Boys A and B, are pulling a heavy cabinet at the same time with 5 N of force

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has

irga5000 [103]

Answer:

The total linear acceleration is approximately 0.246 meters per square second.

Explanation:

The total linear acceleration ( $a$ ) consist in two components, radial ( $a_{r}$ ) and tangential ( $a_{t}$ ), in meters per square second:

$a_{r} = \omega^{2}\cdot r$ (1)

$a_{t} = \alpha \cdot r$ (2)

Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$ (3)

Where:

$r$ - Radius of the wheel, in meters.

$\omega$ - Angular speed, in radians per second.

$\alpha$ - Angular acceleration, in radians per square second.

Given that wheel accelerates uniformly, we use the following kinematic equation:

$\omega = \omega_{o}+ \alpha\cdot t$ (4)

Where:

$\omega_{o}$ - Initial angular speed, in radians per second.

$t$ - Time, in seconds.

If we know that $r = 0.1\,m$ , $\alpha = 2\,\frac{rad}{s^{2}}$ , $\omega_{o} = 0\,\frac{rad}{s}$ and $t = 0.60\,s$ , then the total linear acceleration is:

$\omega = \omega_{o}+ \alpha\cdot t$

$\omega = 1.2\,\frac{rad}{s}$

$a_{r} = \omega^{2}\cdot r$

$a_{r} = 0.144\,\frac{m}{s^{2}}$

$a_{t} = \alpha \cdot r$

$a_{t} = 0.2\,\frac{m}{s^{2}}$

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$

$a \approx 0.246\,\frac{m}{s^{2}}$

The total linear acceleration is approximately 0.246 meters per square second.

3 0

2 years ago

A planned high-speed train between Houston and Dallas will travel a distance of 386 kilometers in 5.40 × 10^3 seconds. What is t

Mazyrski [523]

¡Hellow!

For this problem, first, lets convert the seconds in hours:

5,4x10³ $\rightarrow$ 5400

h = sec / 3600

h = 5400 s / 3600

h = 1,5

Let's recabe information:

d (Distance) = 386 km

t (Time) = 1,5 h

v (Velocity) = ?

For calculate velocity, let's applicate formula:

$\boxed{\boxed{\textbf{d = v * t} } }$

Reeplace according we information:

386 km = v * 1,5 h

v = 386 km / 1,5 h

v = 257,33 km/h

The velocity of the train is of 257,33 kilometers for hour.

Extra:

For convert km/h to m/s, we divide the velocity of km/h for 3,6:

m/s = km/h / 3,6

Let's reeplace:

m/s = 257,33 km/h / 3,6

m/s = 71,48

¿Good Luck?

7 0

3 years ago

The gravity of earth depends on the _____ from its ____ . as this _____increase, the magnitude of the gravity _____.

Anna [14]

The gravity of earth depends on the magnetism from its core. as this magnetism increases, the magnitude of the gravity increases.

3 0

3 years ago

What is the magnitude of the torque about his shoulder due to the weight of the ball and his arm if he holds his arm straight ou

Lubov Fominskaja [6]

Answer:

The torque about his shoulder is 34.3Nm.

The solution approach assumes that the weight of the boy's arm acts at the center of the boy's arm length 35cm from the shoulder.

Explanation:

The solution to the problem can be found in the attachment below.

6 0

3 years ago

Determine the velocity required for a moving object 2.00 x 10^4 m above the surface of Mars to escape from Mars's gravity. The m

Ket [755]

Answer:

Therefore the escape velocity from Mar's gravity is $15.88 \times 10^4$ m/s.

Explanation:

Escape velocity: Escape velocity is a the minimum velocity that a object needs to escape from the gravitational field of massive body.

$V_{escape}=\sqrt{\frac{2GM}{R}}$

$V_{escape}=$ Escape velocity

G=Universal gravitational constant = 6.673×10⁻¹¹N m²/Kg²

M= mass of Mars = 6.42×10²³ kg

R = Radius of the Mars = 3.40×10³m

The escape velocity does not depend on the velocity of a object.

$V_{escape}=\sqrt{\frac{2\times6.673\times 10^{-11}\times 6.42\times 10^{23}}{3.40\times10^3}}$

$=15.88 \times 10^4$ m/s

Therefore the escape velocity from Mar's gravity is $15.88 \times 10^4$ m/s.

3 0

2 years ago