How might the changes in tide affect a fishing community?

A stone is dropped from the upper observation deck of a tower, "500" m above the ground. (Assume g = 9.8 m/s2.) (a) Find the dis

Yuri [45]

Answer:

h₍₁₎ = 495,1 meters

h₍₂₎ = 480,4 m

h₍₃₎ = 455,9 m

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Explanation:

The exercise is "free fall". t = $\sqrt{\frac{2h}{g} }$

Solving with this formula you find the time it takes for the stone to reach the ground (T) = 102,04 s

The heights (h) according to his time (t) are found according to the formula:

h(t) = 500 - 1/2 * g * t²

Remplacing "t" with the desired time.

4 0

3 years ago

Volcanic ash and sulfur dioxide spewed out of Mt. Pinatubo in 1991. These materials can reflect incoming solar radiation. Over t

gogolik [260]

The suspended ash made for some some spectacular sunsets! Sulfuric acid was spread worldwide, increasing acidity of rain. Ash deflected energy from the sun, causing a slight drop on global temps for a few years.

7 0

2 years ago

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A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a

makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

$\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]$ (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

$\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°$

The distance r is:

$r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m$

You replace the values of all parameters in the equation (1):

$\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}$

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0

3 years ago

A disk of radius 10 cm speeds up from rest. it turns 60 radians reaching an angular velocity of 15 rad/s. what was the angular a

stepan [7]

Answer:

a) α = 1.875 $\frac{rad}{s^{2} }$

b) t = 8 s

Explanation:

Given:

ω1 = 0 $\frac{rad}{s}$

ω2 = 15 $\frac{rad}{s}$

theta (angular displacement) = 60 rad

*side note: you can replace regular, linear variables in kinematic equations with angular variables (must entirely replace equations with angular variables)*

a) α = ?

(ω2)^2 = (ω1)^2 + 2α(theta)

$15^{2}$ = $0^{2}$ + 2(α)(60)

225 = 120α

α = 1.875 $\frac{rad}{s^{2} }$

b)

α = (ω2-ω1)/t

t = (ω2-ω1)/α = (15-0)/1.875 = 8

t = 8 s

4 0

2 years ago

Which phrases describe all the outer planets motion? Select two options

11111nata11111 [884]

Answer:

slow revolution and fast rotation

Explanation:

4 0

2 years ago