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Anna11 [10]
2 years ago
10

A 1.0 kg block is attached to an unstretched

Physics
1 answer:
KonstantinChe [14]2 years ago
4 0

Answer:

Change in  potential energy of the block-spring-Earth

system between Figure 1 and Figure 2 = 1 Nm.

Explanation:

Here, spring constant, k  = 50 N/m.

given block comes down eventually 0.2 m below.

here, g = 10 m/s.

let block be at a height h above the ground in figure 1.

⇒In figure 2, potential energy of the block-spring-Earth

system = m×g×(h - 0.2) + 1/2× k × x². where, x = change in spring length.

⇒ Change in  potential energy of the block-spring-Earth

system between Figure 1 and Figure 2 = (m×g×(h - 0.2)) - (1/2× k × x²)

              =  (1×10×0.2) - (1/2×50×0.2×0.2) = 1 Nm.

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1) A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3/s. Determine the minimum power that must be sup
azamat

Answer:

\dot{W} = 339.84 W

Explanation:

given data:

flow Q = 9 m^{3}/s

velocity = 8 m/s

density of air = 1.18 kg/m^{3}

minimum power required to supplied to the fan is equal to the POWER POTENTIAL of the kinetic energy and it is given as

\dot{W} =\dot{m}\frac{V^{2}}{2}

here \dot{m}is mass flow rate and given as

\dot{m} = \rho*Q

\dot{W} =\rho*Q\frac{V^{2}}{2}

Putting all value to get minimum power

\dot{W} =1.18*9*\frac{8^{2}}{2}

\dot{W} = 339.84 W

7 0
2 years ago
In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal
SpyIntel [72]

Answer:

A) F_g = 26284.48 N

B) v = 7404.18 m/s

C) E = 19.19 × 10^(10) J

Explanation:

We are given;

Mass of satellite; m = 3500 kg

Mass of the earth; M = 6 x 10²⁴ Kg

Earth circular orbit radius; R = 7.3 x 10⁶ m

A) Formula for the gravitational force is;

F_g = GmM/r²

Where G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Plugging in the relevant values, we have;

F_g = (6.67 × 10^(-11) × 3500 × 6 x 10²⁴)/(7.3 x 10⁶)²

F_g = 26284.48 N

B) From the momentum principle, we have that the gravitational force is equal to the centripetal force.

Thus;

GmM/r² = mv²/r

Making v th subject, we have;

v = √(GM/r)

Plugging in the relevant values;

v = √(6.67 × 10^(-11) × 6 x 10²⁴)/(7.3 x 10⁶))

v = 7404.18 m/s

C) From the energy principle, the minimum amount of work is given by;

E = (GmM/r) - ½mv²

Plugging in the relevant values;

E = [(6.67 × 10^(-11) × 3500 × 6 × 10²⁴)/(7.3 x 10⁶)] - (½ × 3500 × 7404.18)

E = 19.19 × 10^(10) J

5 0
2 years ago
A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f
sasho [114]

We use the kinematic equations,

v=u+at                                          (A)

S= ut + \frac{1}{2} at^2                  (B)

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given,  u=0, a=0.21 \ m/s^2 and s= 280 m.

Substituting these values in equation (B), we get

280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s.

Therefore from equation (A),

v = 0 + (0.21) \times (51.63 s)= 10.84 \ m/s

Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s



8 0
2 years ago
La altura de un tornillo de banco respecto a la superficie es de 80 cm expresar dicha medida en pies..
Andrej [43]

Answer:

this measurement if feet is: 2.624672 ft

Explanation:

Notice that 80 cm can be expressed as 0.8 meters, and In order to convert from meters to feet, one needs to multiply the meter measurement times 3.28084. Therefore:

0.80 m can be written in feet as: 0.80 * 3.28084 feet = 2.624672 feet

3 0
2 years ago
"As the Voyager spacecraft penetrated into the outer solar system, the illumination from the Sun declined. Relative to the situa
____ [38]

Answer:

\frac{I_{2}}{I_{1}} = 0.04

Explanation:

The intensity of a star noticed at a certain distance is inversely proportional to the square of distance. Then:

I_{1}\cdot r_{1}^{2} = I_{2}\cdot r_{2}^{2}

The intensity of the Sun in Jupiter relative to Earth is:

\frac{I_{2}}{I_{1}} = \frac{r_{1}^{2}}{r_{2}^{2}}

\frac{I_{2}}{I_{1}} = \left(\frac{1\,AU}{5.2\,AU} \right) ^{2}

\frac{I_{2}}{I_{1}} = 0.04

3 0
2 years ago
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