Ask question

Ask question

All categories

2 years ago

9

Un automovil con la velocidad de 5 m/s acelera durante 12 s a 3 m/s2,¿cual es la velocis

1 answer:

harina [27]2 years ago

4 0

Answer:

60 miles

Explanation:

You might be interested in

Which is an example of qualitative data?

LiRa [457]

The answer is C)the rating that the golfers give Callaway clubs

6 0

1 year ago

Where are truss bridges located?

guapka [62]

Basic truss bridge types found in North Carolina (source: HAER) A truss bridge can be characterized by the location of its traffic deck. At a pony truss, the travel surface passes along the bottom chords of trusses standing to either side that are not connected to each other at the top.

7 0

2 years ago

Read 2 more answers

Define mechanics and describe its three major divisions

Andrew [12]

<u>Mechanics</u> is the branch of physics which deals with the study of motion of material objects.

<u><em>Divisions</em></u>

There are three major division of mechanics

Statics

Kinematics

Dynamics.

4 0

2 years ago

A light bulb has a voltage of 36 and a current of 8 A. Calculate the resistance of the light bulb

timurjin [86]

R = U : I. U is in Voltage and I is in Ampère. That gives you R = 36 : 8 = 4,5 Ohm

4 0

2 years ago

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:

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

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

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

$K_f = 19.61 J$

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0

2 years ago