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Vikki [24]
2 years ago
5

Which phrases describe NASA's goals in the coming years? Check all that apply.

Physics
2 answers:
Butoxors [25]2 years ago
4 0

Answer:

true, false, true, true, false

Explanation:

I did the test

Serggg [28]2 years ago
3 0

Answer:

1, 3 and 4

Explanation:

Did the test and saw the answers.

You might be interested in
To measure moderately low pressures, oil with a density of 9.0 x 102 kg/m3 is used in place of mercury in a barometer. A change
Yuki888 [10]

Answer:

Δ P =  13.24 Pa

Explanation:

Given that

Density of oil ,ρ₁ = 9 x 10² kg/m³

We know that density for mercury ,ρ₂  = 13.6 x 10³ kg/m³

The change in the height of column ,Δh = 1.5 mm

The pressure given as

P = ρ g h

Change in the pressure

Δ P =  ρ₁ g Δh

Now by putting the values

Δ P =  9 x 10² x 9.81 x 1.5 x 10⁻³    Pa

Δ P =  13.24 Pa

Therefor the change in the pressure will be 13.24 Pa.  

       

3 0
2 years ago
Greg walks on a straight road from his home to a convenience store 3.0 km away with a speed of 6.0 km/h. On reaching the store h
VladimirAG [237]

This question was apprently selected from the "Sneaky Questions" category.

The store is 3 km from his home, and he walks there with a speed of 6 km/hr.  So it takes him (3 km) / (6 km/hr)  =  1/2 hour to get to the store.

That's 30 minutes.  So the whole part-(a.) of the question refers to only that part of the trip, and we don't care what happens once he reaches the store.  

a). Over the first 30 minutes of his travel, Greg walks 3.0 km on a straight road, and he ends up 3.0 km away from where he started.

Average speed = (distance/time) = (3.0 km) / (1/2 hour) = <em>6.0 km/hr</em>

Average velocity = (displacement/time) = (3.0 km) / (1/2 hour) = <em>6.0 km/hr</em>

There's probably some more questions in part-(b.) where you'd need to use Greg's return trip to find the answers, but johnaddy210 is only asking us for part-(a.).

8 0
2 years ago
Read 2 more answers
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
Not too long ago houses were protected from excessive currents by fuses rather than circuit breakers. sometimes a fuse blew out
olganol [36]

Answer: Resistance = 8.21 \times 10^{-8} \Omega

The approximate diameter of a penny is, <em>d</em> = 20 mm

thickness of penny is, <em>L = </em> 1.5×10^{-3} mm

The area of penny along circular face is,A = \frac{\pi d^2}{4} =\frac{\pi (20 mm\frac{1 m}{1000 m})^2}{4}

= 3.14×10^{-4} m²

The resistivity of copper is <em>ρ</em> = 1.72 x 10-8 Ωm.

Resistance,

R = \rho \frac{L}{A}  = (1.72 \times 10^{-8} \Omega m)\frac{1.5 \times 10^{-3} m}{3.14 \times 10^{-4} m^2} = 8.21 \times 10^{-8} \Omega

5 0
2 years ago
Block A, mass 250 g , sits on top of block B, mass 2.0 kg . The coefficients of static and kinetic friction between blocks A and
masha68 [24]

Answer:

  F = 69.3 N

Explanation:

For this exercise we use Newton's second law, remembering that the static friction force increases up to a maximum value given by

               fr = μ N

We define a reference system parallel to the floor

block B  ( lower)

Y axis  

            N - W₁-W₂ = 0

            N = W₂ + W₂

            N = (M + m) g

X axis

              F -fr = M a

for block A (upper)

X axis

              fr = m a                 (2)

so that the blocks do not slide, the acceleration in both must be the same.

Let's solve the system by adding the two equations

             F = (M + m) a          (3)

             a =\frac{F}{ M+m}

the friction force has the formula

            fr = μ N

             fr = μ (M + m) g

let's calculate

            fr = 0.34 (2.0 + 0.250) 9.8

            fr = 7.7 N

we substitute in equation 2

             fr = m a

             a = fr / m

             a = 7.7 / 0.250

             a = 30.8 m / s²

we substitute in equation 3

             F = (2.0 + 0.250) 30.8

             F = 69.3 N

5 0
2 years ago
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