Which of earth sphere have the most mass

A golfer hits her tee-shot due north towards the fairway. Her shot has an initial velocity of 60 m/s. A 15 m/s wind is blowing i

vovangra [49]

Answer:

$c=71.4m/s$

$\theta=8.54\textdegree$

Explanation:

From the question we are told that

Initial velocity of 60 m/s

Wind speed $V_w= 15 m/s \angle 45 \textdegree$

Generally Resolving vector mathematically

$sin(45\textdegree)15=10.6\\cos(45\textdegree)15=10.6$

Generally the equation Pythagoras theorem is given mathematically by

$c^2=a^2+b^2$

$c^2=10.6^2 +(10.6+60)^2$

$c=\sqrt{10.6^2 +(10.6+60)^2}$

Therefore Resultant velocity (m/s)

$c=71.4m/s$

b)Resultant direction

Generally the equation for solving Resultant direction

$\theta=tan^-1(\frac{y}{x})$

Therefore

$\theta=tan^-1(\frac{10.6}{70.6})$

$\theta=8.54\textdegree$

7 0

2 years ago

7. Un bloque de 700 N se encuentra sobre una viga uniforme de 200 N y 6 m de longitud. El bloque está a una distancia de 1 m del

GrogVix [38]

Answer:

x = 0.176 m

Explanation:

For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.

Let's use trigonometry to decompose the tension

sin 60 = $T_{y}$ / T

T_{y} = T sin 60

cos 60 = Tₓ / T

Tₓ = T cos 60

we apply the equation

∑ τ = 0

-W L / 2 - w x + T_{y} L = 0

the length of the bar is L = 6m

-Mg 6/2 - m g x + T sin 60 6 = 0

x = (6 T sin 60 - 3 M g) / mg

let's calculate

let's use the maximum tension that resists the cable T = 900 N

x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)

x = (4676 - 5880) / 6860

x = - 0.176 m

Therefore the block can be up to 0.176m to keep the system in balance.

5 0

3 years ago

A compact car has a mass of 1380 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that

astraxan [27]

Answer:

A) $k=34867.3384\ N.m^{-1}$

B) $\omega'\approx84\ Hz$

Explanation:

Given:

mass of car, $m=1380\ kg$

A)

frequency of spring oscillation, $f=1.6\ Hz$

We knkow the formula for spring oscillation frequency:

$\omega=2\pi.f$

$\Rightarrow \sqrt{\frac{k_{eq}}{m} } =2\pi.f$

$\sqrt{\frac{k_{eq}}{1380} } =2\times \pi\times 1.6$

$k_{eq}=139469.3537\ N.m^{-1}$

Now as we know that the springs are in parallel and their stiffness constant gets added up in parallel.

<u>So, the stiffness of each spring is (as they are identical):</u>

$k=\frac{k_{eq}}{4}$

$k=\frac{139469.3537}{4}$

$k=34867.3384\ N.m^{-1}$

B)

given that 4 passengers of mass 70 kg each are in the car, then the oscillation frequency:

$\omega'=\sqrt{\frac{k_{eq}}{(m+70\times 4)} }$

$\omega'=\sqrt{\frac{139469.3537}{(1380+280)} }$

$\omega'\approx84\ Hz$

7 0

3 years ago

An automobile engine delivers 55.0 hp. How much time will it take for the engine to do 6.22 × 105 J of work? One horsepower is e

Gennadij [26K]

Answer:

15.2 s

Explanation:

Convert hp to W:

55.0 hp × 746 W/hp = 41,030 W

Power = energy / time

41030 W = 6.22×10⁵ J / t

t = 15.2 s

8 0

2 years ago

Item 19 Question 1 A Ferris wheel has a radius of 75 feet. You board a car at the bottom of the Ferris wheel, which is 10 feet a

sergey [27]

Answer:

$y = 104.4 ft$

Explanation:

As we know that we board in the car of ferris wheel at the bottom position

So we will have

final height of the car at angular displacement given as

$y = y_o + R + R sin(270 - 255)$

$y = y_o + R + R sin 15$

here we know that

$y_o = 10 ft$

$R = 75 ft$

so we have

$y = 10 + 75 + 75 sin15$

$y = 104.4 ft$

4 0

2 years ago