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3 years ago

The filament is broken in a light bulb. The circuit is now

Physics

1 answer:

Zina [86]3 years ago

3 0

Answer:

It is A. Open Circuit.

Explanation:

Filament heats up and produces the light. If the filament breaks, the current does not flow and the circuit will be open. The ammeter will reach zero when this happens.

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Calculate the tension (in N) in a vertical strand of spiderweb if a spider of mass 5.00 ✕ 10-5 kg hangs motionless on it.

Mandarinka [93]

Answer:

4.9 x 10⁻⁴N

Explanation:

Given parameters:

Mass of the vertical strand of spiderweb = 5 x 10⁻⁵kg

Unknown:

Tension in the vertical strand = ?

Solution:

The tension is the vertical strand will be the weight of the strand.

Weight of strand = mg

m is the mass

g is the acceleration due to gravity = 9.8m/s²

Tension in the vertical strand = 5 x 10⁻⁵kg x 9.8m/s²

Tension in the vertical strand = 4.9 x 10⁻⁴N

3 0

3 years ago

If a proton and an electron are released when they are 2.50×10^-10m apart (typical atomic distances), find the initial accelerat

katrin [286]

To solve this exercise, we will first proceed to calculate the electric force given by the charge between the proton and the electron (it). From the Force we will use Newton's second law that will allow us to find the acceleration of objects. The Coulomb force between two charges is given as

$F = k \frac{q_1q_2}{r^2}$

Here,

k = Coulomb's constant

q = Charge of proton and electron

r = Distance

Replacing we have that,

$F = (9*10^9)(\frac{(1.602*10^{-19})^2}{2.5*10^{-10}})$

$F = 3.6956*10^{-9}N$

The force between the electron and proton is calculated. From Newton's third law the force exerted by the electron on proton is same as the force exerted by the proton on electron.

The acceleration of the electron is given as

$a_e = \frac{F}{m_e}$

$a_e = \frac{3.6956*10^{-9}}{9.11*10^{-31}}$

$a_e = 4.0566*10^{21}m/s^2$

The acceleration of the proton is given as,

$a_p = \frac{F}{m_p}$

$a_p = \frac{3.6956*10^{-9}}{1.672*10^{-27}}$

$a_p = 2.21*10^{18}m/s^2$

3 0

2 years ago

Part C: Quantitative Problems when vf is not 0

Alina [70]

Answer:

(a)

$\triangle v=-8\ m/s\\\triangle mv=-56\ kg.m/s$

(b)

1120 N

Explanation:

Change in velocity, $\triangle v$ is given by subtracting the initial velocity from the final velocity and expressed as $\triangle v= v_f -v_i$

Where v represent the velocity and subscripts f and i represent final and initial respectively. Since the ball finally comes to rest, its final velocity is zero. Substituting 0 for final velocity and the given figure of 8 m/s for initial velocity then the change in velocity is given by

$\triangle v=0-8=-8\ m/s$

To find $m\triangle v$ then we substitute 7 kg for m and -8 m/s for $\triangle v$ therefore $\triangle\ v=7 Kg\times -8 m/s=-56\ Kg.m/s$

(b)

The impact force, F is given as the product of mass and acceleration. Here, acceleration is given by dividing the change in velocity by time ie

$a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}$

Substituting t with 0.05 s then $a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}=\frac {-8}{0.05}=-160 m/s^{2}$

Since F=ma then substituting m with 7 Kg we get that F=7*-160=-1120 N

Therefore, the impact force is equivalent to 1120 N

3 0

3 years ago

A ball is thrown horizontally from the top of a 60 m building and lands 100 m from the base of the building. How long is the bal

zhannawk [14.2K]

Answer:

The ball is in the air for 3.5 seconds

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

Explanation:

A ball is thrown horizontally

That means the vertical component of the initial velocity $u_{y}=0$

The initial velocity is the horizontal component $u_{x}$

The ball is thrown from the top of a 60 m

That means the vertical displacement component y = 60 m

→ y = $u_{y}$ t + $\frac{1}{2}$ gt²

where g is the acceleration of gravity and t is the time

y = -60 m , g = -9.8 m/s² , $u_{y}=0$

Substitute these values in the rule

→ -60 = 0 + $\frac{1}{2}$ (-9.8)t²

→ -60 = -4.9t²

Divide both sides by -4.9

→ 12.2449 = t²

Take √ for both sides

∴ t = 3.5 seconds

* The ball is in the air for 3.5 seconds 

The initial velocity is the horizontal component $u_{x}$

The ball lands 100 meter from the base of the building

That means the horizontal displacement x = 100 m

→ x = $u_{x}$ t

→ t = 3.5 s , x = 100 m

Substitute these values in the rule

→ 100 = $u_{x}$ (3.5)

Divide both sides by 3.5

→ $u_{x}$ = 28.57 m/s

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is $v_{y}$

→ $v_{y}$ = $u_{y}$ + gt

→ $u_{y}$ = 0 , g = -9.8 m/s² , t = 3.5 s

Substitute these values in the rule

→ $v_{y}$ = 0 + (-9.8)(3.5)

→ $v_{y}$ = -34.3 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity v is the resultant vector of $v_{x}$ and $v_{y}$

→ Its magnetude is $v=\sqrt{(v_{x})^{2}+(v_{y})^{2}}$

→ Its direction $tan^{-1}\frac{v_{y}}{v_{x}}$

→ $v_{y}$ = 28.6 , $v_{y}$ = -34.3

Substitute this values in the rules above

→ $v=\sqrt{(28.6)^{2}+(-34.3)^{2}}=44.66$

→ Its direction $tan^{-1}\frac{-34.3}{28.6}=-50.18$

The negative sign means the direction is below the horizontal

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

7 0

2 years ago

Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

The shear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

b) If the railing edge is 2 ft, the Reynold number is:

$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

The friction drag is equal to:

$F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf$

7 0

3 years ago