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3 years ago

What is cell in electricity?

Physics

1 answer:

agasfer [191]3 years ago

7 0

Answer:

An electrical cell is a device used to generate electricity, or to make chemical reactions by applying electricity.

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The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 3 months. The line

Sladkaya [172]

Answer:

Explanation:

given

T = 3months = 7.9 × 10⁶s

orbital speed = 88 × 10³m/s

V= 2πr÷T

∴ r = (V×T) ÷ 2π

r = (88km × 7.9 × 10⁶s) ÷ 2π

r = 1.10 × 10⁸km

using kepler's 3rd law

mass of both stars = (seperation diatance)³/(orbital speed)²

M₁ + M₂ = (2r)³/( $\frac{1}{4}$ year)²

= (1.06 × 10²⁵)/(6.2×10¹³)

1.71×10¹²kg

since M₁ = M₂ =1.71×10¹²kg ÷ 2

M₁ = M₂ = 8.55×10¹¹kg

6 0

2 years ago

What is a scientific theory

Tomtit [17]

Answer:

They arn't usually guesses, but they are well made theorys or explanations. Its a well-substantiated explanation of facts that have been confirmed in expirements.

3 0

2 years ago

A cyclist is traveling 10 m/s with an acceleration of 6 m/s2. How fast is the cyclist traveling at the end of 20 seconds

krok68 [10]

Answer:

The cyclist is traveling at 130 m/s

Explanation:

Constant Acceleration Motion

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

$v_f=v_o+at$

The cyclist initially travels at 10 /s and it's accelerating at a=6m/s^2. We need to know the new speed when t= 20 seconds have passed.

Apply the above equation:

$v_f=10+6\cdot 20$

$v_f=10+120$

$v_f=130\ m/s$

The cyclist is traveling at 130 m/s

7 0

3 years ago

When water flows in to the cracks of a rock freezes and breaks into smaller pieces is it a chemical or physical change?

prisoha [69]

Most probably physical, even though the change of liquids to solids is directly related to chemistry too

7 0

3 years ago

While unrealistic, we will examine the forces on a leg when one falls from a height by approximating the leg as a uniform cylind

Leno4ka [110]

Answer:

Part A: 7.75 m/s

Part B: 2330.8 kN

Part C: 24.03 kN

Part D: 4.8 kN

Part E: $1.7\times 10^{9} Dyn/cm^{2}$

Part F: Option D

Bending his legs increases the time over which the ground applies force, thus decreasing the force applied by the ground.

Explanation:

Part A

From the fundamental kinematic equation

$v^{2}=u^{2}+2gh$ where v is the velocity of the man just before hitting the ground, g is acceleration due to gravity, u is initial velocity, h is the height.

Since the initial velocity is zero hence

$v^{2}=2gh$

$v=\sqrt 2gh$

Substituting 10 m/s2 for g and 3 m for h we obtain

$v=\sqrt 2\times 10\times 3 =\sqrt 60= 7.745967\approx 7.75 m/s$

Part B 

Force exerted by the leg is given by

F=PA where P is pressure, F is force, A is the cross-section of the bone

$A=\frac {\pi d^{2}}{4}$

Substituting 2.3 cm which is equivalent to 0.023m for d and $1.7\times10^{8} N/m2$ for P we obtain the force as

$F=PA=1.7\times10^{8}*\frac {\pi (0.023)^{2}}{4}= 2330818.276\approx 2330.8 kN$

Part C

The fundamental kinematic equation is part (a) can also be written as

$v^{2}=u^{2}+2a\triangle x$ and making a the subject then

$a=\frac {v^{2}-u^{2}}{2\triangle x}$ where a is acceleration and $\triangle x$ is the change in length

Substituting the value obtained in part a, 7.75 m/s for v, u is zero and 1cm which is equivalent to 0.01 m for $\triangle x$ then

$a=\frac {7.75^{2}-0^{2}}{2\times 0.01}= 3003.125 m/s^{2}$

Force exerted on the man is given by

$F=ma=80\times 3003.125= 240250 N\approx 24.03 kN$

Part D

The fundamental kinematic equation is part (a) can also be written as

$v^{2}=u^{2}+2a\triangle h$ and making a the subject then

$a=\frac {v^{2}-u^{2}}{2\triangle h}$ where a is acceleration and $\triangle h$ is the change in height

Also, force exerted on the man is given by $F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}$

Substituting 80 Kg for m, 50 cm which is equivalent to 0.5m for \triangle h and other values as used in part c

$F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}=80\times \frac {7.75^{2}-0^{2}}{2\times 0.5}= 4805 N\approx 4.8 kN$

Part E

$P=1.7\times 10^{8}=1.7\times 10^{8}\times (\frac {10^{5} Dyn}{10^{4} cm^{2}}=1.7\times 10^{9} Dyn/cm^{2}$

Part F

Bending his legs increases the time over which the ground applies force, thus decreasing the force applied by the ground

7 0

2 years ago