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3 years ago

Please verify these are correct. No work needs to be shown unless I've made a mistake. Thank you.

Physics

2 answers:

nevsk [136]3 years ago

8 0

Answer:

Those are correct, well done.

Tema [17]3 years ago

8 0

Answer:

your workings are correct

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Achargingbullelephantwithamassof5240kgcomesdi- rectly toward you with a speed of 4.55 m/s. You toss a 0.150-kg rubber ball at th

Ierofanga [76]

Answer:

Given:

Mass of elephant = 5240 kg

The initial speed of the elephant = 4.55 m/s

Mass of the rubber ball, m, = 0.15 kg

Inital speed of the rubber ball, v = 7.81 m/s

On substitution in $V_{2f}$ = $[\frac{2m_{1} }{m_{1} +m_{2} } ]V_{1f}$ + $[\frac{m_{2}-m_{1}}{m_{1}+m_{2} } ] v_{2f}$

$V_{2f}$ = $[\frac{2*5240_{} }{5240_{} +0.15_{} } ](-4.55_{})$ + $[\frac{0.15_{}-5240_{}}{5240_{}+0.15_{} } ] (7.81_{})$

a) The negatıve sign shows that the ball bounces back in the direction opposite to the incident

b) it is clear that the velocity of the ball increases and therefore it is kinetic energy . The ball gains kinetic energy from the elephant.

8 0

3 years ago

A jewellery melts 500g of Silver to pour into a mould. Calculate how much energy was released as the silver solidified.

irga5000 [103]

When silver is poured into the mould the it will solidify

In this process the phase of the Silver block will change from liquid to solid.

This phase change will lead to release in heat and this heat is known as latent heat of fusion.

The formula to find the latent heat of fusion is given as

$Q = mL$

here given that

$m = mass = 500 g$

$L = 111 kJ/kg$

now we can find the heat released

$Q = 0.5 * 111 kJ$

$Q = 55.5 kJ$

So it will release total heat of 55.5 kJ when it will solidify

8 0

4 years ago

20 points!

MariettaO [177]

Do they give answer choices? or is it free write? i’ll help if you tell me!!

8 0

3 years ago

An individual white LED (light-emitting diode) has an efficiency of 20% and uses 1.0 W of electric power. a. How many LEDs must

Neporo4naja [7]

Answer:

8, 8 W

Explanation:

The useful power of 1 Light Emitting Diode is

$0.2\times 1=0.2\ W$

Total power required is 1.6 W

Number of Light Emitting Diodes would be

$n=\dfrac{1.6}{0.2}\\\Rightarrow n=8$

The number of Light Emitting Diodes is 8.

Power would be

$P=8\times 1=8\ W$

The power that is required to run the Light Emitting Diodes is 8 W

7 0

4 years ago

A mass of 4.10 kg is suspended from a 1.69 m long string. It revolves in a horizontal circle as shown in the figure.

nikklg [1K]

The horizontal component of the tension in the string is a centripetal force, so by Newton's second law we have

• net horizontal force

$F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R$

where $m=4.10\,\rm kg$ , $v=2.85\frac{\rm m}{\rm s}$ , and $R$ is the radius of the circular path.

As shown in the diagram, we can see that

$\sin(\theta) = \dfrac Rr \implies R = r\sin(\theta)$

where $r=1.69\,\rm m$ , so that

$F_{\rm tension} \sin(\theta) = \dfrac{mv^2}R \\\\ \implies F_{\rm tension} = \dfrac{mv^2}{r\sin^2(\theta)}$

The vertical component of the tension counters the weight of the mass and keeps it in the same plane, so that by Newton's second law we have

• net vertical force

$F_{\rm \tension} \cos(\theta) - mg = 0 \\\\ \implies F_{\rm tension} = \dfrac{mg}{\cos(\theta)}$

Solve for $\theta$ :

$\dfrac{mv^2}{r\sin^2(\theta)} = \dfrac{mg}{\cos(\theta)} \\\\ \implies \dfrac{\sin^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \dfrac{1-\cos^2(\theta)}{\cos(\theta)} = \dfrac{v^2}{rg} \\\\ \implies \cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) - 1 = 0$

Complete the square:

$\cos^2(\theta) + \dfrac{v^2}{rg} \cos(\theta) + \dfrac{v^4}{4r^2g^2} = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \left(\cos(\theta) + \dfrac{v^2}{2rg}\right)^2 = 1 + \dfrac{v^4}{4r^2g^2} \\\\ \implies \cos(\theta) + \dfrac{v^2}{2rg} = \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}} \\\\ \implies \cos(\theta) = -\dfrac{v^2}{2rg} \pm \sqrt{1 + \dfrac{v^4}{4r^2g^2}}$

Plugging in the known quantities, we end up with

$\cos(\theta) \approx 0.784 \text{ or } \cos(\theta) \approx -1.27$

The second case has no real solution, since $-1\le\cos(\theta)\le1$ for all $\theta$ . This leaves us with

$\cos(\theta) \approx 0.784 \implies \theta \approx \cos^{-1}(0.784) \approx \boxed{38.3^\circ}$

7 0

2 years ago