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AnnyKZ
20 days ago
7

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

ry itself just below the surface.
Physics
1 answer:
ValentinkaMS [3.4K]20 days ago
6 0

Complete Question:

Imagine an asteroid on a direct collision course with the Earth at the equator, embedding itself just beneath the surface. What mass must this asteroid possess, in terms of Earth's mass M, to extend the length of the day by 25.0% due to the collision? Assume the asteroid's mass is negligible relative to the Earth and that the Earth is homogenous throughout.

Answer:

m = 0.001 M

For detailed calculations, refer to 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/[[TAG_14]]

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The magnitude of the net force versus time graph has a rectangular shape. Often in physics geometric properties of graphs have p
Maru [3345]
True. Explanation: In this instance, the area of the graph represents the impulse. Impulse is defined as the change in an object's momentum. Moreover, it is also expressed as the product of the force acting on an object and the duration of the impact. When we graph the force against time, if the force remains constant, the resultant graph will take on a rectangular shape, and the area under that graph will equal the impulse's definition.
8 0
28 days ago
Calculate the minimum average power output necessary for a person to run up a 12.0 m long hillside, which is inclined at 25.0° a
Ostrovityanka [3204]

Answer:

Power output, P = 924.15 watts

Explanation:

We have the following parameters:

Length of the ramp, l = 12 m

Weight of the individual, m = 55.8 kg

Incline angle with respect to the horizontal, \theta=25^{\circ}

Elapsed time, t = 3 s

Let h represent the vertical height of the hill:

h=l\ sin\theta

h=12\times \ sin(25)

h = 5.07 m

Power P required for a person to ascend the hill can be expressed as:

P=\dfrac{E}{t}

P=\dfrac{mgh}{t}

P=\dfrac{55.8\times 9.8\times 5.07}{3}

P = 924.15 watts

This indicates that a minimum average power output of 924.15 watts is essential for an individual to ascend this elevation. Thus, this is the answer sought.

3 0
1 month ago
Steel blocks A and B, which have equal masses, are at TA = 300 oC and T8 = 400 oC. Block C, with mc - 2mA, is at TC = 350 oC. Bl
serg [3582]

Answer:

b) TA = TB = TC

Explanation:

  • When the blocks are brought into contact and isolated from the environment, they will exchange heat until they achieve thermal equilibrium.
  • During this exchange, the hotter body will lose heat, which will be gained by the cooler body.
  • The equilibrium state will be established once this equation is satisfied:

       \Delta Q = c_{st}* m_{A} * (T_{fin} - T_{0A} ) = c_{st}* m_{B} * (T_{0B} - T_{fin} )

  • Substituting the initial temperatures T₀A = 300º C and T₀B = 400ºC, while simplifying for equal block masses mA = mB, enables us to solve for the final temperature, Tfin:

       (400 \ºC - T_{fin}) = (T_{fin} - 300 \ºC) \\ \\ 2* T_{fin} = 700\ºC\\ \\ T_{fin} = 350\ºC

  • At equilibrium, when both blocks combine, they will yield a uniform final temperature of 350ºC.
  • When block C, also at this temperature, makes contact, all three blocks will simultaneously reflect this final temperature of 350 ºC.
  • Therefore, option b) is correct.
8 0
1 month ago
A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'
Ostrovityanka [3204]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half the diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Cross-sectional area of the wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3

= 7.85 x 10^-7 m^2

Let the change in length of the wire be ΔL.

The equation for Young's modulus is given by

Y =\frac{mgL}{A\Delta L}

\Delta L =\frac{mgL}{A\times Y}

ΔL = 0.035 m = 3.5 cm

Thus, the wire stretches by 3.5 cm.

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2 months ago
Planet X has a magnetic field with strength 0.80 G in the southern direction. When a probe is placed 30 mm east of a vertical wi
kicyunya [3294]
0.20 G directed southward
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1 month ago
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