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2 years ago

Thanks to me, you can see straight through the wall. What am I?

Physics

2 answers:

Bezzdna [24]2 years ago

5 0

Answer:

A window

planation:

klasskru [66]2 years ago

4 0

Answer:

A window

Explanation:

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Two blocks, 1 and 2, are connected by a rope R1 of negligible mass. A second rope R2, also of negligible mass, is tied to block

alekssr [168]

Answer:

Explanation:

Given

Two block are connected by rope $R_1$

$R_2$ rope is attached to block 2

suppose $F_2$ is a force applied to Rope $R_2$

Applied force $F_2$ =Tension in Rope 2

$F_2=(m_1+m_2)a---1$

where a=acceleration of system

Tension in rope $R_1$ is denoted by $F_1$

$F_1=m_1a---2$

divide 1 and 2 we get

$\frac{F_2}{F_1}=\frac{(m_1+m_2)a}{m_1a}$

also $m_1=2.11\cdot m_2$

$\frac{F_2}{F_1}=\frac{2.11m_2+m_2}{2.11m_2}$

$\frac{F_2}{F_1}=\frac{3.11}{2.11}$

$\frac{F_1}{F_2}=\frac{2.11}{3.11}$

3 0

2 years ago

A Body OF Volume 36cc Floats With 3/4 of its volume submerged in water . The density Of Body is

Radda [10]

Answer:

Density of body = 0.25g/cc

Explanation:

Given:

Volume submerged in water = 3/4

Find:

Density Of Body

Computation:

Density of body = fraction of body in liquid x density of water

Density of body = [1-3/4]1

Density of body = 0.25g/cc

8 0

1 year ago

Consider the Bohr energy expression (Equation 30.13) as it applies to singly ionized helium He+ (Z = 2) and an ionized atom with

ella [17]

Answer:

Explanation:

Bohr's energy expression is as follows

E_n = 13.6 z² /n² where z is atomic no and n is principal quantum no of the atom .

z for helium is 2 and for ionised atom is 5 . Let energy of n₁ level of He is equal to energy level n₂ of ionised atom

13.6 x 2² / n₁² = 13.6 x 5² / n₂²

n₁ / n₂ = 2/5 , ie 2nd energy level of He matches 5 th energy level of ionised atom .

For quantum numbers less than or equal to 9 , If we take n₁ = 8 for He

Putting it in the equation above

2² / 8² = 5² / n₂²

n₂ = 5 x 8 / 2

= 20 .

energy

= - 13.6 x2² / 8²

= - 0.85 eV .

3 0

2 years ago

Stu wanted to calculate the resistance of a light bulb connected to a 4.0 V battery, with a resulting current of 0.5 A. He used

Dimas [21]

His answer was incorrect because according to ohm's law the formula used should have been R=V/I instead of multiplying and the answer should be 8ohms

4 0

2 years ago

Read 2 more answers

Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller

kobusy [5.1K]

Answer:

σ₁ = $3.167 * 10^{-6}$ C/m²

σ₂ = $7.6 * 10 ^{-6}$ C/m²

Explanation:

The given data :-

i) The radius of smaller sphere ( r ) = 5 cm.

ii) The radius of larger sphere ( R ) = 12 cm.

iii) The electric field at of larger sphere ( E₁ ) = 358 kV/m. = 358 * 1000 v/m

$E_{1} = (\frac{1}{4\pi\epsilon }) (\frac{Q_{1} }{R^{2} } )$

$358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }$

Q₁ = 572.8 $* 10^{-9}$ C

Since the field inside a conductor is zero, therefore electric potential ( V ) is constant.

V = constant

∴ $\frac{Q_{1} }{R} = \frac{Q_{2} }{r}$

$Q_{2} = \frac{r}{R} *Q_{1}$

$Q_{2} = \frac{5}{12} *572.8*10^{-9}$ = $238.666 *10^{-9}$ C

Surface charge density ( σ₁ ) for large sphere.

Area ( A₁ ) = 4 * π * R² = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁ = $\frac{Q_{1} }{A_{1} }$ = $\frac{572.8 *10^{-9} }{0.180864}$ = $3.167 * 10^{-6}$ C/m².

Surface charge density ( σ₂ ) for smaller sphere.

Area ( A₂ ) = 4 * π * r² = 4 * 3.14 * 0.05² =0.0314 m².

σ₂ = $\frac{Q_{2} }{A_{2} }$ = $\frac{238.66 *10^{-9} }{0.0314}$ = $7.6 * 10 ^{-6}$ C/m²

8 0

2 years ago