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3 years ago

Conservation of Energy:

Physics

1 answer:

e-lub [12.9K]3 years ago

5 0

The new speed of the glider is 42.2 m/s.

The new speed of the glider can be calculated using the principle of the law of conservation of energy.

Note: Total energy of the glider before the dive = Total energy of the glider after the dive.

<h3> Formula:</h3>

mv²/2+mgh = mV²/2 +mgH................ Equation 1

Simplifying the equation above,

v²/2+gh = V²+gH

make V the subject of the equation.

V = √[(v²/2)+gh]-gH]............... Equation 2

<h3>Where:</h3>

V = New speed of the glider
v = initial speed of the glider
h = initial height of the glider
H = New height of the glider
g = acceleration due to gravity.

From the question,

<h3>Given:</h3>

v = 40 m/s
h = 300 meters
H = 200 meters
g = 9.8 m/s²

Substitute these values into equation 2.

V = √[(40²/2)+(9.8×300)-(9.8×200)]
V = √(800+2940-1960)
V = √1780
V = 42.2 m/s.

Hence, The new speed of the glider is 42.2 m/s.

Learn more about speed here: brainly.com/question/4931057

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What is a period in physics?<br>

Citrus2011 [14]

Answer:

The period of a wave is the time for a particle on a medium to make one complete vibrational cycle. Period, being a time, is measured in units of time such as seconds, hours, days or years. The period of orbit for the Earth around the Sun is approximately 365 days; it takes 365 days for the Earth to complete a cycle.

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3 years ago

How many meters are in 32 kilometers

lukranit [14]

Answer:

32,000 m

Explanation:

Conversion

1km ===> 1000m

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x = (32 × 1000) ÷ 1

5 0

3 years ago

A glass capillary tube with a diameter of 8.5 mm and length 8 cm is filled with a salt solution with a resistivity of 2.5 ?m. Wh

MA_775_DIABLO [31]

Answer:

The resistance is $3.5\times10^{-4}\ \Omega$

Explanation:

Given that,

Diameter of tube = 8.5 mm

Length = 8 cm

Resistivity = 2.5 m

We need to calculate the resistance

The resistance is equal to the product of the resistivity and length divided by the area of cross section .

In mathematical form,

$R = \dfrac{\rho\times l}{A}$

Where, $\rho$ =resistivity

l = length

A = area of cross section

Put the value into the formula

$R = \dfrac{2.5\times8\times10^{-2}}{3.14\times(\dfrac{8.5}{2}\times10^{-3})^2}$

$R=3526.32\ \Omega$

$R=3.5\times10^{-4}\ \Omega$

Hence, The resistance is $3.5\times10^{-4}\ \Omega$

6 0

3 years ago

A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre

Brut [27]

Answer:

The time taken is $t = 0.356 \ s$

Explanation:

From the question we are told that

The length of steel the wire is $l_1 = 31.0 \ m$

The length of the copper wire is $l_2 = 17.0 \ m$

The diameter of the wire is $d = 1.00 \ m = 1.0 *10^{-3} \ m$

The tension is $T = 122 \ N$

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

$t = t_s + t_c$

Where $t_s$ is the time taken to transverse the steel wire which is mathematically represented as

$t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]$

here $\rho_s$ is the density of steel with a value $\rho_s = 8920 \ kg/m^3$

$t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_s = 0.235 \ s$

And

$t_c$ is the time taken to transverse the copper wire which is mathematically represented as

$t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]$

here $\rho_c$ is the density of steel with a value $\rho_s = 7860 \ kg/m^3$

$t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_c =0.121$

$t = t_c + t_s$

$t = 0.121 + 0.235$

$t = 0.356 \ s$

4 0

3 years ago

How much work would have to be done to bring a 1150kg automobile traveling at 86km/h to a stop?

krok68 [10]

Explanation:

We have,

Mass of an automobile is 1150 kg

The automobile traveling at 86 km/h and then it comes to stop.

86 km/h = 23.88 m/s

It is required to find work done by the automobile.

Concept used : Work energy theorem

Th change in kinetic energy of an object is equal to the work done by it. The work done is then given by :

$W=\dfrac{1}{2}m(v^2-u^2)$

Here, v = 0

$W=-\dfrac{1}{2}mu^2\\\\W=-\dfrac{1}{2}\times 1150\times (23.88)^2\\\\W=-327896.28\ J$

$W=3.27\times 10^5\ J$

Therefore, the work done by the automobile is $-3.27\times 10^5\ J$ .

7 0

3 years ago