¿Es posible determinar el valos de las variables que intervienen en los diferentes movimientos en dos dimensiones?

A boy pulls a 28.0-kg box with a 230-N force at 35° above a horizontal surface. If the coefficient of kinetic friction between t

ddd [48]

Answer:

1977.696 J

Explanation:

Given;

Weight of the box = 28.0 kg

Force applied by the boy = 230 N

angle between the horizontal and the force = 35°

Therefore,

the horizontal component of the force = 230 × cosθ

= 230 × cos 35°

= 188.405 N

Coefficient of kinetic friction, μ = 0.24

Force by friction, f = μN

here,

N = Normal force = Mass × acceleration due to gravity

or

N = 28 × 9.81 = 274.68 N

therefore,

f = 0.24 × 274.68

or

f = 65.9232 N

Now,

work done by the boy, W₁ = 188.405 N × Displacement

= 188.405 N × 30

= 5652.15 J

and,

the

work done by the friction, W₂ = - 65.9232 N × Displacement

= - 65.9232 N × 30 m

= - 1977.696 J

[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]

8 0

3 years ago

According to the big bang theory, the universe is continually getting

oee [108]

The answer on Edge would be (A.)= Larger and Cooler ! I'm doing the same thing as y'all. Good luck everyone.

7 0

3 years ago

Read 2 more answers

In a circus act, a 70 kg clown is shot from a cannon with an initial velocity of 17 m/s at some unknown angle above the horizont

EleoNora [17]

Answer:

$K_{2}=7302.4J$

Explanation:

Given the initial velocity of the clown, his mass and final height we can calculate the final kinetic energy using the <em><u>conservation of total mechanical energy</u></em>

$K_{1}+U_{1}=K_{2}+U_{2}$

$K_{2}=K_{1}+U_{1}-U_{2}$

$K_{2}=\frac{1}{2}mv_{1}^{2}+mgh_{1}-mgh_{2}$

Since $h_{1}=0$

$K_{2}=\frac{1}{2}mv_{1}^{2}-mgh_{2}$

$K_{2}=\frac{1}{2}(70kg)(17m/s)^2-(70kg)(9.8m/s^2)(4.1m)=7302.4J$

$K_{2}=7302.4J$

7 0

3 years ago

Estimate the length of the neptunian year using the fact that the earth is 1.50Ã108km from the sun on average.

Ksenya-84 [330]

Answer:

Explanation:

We know that,

Neptune is 4.5×10^9 km from the sun

And given that,

Earth is 1.5×10^8km from sun

Then,

Let P be the orbital period and

Let a be the semi-major axis

Using Keplers third law

Then, the relation between the orbital period and the semi major axis is

P² ∝ a³

Then,

P² = ka³

P²/a³ = k

So,

P(earth)²/a(earth)³ = P(neptune)² / a(neptune)³

Period of earth P(earth) =1year

Semi major axis of earth is

a(earth) = 1.5×10^8km

The semi major axis of Neptune is

a (Neptune) = 4.5×10^9km

So,

P(E)²/a(E)³ = P(N)² / a(N)³

1² / (1.5×10^8)³ = P(N)² / (4.5×10^9)³

Cross multiply

P(N)² = (4.5×10^9)³ / (1.5×10^8)³

P(N)² = 27000

P(N) =√27000

P(N) = 164.32years

The period of Neptune is 164.32years

6 0

4 years ago

Name the four phases of the lunar cycle

Serggg [28]

Answer:

Hello there!

This is a picture that helped me learn the phases in an easy way. I hope it helps you too.

Explanation:

4 0

3 years ago