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Jobisdone [24]
3 years ago
13

 If the gauge pressure of a gas is 114 kPa, what is the absolute pressure?

Physics
2 answers:
Anastasy [175]3 years ago
4 0

Answer:

D. 214 kPa

Explanation:

The absolute pressure is given by:

p = p_a + p_g

where

p is the absolute pressure

p_a \sim 100 kPa is the atmospheric pressure

p_g is the gauge pressure

In this problem, we have

p_g = 114 kPa

So, the atmospheric pressure is

p = 100 kPa + 114 kPa = 214 kPa

dalvyx [7]3 years ago
3 0

Answer:

your answer is 214kPa

Explanation:

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a motorcycle starts from rest covers 200 meter distance in 6 second calculate final velocity and acceleration​
Ivanshal [37]

Explanation:

s = ut + 1/2 a t^2

200 = 0 * 6 + 1/2 * a * (6)^2

200 = 1/2 * a * 36

200 = 18 a

a = 200/18

a= 11.1m/sec^2

v = u + at

v = 0 + 11.1 * 6

v = 66.6m/s

hope it helps you

3 0
3 years ago
Jamie is a hairstylist who works in a salon. He noticed that when many of the stylist or blow drying hair the power goes out wha
Mashcka [7]
The people are using a lot of electricity blow drying to many peoples hair so i would make a schedule so it dosent get to busy with costumers
6 0
3 years ago
Read 2 more answers
What is the energy in joules of a mole of photons associated with visible light of wavelength 486 nm?
ivann1987 [24]

Answer:

2.46\cdot 10^5 J

Explanation:

The energy of a single photon is given by:

E=\frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda is the wavelength

For the photon in this problem,

\lambda=486 nm=4.86\cdot 10^{-7}m

So, its energy is

E_1=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.86\cdot 10^{-7}m}=4.09\cdot 10^{-19} J

One mole of photons contains a number of photons equal to Avogadro number:

N_A = 6.022\cdot 10^{23}

So, the total energy of one mole of photons is

E=N_A E_1 = (6.022\cdot 10^{23})(4.09\cdot 10^{-19} J)=2.46\cdot 10^5 J

7 0
4 years ago
calculate the spring constant if a weight of 250N is added to a spring which increases in length by 20cm
ZanzabumX [31]
Since, F = k . ∆x

Therefore, k = F / ∆x = 250 / 0.2 = 1250 N/m

(ps: convert 20 cm into 0.2 m)
8 0
3 years ago
Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initia
sergejj [24]

Answer:

0.176m from the flagpole, westward.

Explanation:

Let the Eastward be the positive direction. So initially runner A is at position -6km, running with velocity of 9km/h while runner B is at position 5km running at a velocity of -8km/h. We can conduct the following equation for their distances over the same time t

s_A = -6 + 9t

s_B = 5 - 8t

When A an B meets, they are at the same position and at the same time. So

s_A = s_B

-6 +9t = 5 - 8t

17t = 5 + 6 = 11

t = 11/17 = 0.647 s

s_A = -6 + 9*0.647 = -0.176 m

So where they meet is 0.176m from the flagpole, westward.

5 0
3 years ago
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