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4 years ago

What is the cell cycle?

Physics

2 answers:

dimulka [17.4K]4 years ago

7 0

I believe the answer is A it seems most likely

In-s [12.5K]4 years ago

4 0

The answer was A on my test, please let me know if it's the same for yours :)

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What is a phone message?

malfutka [58]

A message telephone messagemessage a communication usually brie that is written or spoken or signaled; "hesent a three-word <span>message"</span>

8 0

3 years ago

Read 2 more answers

A potential energy function is given by U(x)=(3.00J)x+(1.00J/m^2)x^3. What is the force function F(x) (in newtons) that is assoc

snow_lady [41]

Answer:

$F(x)=-3 N -(3N/m^2) x^2$

Explanation:

The force is equal to the negative of the derivative of the potential energy function:

$F=-\frac{dU}{dx}$

In this problem, the potential energy function is

$U(x)=3 x + 1 x^3$

Therefore, the force function is:

$F(x)=-\frac{dU}{dx}=-\frac{d}{dx}(3x+1x^3)=-3-3x^2$

So, adding the units of measurement,

$F(x)=-3 N -(3N/m^2) x^2$

4 0

4 years ago

An elevator cab is pulled directly upward by a single cable. The elevator cab and its single occupant have a mass of 2300 kg. Wh

Murrr4er [49]

Answer:

15.64 KN

Explanation:

mass of the elevator cab with a single occupant= 2300 kg

acceleration relative to the cab $a_{ce}$ = 6.80 m/s^2

acceleration of the coin relative to the cab in the opposite direction of motion of cab so we can consider it as a= -6.80 m/s^2

The acceleration of elevator cab relative to the ground $a_{cg}$

now we can say that

$a_{ce} +a_{eg} =a_{cg}$

=-6.80+ a_{eg}= -9.8

[tex]a_{eg}=-9.8+6.80=-3.8

The forces that act on elevator cab are tension and gravitational, applying newtons second law

T- mg= ma_{eg}

Then the tension in the cable is

T= 2300(-3.8)+2300×9.8= 15640 N= 15.64 KN

therefore tension in the string will be 15.64 KN

3 0

4 years ago

What happens to the kinetic energy of water when it changes from the liquid state to the solid state

d1i1m1o1n [39]

k e is energy of movement. When thing solidifies, atoms stop moving ke is less.

8 0

3 years ago

There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C

liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

$V_{tot} = V_{q1} + V_{q2}$

$V_q = \dfrac{k \cdot q}{r}$

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

$V_{tot} = \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05} = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0$

The electric field at the point exactly midway between the plates, $V_{tot}$ = 0

3) The electric field, 'E', between plates is given as follows;

$E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C$

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, $F_e$ = E × e

∴ $F_e$ = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, $F_e$ ≈ 1.807 × 10⁻⁷ N

4 0

3 years ago