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2 years ago

10

an automobile has a tire pressure of 325 kpa when the temperature is 10°C of the temperature rises to 50°C what is the new press

ure

1 answer:

Finger [1]2 years ago

8 0

Old temperature = 283 K.
New temp = 323 K.

(323/283) x (325 kPa) = 371 kPa.

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A student sects a leaf of length 7.2 cm to draw. Her drawing is 28.8 cm in length. What is the magnification of the drawing?

Alina [70]

Answer:

A) x4

Explanation:

Magnification is equal to image size divided by the actual size, or M = I/A.

The image size is the student's drawing, which is 28.8 cm, and the actual size is 7.2 cm. Divide them, and cancel out the units, and you should get:

28.8 cm/7.2 cm = 4

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2 years ago

My teacher didnt explain this well and im lost

Vedmedyk [2.9K]

Explanation:

(a) The given figure is a convex lens.

(b) In this figure, the object is placed between F and optical center of a lens. Convex lens is a converging lens. It converges the beam of light falling on it after reflection. The image is formed on the same side of the lens as the object.

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2 years ago

Linus builds an electrical circuit with a battery and with wires that carry the current. As the battery weakens, the current als

pantera1 [17]

Answer:

The field gets weaker

Explanation:

I’m taking the test right now, hope this helps!!

7 0

3 years ago

Read 2 more answers

An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see

siniylev [52]

Answer:

3.88 * 10^(-15) J

Explanation:

We know that the Potential energy of the electron at the beginning of its motion is equal to the Kinetic energy at the end of its motion, when it reaches the plates.

First, we get the potential and potential energy:

Electric potential = E * r

E = electric field

r = distance between plates

Potential = 2.2 * 10^6 * 0.011

= 2.42 * 10^4 V

The relationship between electric potential and potential energy is:

P. E. = q*V

q = charge of electron = 1.602 * 10^(-19) C

P. E. = 2.42 * 10^4 * 1.602 * 10^(-19)

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6 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