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Black_prince [1.1K]

3 years ago

11

If an applied force of 20 N to the East on a 4 kg object results in change in velocity of 5 m/s per second which equation is nee

ded to find the acceleration

1 answer:

jekas [21]3 years ago

7 0

Answer:

Explanation

F=ma

20=4*a

20/4 = a

5 ms⁻² = a

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A certain nuclear power plant is capable of producing 1.2×10^9 W of electric power. During operation of the reactor, mass is con

alukav5142 [94]

Answer:

0.00016 kg

Explanation:

Given:

Power = P = 1.2 × 10⁹ Watts

Power = work done / Time

efficiency = 0.30

Input power = 1.2 × 10⁹ / 0.30 = 4 × 10⁹ W

Energy = 4 × 10⁹ x 60 x 60 = 1.44 x 10¹³ joules

E = m c² , where c is the speed of light and m is the mass.

⇒ mass = m = E / c² = (1.44 x 10¹³) / (3 × 10⁸ )²

= 0.00016 kg

6 0

4 years ago

The wavelength of sound wave is 30 m. Frequency is 7 Hz. What is the wave speed

MaRussiya [10]

Answer:23hz

Explanation:

6 0

3 years ago

Elliot jumps up and down on a pogo stick. He weighs 600.N, and his pogo stick has a spring with spring constant 1100N/m. What is

tia_tia [17]

From conservation of energy, the height he will reach when he has gravitational potential energy 250J is 0.42 meters approximately

The given weight of Elliot is 600 N

From conservation of energy, the total mechanical energy of Elliot must have been converted to elastic potential energy. Then, the elastic potential energy from the spring was later converted to maximum potential energy P.E of Elliot.

P.E = mgh

where mg = Weight = 600

To find the height Elliot will reach, substitute all necessary parameters into the equation above.

250 = 600h

Make h the subject of the formula

h = 250/600

h = 0.4167 meters

Therefore, the height he will reach when he has gravitational potential energy 250J is 0.42 meters approximately

Learn more about energy here: brainly.com/question/24116470

4 0

3 years ago

Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

The shear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

b) If the railing edge is 2 ft, the Reynold number is:

$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

The friction drag is equal to:

$F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf$

7 0

3 years ago

A baby born more than 3 weeks before the due date is considered

natulia [17]

Fetal because you are still in the womb and not fully developed

6 0

3 years ago

Read 2 more answers