A skydiver uses a parachute to:

A passenger in a train accelerating smoothly away from a station observes that a child’s yo-yo hanging by its string from a lugg

olchik [2.2K]

Answer:

1.73 m/s²

3.0 cm

Explanation:

Draw a free body diagram of the yo-yo. There are two forces: weight force mg pulling down, and tension force T pulling up 10° from the vertical.

Sum of forces in the y direction:

∑F = ma

T cos 10° − mg = 0

T cos 10° = mg

T = mg / cos 10°

Sum of forces in the x direction:

∑F = ma

T sin 10° = ma

mg tan 10° = ma

g tan 10° = a

a = 1.73 m/s²

Draw a free body diagram of the sphere. There are two forces: weight force mg pulling down, and air resistance D pushing up. At terminal velocity, the acceleration is 0.

Sum of forces in the y direction:

∑F = ma

D − mg = 0

D = mg

½ ρₐ v² C A = ρᵢ V g

½ ρₐ v² C (πr²) = ρᵢ (4/3 πr³) g

3 ρₐ v² C = 8 ρᵢ r g

r = 3 ρₐ v² C / (8 ρᵢ g)

r = 3 (1.3 kg/m³) (100 m/s)² (0.47) / (8 (7874 kg/m³) (9.8 m/s²))

r = 0.030 m

r = 3.0 cm

3 0

3 years ago

9. A 5.0 kg block on an inclined plane is acted upon by a horizontal force of 100 N shown in the figure below. The coefficient o

Helga [31]

Answer:

A: The acceleration is 7.7 m/s up the inclined plane.

B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane

Explanation:

Let us work with variables and set

$m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.$

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

$F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).$

Now the force of friction is $F_s=\mu*N,$ where $N$ is the normal force and from the diagram it is $F_y=mg*cos(\theta).$

Thus $F_s=\mu*N=\mu*mg*cos(\theta).$

Therefore,

$F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).$

Substituting the value for $F_H,m,\mu, and \:\theta$ we get:

$F_{tot}= -38.63N.$

Now acceleration is simply

$a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.$

The negative sign indicates that the acceleration is directed up the incline.

Part B:

$d=\frac{1}{2} at^2$

Which can be rearranged to solve for t:

$t=\sqrt{\frac{2d}{a} }$

Substitute the value of $d=0.50m$ and $a=7.7m/s$ and we get:

$t=0.36s.$

which is our answer.

Notice that in using the formula to calculate time we used the positive value of $a$ , because for this formula absolute value is needed.

5 0

4 years ago

Someone help please by providing work and answers please :)

Nastasia [14]

First we gotta use an equation of motion:

$d = ut + \frac{1}{2} a {t}^{2}$

Our vertical distance d= 100 m, initial vertical speed u = 0 m/s (because velocity is fully horizontal), and vertical acceleration a = 9.8 m/s2 because of gravity. Let's plug it all in!

$100 = 0 + \frac{1}{2} (9.8) {t}^{2}$

Now we just need to solve for t:

${t}^{2} = \frac{2(100)}{9.8} \\ \\ t = \sqrt{\frac{2(100)}{9.8}}$

Hit the calculators, and you'll get 4.5 seconds!

5 0

3 years ago

a radio antenna broadcasts a 1.0 MHz radio wav e with 21 kW of power. Assume that the radiation is emitted uniformly in all dire

My name is Ann [436]

Answer:

$I=2.67\times 10^{-6}\ W/m$

Explanation:

Given that,

Frequency of a radio antenna is 1 MHz

Power, P = 21 kW

We need to find the the waves intensity 25 km from the antenna . The object emits intenisty evenly in all direction. It can be given by :

$I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{21\times 10^3}{4\pi (25000)^2}\\\\I=2.67\times 10^{-6}\ W/m$

So, the wave intensity 25 km from the antenna is $2.67\times 10^{-6}\ W/m$ .

5 0

3 years ago

Mass can be considered concentrated at the center of mass for rotational motion.Select one:TrueFalse

frez [133]

iIn this case the mass of a body cannot be considered to be concentrated at the centre of mass of the body for the purpose of computing the rotational motion

Therefore the answer is False

6 0

1 year ago