Which of the following is true about "natural" steroids?

A force of 200 N is applied on small piston of a pascal press. What would be the

VladimirAG [237]

Answer:

The force applied on the big piston is 1306.67 N

Explanation:

Given;

force applied on small piston, F₁ = 200 N

diameter of the small piston, d₁ = 4.37 cm

radius of the small piston, r₁ = d₁/2 = 2.185 cm

Area of the small piston, A₁ = πr₁² = π(2.185 cm)² = 15 cm²

Area of the big piston, A₂ = 98 cm²

The pressure of the piston is given by;

$P = \frac{F}{A} \\\\\frac{F_1}{A_1} = \frac{F_2}{A_2}\\\\ F_2 = \frac{F_1A_2}{A_1}$

Where;

F₂ is the force on big piston

$F_2 = \frac{200*98}{15} \\\\F_2 = 1306.67 \ N$

Therefore, the force applied on the big piston is 1306.67 N

3 0

3 years ago

Three guns are aimed at the center of a circle, and each fires a bullet simultaneously. The directions in which they fire are 12

ikadub [295]

Answer:

The unknown mass of the bullet is $m1=3.751x10^{-3} kg$

Explanation:

According to Newton's laws of motion, when a net external force acts on a body of mass <u><em>m</em></u> , it results in change in momentum of the body and is given by:

$F=\frac{P}{dt}$

Where:

P

is the linear momentum of the body

As a consequence, when there are no external forces acting on the body the total momentum remains conserved i.e.

Given:

$m_{2}=5.79x10^{-3}kg \\m_{3}=5.79x10^{-3}kg\\v_{2}=v_{3}=392 \frac{m}{s}\\$

For momentum along the y-direction to be zero, it is achieved when the equal masses are moving at angles of

θ1=180°, θ2=60°, θ3=-60°

Therefore, from conservation of momentum along x - direction:

$m_{1}*v_{1}*cos(180)+m_{2}*v_{2}*cos(60)+m_{3}*v_{3}*cos(-60)=0\\$ $m_{1}*605\frac{m}{s}*cos(180)+5.79x10^{-3}kg *392\frac{m}{s}*cos(60)+5.79x10^{-3}kg*392\frac{m}{s}*cos(-60)=0\\$

$-m_{1}*605+5.79x10^{-3}kg*196\frac{m}{s}+5.79x10^{-3}kg*196\frac{m}{s}=0\\m_{1}*605kg= 2.26968\frac{kg*m}{s}\\m_{1}=3.75 x10^{-3} kg$

3 0

3 years ago

An airplane is traveling at a fixed altitude with an outside wind factor. The airplane is headed N 40° W at a speed of 600 miles

bija089 [108]

Answer: $\vec{v_R}=\hat{i}[-329.11]+\hat{j}[516.18]$

Explanation:

Given

Plane is initially flying with velocity of magnitude $v=600\ mph$

at angle of $40^{\circ}$ with North towards west

Velocity of plane airplane can be written as

$v_a=600(-\sin 40\hat{i}+\cos 40\hat{j})$

Now wind is encountered with speed of $v=80\ mph$ at angle of $N45^{\circ}E$

$v_w=80(\cos 45\hat{i}+\sin 45\hat{j})$

resultant velocity

$\vec{v_R}=\vec{v_a} +\vec{v_w}$

$\vec{v_R}=600(-\sin 40\hat{i}+\cos 40\hat{j})+ 80(\cos 45\hat{i}+\sin 45\hat{j})$

$\vec{v_R}=\hat{i}[-385.67+56.56]+\hat{j}[459.62+56.56]$

$\vec{v_R}=\hat{i}[-329.11]+\hat{j}[516.18]$

for direction $\tan \theta =\frac{516.18}{329.11}$

$\tan \theta =1.568$

$\theta =57.47^{\circ}$ west of North

3 0

3 years ago

What is a metallic bond and how does it hold metals together?

Marysya12 [62]

Answer:

Force that holds atoms together in a metallic substance.

Explanation:

Hope this helps? C:

~Chiena

4 0

2 years ago

Suppose a 350-g kookaburra (a large kingfisher bird) picks up a 75-g snake and raises it 2.5 m from the ground to a branch. (a)

mafiozo [28]

(a) The work done by the bird to raise the snake on the branch is equal to the product between the weight of the snake and the height of the branch above the ground:

$W_1 = mg h$

where $m=75 g=0.075 kg$ is the mass of the snake and $h=2.5 m$ is the height of the branch with respect to the ground. Substituting numbers into the equation, we find

$W_1 = (0.075 kg)(9.81 m/s^2)(2.5 m)=1.84 J$

(b) The work the bird did to raise its own centre of mass from the ground to the branch is equal to the product between the bird's weight and the height of the branch:

$W_2 = mgh$

where $m=350 g=0.35 kg$ is the mass of the bird. Substituting numbers into the equation, we find

$W_2 = (0.35 kg)(9.81 m/s^2)(2.5 m)=8.58 J$

6 0

3 years ago