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3 years ago

In which type of collision is no kinetic energy converted to heat or sound energy?

Physics

1 answer:

mezya [45]3 years ago

5 0

Answer:

elastic collision

An elastic collision occurs when the two objects "bounce" apart when they collide. Two rubber balls are a good example. In an elastic collision, both momentum and kinetic energy are conserved. Almost no energy is lost to sound, heat, or deformation.

I hope it's helpful!

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What is the primary outgoing radiation put off by the sun?

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3 years ago

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marusya05 [52]

Answer:

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Explanation:

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8 0

3 years ago

Read 2 more answers

If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.41 m

AveGali [126]

Answer:

6.88 m/s

Explanation:

The Conservation of Energy states that:

Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy + Final Potential Energy

So we can write

$mgh_{i}+\frac{1}{2}mv_{i} ^{2}=mgh_{f}+\frac{1}{2}mv_{f} ^{2}$

We can cancel the common factor of $m$ which leaves us with

$gh_{i}+\frac{1}{2}v_{i} ^{2}=gh_{f}+\frac{1}{2}v_{f} ^{2}$

Lets solve for $v_f$

$gh_{i}+\frac{v_{i} ^{2}}{2}=gh_{f}+\frac{v_{f} ^{2}}{2}$

Subtract $gh_f$ from both sides of the equation.

$gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f}=\frac{v_{f} ^{2}}{2}$

Multiply both sides of the equation by 2.

$2(gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f})={v_{f} ^{2}$

Simplify the left side.

Apply the distributive property.

$2(gh_{i})+2\frac{v_{i} ^{2}}{2}+2(-gh_{f})={v_{f} ^{2}$

Cancel the common factor of 2.

$2gh_{i}+v_{i} ^{2}-2gh_{f}={v_{f} ^{2}$

Take the square root of both sides of the equation to eliminate the exponent on the right side.

${v_{f}=\sqrt{2gh_{i}+v_{i} ^{2}-2gh_{f}}$

We are given $g,v_{i},h_{i},h_{f}$ .

We can now solve for the final velocity.

${v_{f}=\sqrt{(2*9.81*2.41)+(0^{2})-(2*9.81*0)$

Anything multiplied by 0 is 0.

${v_{f}=\sqrt{2*9.81*2.41$

${v_{f}=\sqrt{47.2842$

$v_f=6.88$

7 0

1 year ago

How does the work required to accelerate a particle from 10 m/s to 20 m/s compare to that required to accelerate it from 20 m/s

poizon [28]

To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore

$W = \Delta KE$

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

Here,

m = mass

$v_{f,i}$ = Velocity (Final and initial)

First case) When the particle goes from 10m/s to 20m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(20)^2 -\frac{1}{2} (m)(10)^2$

$W_1 = 150(m) J$

Second case) When the particle goes from 20m/s to 30m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(30)^2 -\frac{1}{2} (m)(20)^2$

$W_1 = 250(m) J$

As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.

5 0

3 years ago

An aluminum clock pendulum having a period of 1.00 s keeps perfect time at 20.0°C.

amm1812

Answer:

a) T ’= 0.999 s , b) t = 3596.4 s

Explanation:

The angular velocity of a simple pendulum is

w = √g / L

The angular velocity, frequency and period are related

w = 2π f = 2π / T

2π / T = √ g / L

T = 2π √ L / g

L = T² g / 4π²

L = 1² 9.8 / 4π²

L = 0.248 m

To know the effect of the temperature change let's use the thermal expansion ratios

ΔL = α L ΔT

ΔL = 24 10⁻⁶ 0.248 (-4 - 20)

ΔL = 142.8 10⁻⁶ m

Lf - L = -142. 8 10⁻⁶

Lf = 142.8 10⁻⁶ + 0.248

Lf = 0.2479 m

Let's calculate new period

T ’= 2π √ L / g

T ’= 2π √ (0.2479 / 9.8)

T ’= 0.999 s

We can see that the value of the period is reduced so that the clock is delayed

b) change of time in 1 hour

When the clock is at 20 ° C in one hour it performs 3600 oscillations, for the new period the time of this number of oscillations is

t = 3600 0.999

t = 3596.4 s

Therefore the clock is delayed almost 4 s

6 0

4 years ago