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3 years ago

A car travels 20 km southwards and then travel another 20km westward what is the displacement from the rest position

1 answer:

7 0

$\text{Hello there!}\\\\\text{We're trying to find displacement}\\\\\text{We would use the Pythagorean theorem to find the displacement:}\\a^2+b^2=c^2\\\\\text{We know that the car goes 20 km south and 20 km west}\\\\\text{When you draw it out, it should look like a triangle, you would be finding}\\\text{the length of the hypotenuse}\\\\\text{Plug in 20 to} \,\,a^2\,\, \text{and}\,\, b^2}\\\\20^2+20^2=c^2\\\\\text{Solve:}\\\\20^2+20^2=c^2\\\\400+400=c^2\\\\800=c^2\\\\\sqrt800=\sqrt c\\$

$\large\boxed{28.28}\\\\\normalsize\text{The displacment would be 28.28}$

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In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin

alexandr402 [8]

Answer:

a) $v=0.999124c$

b) $E=7.566*10^{22}$

c) $E_a=760 times\ larger$

Explanation:

From the question we are told that

Distance to Betelgeuse $d_b=430ly$

Mass of Rocket $M_r=20000$

Total Time in years traveled $T_d=36years$

Total energy used by the United States in the year 2000 $E_{2000}=1.0*10^20$

Generally the equation of speed of rocket v mathematically given by

$v=\frac{2d}{\triangle t}$

$v=860ly/ \triangle t$

where

$\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\sqrt{(860)^2+(36)^2}$

$\triangle t=860.7532$

Therefore

$v=\frac{860ly}{ 860.7532}$

$v=0.999124c$

Generally the equation of the energy E required to attain prior speed mathematically given by

$E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2$

$E=7.566*10^{22}$

c)Generally the equation of the energy $E_a$ required to accelerate the rocket mathematically given by

$E_a=\frac{E}{E_{2000}}$

$E_a=\frac{7.566*10^{22}}{1.0*10^{20}}$

$E_a=760 times\ larger$

8 0

3 years ago

Suppose a free-fall ride at an amusement park starts at rest and is in free fall. What is the velocity of the ride after 2.3 s?

horsena [70]

Answer:

V = a * t = 9.8 m/s^2 * 2.3 s = 22.5 m/s velocity after 2.3 s

S = 1/2 g t^2 since initial speed is zero

S = 1/2 * 9.8 m/s^2 * 5.29 s^2 = 25.9 m

8 0

3 years ago

Which statement describes the kinetic energy of a particle?

worty [1.4K]

Kinetic energy has nothing to do with anything other than motion of the particle.
When a particle with velocity v collides another particle(suppose it is at rest for simplication), assuming that there is perfectly elastic collision between them, the velocity of particle which was at rest becomes mv/M ( assuming mass of particle in motion to be m and at rest to be M) from convervation of linear momentum. And all this transfer of energy happens in a fraction of seconds which is not visible to naked eyes.

Hence 1st option is correct!

5 0

3 years ago

Work out the kinetic energy of a 2.5 kg remote-controlled car that is moving at 2 m/s.

lbvjy [14]

Answer: 5 joules

Explanation:

mass=m=2.5kg

Velocity=v=2m/s

Kinetic energy=ke

ke=(m x v x v)/2

ke=(2.5 x 2 x 2)/2

Ke=10/2

Ke=5

Kinetic energy=5 joules

8 0

3 years ago

A 5.50-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical

iris [78.8K]

Answer:

17.71N/m

Explanation:

The period of the spring is expressed according to the expression;

$T = 2 \pi \sqrt{\frac{m}{k} } \\$

m is the mass of the object

k is the force constant

Given

m = 5.50kg

T = 3.50s

Substitute into the formula;

$T = 2 \pi \sqrt{\frac{m}{k} } \\3.5 = 2 (3.14) \sqrt{\frac{5.5}{k} } \\3.5 = 6.28 \sqrt{\frac{5.5}{k} } \\\frac{3.5}{6.28} = \sqrt{\frac{5.5}{k} } \\0.557 = \sqrt{\frac{5.5}{k} } \\square \ both \ sides\\0.557^2 = (\sqrt{\frac{5.5}{k} })^2 \\0.3106 = \frac{5,5}{k}\\k = \frac{5.5}{0.3106}\\k = 17.71N/m$

Hence the force constant of the spring is 17.71N/m

4 0

3 years ago