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LiRa [457]
3 years ago
13

Why there's a rainbow after the rain?

Physics
1 answer:
Diano4ka-milaya [45]3 years ago
3 0

Answer:

Rainbow is formed due to dispersion of sunlight by tiny water droplets suspended in the atmosphere after a rain shower .....After rain shower the sun is visible to us again and the rain water droplets are suspended in the atmosphere therefore they disappear sunlight and and rainbow is formed....

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How are force, mass and acceleration related?
Zepler [3.9K]

Newton's 2nd law of motion:

F = ma

F is force, m is mass, and a is acceleration.

6 0
3 years ago
A rocket engine uses fuel and oxidizer in a reaction that produces gas particles having a velocity of 1380 ms The desired thrust
bearhunter [10]

Answer:

a. 141.3 kg/s b. 5.49 m/s² c. i. 104228.9 N ii. 8.53 m/s² d. i. 97305.2 N ii. 9.84 m/s²

Explanation:

a. What must be the fuel/oxidizer consumption rate (in kg s1)?

The thrust T = Rv where R = mass consumption rate and v = velocity of rocket. Since T = 195000 N and v = 1380 m/s,

R = T/v = 195000 N/1380 m/s = 141.3 kg/s

b. If the initial weight of the rocket is 125000 N, what is its initial acceleration?

We also know that thrust T - W = ma since the rocket has to move against gravity. where M = mass of rocket = W/g = 125000 N/9.8m/s² = 12755.1 kg, W = weight of rocket = 125000 N, a = acceleration of rocket and T = thrust = 195000 N.

So, T - W = Ma

195000 N - 125000 N = (12755.1 kg)a

70000 N = ma

a = 70000 N/12755.1 kg = 5.49 m/s²

c. What are the weight and acceleration of the rocket at t 15.0 s after ignition?

We know that the loss in mass ΔM = mass consumption rate × time = Rt. Since R = 141.3 kg/s and t = 15 s,

ΔM = 141.3 kg/s × 15 = 2119.5 kg

The new mass is thus M = M - ΔM = 12755.1 kg - 2119.5 kg = 10635.6 kg

i.The weight after 15 seconds is thus W' = M'g = 10635.6 kg × 9.8m/s² = 104228.9 N

ii. Since T - W' = M'a. where M' is our new mass and a our new acceleration,

a = (T - W')/M'

= (195000 N - 104228.9 N)/10635.6 kg

= 90771.1 N/10635.6 kg

= 8.53 m/s²

d. What are the weight and acceleration of the rocket at 20.0 s after ignition?

We know that the loss in mass ΔM" = mass consumption rate × time = Rt. Since R = 141.3 kg/s and t = 20 s,

ΔM" = 141.3 kg/s × 20 = 2826 kg

The new mass is thus M" = M - ΔM" = 12755.1 kg - 2826 kg = 9929.1 kg

i. The weight after 20 seconds is thus W" = M"g = 9929.1 kg × 9.8m/s² = 97305.2 N

ii. Since T - W" = M"a. where M" is our new mass and a our new acceleration,

a = (T - W")/M"

= (195000 N - 97305.2 N)/9929.1 kg

= 97694.8 N/9929.1 kg

= 9.84 m/s²

4 0
3 years ago
A uniform electric field of magnitude 110 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.
Marizza181 [45]

Answer:

1.7×10^5 ms-1

Explanation:

From

qE= qvB

q= charge on the electron

E = electric field

v= velocity

B= magnetic field

E= vB

v= E/B= 110×10^3/0.6

v= 1.7×10^5 ms-1

3 0
4 years ago
Read 2 more answers
why is the earth more attracted to the sun gravitational pull than the gravitational pull of the moon
Ierofanga [76]
First, let's take a look at the equation for the force of gravity between two objects:

F = (GMm)/r², where, 

G = gravitational constant = 6.67 x 10⁻¹¹
M = mass of one object
m = mass of the other object
r = distance between the two objects

From this equation, we can see that the force of gravity is directly proportional to the mass of the two objects and inversely proportional to the distance between them. We can then say that the Earth is <span>more attracted to the sun than the moon because of the massive mass of the Sun (1.9891 x 10</span>³⁰)<span> compared to moon (7.3577 x 10</span>²²<span>). Although, the moon is nearer to the Earth, it has little effect to bring down the gravitational pull of the Sun. </span>
5 0
3 years ago
During a marathon race, a runner’s blood flow increases to 10.0 times her resting rate. Her blood’s viscosity has dropped to 95.
Phoenix [80]

To solve the problem it is necessary to apply the equations related to the Poiseuilles laminar flow law, with which the stationary laminar flow ΦV of an incompressible and uniformly viscous liquid (also called Newtonian fluid) can be determined through a cylindrical tube of constant circular section. Mathematically this can be expressed:

Q = \frac{\Delta P \pi r^4}{8\eta l}

Where:

\eta_i = are the viscosities of the concrete before and after the increase

l = Length of the vessel

r_1, R_2 = Radio of the vessel before and after the increase

\Delta P= Change in the pressure

Q_{1,2} = The rates of flow before and after he increase

Our values are given as:

Q_2 = 10Q_1 \rightarrow 10 times her resting rate

\eta_2 = 0.95\eta_1 95% of its normal value

\Delta P_2 = 1.5\Delta P_1 Increase of 50%

Plugging known information to get

Q_1 = \frac{\Delta P \pi r^4}{8\eta l}

Q_1 8\eta_1 l = \Delta P_1 \pi r_1^4

r_1^4 = \frac{Q_1 8\eta_1 l}{\Delta P_1 \pi}

r_1 = (\frac{Q_1 8\eta_1 l}{\Delta P_1 \pi})^{1/4}

r_2 = (\frac{Q_2 8\eta_2 l}{\Delta P_2 \pi})^{1/4}

r_2 = (\frac{10Q_18 \times 0.95\eta_1 l}{1.5\Delta P_1 \pi})^{1/4}

r_2 = 1.586r_1

Therefore the factor of average radio of her blood vessels increased is 1.589 the initial factor after the increase.

7 0
4 years ago
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