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3 years ago

All known matter is made up of Question 1 options:

Physics

2 answers:

Luda [366]3 years ago

6 0

Answer:

D. Atoms

Explanation:

You are made of matter, everything you touch is made of matter if your at a desk thats matter. Therefore the answer is Atoms

Tatiana [17]3 years ago

4 0

the answer is (a) molecules

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When a bow is stretched back and an arrow is shot, what type of energy conversion has occurred?

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D. mechanical to elastic

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3 years ago

As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i

11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

Explanation:

Given that,

Height = 1.94 m

Bounced height = 1.48 m

Time interval $t=14.86\times10^{-3}\ s$

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$9.8\times1.94=\dfrac{1}{2}\times v^2$

$v=\sqrt{2\times9.8\times1.94}$

$v=6.166\ m/s$

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2'$

Put the value into the formula

$9.8\times1.48=\dfrac{1}{2}\times v^2'$

$v'=\sqrt{2\times9.8\times1.48}$

$v'=5.38\ m/s$

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

$a=\dfrac{v-v'}{t}$

Put the value into the formula

$a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}$

$a=776.98\ m/s^2$

$a=0.77\times10^{3}\ m/s^2$

Hence,The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

6 0

3 years ago

Two steel balls, of masses m1=1.00 kg and m2=2.00 kg, respectively, are hung from the ceiling with light strings next to each ot

Zolol [24]

Answer:

(a) The maximum height achieved by the first ball, m₁ is 0.11 m

(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m

Explanation:

Given;

mass of the first ball, m₁ = 1 kg

mass of the second ball, m₂ = 2 kg

The velocity of the first when released from a height of 1 m before collision;

u₁² = u₀² + 2gh

u₀ = 0, since it was released from rest

u₁² = 2gh

u₁² = 2 x 9.8 x 1

u₁² = 19.6

u₁ = √19.6

u₁ = 4.427 m/s

The velocity of the second ball before collision, u₂ = 0

Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final velocity of the first ball after an elastic collision

v₂ is the final velocity of the second ball after an elastic collision

m₁u₁ + m₂(0) = m₁v₁ + m₂v₂

m₁u₁ = m₁v₁ + m₂v₂

1 x 4.427 = v₁ + 2v₂

v₁ + 2v₂ = 4.427

v₁ = 4.427 - 2v₂ ----- equation (1)

one directional velocity;

u₁ + v₁ = u₂ + v₂

u₂ = 0

u₁ + v₁ = v₂

v₁ = v₂ - u₁

v₁ = v₂ - 4.427 ------ equation (2)

Substitute v₁ into equation (1)

v₂ - 4.427 = 4.427 - 2v₂

3v₂ = 4.427 + 4.427

3v₂ = 8.854

v₂ = 8.854 / 3

v₂ = 2.95 m/s (→ forward direction)

v₁ = v₂ - 4.427

v₁ = 2.95 - 4.427

v₁ = - 1.477 m/s

v₁ = 1.477 m/s ( ← backward direction)

Apply the law of conservation of mechanical energy

$mgh_{max} = \frac{1}{2}mv_{max}^2$

(a) The maximum height achieved by the first ball (v₁ = 1.477 m/s)

$mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max} = \frac{1}{2g}v_{max}^2\\\\ h_{max} = \frac{1}{2*9.8}(1.477^2)\\\\ h_{max} = 0.11 \ m$

(b) The maximum height achieved by the second ball (v₂ = 2.95 m/s)

$mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max} = \frac{1}{2g}v_{max}^2\\\\ h_{max} = \frac{1}{2*9.8}(2.95^2)\\\\ h_{max} = 0.44 \ m$

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3 years ago

Which change indicates that the universe is expanding?

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