A water-balloon launcher with mass 5 kg fires a 1 kg balloon with a velocity of

A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the sur

emmasim [6.3K]

• Net vertical force on the block:

∑ F = n - w = 0

(n = magnitude of normal force, w = weight)

n = w = m g

(m = mass, g = 9.8 m/s²)

n = (4 kg) (9.8 m/s²) = 39.2 N

• Net horizontal force:

∑ F = -f = m a

(f = mag. of friction, a = acceleration)

We have f = µ n = 0.5 (39.2 N) = 19.6 N, so

-19.6 N = (4 kg) a

a = -4.9 m/s²

With this acceleration, the block comes to a rest from an initial speed of 5 m/s, so that it travels a distance ∆x in this time such that

0² - (5 m/s)² = 2 (-4.9 m/s²) ∆x

∆x = (25 m²/s²) / (9.8 m/s²) ≈ 2.55 m ≈ 2.6 m

8 0

3 years ago

The oceans level is currently rising at about 2 mm per year at this rate in how many years will sea level be 3 m higher than now

lesantik [10]

(3 m) / (2 mm/yr) = (3,000mm)/(2mm/yr) = 1,500 yrs.
This is the time required to age approx 75 generations of the best wine.

5 0

4 years ago

Which of the following objects has more kinetic energy than it does potential energy? Select the correct answer below:______1.a

Lilit [14]

Answer:

The pendulum of the clock.

Explanation:

Hi there!

The kinetic energy is the energy associated with the velocity of the object. The potential energy is the energy associated with the position of the object. In the objects listed in the question, only one object is moving: the pendulum of the clock (assuming that the clock is functioning). If the clock functions, the pendulum is moving when it is at the lowest point of its arc of motion and with maximum velocity. All potential energy that the pendulum stored when it reached the highest height, is transformed into kinetic energy at the lowest point. Thus, at that point, the object has more kinetic energy than potential energy.

6 0

3 years ago

A 47.2 kg girl is standing on a 177 kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,

Softa [21]

Answer:

v_g,i = 1.208 m/s

Explanation:

We are given;

Mass of girl; m_g = 47.2 kg

Mass of plank; m_p = 177 kg

Let the velocity of girl to ice be v_g,i

Let the velocity of plank to ice be v_p,i

Since the velocity of the girl is 1.53 m/s relative to the plank, then;

v_g,i + v_p,i = 1.53

From conservation of momentum;

m_g × v_g,i = m_p × v_p,i

Thus;

47.2(v_g,i) = 177(v_p,i)

Dividing both sides by 47.2 gives;

v_g,i = 3.75(v_p,i)

v_pi = (v_g,i)/3.75

Thus, from v_g,i + v_p,i = 1.53, we have;

v_g,i + ((v_g,i)/3.75) = 1.53

v_g,i(1 + 1/3.75) = 1.53

1.267v_g,i = 1.53

v_g,i = 1.53/1.267

v_g,i = 1.208 m/s

5 0

3 years ago

A cannon fires a shell straight upward; 2.3 s after it is launched, the shell is moving upward with a speed of 17 m/s. Assuming

PtichkaEL [24]

Answer:

The speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.

Explanation:

Let u is the initial speed of the launch. Using first equation of motion as :

$u=v-at$

a=-g

$u=v+gt\\\\u=17+9.8\times 2.3\\\\u=39.54\ m/s$

The velocity of the shell at launch and 5.4 s after the launch is given by :

$v=u-gt\\\\v=39.54-9.8\times 5.4\\\\v=-13.38\ m/s$

So, the speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.

6 0

3 years ago