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3 years ago

How much kinetic energy the truck still has

Physics

1 answer:

seropon [69]3 years ago

8 0

Answer:

47.5kJ

Explanation:

Before climbing the cliff

$E_t = E_k$

$E_k=\frac{1}{2} mv^2\\E_k=\frac{1}{2} 500*15^2\\\\E_k = 56250J\\$

At 2.

$E_t = E_k+E_p$

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Which transformation will map figure L onto figure L'? (1 point) Two congruent triangles Figure L and Figure L prime are drawn o

Ghella [55]

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Horizontal translation of six units

Explanation:

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2 years ago

The pilot of an airplane reads the altitude 6400 m and the absolute pressure 46 kPa when flying over a city. Calculate the local

olga nikolaevna [1]

Answer:

1. 6.672 kPa

2. 49.05 mm of mercury

Explanation:

h = 6400 m

Absolute pressure, p = 46 kPa = 46000 Pa

density of air, d = 0.823 kg/m^3

density of mercury, D = 13600 kg/m^3

(a) Absolute pressure = Atmospheric pressure + pressure due to height

46000 = Atmospheric pressure + h x d x g

Atmospheric pressure = 46000 - 6400 x 0.823 x 10 = 6672 Pa = 6.672 kPa

(b) To convert the pressure into mercury pressure

Atmospheric pressure = H x D x g

Where, H is the height of mercury, D be the density of mercury, g be the acceleration due to gravity

6672 = H x 13600 x 10

H = 0.04905 m

H = 49.05 mm of mercury

4 0

3 years ago

Physicists and engineers from around the world have come together to build the largest accelerator in the world, the Large Hadro

elena-14-01-66 [18.8K]

Solution :

Energy of photon, E = 6.7 eV

E = $6.7 \times 1.602 \times 10^{-7}$ joule

Kinetic energy, $K.E. =\frac{1}{2} mv^2 = 1.602 \times 6.7 \times 10^{-7}$

$v^2=\frac{2 \times 1.602 \times 6.7 \times 10^{-7}}{1.6726 \times 10^{-27}}$

$=12.834 \times 10^{-20}$

Kinetic energy at high speeds

$(r-1)\times mc^2 = 6.7 \ eV$

$(r-1)=\frac{6.7 \times 1.602 \times 10^{-7}}{1.6726 \times 10^{-27} \times 9 \times 10^{16}}$

r - 1 = 7130

r = 7130 + 1

r = 7131

$\frac{1}{\sqrt{1-\frac{v^2}{C^2}}}=7131$

$1-\frac{v^2}{C^2} = \left(\frac{1}{7131}\right)^2$

$v^2=C^2\left[1-\left(\frac{1}{7131}\right)^2\right]$

$v=0.99999999017C$

Δ = 1 - 0.99999999017

= 0.00000000933

Relative mass, $m_{rel}=r.m$

$=7131 \times 1.6728 \times 10^{-27}$

$=1.1927 \times 10^{-23}$ kg

6 0

2 years ago

An experimental rocket designed to land upright falls freely from a height of 2.59 102 m, starting at rest. At a height of 86.9

aleksandr82 [10.1K]

Answer:

The acceleration required by the rocket in order to have a zero speed on touchdown is 19.96m/s²

The rocket's motion for analysis sake is divided into two phases.

Phase 1: the free fall motion of the rocket from the height 2.59*102m to a height 86.9m

Phase 2: the motion of the rocket due to the acceleration of the rocket also from the height 86.9m to the point of touchdown y = 0m.

Explanation:

The initial velocity of the rocket is 0m/s when it started falling from rest under free fall. g = 9.8m/s² t1 is the time taken for phase 1 and t2 is the time taken for phase2.

The final velocity under free fall becomes the initial velocity for the accelerated motion of the rocket in phase 2 and the final velocity or speed in phase 2 is equal to zero.

The detailed step by step solution to the problems can be found in the attachment below.

Thank you and I hope this solution is helpful to you. Good luck.

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2 years ago

Please help ASAP WILL GIVE BRAINLIEST

Westkost [7]

Answer:

Explanation: It makes the most sense. Plz mark brainliest

5 0

2 years ago

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