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4 years ago

Please help! How did Jessica Kusher create her new material?

Physics

2 answers:

zimovet [89]4 years ago

7 0

Answer:

she developed something to put on roofs to reduce the amount of ozone produced when heat hit the shingles

Explanation:

Arte-miy333 [17]4 years ago

7 0

Answer:

Explanation:

Jesseca Kusher, an 18-year-old researcher from Spartansburg, S.C., invented a paint-on coating for roofing shingles. Her formula could reduce a home's cooling costs and possibly cut ozone pollution in urban areas. ... One coating got graphite, the same material in pencil lead. Another recipe included gypsum

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A 10 kilogram sled is pulled across a frictionless surface with a force of 50 newtons for a distance of 10 meters. The pull is a

Gala2k [10]

Answer:

The power will be "250 watt". A further explanation is given below.

Explanation:

The given values are:

Force,

F = 50 N

Displacement,

d = 20 m

Time,

t = 2.0 seconds

Whenever the block is pulled, the angle will be "0" i.e., Cos0° = 1

Now,

The work will be:

= $Force\times displacement\times \Theta$

On substituting the given values, we get

= $50\times 10\times Cos0^{\circ}$

= $50\times 10\times 1$

= $500 \ Newton$

Now,

The Power will be:

= $\frac{Work \ done}{time}$

= $\frac{500}{2.0}$

= $250 \ watt$

8 0

3 years ago

Two point charges totaling 8 μC exert a repulsive force of 0.15 N on one another when separated by 0.5 m. What is the charge on

nexus9112 [7]

Answer:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

Explanation:

From Coulomb's Law the electrostatic repulsive force is given by the following formula:

F = kq₁q₂/r²

where,

F = Repulsive Force = 0.15 N

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = Magnitude of 1st Charge = ?

q₂ = Magnitude of 2nd Charge = ?

r = Distance between Charges = 0.5 m

Therefore,

0.15 N = (9 x 10⁹ N.m²/C²)q₁q₂/(0.5 m)²

q₁q₂ = (0.15 N)(0.5 m)²/(9 x 10⁹ N.m²/C²)

q₁q₂ = 4.17 x 10⁻¹²

q₁ = (4.17 x 10⁻¹²)/q₂ -------------------- equation (1)

The sum of charges is given as:

q₁ + q₂ = 8 μC

q₁ + q₂ = 8 x 10⁻⁶

using equation (1):

(4.17 x 10⁻¹²)/q₂ + q₂ = 8 x 10⁻⁶

(4.17 x 10⁻¹²) + q₂² = 8 x 10⁻⁶ q₂

q₂² - (8 x 10⁻⁶) q₂ + (4.17 x 10⁻¹²) = 0

Solving this quadratic equation:

q₂ = 7.4 x 10⁻⁶ C (OR) q₂ = 0.56 x 10⁻⁶ C

q₂ = 7.4 μC (OR) q₂ = 0.6 μC

Therefore,

q₁ = (4.17 x 10⁻¹² C)/(7.4 x 10⁻⁶ C)

q₁ = 0.6μC

Now, if we solve with q₂ = 0.6 μC, we will get q₁ = 7.4 μC.

Therefore, the correct option will be:

8 0

3 years ago

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0° below the horizontal. The negligent

anzhelika [568]

Answer:

The position is $P = 47.4 \ m$ relative to the base of the ocean

Explanation:

From the question we are told that

The angle made by the incline with the horizontal is $\theta = 24.0 ^o$

The constant acceleration is $a = 3.82 \ m/s^2$

The distance covered is $d = 60.0 \ m$

The height of the cliff is $h = 50 .0 \ m$

The velocity of the car is mathematically represented as

$v^2 = u^2 + 2ad$

The initial velocity of the car is u= 0

$v^2 = 2ad$

substituting values

$v^2 = 2 * 3.82 * 60$

$v = 21.4 \ m/s$

The vertical component of this velocity is

$v_v = -v * sin(\theta )$

substituting values

$v_v = -21.4 * sin(24.0)$

$v_v = -8.7 \ m/s$

The negative sign is because is moving in the negative direction of the y-axis

The horizontal component of this velocity is

$v_h = v * cos (\theta)$

$v_h = 21.4 * cos (24.0)$

$v_h = 19.5 \ m/s$

Now according to equation of motion we have

$h = v_v*t - \frac{1}{2} * g t^2$

substituting values

$50 = -8.7 t - \frac{1}{2} * 9.8 t^2$

$4.9t^2 +8.7t -50 = 0$

using quadratic equation we have that

$t_1 = 2.42\ s \ and\ t_2 = -4.20\ s$

given that time cannot be negative

$t = 2.42 \ s$

The car’s position relative to the base of the cliff when the car lands in the ocean is mathematically evaluate as

$P = v_h * t$

substituting values

$P = 19.5 * 2.43$

$P = 47.4 \ m$

5 0

3 years ago

What is the science definition of rounding

MrRa [10]

Alter (a number) to one less exact but more convenint calculations.

3 0

3 years ago

2. A small piece of dust is located in between two oppositely charged parallel plates where there exists a uniform electric fiel

Alla [95]

Answer:

The total charge is $q = 1.6*10^{-4}C$

The change in potential energy is $\Delta U = -4*10^{-2}J$

The height is $h= 8.2cm$

Explanation:

From the question we are told that the

The magnitude of electric field is $E = 5000N/C$

The number of proton is $N_p = 1*10^{15}$

The distance between the plates is $d = 5cm = \frac{5}{100} = 0.05m$

The total charge can be mathematically represented as

$q = N_p * e$

Where e is the charge on one proton which has a value of $=1.6*10^{-19}C$

Substituting values

$q = 1 *10^{15} * 1.6*10^{-19} = 1.6 *10^{-4}C$

The change in potential energy is mathematically represented ads

$\Delta U = -(qE)d$

where the negative sign shows that the work done by the electric force is against the electric field

Substituting values

$\Delta U = - 1.6*10^{-4} * 5000 * 0.05$

$= -4*10^{-2}J$

The mass of the bead is given as 0.05kg

The change in potential due to gravity is mathematically given as

$\Delta U = -mgh$

the negative sign is due to the fact that the height is decreasing

And $g =9.8m/s^2$

Making h the subject

$h = \frac{\Delta U}{mg}$

Substituting values

$h = \frac{4*10^{-2}}{0.05 * 9.8}$

$=0.081m = 0.082 *100 = 8.2cm$

4 0

4 years ago