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3 years ago

What is indirect sunlight? how does the tilt of the earth affect which type of sunlight the earth receives?

1 answer:

6 0

Indirect sunlight is sunlight that doesn't shine onto a plant at full strength and is weakened by something coming between it and the plant. the tilt of the earth on it's axis allows more direct sunlight on certain parts of the earth at particular times

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A uniform, 0.0300-kg rod of length 0.400 m rotates in a horizontal plane about a fixed axis through its center and perpendicular

AURORKA [14]

Answer:

A) ω₂ = 28 rev/min

B) ω_3 = 28 rev/min

Explanation:

A) Initial moment of inertia (I₁) is the rod's own ((1/12)ML²) plus that of the two rings (two lots of mr² if we treat each ring as a point mass).

Thus;

I₁ = ((1/12)ML²) + 2(mr²)

Where;

M is mass of rod = 0.03 kg

L is length of rod = 0.4m

m is mass of small rings = 0.02 kg

r is radius of small rings = 0.05 m

Thus;

I₁ = ((1/12) x 0.03 x 0.4²) + 2(0.02 x 0.05²)

I₁ = 0.0021 kg.m²

Now, when the rings slide and reach the ends of the rod i.e at r = L/2 = 0.4/2 = 0.2m from axis, the new Moment of inertia is;

I₂ = ((1/12) x 0.03 x 0.4²) + 2(0.02 x 0.2²)

I₂ = 0.0036 kg.m²

From conservation of angular momentum, we know that:

I₁ω₁ = I₂ω₂

We are given ω₁ = 48 rev/min.

Thus; plugging in the relevant values;

0.0021 x 48 = 0.0036ω₂

0.1008 = 0.0036ω₂

ω₂ = 0.1008/0.0036

ω₂ = 28 rev/min

b) For this, we have the same scenario as the case above where the ring just reaches the ends of the rods.

Thus,

I₂ω₂ = I_3•ω_3

So,0.0021 x 48 = 0.0036ω_3

ω_3 = 28 rev/min

7 0

3 years ago

A 5.0 g coin is placed 15 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the t

Kaylis [27]

Answer:

Angular speed will reach 6.833rad/s before the coin starts slipping

Explanation:

There is no question but I'll asume the common one: Calculate the speed of the turntable before the coin starts slipping.

With a sum of forces:

$Ff = m*a$

$Ff=m*V^2/R$

At this point, friction force is maximum, so:

$\mu*N=m*V^2/R$

$\mu*m*g=m*V^2/R$

Solving for V:

$V=\sqrt{\mu*g*R}$

V=1.025 m/s

The angular speed of the turntable will be:

ω = V/R = 6.833 rad/s This is the maximum speed it can reach before the coin starts slipping.

7 0

3 years ago

An object with a mass of 3 kg has a momentum of 75 kg.m/s. What is its velocity?

tiny-mole [99]

Answer:

25m/s

Explanation:

p=mv

75 = 3v

v = 75/3

v = 25 m/s

If you want to verify your answer, just insert the value of v to the equation.

8 0

3 years ago

I don’t understand can someone break it down for me

Westkost [7]

Answer:

a = (v² – v₀²)/ 2(s – s₀)

Explanation:

v² = v₀² + 2a (s – s₀)

We can make 'a' the subject of the above expression as follow:

v² = v₀² + 2a (s – s₀)

Subtract v₀² from both side

v² – v₀² = v₀² + 2a (s – s₀) – v₀²

v² – v₀² = v₀² – v₀² + 2a (s – s₀)

v² – v₀² = 2a (s – s₀)

Divide both side by (s – s₀)

(v² – v₀²)/ (s – s₀) = 2a

Divide both side by 2

(v² – v₀²)/ (s – s₀) ÷ 2 = a

(v² – v₀²)/ (s – s₀) × 1/2 = a

(v² – v₀²)/ 2(s – s₀) = a

a = (v² – v₀²)/ 2(s – s₀)

6 0

3 years ago

A car of mass 1400 kg travelling at 15 m/s goes over a circular arc-shaped bump in the road. if the radius of the arc is 40 m, w

kari74 [83]

The car on the top of the arc;
m = 1,400 kg, v = 15 m/s, r = 40 m;
The normal force:
F n = m g - m v ² / r =
= 1,400 kg · 9.8 m / s² - (1,400 kg · ( 15 m/s )² : 40 m ) =
= 13,720 N - 7,875 N = 5,845 N
Answer:
B ) 5,800 N

8 0

4 years ago