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Liula [17]
2 years ago
6

I need help with one through six please

Physics
1 answer:
dybincka [34]2 years ago
7 0

Answer:

1] 8500000 = <u>8.5 × 10⁶</u>

2] .000072 = <u>7.2 × 10⁻⁵</u>

3] 5.3 × 10⁴ = <u>53000</u>

4] 2.8 × 10⁻³ = <u>0.0028</u>

5] Velocity = \frac{distance}{time}

 V = \frac{50}{10}

 <u>V = 5 m/s</u>

6] Acceleration = \frac{V1-V2}{time}

 A = \frac{30-15}{3}

 A = \frac{15}{3}

 <u>A = 3 m/s²</u>

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1.2miles=__________km
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1.931 kilometres is the answer of 1.2 miles

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The volume of water in the Pacific Ocean is about 7.0 × 10 8 km 3 . The density of seawater is about 1030 kg/m3. (a) Determine t
Novay_Z [31]

To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.

Potential Energy because of a planet would be given by the equation,

PE=\frac{GMm}{r}

Where,

G = Gravitational Universal Constant

M = Mass of Ocean

M = Mass of Moon

r = Radius

From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,

m = \rho V

m = (1030Kg/m^3)(7*10^8m^3)

m = 7.210*10^{11}Kg

It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.

When it is far away, it will be the total diameter from the center of the earth to the center of the moon.

r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m

When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,

r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m

PART A) Potential energy when the ocean is at its furthest point to the moon,

PE_1 = \frac{GMm}{r_1}

PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}

PE_1 = 9.05*10^{15}J

PART B) Potential energy when the ocean is at its closest point to the moon

PE_2 = \frac{GMm}{r_2}

PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}

PE_2 = 9.361*10^{15}J

PART C) The maximum speed. This can be calculated through the conservation of energy, where,

\Delta KE = \Delta PE

\frac{1}{2}mv^2 = PE_2-PE_1

v=\sqrt{2(PE_2-PE_1)/m}

v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}

v = 29.4m/s

8 0
3 years ago
Starting from rest, a wheel with constant angular acceleration turns through an angle of 25 rad in a time t. What will its angul
alexandr402 [8]

Answer:

θ = 225 rad

Explanation:

given data

angle = 25 rad

to find out

angular velocity after 3t?

solution

let angular acceleration α in t

θ = ω × t + 0.5 × α × t²        ........................1

here ω  = 0 (initial velocity )

so put this value here

25 = 0 + 0.5 × α × t²             ..........................2

α = 25 ÷ (0.5 t²)

α = 50 ÷ t²                            .........................3

now here we take in 3t

θ = ω × 3t + 0.5 × α × (3t)²

for ω  = 0

θ = 0 + 0.5 × α × 9t²

now put value in eq 2

so

θ = (0.5) ×  (50 ÷ t²) ×  (3t)²

θ = 25 × 9

θ = 225 rad

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2 years ago
What is science explain?​
slava [35]

Answer:

science is a subject

nice to meet you

8 0
3 years ago
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