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iragen [17]
2 years ago
9

You drop your keys in a high-speed elevator going up at a constant speed. Part APart complete Do the keys accelerate faster towa

rd the elevator floor than they would if the elevator were not moving? Do the keys accelerate faster toward the elevator floor than they would if the elevator were not moving? Yes No Previous Answers Correct Part B Do the keys accelerate faster toward the elevator floor than they would if the elevator were accelerating downward? Do the keys accelerate faster toward the elevator floor than they would if the elevator were accelerating downward? No Yes
Physics
1 answer:
anzhelika [568]2 years ago
5 0

Answer:

Explained

Explanation:

a) No, the keys were initially moving upward in the elevator only effects the initial velocity of the key and not the rate of change of velocity that is acceleration. So, the keys accelerate with the same acceleration as before.

b)Yes, keys will accelerate towards the floor faster if it is a constant speed than it is moving downward because if the elevator is accelerating downward, the downward change in velocity of the keys is at least partially matched by a downward change in the velocity of the of the elevator.

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g The international space station has an orbital period of 93 minutes at an altitude (above Earth's surface) of 410 km. A geosyn
krok68 [10]

Answer:

r = 4.21 10⁷ m

Explanation:

Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining

            T² = (\frac{4\pi }{G M_s} ) r³             (1)

           

in this case the period of the season is

            T₁ = 93 min (60 s / 1 min) = 5580 s

            r₁ = 410 + 6370 = 6780 km

            r₁ = 6.780 10⁶ m

for the satellite

           T₂ = 24 h (3600 s / 1h) = 86 400 s

if we substitute in equation 1

            T² = K r³

            K = T₁²/r₁³

            K = \frac{ 5580^2}{ (6.780 10^6)^2}

            K = 9.99 10⁻¹⁴ s² / m³

we can replace the satellite values

            r³ = T² / K

            r³ = 86400² / 9.99 10⁻¹⁴

            r = ∛(7.4724 10²²)

            r = 4.21 10⁷ m

this distance is from the center of the earth

7 0
2 years ago
1. For each of the following scenarios, describe the force providing the centripetal force for the motion: a. a car making a tur
GenaCL600 [577]

Complete Question

For each of the following scenarios, describe the force providing the centripetal force for the motion:

a. a car making a turn

b. a child swinging around a pole

c. a person sitting on a bench facing the center of a carousel

d. a rock swinging on a string

e. the Earth orbiting the Sun.

Answer:

Considering a

    The force providing the centripetal force is the frictional force on the tires \

          i.e  \mu mg  =  \frac{mv^2}{r}

    where \mu is the coefficient of static friction

Considering b

   The force providing the centripetal force is the force experienced by the boys  hand on the pole

Considering c

     The force providing the centripetal force is the normal from the bench due to the boys weight

Considering d

     The force providing the centripetal force is the tension on the string

Considering e

      The force providing the centripetal force is the force of gravity between the earth and the sun

Explanation:

6 0
2 years ago
A pair of glasses is dropped from the top of a 32.0m stadium. A pen is dropped 2.Os later. How high above the ground is the pen
Svetllana [295]

Answer:

h_p = 30.46\ m

Explanation:

<u>Free Fall Motion</u>

A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.

The speed vf of the object when a time t has passed is given by:

v_f=g\cdot t

Where g = 9.8 m/s^2

Similarly, the distance y the object has traveled is calculated as follows:

\displaystyle y=\frac{g\cdot t^2}{2}

If we know the height h from which the object was dropped, we can solve the above equation for t:

\displaystyle t=\sqrt{\frac{2\cdot y}{g}}

The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:

\displaystyle t_1=\sqrt{\frac{2\cdot 32}{9.8}}=2.56\ sec

The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:

t_2=2.56 - 2 = 0.56\ sec

Therefore, it has traveled down a distance:

\displaystyle y=\frac{9.8\cdot 0.56^2}{2} = 1.54\ m

Thus, the height of the pen is:

h_p = 32 - 1.54\Rightarrow h_p=30.46\ m

8 0
2 years ago
Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The
Eddi Din [679]

Answer:

(a) Steel rod: 1.1 * 10^{-4}

    Copper rod: 1.88 * 10^{-4}

(b) Steel rod: 8.3 * 10^{-5} m

Copper rod: 1.41 * 10^{-4} m

Explanation:

Length of each rod = 0.75 m

Diameter of each rod = 1.50 cm = 0.015 m

Tensile force exerted = 4000 N

(a) Strain is given as the ratio of change in length to the original length of a body. Mathematically, it is given as

Strain = \frac{1}{Y} * \frac{F}{A}

where Y = Young modulus

F = Fore applied

A = Cross sectional area

For the steel rod:

Y =  200 000 000 000 N/m^{2}

F = 4000N

A = \pi r^{2}      (r = d/2 = 0.015/2 = 0.0075 m)

=> A = \pi * (0.0075)^{2}

=> A = 0.000177 m^{2}

∴ Strain = \frac{4000}{200000000000 * 0.000177} \\\\Strain = \frac{4000}{35400000}\\ \\Strain = 0.000113 = 1.13 * 10^{-4}

For the copper rod:

Y =  120 000 000 000 N/m²

F = 4000N

A = \pi r^{2}      (r = d/2 = 0.015/2 = 0.0075 m)

=> A = \pi * (0.0075)^{2}

=> A = 0.000177 m^{2}

Strain = \frac{4000}{120 000 000 000 * 0.000177} \\\\Strain = \frac{4000}{21240000}\\ \\Strain =  = 1.88 * 10^{-4}

(b) We can find the elongation by multiplying the Strain by the original length of the rods:

Elongation = Strain * Length

For the steel rod:

Elongation = 1.1 * 10^{-4} * 0.75 = 8.3 * 10^{-5} m

For the copper rod:

Elongation = 1.88 * 10^{-4} * 0.75 = 1.41 * 10^{-4} m

6 0
2 years ago
Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu
MArishka [77]

Complete Question:

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Answer:

m = 0.001 M

For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/

6 0
2 years ago
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