If an acid is combined with a base of equal strength, the result will most likely be

A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra

kkurt [141]

Answer:

$V_{ft}= 317 cm/s$

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

$P_i = P_f$

Where:

$P_i=M_cV_{ic} + M_tV_{it}$

$P_f = M_cV_{fc} + M_tV_{ft}$

Now:

$M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}$

Where $M_c$ is the mass of the car, $V_{ic}$ is the initial velocity of the car, $M_t$ is the mass of train, $V_{fc}$ is the final velocity of the car and $V_{ft}$ is the final velocity of the train.

Replacing data:

$(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}$

Solving for $V_{ft}$ :

$V_{ft}= 3.17 m/s$

Changed to cm/s, we get:

$V_{ft}= 3.17*100 = 317 cm/s$

b) The kinetic energy K is calculated as:

K = $\frac{1}{2}MV^2$

where M is the mass and V is the velocity.

So, the initial K is:

$K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2$

$K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2$

$K_i = 22.06 J$

And the final K is:

$K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = 19.61 J$

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0

3 years ago

Your friend tells you that a shiny element has to be a metal. Do you agreed? Explain your reasoning

Svetradugi [14.3K]

no i dont agree, because a glass bowl is shiny but its not a metal. just because some metals are shiny doesnt make all metals shiny

3 0

2 years ago

Read 2 more answers

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

2 years ago

Use newton's law to explain the vertical acceleration of a projectile

NISA [10]

Answer:

Explained below

Explanation:

Newton's first law of motion: This law states that an object will remain at rest or continue in constant motion except it's acted upon by an external force. In projectile motion, the horizontal component of velocity will remain unchanged because we ignore air resistance since no force is acting in that horizontal direction.

Newton's second law of motion: This law states that force is the product of mass and acceleration. In projectile the force acts downwards, thus f = mg.

But g = a since internal forces will cancel out.

Thus, F = ma

5 0

3 years ago

What constant acceleration (in ft/s2) is required to increase the speed of a car from 27 mi/h to 50 mi/h in 5 seconds

SIZIF [17.4K]

Answer:

6.746 ft/s^2

Explanation:

v(t)=50

v(0)=27

t=5/3600 = 1/720 hours

v(t)-v(0)= a(t-0)

50-27= a(1/720)

a= 23*720= 16560 mi/h^2

16560mi/h^2 * 5280/3600^2 (ft/s^2) =6.746 ft/s^2

4 0

2 years ago