According to your textbook, the closest match for “nervous breakdown” would be

Which property of potential energy distinguishes it from kinetic energy

Bas_tet [7]

Answer:

Shape and position

Explanation:

Hope this helps! :)

7 0

2 years ago

A ball is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the wa

vovangra [49]

Answer:

a) 48.5 ft/s

b) 36.5 ft

c) -80.3 ft/s

Explanation:

a)

The equation of motion of the ball is :

y(t) = -16.1 ft/s^2 * t^2 + Vo*t

Where Vo is the initial velocity

If y(5s) = - 160 ft:

-160 ft = -16.1 ft/s^2 * (5 s)^2 + Vo*(5s)

Solving for Vo

Vo = (16.1*25- 160) ft / 5s = 48.5 ft/s

b)

To answer this question we must first know when the velocity became zero, at this time is when the ball was at its highest point.

v(t) = -32.2 ft/s^2 * t + Vo

t = Vo/32.2ft/s^2 = 1.5 s

And now, the highest point which the ball reached is given by:

y(1.5s) = -16.1 ft/s^2 * (1.5)^2 + Vo*(1.5s)

y(1.5s) = 36.52 ft

c)

We now need the time at which y(t') = -64 ft

-64 = -16.1*t'^2 + 48.5*t'

By means of the quadratic formula, we find that

t' = 4.00498 s ≈ 4 s

And the velocity at t = 4s is:

v(4s) = -32.2 ft/s^2 * 4s +48.5 ft/s = -80.3 ft/s

3 0

3 years ago

Acceleration of a free-falling object in a frictionless environment increases as a function of time.

Digiron [165]

A free-falling object is an object moving under the effect of gravitational forces alone

The correct option to select for the True or False question is False

The reason the above selected option is correct is as follows:

According to Newton's second law of motion, we have;

Force = Mass × Acceleration

The force of gravity is $F_{g} =G \cdot \dfrac{M \cdot m}{r^{2}}$

Where;

$G \cdot \dfrac{M }{r^{2}} = Acceleration \ due \ to \ gravity , \ g \approx 9.81 m/s^2$

m = The mass of the object

∴ The force acting on an object in free fall, $F_g$ = m × g

Therefore the acceleration of an object in free fall is the constant acceleration due to gravity, and it therefore, does not change with time

The correct option for the question, acceleration of a free-falling object in a frictionless environment increases as a function of time is <u>False</u>

<u></u>

Learn more about object in free fall here:

brainly.com/question/13712424

brainly.com/question/11698474

6 0

2 years ago

A 0.150 kg baseball has 118 j of KE. how fast is the ball moving?(unit=m/s)

MrRissso [65]

Answer:

Explanation 118 = (1/2) * 0.15 * v² 118 = 0.075 * v² v² = 1573.33 m/s ... since KE = m/2*V^2 , then : V = √2KE/m = √20*118/1.5 = 39.67 m//sec ( 142.8 km/h ; 88.75 mph).:

4 0

3 years ago

A hockey puck with a mass of 0.16 kg travels at a velocity of 40 m/s toward a goalkeeper. The goalkeeper has a mass of 120 kg an

Ainat [17]

Answer:

Explanation:

Momentum is the product of mass of a body and its velocity.

Given the mass of the puck m1 = 0.16kg

velocity of the puck v1 = 40m/s

Given the mass of the goalkeeper m2 = 120kg

velocity of the goalkeeper v2= 0m/s (goal keeper at rest)

The total momentum of the goalkeeper and puck after the puck is caught by the goalkeeper is expressed as:

m1v1 + m2v2 (their momentum will be added since they collide)

= 0.16(40) + 120(0)

= 0.16(40) + 0

= 6.4kgm/s

Let us calculate their common velocity using the conservation of momentum formula;

m1u1 + m2u2 = (m1+m2)v

6.4 = (0.16+120)v

6.4 = 120.16v

v = 6.4/120.16

v = 0.053m/s

Hence after collision, both objects move at a velocity of 0.053m/s

Momentum of the puck after collision = m1v

Momentum of the puck after collision = 0.16*0.053m/s

Momentum of the puck after collision = 0.0085kgm/s

Momentum of the keeper after collision = m2v

Momentum of the keeper after collision = 120*0.053m/s

Momentum of the keeper after collision = 6.36kgm/s

From the calculation above, it can be seen that the keeper has the greater momentum after the puck was caught since the momentum of the keeper after collision is greater than that of the puck

4 0

2 years ago