r=2.1932 m

r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m

So

a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\

The transverse component of acceleration is 26.32 m/s^2

The radial component is given as

a_r=r''-r\theta'^2

Here

r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m

So

a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2

The radial component of acceleration is 8.77 m/s^2

6 0
3 years ago
A 1.2 KG rubber ball is being thrown in the air if the ball is traveling at 2.0 M/S when it is 3.0 M off the ground what is the
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Rocket engineers use newton's third law during launch. identify the action force.
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The answer is A. Newton's third law of motion states that for every action, there is an equal and opposite reaction. A rocket exerts a large force on the gas that is in the rocket chamber (action). The gas thus exerts a large reaction force forward on the rocket (reaction). The large reaction force is called thrust.
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3 years ago
Read 2 more answers
To win the game, a placekicker must kick a football from a point 44 m (48.1184 yd) from the goal, and the ball must clear the cr
Ganezh [65]

Answer:

The distance by the ball clear the crossbar is 1.15 m

Explanation:

Given that,

Distance = 44 m

Speed = 24 m/s

Angle = 31°

Height = 3.05 m

We need to calculate the horizontal velocity

Using formula of horizontal velocity

u_{x}=u\cos\theta

Put the value into the formula

u_{x}=24\cos(31)

u_{x}=20.5\ m/s

We need to calculate the vertical velocity

Using formula of vertical velocity

u_{y}=u\sin\theta

Put the value into the formula

u_{y}=24\sin(31)

u_{y}=12.3\ m/s

We need to calculate the time

Using formula of time

t=\dfrac{d}{u_{x}}

Put the value into the formula

t=\dfrac{44}{20.5}

t=2.1\ sec

We need to calculate the vertical height

Using equation of motion

h=u_{y}t+\dfrac{1}{2}at^2

Put the value into the formula

h=12.3\times2.1-\dfrac{1}{2}\times9.8\times(2.1)^2

h=4.2\ m

We need to calculate the distance by the ball clear the crossbar

Using formula for vertical distance

d=h-3.05

Put the value of h

d=4.2-3.05

d=1.15\ m

Hence, The distance by the ball clear the crossbar is 1.15 m

7 0
3 years ago
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