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Sergio [31]
3 years ago
7

State 1 pascal pressure​

Physics
1 answer:
cricket20 [7]3 years ago
6 0

Answer:

1 pascal pressure is define as the one newton force is given which covers 1 square metre area

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a car moving on a road passes by kilometer 218 at 10:15 am and milestone 236 at 10:30 am. determine an average scalar speed in k
Leno4ka [110]

Answer:

v?

Explanation:

v = ∆s / ∆t

v =?

∆s = 236 - 218 = 18km

=t = 10: 30 - 10:15 = 15 minutes to hour 15/60 = 0.25h

v = 18 / 0.25

v = 72km / h

speed of 72km / h

5 0
3 years ago
Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect
Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a)  0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b)  r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 \pi r^{2}  = \frac{Q1}{E_{0} }

So,

Rearranging the above equation to get Electric field, we will get:

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   }

Multiply and divide by r1^{2}

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } x \frac{r1^{2} }{r1^{2} }

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x r1^{2}) /(E_{0} x r^{2})

c) r > r2 :

Electric Field = ?

E x 4 \pi r^{2}  = \frac{Q1 + Q2}{E_{0} }

Rearranging the above equation for E:

E = \frac{Q1+Q2}{E_{0} . 4 \pi. r^{2}   }

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

As we know from above, that:

\frac{Q1}{E_{0} . 4 \pi. r^{2}   } =  (σ1 x r1^{2}) /(E_{0} x r^{2})

Then, Similarly,

\frac{Q2}{E_{0} . 4 \pi. r^{2}   } = (σ2 x r2^{2}) /(E_{0} x r^{2})

So,

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

Replacing the above equations to get E:

E = (σ1 x r1^{2}) /(E_{0} x r^{2}) + (σ2 x r2^{2}) /(E_{0} x r^{2})

Now, for

d) Under what conditions,  E = 0, for r > r2?

For r > r2, E =0 if

σ1 x r1^{2} = - σ2 x r2^{2}

4 0
3 years ago
A 5 meter long ladder leans against a wall. The bottom of the ladder slides away from the wall at the constant rate of 1 3 m/s.
Oksana_A [137]

Answer:9.75 m/s

Explanation:

Given

Length of ladder (L)=5 m

Foot the ladder is moving away with speed of \frac{\mathrm{d} x}{\mathrm{d} t}=13 m/s

From diagram

x^2+y^2=L^2------1

at x=3

y^2=25-9=16

y=4 m

Now differentiating equation 1 w.r.t time

2x\frac{\mathrm{d} x}{\mathrm{d} t}+2y\frac{\mathrm{d} y}{\mathrm{d} t}=0

x\frac{\mathrm{d} x}{\mathrm{d} t}=-y\frac{\mathrm{d} y}{\mathrm{d} t}

3\times 13=-4\times \frac{\mathrm{d} y}{\mathrm{d} t}

\frac{\mathrm{d} y}{\mathrm{d} t}=-\frac{3\times 13}{4}=-9.75 m/s

negative indicates distance is decreasing with time

5 0
3 years ago
An 81.5-kg man stands on a horizontal surface.
OLga [1]

Answer:

a)V= 0.0827 m³

b)P=181.11 x 10²  N/m²

Explanation:

Given that

m = 81.5 kg

Density ,ρ = 985 kg/m³

As we know that

Mass = Volume x Density

81.5 = V x 985

V= 0.0827 m³

The force exerted by weight = m g

 F= m g= 81.5 x 10 = 815 N      ( Take ,g= 10 m/s²)

Area ,A= 4.5 x 10⁻² m²

The Pressure P

P=\dfrac{F}{A}

P = \dfrac{815}{4.5\times 10^{-2}}\ N/m^2

P=181.11 x 10²  N/m²

7 0
3 years ago
so i had to work for halloween, my first Halloween working. When i got home my lil bro showed me all the candy he collected so i
Mrrafil [7]

Answer:

now that's a sweet little bro

4 0
2 years ago
Read 2 more answers
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