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4 years ago

How to do this question?

Physics

2 answers:

marissa [1.9K]4 years ago

7 0

Answer:

i3 =11.014A

i5 = 3.15A

Explanation:

Here according to k'chofs first law

i1 =i2 + i3

i3 = i4 + i5

For determine the i1 you have to consider the resultant resistor of the system

4 , 1 and 3 resistors are in pararel

Then, Resultant is

1/4 + 1/1 + 1/3 = 1/ R

R = 12/19

For get total we have to add another remaining 3 resistor because of serious

Then Resultant is = 12/19 + 3

= 69/19

Then using V = IR

40 =i3* 69/19

i3 = 11.014 A

Other 3 resistors are parrarel because of this voltage of those resistors are same.

Then i inversely propotional to its resistor

Then ,

i5 * 2 = (i3-i5)*4/5

i 5 = 3.15 A

Diano4ka-milaya [45]4 years ago

6 0

Answer:

I₁ = 11.2 A

I₂ = 1.6 A

I₃ = 9.6 A

I₄ = 6.4 A

I₅ = 3.2 A

Explanation:

Find the equivalent resistances.

Resistors 4 and 5 are in parallel:

R₄₅ = 1 / (1/R₄ + 1/R₅)

R₄₅ = 1 / (1/1 + 1/2)

R₄₅ = 2/3

Resistor 2 is in parallel with that:

R₂₄₅ = 1 / (1/R₂ + 1/R₄₅)

R₂₄₅ = 1 / (1/4 + 3/2)

R₂₄₅ = 4/7

Resistor 1 is in series with that:

R = R₁ + R₂₄₅

R = 3 + 4/7

R = 3 4/7

Now use Ohm's law to find I₁.

V = I₁ R

40 V = I₁ (3 4/7)

I₁ = 11.2 A

Find the voltage drop across R₁:

V₁ = I₁ R₁

V₁ = (11.2 A) (3 Ω)

V₁ = 33.6 V

So the voltage after is 40 V − 33.6 V = 6.4 V.

Find I₂:

6.4 V = I₂ (4 Ω)

I₂ = 1.6 A

Find I₃:

6.4 V = I₃ (2/3 Ω)

I₃ = 9.6 A

Find I₄:

6.4 V = I₄ (1 Ω)

I₄ = 6.4 A

Find I₅:

6.4 V = I₅ (2 Ω)

I₅ = 3.2 A

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A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

3 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

4 years ago

A box with a total surface area of 1.88 m^2

Veseljchak [2.6K]

Answer:9.8x10^(-3) w/m°C

Explanation:

Quantity of heat(Q)=14.1w

Distance(d)=2.44cm=2.44/100=0.0244m

Area(a)=1.88m^2

Temperature change(t)=18.6°C

thermal conductivity(k)=?

K=(Qxd) ➗ (a x t)

K=(14.1x0.0244) ➗ (1.88x18.6)

K=0.34404 ➗ 34.968

K=9.8x10^(-3) w/m°C

3 0

3 years ago

A balloon has a volume of 8.5 L at a pressure of 150 kPa. What will be the new volume when the pressure drops to 55.0 kPa?

lukranit [14]

Answer:

V₂ = 23.18 L

Explanation:

Given that,

Initial volume of a balloon, V₁ = 8.5 L

Initial pressure, P₁ = 150 kPa

We need to find new volume when the pressure drops to 55 kPa.

It is based on the concept of Boyle's law. The mathematical form of the Boyle's law is given by :

$P_1V_1=P_2V_2\\\\V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{150\times 8.5}{55}\\\\=23.18\ L$

So, the new volume is 23.18 L.

6 0

3 years ago

Which of these waves would have the greatest amount of energy?

alexdok [17]

Answer:

The answer is Gamma Rays

Hopes it Helps!

4 0

3 years ago

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