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4 years ago

Physical change of a solid to a liquid at the melting point

1 answer:

5 0

This occurs when the internal energy of the solid increases, typically by the application of heat or pressure, which increases the substance's temperature to the melting point. At the melting point, the ordering of ions or molecules in the solid breaks down to a less ordered state, and the solid melts to become a liquid.

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After Thanksgiving, Kevin and Gamal use the turkey’s wishbone to make a wish. If Kevin pulls on it with a force 0.17 N larger th

Sholpan [36]

Answer:

Explanation:

2F-F=ma, so F=ma or .17=(13x10-3kg)a then a=13.07m/s2

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3 years ago

Nina is doing a workout where she runs at 50% speed for 3 minutes, 75% speed for 2 minutes, and 100% speed for 30 seconds. This

Dovator [93]

D. Interval training

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2 years ago

Read 2 more answers

An insulating sphere is 8.00 cm in diameter and carries a 6.50 ÂµC charge uniformly distributed throughout its interior volume.

Kobotan [32]

Explanation:

(a) Formula to calculate the density is as follows.

$\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}$

= $2.42 \times 10^{-2} C/m^{3}$

Now, calculate the charge as follows.

$q_{in} = \rho(\frac{4}{3} \pi r^{3})$

= $2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}$

= $10.106 \times 10^{-8}$ C

or, = 101.06 nC

(b) For r = 6.50 cm, the value of charge will be calculated as follows.

$q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}$

= 7.454 $\mu C$

7 0

3 years ago

A car company wants to ensure its newest model can stop in less than 450 ft when traveling at 60 mph. If we assume constant dece

seraphim [82]

Answer:

The value of acceleration that accomplishes this is 8.61 ft/s² .

Explanation:

Given;

maximum distance to be traveled by the car when the brake is applied, d = 450 ft

initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s

final velocity of the car when it stops, v = 0

Apply the following kinematic equation to solve for the deceleration of the car.

v² = u² + 2as

0 = 88.02² + (2 x 450)a

-900a = 7747.5204

a = -7747.5204 / 900

a = -8.61 ft/s²

|a| = 8.61 ft/s²

Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .

4 0

3 years ago

A rock is thrown upward from level ground in such a way that the maximumheight of its flight is equal to its horizontal rangeR.

Ne4ueva [31]

The range of a projectile motion is given by:

$\frac{u_o^2 sin2\theta}{g}$

where, $u$ is the initial speed of the projectile, $\theta$ is the angle of the projectile and $g$ is the acceleration due to gravity.

The maximum height reached is given by:

$\frac{u_o^2 sin^2\theta}{2g}$

Part a

It is given that the maximum height reached is equal to the horizontal range. we need to find the angle of the projectile.

Equating the two:

$\frac{u_o^2 sin2\theta}{g}=\frac{u_o^2 sin^2\theta}{2g}\\ \Rightarrow 2 sin2\theta =sin^2\theta\\ \Rightarrow 2\times 2 sin\theta cos\theta=sin^2 \theta\\ \Rightarrow tan\theta =4\\ \Rightarrow \theta=tan^{-1}4=75.96^o$

Hence, the projectile was thrown at an initial angle of $\theta=75.96^o$ .

Part b

we need to find the angle for which range would be maximum and then write this maximum range in terms of original range.

So, we know that range is given by:

$R= \frac{u_o^2sin2\theta}{g}$

It would be maximum when $sin2\theta=1\Rightarrow 2\theta=90^o\Rightarrow \theta=45^o$

Hence, $R_{max}=\frac{u_o^2}{g}$

Original range,

$R=\frac{u_o^2sin2\times75.96^o}{g}\\ \Rightarrow R=\frac{u_o^2}{g}\times 0.47\\ \Rightarrow R=R_{max}0.47\\ \Rightarrow R_{max}=2.125R$

Part c:

In the part a, we know that the angle of the projectile is independent of the $g$ i.e. the acceleration due to gravity and this is the only factor that varies with the different planets. Hence, the answer would remain same.

7 0

3 years ago