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soldi70 [24.7K]
3 years ago
12

A person drops a brick from the top of a building. The height of the building is 400 m and the mass of the brick is 2.00 kg. Wh

at will be the speed of the brick right before it touches the ground? Use g = 10.0 m/s2
Physics
1 answer:
DaniilM [7]3 years ago
5 0
-- It takes the brick 8.9 seconds to reach the ground. 

-- At the instant of the "splat", it's falling at 89 m/s.

-- The mass doesn't matter. If not for air resistance, every object
    would fall at the same rate.  The answer is the same for a feather,
    a rubber chicken, a brick, or a school bus.
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I need help ASAP please and thank u very much if u can answer i’m giving a brainliest to btw
NemiM [27]

Answer:

A I Think

Explanation:

8 0
3 years ago
A star rotates with a period of 37 days about an axis through its center. The period is the time interval required for a point o
vlabodo [156]

Answer:

T = 1.1285 10⁻² day

Explanation:

For this exercise the forces in the premiere are internal, so the angular momentum is conserved

         L₀ = I₀ w₀

         L  = I w

         L₀ = L

         I₀ w₀ = I w

Angular velocity and period are related

         w₀ = 2π / T₀

         w = 2π  / T

         

The moment of inertia of a sphere is

       I₀ = 2/5 M R²

       I = 2/5 m r²

If we assume that the mass of the star does not change in the transformation

We substitute

         2/5 M R² 2π/T₀ = 2/5 M r² 2π/ T

          R² /T₀ = r² / T

          T = (r / R)² T₀

          T = (6.1 / 2.0 104) 37

          T = 1.1285 10⁻² day

5 0
3 years ago
Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

6 0
3 years ago
1. explain why nichrome wire used in the heater.
Monica [59]

Answer:

1. The reason why nichrome wire is used in a heater is due to the properties of nichrome which makes it a material of choice for heating application including;

i) The cost of manufacture of nichrome is low

ii) Nichrome has both high mechanical and creep strength, and therefore is suitable for installation and use in a wide setting

iii) Nichrome has a high resistance to electron flow, making it a good electrical heating element

iv) Nichrome has a high resistance to oxidation, such that it can be reused several times as a heater

v) The high ductility of nichrome allows it to be stretched and drawn around different vessels with contents to which heat is to be supplied

2. Copper wire is not used in heaters because according to Joule's law, the heat produced from the flow of electric current, 'I', through a material having a resistance, 'R', for a time, 't', is proportional to I²·R·t, and copper has a very low resistance, 'R', to the flow of electric current, and therefore, will not produce sufficient heat required at a given current level

Explanation:

3 0
3 years ago
What is the smallest particle that relains<br> the properties of an element
VLD [36.1K]

Answer:

atom

Explanation:

Smallest particle of an element that identifies that element is an atom.

8 0
2 years ago
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