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3 years ago

3. Two bullets have masses of 0.003 kg and 0.006 kg, respectively. Both are fired with a speed of 40.0 m/s.

Physics

1 answer:

Novay_Z [31]3 years ago

6 0

Answer:

A. The bullet with 0.006kg has more energy

B. When the mass is doubled the kinetic energy increases

Explanation:

Kinetic energy increases when mass increases

kinetic energy increases when velocity increases

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If I work out rotational energy to be 102.2J which equals kg.M/s^2, and I hadn't factored time into it, would that be Joules per

Marina CMI [18]

Answer:

0.057 joules is needed to create the total rotational energy each second.

Explanation:

The energy rate is the ratio of total energy to time, which coincides with the definition of power at constant rate:

$\dot W = \frac{\Delta E}{\Delta t}$

$\dot W = \frac{102.2\,J}{\left(30\,min\right)\cdot \left(60\,\frac{s}{min} \right)}$

$\dot W = 0.057\,\frac{J}{s}$

$\dot W = 0.057\,W$

0.057 joules is needed to create the total rotational energy each second.

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3 years ago

If you have 80g of a radioactive substance whose half life is 2 days, how long will it take for the substance to decay to the po

Ber [7]

Answer:

6 days.

Explanation:

From radioactivity, The expression for half life is given as,

R/R' = 2⁽ᵃ/ᵇ)................... Equation 1

Where R = original mass of the radioactive substance, R' = Remaining mass of the radioactive substance after decay, a = Total time taken to decay, b = half life.

Given: R = 80 g, R' = 10 g, b = 2 days.

Substitute into equation 1

80/10 = 2⁽ᵃ/²⁾

8 = 2⁽ᵃ/²⁾

2³ = 2⁽ᵃ/²)

Equating the base and solving for a

3 = a/2

a = 2×3

a = 6 days.

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3 years ago

What is the speed of a bobsled whose distance-time graph indicates that it traveled 122m in 27s?

Stolb23 [73]

Speed = distance/time
speed= 122÷27=4.52m/s (3sf)

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3 years ago

Calculate the acceleration if you push with a 20-N horizontal force on a 2-kg block on a horizontal friction-free air table

Afina-wow [57]

Acceleration = force / mass = 20 / 2 = 10 m/s^2

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3 years ago

A 2.6 kg mass attached to a light string rotates on a horizontal,

Ainat [17]

The maximum speed the mass can have before it breaks is 2.27 m/s.

The given parameters:

<em>maximum mass the string can support before breaking, m = 17.9 kg</em>
<em>radius of the circle, r = 0.525 m</em>

The maximum speed the mass can have before it breaks is calculated as follows;

$T = ma_c\\\\Mg = \frac{Mv^2}{r} \\\\v^2 = rg\\\\v = \sqrt{rg} \\\\v_{max} = \sqrt{0.525 \times 9.8} \\\\v_{max} = 2.27 \ m/s$

Thus, the maximum speed the mass can have before it breaks is 2.27 m/s.

Learn more about maximum speed of horizontal circle here:brainly.com/question/21971127

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2 years ago