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Oduvanchick [21]
3 years ago
12

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

Physics
1 answer:
Juli2301 [7.4K]3 years ago
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

b)

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
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