A source is more reliable if ____

Formula for the distance (d) is given by d = rate*time. For example if you are traveling at 60 mph for 3 hours the distance trav

babunello [35]

Explanation:

Distance covered by the particle is given by:

Distance (d) = rate (v) × time (t)

Speed of Mary, v₁ = 50 mph

Speed of Jim, v₂ = 60 mph

It is assumed that, Mary and Jim leave at the same time. After one hour, Jim is 10 miles ahead.

Distance travelled by Jim, d₁ = (60t + 10)

Distance travelled by Mary, d₂ = 50t

The distance between Mary and Jim is greater than or equal to 100 miles.

$60t+10-50t\ge100$

$10t\ge90$

$t\ge9\ h$

So, Jim takes is 9 hours more than Mary to cover same distance. Hence, this is the required solution.

7 0

3 years ago

What are the inner planets relative distance from the Sun

Inessa [10]

Closer than the outer planets, inside the Asteroid Belt between Mars and Jupiter.

5 0

2 years ago

Read 2 more answers

The volume electric charge density of a solid sphere is given by the following equation: The variable r denotes the distance fro

qwelly [4]

Answer:

62.8 μC

Explanation:

Here is the complete question

The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?

Solution

The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ

So, Q = ∫∫∫ρdV

Q = ∫∫∫ρr²sinθdθdrdΦ

Q = ∫∫∫(0.2r²)r²sinθdθdrdΦ

Q = ∫∫∫0.2r⁴sinθdθdrdΦ

We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π

So, Q = ∫∫∫0.2r⁴sinθdθdrdΦ

Q = ∫∫∫0.2r⁴[∫sinθdθ]drdΦ

Q = ∫∫0.2r⁴[-cosθ]drdΦ

Q = ∫∫0.2r⁴-[cosπ - cos0]drdΦ

Q = ∫∫∫0.2r⁴-[-1 - 1]drdΦ

Q = ∫∫0.2r⁴-[- 2]drdΦ

Q = ∫∫0.2r⁴(2)drdΦ

Q = ∫∫0.4r⁴drdΦ

Q = ∫0.4r⁴dr∫dΦ

Q = ∫0.4r⁴dr[Φ]

Q = ∫0.4r⁴dr[2π - 0]

Q = ∫0.4r⁴dr[2π]

Q = ∫0.8πr⁴dr

Q = 0.8π∫r⁴dr

Q = 0.8π[r⁵/5]

Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]

Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]

Q = 0.8π[0.025 m⁵ - 0 m⁵]

Q = 0.8π[0.025 m⁵]

Q = (0.02π mC/m⁵) m⁵

Q = 0.0628 mC

Q = 0.0628 × 10⁻³ C

Q = 62.8 × 10⁻³ × 10⁻³ C

Q = 62.8 × 10⁻⁶ C

Q = 62.8 μC

3 0

2 years ago

A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th

Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(4)}{2}(2(0)+4)$

$F = 25000 lb$

b) Force on deep end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(10)}{2} (2(0)+10)$

$F = 187500 lb$

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

$F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)$