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Karolina [17]
3 years ago
7

True or false, only a radioactive isotope will have a half-life

Physics
1 answer:
maks197457 [2]3 years ago
5 0
False not radioactive isotope will have a half-life
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Write one to two sentences explaining whether this passage from the story is exposition, rising action, climax, or resolution: “
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Do houses use parallels or series circuits
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Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic
VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

m_Agh=\frac{1}{2}m_Av_A^2

where

g=9.8 m/s^2 is the acceleration due to gravity

m_A=0.5 kg is the mass of the block

v_A is the speed of the block A just before touching block B

Solving for the speed,

v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

e=\frac{v'_B-v'_A}{v_A-v_B}

where:

e = 0.7 is the coefficient of restitution in this case

v_B' is the final velocity of block B

v_A' is the final velocity of block A

v_A=3.83 m/s

v_B=0 is the initial velocity of block B

Solving,

v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s

Re-arranging it,

v_A'=v_B'-2.68 (1)

Also, the total momentum must be conserved, so we can write:

m_A v_A + m_B v_B = m_A v'_A + m_B v'_B

where

m_B=2 kg

And substituting (1) and all the other values,

m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m

5 0
3 years ago
A solid sphere is released from the top of a ramp that is at a height h1 = 2.30 m. It rolls down the ramp without slipping. The
Oksi-84 [34.3K]

Answer:

The horizontal distance d does the ball travel before landing is 1.72 m.

Explanation:

Given that,

Height of ramp h_{1}=2.30\ m

Height of bottom of ramp h_{2}=1.69\ m

Diameter = 0.17 m

Suppose we need to calculate the horizontal distance d does the ball travel before landing?

We need to calculate the time

Using equation of motion

h_{2}=ut+\dfrac{1}{2}gt^2

t=\sqrt{\dfrac{2h_{2}}{g}}

t=\sqrt{\dfrac{2\times1.69}{9.8}}

t=0.587\ sec

We need to calculate the velocity of the ball

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times(\dfrac{2}{5}mr^2)\times(\dfrac{v}{r})^2

K.E=\dfrac{7}{10}mv^2

Using conservation of energy

K.E=mg(h_{1}-h_{2})

\dfrac{7}{10}mv^2=mg(h_{1}-h_{2})

v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})

Put the value into the formula

v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}

v=2.922\ m/s

We need to calculate the horizontal distance d does the ball travel before landing

Using formula of distance

d =vt

Where. d = distance

t = time

v = velocity

Put the value into the formula

d=2.922\times 0.587

d=1.72\ m

Hence, The horizontal distance d does the ball travel before landing is 1.72 m.

8 0
3 years ago
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