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2 years ago

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A robot probe drops a camera off the rim of a 278 m high cliff on Mars, where the free-fall acceleration is 3.7 m/s2 . Find the

velocity with which it hits the ground. Answer in units of m/s.

1 answer:

FromTheMoon [43]2 years ago

4 0

S = u + at u = 0 278 = 3.7t t = 278/3.7 = 75.135.. v = ut + 0.5at^2 u = 0 v = 0.5 * 3.7 * 75.135^2 = 10,443 m/sec

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A 2000 kg truck traveling north at 34 km/h turns east and accelerates to 58 km/h. (a) What is the change in the truck's kinetic

barxatty [35]

Explanation:

It is given that,

Mass of the truck, m = 2000 kg

Initial velocity of the truck, u = 34 km/h = 9.44 m/s

Final velocity of the truck, v = 58 km/h = 16.11 m/s

(a) Change in truck's kinetic energy, $\Delta E=\dfrac{1}{2}m(v^2-u^2)$

$\Delta E=\dfrac{1}{2}\times 2000\ kg\times (16.11^2-9.44^2)$

$\Delta E=170418.5\ J$

$\Delta E=1.7\times 10^5\ J$

(b) Change in momentum of the truck, $\Delta p=m(v-u)$

$\Delta p=2000\ kg\times (16.11-9.44)$

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3 years ago

Two 10-N weights are hanging from two Springs A and B. Spring A has a greater spring constant and therefore will ____. a. have a

arsen [322]

The spring should have a greater displacement .
Greater spring constant = the more difficult it is to stretch a spring , due to stiffness.

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1 year ago

NaCl is common table salt and is an ionic compound. Table salt can be stirred into water to create a salt-water solution. What h

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a force of 5 N extends a spring of natural length 0.5m by 0.01, what will be the length of the spring when the applied force is

vova2212 [387]

Answer:

L_new =L+x^2 = L_new = 0.54_m.

Explanation:

Given data:

Force in the first case,

F_1 = 5N

Force in the second case,

F_2 = 20 N

Natural length of spring,

L= 0.5

Extension in the first case,

x_1 = 0.01m

Let the force constant of the spring be k.

Thus,

F_1=kx_1

5 = k × 0.01

⇒ k = 500 N/m.

The extension in the spring in the second case can be given as,

F_2=kx_2

20 = 500x_2

⇒ x_2 = 0.04 m.

Thus, the effective length of the spring would be,

L_new =L+x^2

L_new = 0.5+0.04

L_new = 0.54_m.

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2 years ago

A 2010 kg space station orbits Earth at an altitude of 5.35×105 m. Find the magnitude of the force with which the space station

Shkiper50 [21]

Answer:

Force, F = 16814.95 N

Explanation:

It is given that,

Mass of space station, m = 2010 kg

Altitude, $d=5.35\times 10^5\ m$

Mass of earth, $m=5.98\times 10^{24} kg$

Mean radius of earth, $r=6.37\times 10^6\ m$

Magnitude of force is given by :

$F=G\dfrac{Mm}{R^2}$

R = r + d

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$F=6.67\times 10^{-11}\times \dfrac{2010\ kg\times 5.98\times 10^{24} kg}{(6905000\ m)^2}$

F = 16814.95 N

So, the force between the space station and the Earth is 16814.95 N. Hence, this is the required solution.

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3 years ago