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daser333 [38]
3 years ago
9

Which parameter of the present universe, more than any other, is considered to be critical in determining the ultimate fate of t

he universe?
Physics
1 answer:
pochemuha3 years ago
8 0
The slope and curvature of space time which is being derived from the Einstein's law of gravitation which was modified later it gives three slopes value (-,0,+ )
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What is the maximum speed with which a 1200-kg car can round a turn of radius 88.0 m on a flat road if the coefficient of static
boyakko [2]

Answer:

18.6 m/s

Explanation:

The frictional force acting on the car is given by:

F_f = \mu m g=(0.40 )(1200 kg)(9.81 m/s^2)=4709 N

But the frictional force also corresponds to the centripetal force that keeps the car in circular motion in the turn:

F_f = F_c = m \frac{v^2}{r}

Where v is the maximum speed the car can achieve remaining in the turn. Substituting r=88.0 m and re-arranging the formula, we can find the value of v:

v=\sqrt{\frac{Fr}{m}}=\sqrt{\frac{(4709 N)(88.0 m)}{1200 kg}}=18.6 m/s

6 0
3 years ago
A cylindrical rod with length (1.7 m) and radius (2 cm) is fixed from one end and a mass of (20 kg) is attached to the other fre
belka [17]

Answer:

0.44

Explanation:

Poisson's ratio is:

ν = (3K − E) / 6K

where K is the bulk modulus and E is Young's modulus.

Young's modulus is:

E = FL / (AΔL)

where F is the force, L is the initial length, A is the cross sectional area, and ΔL is the change in length.

E = (20 kg × 9.8 m/s²) (1.7 m) / (π (0.02 m)² × 0.0005 m)

E = 0.530×10⁹ Pa

Bulk modulus is:

K = -ΔP / (ΔV/V)

where ΔP is the change in pressure, ΔV is the change in volume, and V is the initial volume.

K = -(180 atm × 101325 Pa/atm) / (-0.012)

K = 1.52×10⁹ Pa

Therefore, the Poisson's ratio is:

ν = (3(1.52×10⁹ Pa) − 0.530×10⁹ Pa) / 6(1.52×10⁹ Pa)

ν = (3(1.52) − 0.530) / 6(1.52)

ν = 0.442

Rounded to 2 significant figures, the Poisson's ratio is 0.44.

3 0
3 years ago
A bus is moving at a constant speed of 40 m/s. How many hours will it takes to travel 260 miles?
lawyer [7]

Explanation:

1 mile = 1609m

Distance=260 miles= 418340 m

Speed= 40 m/s

Time = distance/speed

= 418340/40

=10458.5 seconds

= 2.9 hours

8 0
3 years ago
what is the rotational kinetic energy of the earth? use the moment of inertia you calculated in part a rather than the actual mo
Ivenika [448]

The Earth's rotational kinetic energy is the kinetic Energy that the Earth

has due to rotation.

The rotational kinetic energy of the Earth is approximately <u>3.331 × 10³⁶ J</u>

Reasons:

<em>The parameters required for the question are;  </em>

<em>Mass of the Earth, M = </em><em>5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = </em><em>6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = </em><em>24.0 hrs</em><em>.</em>

The \ moment  \ of \  inertia \  of \  uniform \  sphere \  is \ I =   \mathbf{\dfrac{2}{5} \cdot M \cdot R^2}

Which gives;

\mathbf{I_{Earth}} =   \dfrac{2}{5} \times 5.97 \times 10 ^{24} \cdot \left(6.38 \times 10^6 \right)^2 = 9.7202107 \times 10^{37}

\mathrm{The \ rotational \  kinetic  \ energy \  is} \   E_{rotational} = \mathbf{\dfrac{1}{2} \cdot I \cdot \omega^2}

\mathrm{The \ angular \ speed, \ \omega} = \mathbf{\dfrac{2 \dcdot \pi}{T}}

Therefore;

\omega = \dfrac{2 \cdot \pi}{24}  = \dfrac{\pi}{24}

Which gives;

\mathbf{E_{rotational}} = \dfrac{1}{2} \times  9.7202107 \times 10^{37} \times  \left(  \dfrac{\pi}{12} \right)^2 = 3.331 \times 10^{36}

The rotational kinetic energy of the Earth, E_{rotational} = <u>3.331 × 10³⁶ Joules</u>

Learn more here:

brainly.com/question/13623190

<em>The moment of inertia from part A  of the question (obtained online) is that of the Earth approximated to a perfect sphere</em>.

<em>Mass of the Earth, M = 5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = 6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = 24.0 hrs</em>

3 0
3 years ago
50 POINTS! Chipping from the rough, a golfer sends the ball at an angle of 54° with an initial velocity of 13m/s, how far down t
docker41 [41]

Range equation:

x=\dfrac{{v_i}^2\sin2\theta}g

This ball travels a distance of

x=\dfrac{\left(13\frac{\rm m}{\rm s}\right)^2\sin108^\circ}{9.8\frac{\rm m}{\mathrm s^2}}=16\,\mathrm m

5 0
3 years ago
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