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3 years ago

12

On a distant planet a freely falling object has an acceleration of 20m/s. what vertical distance will an object dropped from res

t on this planet cover in 1.8

1 answer:

AveGali [126]3 years ago

3 0

<span>Use s = ut + 1/2 at^2, so

s = 0 + 1/2 x20 x 1.8 x 1.8

s = 32.4 m.</span>

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A rider on a bike with the combined mass of 100kgattains aterminal speed of 15m/son a 12% slope. Assuming that the only forces a

Semmy [17]

Answer:

C_d = 0.942

Explanation:

Let's first calculate the angle of inclination.

Formula is;

tan θ = (%slope)

% Slope is given as 12%

Thus;

θ = tan^(-1) (12/100)

θ = 6.843°

Let's now calculate the force due to the weight of the rider and bike combined from;

F = mg sin θ

We are given; m = 100 kg.

Thus;

F = 100 × 9.81 × sin 6.843

F = 116.885 N

The drag force will also be the same as the force due to the weight of the body. Thus;

Drag force; F = C_d(½ρu²A)

Where;

C_d is drag coefficient

ρ is density

U is terminal speed

A is area

We are given;

A = 0.9 m²

U = 15 m/s

From online tables, density of air is approximately 1.225 kg/m³

Thus;

116.885 = C_d(½ × 1.225 × 15² × 0.9)

116.885 = 124.03125C_d

C_d = 116.885/124.03125

C_d = 0.942

4 0

3 years ago

Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p

worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

$U1/U2=6$
$U1/U2=6$

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy is mathematically given as

$U=\frac{-k*qQ}{r}$

Therefore

The Equations of U1 and U2 is

For U1

$U1=\frac{-k*q_1Q}{r}$

For U2

$U2=\frac{-k*q_1Q}{3*2r}$

Since

U is a function of q and q2=q1/3

Therefore

$U1/U2=6$

For Question 2

For U1

$U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}$

Therefore

$U1/U2=6$

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7 0

3 years ago

What is the temperature if the peak of a blackbody spectrum is at 19.0 nm ?

nlexa [21]

Answer:

1.52 * 10⁵ K

Explanation:

When the temperature of a blackbody radiator increases, the overall radiated energy increases and the peak of the radiation curve moves to shorter wavelengths. When the maximum is evaluated from the Planck radiation formula, the product of the peak wavelength and the temperature is found to be a constant.

By Wien's Law,

λ * T = 2.898 * 10⁻³ mK

∴ T = 2.898 * 10⁻³/λ

Given λ = 19 nm = 19 * 10⁻⁹

T = 2.898 * 10⁻³ / 19 * 10⁻⁹

=1.52 * 10⁵ K

6 0

3 years ago

At one instant of time a rocket is traveling in outer space at 2500m/s and is exhausting fuel at a rate of 100 kg/s. If the spee

lawyer [7]

Thrust is a reaction force described quantitatively by Newton's third law. When a system expels or accelerates mass in one direction, the accelerated mass will cause a force of equal magnitude but opposite direction on that system. Mathematically can be written as,

$T = v\frac{dm}{dt}$

Here,

v = speed of the exhaust gases measured relative to the rocket.

$\frac{dm}{dt}$ = Rate of change of mass with respect to time

Our values are given as,

$v = 1500m/s$

$\frac{dm}{dt} = 100kg/s$

Replacing we have that

$T = 1500*100$

$\therefore T = 1.5*10^5N$

7 0

3 years ago

Please help ASAP!

Softa [21]

Answer:

100 newton

Explanation:

newton third law of motion says to every action there is an always an equal and opposite reaction so the magnitude will stay equal but opposite direction

8 0

3 years ago

Read 2 more answers