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victus00 [196]
4 years ago
10

2. Unless an object at rest is acted upon by a force, it stays at rest due to it's __________.

Physics
1 answer:
Leokris [45]4 years ago
5 0

Inertia. (Newton's Laws of Motion)

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A projectile is launched with an initial speed of 40 mys from the foor of a tunnel whose height is 30 m. What angle of elevation
KiRa [710]

Answer:

We are given the trajectory of a projectile:

y=H+xtan(θ)−g2u2x2(1+tan2(θ)),

where H is the initial height, g is the (positive) gravitational constant and u is the initial speed. Since we are looking for the maximum range we set y=0 (i.e. the projectile is on the ground). If we let L=u2/g, then

H+xtan(θ)−12Lx2(1+tan2(θ))=0

Differentiate both sides with respect to θ.

dxdθtan(θ)+xsec2(θ)−[1Lxdxdθ(1+tan2(θ))+12Lx2(2tan(θ)sec2(θ))]=0

Solving for dxdθ yields

dxdθ=xsec2(θ)[xLtan(θ)−1]tan(θ)−xL(1+tan2(θ))

This derivative is 0 when tan(θ)=Lx and hence this corresponds to a critical number θ for the range of the projectile. We should now show that the x value it corresponds to is a maximum, but I'll just assume that's the case. It pretty obvious in the setting of the problem. Finally, we replace tan(θ) with Lx in the second equation from the top and solve for x.

H+L−12Lx2−L2=0.

This leads immediately to x=L2+2LH−−−−−−−−√. The angle θ can now be found easily.

7 0
2 years ago
Arctic sea ice has declined over the past few decades causing water levels to increase. This is an interaction of which two sphe
konstantin123 [22]

Answer:

B

Explanation:

4 0
3 years ago
Air enters an adiabatic compressor at 104 kPa and 292 K and exits at a temperature of 565 K. Determine the power (kW) for the co
ladessa [460]

Answer:

\dot W_{in} = 49.386\,kW

Explanation:

An adiabatic compressor is modelled as follows by using the First Law of Thermodynamics:

\dot W_{in} + \dot m \cdot c_{p}\cdot (T_{1}-T_{2}) = 0

The power consumed by the compressor can be calculated by the following expression:

\dot W_{in} = \dot m \cdot c_{v}\cdot (T_{2}-T_{1})

Let consider that air behaves ideally. The density of air at inlet is:

P\cdot V = n\cdot R_{u}\cdot T

P\cdot V = \frac{m}{M}\cdot R_{u}\cdot T

\rho = \frac{P\cdot M}{R_{u}\cdot T}

\rho = \frac{(104\,kPa)\cdot (28.02\,\frac{kg}{kmol})}{(8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K} )\cdot (292\,K)}

\rho = 1.2\,\frac{kg}{m^{3}}

The mass flow through compressor is:

\dot m = \rho \cdot \dot V

\dot m = (1.2\,\frac{kg}{m^{3}})\cdot (0.15\,\frac{m^{3}}{s} )

\dot m = 0.18\,\frac{kg}{s}

The work input is:

\dot W_{in} = (0.18\,\frac{kg}{s} )\cdot (1.005\,\frac{kJ}{kg\cdot K})\cdot (565\,K-292\,K)

\dot W_{in} = 49.386\,kW

5 0
3 years ago
If a displacement vector has a negative x component and a positive y component,
bezimeni [28]

Answer:

The direction will be NE i.e North east.

Explanation:

From the question, it will be travelling in North east direction because the the negative X components in vector is in horizontal direction which is the Eastern direction when using the four cardinal points and the Y components, the positive is in North direction if you use the four cardinal points. Therefore, the vector with both X components and Y components will be travelling in Northeast direction.

8 0
3 years ago
What is the maximum number of electrons that can occupy each of the following subshells?
aleksandr82 [10.1K]

Answer:

s (l = 0) only two electrons

p (l = 1) sublevel six (6) electrons fit

d (l = 2) sub-level has 10 electrons

f (l = 3) has 7 has 14 electrons

Explanation:

The electrons have an odd spin, so they must comply with the Pauli principle, therefore only two electrons fit in a given sublevel.

Sub level s (l = 0) only two electrons

Sub level p (l = 1) has three orbitals in which two electrons fit, therefore in the sublevel six (6) electrons fit

Sub-level d (l = 2) has 5 orbitals each with two electrons, the sub-level has 10 electors

Sub-level f (l = 3) has 7 orbitals and two electrons each, the sub-level has 14 electrons

3 0
3 years ago
Read 2 more answers
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