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4 years ago

7

if a gas has a gage pressure of 156 kpa its absolute pressure is approximately .....A... 56 kPa ....B....100 kPa.......C......25

6 kPa....D...300 kPa.....

1 answer:

NeX [460]4 years ago

7 0

256 kPa because p-guage + p-absolute + p-atmospheric = 256

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nata0808 [166]

Answer:

?

Explanation:

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3 years ago

if a man hits a golf ball (0.2 kg) which accelerates at a rate of 20 m/s squared. what amount of force acted on the ball?

kkurt [141]

F = ma

We have mass = 0.2kg

and acceleration = 20 m/s^2

So..

F = (0.2)(20)

F = 4 N

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3 years ago

A plastic rod 1.3 m long is rubbed all over with wool, and acquires a charge of -3e-08 coulombs. we choose the center of the rod

Anestetic [448]

(a) The plastic rod has a length of L=1.3m. If we divide by 8, we get the length of each piece:
$L/8=1.3m/8=0.1625 m$

(b) The center of the rod is located at x=0. This means we have 4 pieces of the rod on the negative side of x-axis, and 4 pieces on the positive side. So, starting from x=0 and going towards positive direction, we have: piece 5, piece 6, piece 7 and piece 8. Each piece is 0.1625 m long. Therefore, the center of piece 5 is at 0.1625m/2=0.0812 m. And the center of piece 6 will be shifted by 0.1625m with respect to this:
$c_6 = 0.0812m+0.1625m=0.2437 m$

(c) The total charge is $Q=-3 \cdot 10^{-8}C$ . To get the charge on each piece, we should divide this value by 8, the number of pieces:
$Q/8=-3\cdot 10^{-8}C/8=-3.75\cdot 10^{-9}C$

(d) We have to calculate the electric field at x=0.7 generated by piece 6. The charge on piece 6 is the value calculated at point (c):
$q= -3.75\cdot 10^{-9}C$
If we approximate piece 6 as a single charge, the electric field is given by
$E=k_e \frac{q}{d^2}$
where $k_e=8.99\cdot 10^9Nm^2C^{-2}$ and d is the distance between the charge (center of piece 6, located at 0.2437m) and point a (located at x=0.7m). Therefore we have
$E= 8.99\cdot 10^9 Nm^2C^{-2} \frac{-3.75\cdot 10^9 C}{(0.2437m-0.7m)^2} =-161.9 V/m$
poiting towards the center of piece 6, since the charge is negative.

(e) missing details on this question.

5 0

3 years ago

An event occurs in system K' at x' = 2 m, y' = 3.7 m, z' = 3.7 m, and t' = 0. System K' and K have their axes coincident at t =

fiasKO [112]

Answer:

Coordinates of event in system K are (x,y,z,t)=(5.103m , 3.7m , 3.7m , 1.57×10⁻⁸s)

Explanation:

To find the coordinates of event in system K ,we have to use inverse Lorentz transformation

So

$x=\frac{x^{|}+vt^{|} }{\sqrt{1-\frac{v^{2} }{c{2} } } } \\x=\frac{2m+0.92c(0) }{\sqrt{1-\frac{(0.92c)^{2} }{c{2} } } }\\x=5.103m\\y=y^{|}\\ y=3.7m\\z=z^{|}\\ z=3.7m$

for t

$r=\frac{1}{\sqrt{1-v^{2} } } \\r=\frac{1}{\sqrt{1-(0.92)^{2} } } \\r=2.551\\t=r(t^{|}+vx^{|}/c^{2} )\\t=2.551(0s+(0.92c)(2)/c^{2} )\\t=1.57*10^{-8}s$

Coordinates of event in system K are (x,y,z,t)=(5.103m , 3.7m , 3.7m , 1.57×10⁻⁸s)

6 0

3 years ago

Austin performed an experiment. He put 100mL of vegetable oil (density 0.9 g/ml) in a graduated cylinder. He put a 100g mass of

Grace [21]

Answer:

Explanation:

mass of displaced oil = 11 x .9

= 9.9 gm

9.9 x 10⁻³ kg

weight of displaced oil = 9.9 x 9.81 x 10⁻³ N

= .097 N .

buoyant force by oil = .097 N

weight of unknown metal = .1 x 9.8

= .98 N .

weight of metal in oil = .98 - .097

= .883 N .

=

6 0

3 years ago