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4 years ago

10

The illustration above depicts a string being stretched and then released. Changes in both potential and kinetic Energy occur du

ring this process. Which statement most accurately describes the change in potential or kinetic energy

1 answer:

Masja [62]4 years ago

3 0

The answer is C -<span> The potential energy is increasing through steps A, B, & C and then decreases </span>

You might be interested in

1. Which statement is true about natural

lara [203]

Answer:

1. D. They can be a substance, material, object or source of energy.

2. B. The properties will be different.

3. C. When two reactants form one product, the reaction is spontaneous.

4 0

3 years ago

Two balloons (m = 0.012 kg) are separated by a distance of d = 15 m. They are released from rest and observed to have an instant

Evgesh-ka [11]

Answer:

1.492*10^14 electrons

Explanation:

Since we know the mass of each balloon and the acceleration, let’s use the following equation to determine the total force of attraction for each balloon.

F = m * a = 0.012 * 1.9 = 0.0228 N

Gravitational forces are negligible

Charge force = 9 * 10^9 * q * q ÷ 225

= 9 * 10^9 * q^2 ÷ 225 = 0.0228

q^2 = 5.13 ÷ 9 * 10^9

q = 2.387 *10^-5

This is approximately 2.387 *10^-5 coulomb of charge. The charge of one electron is 1.6 * 10^-19 C

To determine the number of electrons, divide the charge by this number.

N =2.387 *10^-5 ÷ 1.6 * 10^-19 = 1.492*10^14 electrons

3 0

3 years ago

Scientists are working on a new technique to kill cancer cells by zapping them with ultrahigh-energy (in the range of 1012 W) pu

Tcecarenko [31]

1. $7.95\cdot 10^6 J$

The total energy given to the cells during one pulse is given by:

$E=Pt$

where

P is the average power of the pulse

t is the duration of the pulse

In this problem,

$P=1.59\cdot 10^{12}W$

$t=5.0 ns = 5.0\cdot 10^{-9} s$

Substituting,

$E=(1.59\cdot 10^{12}W)(5.0 \cdot 10^{-6}s)=7.95\cdot 10^6 J$

2. $1.26\cdot 10^{21}W/m^2$

The energy found at point (1) is the energy delivered to 100 cells. The radius of each cell is

$r=\frac{4.0\mu m}{2}=2.0 \mu m = 2.0\cdot 10^{-6}m$

So the area of each cell is

$A=\pi r^2 = \pi (2.0 \cdot 10^{-6}m)^2=1.26\cdot 10^{-11} m^2$

The energy is spread over 100 cells, so the total area of the cells is

$A=100 (1.26\cdot 10^{-11} m^2)=1.26\cdot 10^{-9} m^2$

And so the intensity delivered is

$I=\frac{P}{A}=\frac{1.59\cdot 10^{12}W}{1.26\cdot 10^{-9} m^2}=1.26\cdot 10^{21}W/m^2$

3. $9.74\cdot 10^{11} V/m$

The average intensity of an electromagnetic wave is related to the maximum value of the electric field by

$I=\frac{1}{2}c\epsilon_0 E^2$

where

c is the speed of light

$\epsilon_0$ is the vacuum permittivity

E is the amplitude of the electric field

Solving the formula for E, we find:

$E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(1.26\cdot 10^{21} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12}F/m)}}=9.74\cdot 10^{11} V/m$

4. 3247 T

The magnetic field amplitude is related to the electric field amplitude by

$E=cB$

where

E is the electric field amplitude

c is the speed of light

B is the magnetic field

Solving the equation for B and substituting the value of E that we found at point 3, we find

$B=\frac{E}{c}=\frac{9.74\cdot 10^{11} V/m}{3\cdot 10^8 m/s}=3247 T$

3 0

3 years ago

Which element has the same number of valence electrons as a krypton

AleksandrR [38]

Arsenic in period 4 group 5.

3 0

3 years ago

Q.1) The crane arm is pinned at point A and has a mass of 200 kg whose weight is acting at a point 2 m to the right of point A.

Sveta_85 [38]

Answer:

a) Fbc = 22692.23 N

b) Ax = 10148.276 N (→)

Ay = 10486.552 N (↓)

Explanation:

Given

Mass of the crane arm: M = 200 Kg

rM = 2.00 m

Mass of the block: m = 800 Kg

rm = 7.00 m

Fbc = ?

rbcx = (1.80 + 1.20) m = 3.00 m

rbcy = (2.40 - 1.00) m = 1.40 m

a) We apply rotational equilibrium around point A as follows

∑τ = 0 (Counterclockwise is the positive rotation direction)

rbcx*Fbcx + rbcy*Fbcy - rM*WM - rm*Wm = 0 (I)

From the pic we have the Right triangle where

BC² = 1.2²+2.4² = 7.2 ⇒ BC = √7.2 = 2.683

then

Cos ∅ = 1.2/2.683 = 0.447

Sin ∅ = 2.4/2.683 = 0.894

If

Fbcx = Fbc*Cos ∅

⇒ Fbcx = 0.447*Fbc

Fbcy = Fbc*Sin ∅

⇒ Fbcy = 0.894*Fbc

Now, we obtain

3.00*(0.447*Fbc) + 1.40*(0.894*Fbc) - 2.00*(200*9.81) - 7.00*(800*9.81) = 0

⇒ 1.342*Fbc + 1.252*Fbc - 3924 - 54936 = 0

⇒ 2.594*Fbc = 58860

⇒ Fbc = 22692.23 N

b) We apply

∑Fx = 0 (+→)

⇒ Ax - Fbcx = 0

⇒ Ax = Fbcx = 0.447*Fbc = 0.447*(22692.23 N)

⇒ Ax = 10148.276 N (→)

∑Fy = 0 (+→)

⇒ Ay + Fbcy - WM - Wm = 0

⇒ Ay = - Fbcy + WM + Wm

⇒ Ay = - 0.894*Fbc + m*g + M*g

⇒ Ay = - 0.894*(22692.23 N) + (800 Kg*9.81 m/s²) + (200 kg*9.81 m/s²)

⇒ Ay = 10486.552 N (↓)

3 0

3 years ago