All categories

3 years ago

Un automóvil lleva una velocidad inicial de 20 km/h al norte y a los 4 segundos su velocidad es de 50 km/h

Physics

2 answers:

Nookie1986 [14]3 years ago

7 0

Answer:

Desplazamiento

Explanation:

oksian1 [2.3K]3 years ago

6 0

La unidad de aceleración es en metros por segundo al cuadrado m/s2.

Asi que, primero convertí los km/h a m/s:

50 km/h = 13.889 m/s
20 km/h = 5.556 m/s

Y los sustituís en la fórmula de aceleración:

(a = Δv / Δt). En la misma, “a” es la aceleración, “Av” es la variación de la velocidad y “At” es el tiempo que hay en esa variación.

a = (13.889 - 5.556) / 4 segs
a = 8.333 / 4
a = 2.08325 m/s2

Espero te funcione!

Marca mi respuesta como brainliest porfa!!!

Suerte :)

You might be interested in

What is your favorite thing to eat? is it healthy or unhealthy ?

denpristay [2]

Answer:

sometimes I like to eat unhealthy and healthy my healthy are fruits granola a grapes, strawberries, blueberries or make smoothie out fruits my unhealthy is pizza, streak fries, with everything on the side. so ways eaanation:

4 0

3 years ago

Which of the following is responsible for the overall integration of the autonomic nervous system (ANS)?

Dafna1 [17]

Answer:

D. hypothalamus

Explanation:

The autonomic nervous system (ANS) is a component of the peripheral nervous system that is in charge of the involuntary functions, this means it regulates the flight-or-fight response, heart and respiratory rate, sexual arousal, urination, etc. The hypothalamus is the integrator of these functions, it receives the information from the nervous system and responds to it.

I hope you find this information useful and interesting! Good luck!

3 0

3 years ago

A population _____ follows a period of

Nostrana [21]

A population decline follows a period of overshooting.

8 0

3 years ago

Read 2 more answers

3. Consider a locomotive and the rest of a freight train to be a single object. Suppose the locomotive is pulling the train up a

34kurt

Answer:

The friction force and the x component for the weight should be the reaction forces that are opposite and equal to the action force, which causes the locomotive to move up the hill if the velocity of the locomotive remains constant.

Explanation:

When the locomotive starts to pull the train up, appears two reaction forces opposed to the action force in the direction of the move.

The first one is due to the friction between the wheels and the ground, it will be the friction force (Fr):

Fr = μ*Pₓ =μmg*sin(φ)

where μ: friction dynamic coefficient, Pₓ: is the weight component in the x-axis, m: total mass = train's mass + locomotive's mass, g: gravity, and sin(φ): is the angle respect to the x-axis.

And the second one is the x component for the weight (Wₓ):

Wₓ = mg*cos(φ)

where cos(φ): is the angle respect to the y-axis.

These two forces should be the same as the action force, which causes the locomotive to move up the hill if the velocity of the locomotive remains constant.

3 0

3 years ago

While unrealistic, we will examine the forces on a leg when one falls from a height by approximating the leg as a uniform cylind

Leno4ka [110]

Answer:

Part A: 7.75 m/s

Part B: 2330.8 kN

Part C: 24.03 kN

Part D: 4.8 kN

Part E: $1.7\times 10^{9} Dyn/cm^{2}$

Part F: Option D

Bending his legs increases the time over which the ground applies force, thus decreasing the force applied by the ground.

Explanation:

Part A

From the fundamental kinematic equation

$v^{2}=u^{2}+2gh$ where v is the velocity of the man just before hitting the ground, g is acceleration due to gravity, u is initial velocity, h is the height.

Since the initial velocity is zero hence

$v^{2}=2gh$

$v=\sqrt 2gh$

Substituting 10 m/s2 for g and 3 m for h we obtain

$v=\sqrt 2\times 10\times 3 =\sqrt 60= 7.745967\approx 7.75 m/s$

Part B 

Force exerted by the leg is given by

F=PA where P is pressure, F is force, A is the cross-section of the bone

$A=\frac {\pi d^{2}}{4}$

Substituting 2.3 cm which is equivalent to 0.023m for d and $1.7\times10^{8} N/m2$ for P we obtain the force as

$F=PA=1.7\times10^{8}*\frac {\pi (0.023)^{2}}{4}= 2330818.276\approx 2330.8 kN$

Part C

The fundamental kinematic equation is part (a) can also be written as

$v^{2}=u^{2}+2a\triangle x$ and making a the subject then

$a=\frac {v^{2}-u^{2}}{2\triangle x}$ where a is acceleration and $\triangle x$ is the change in length

Substituting the value obtained in part a, 7.75 m/s for v, u is zero and 1cm which is equivalent to 0.01 m for $\triangle x$ then

$a=\frac {7.75^{2}-0^{2}}{2\times 0.01}= 3003.125 m/s^{2}$

Force exerted on the man is given by

$F=ma=80\times 3003.125= 240250 N\approx 24.03 kN$

Part D

The fundamental kinematic equation is part (a) can also be written as

$v^{2}=u^{2}+2a\triangle h$ and making a the subject then

$a=\frac {v^{2}-u^{2}}{2\triangle h}$ where a is acceleration and $\triangle h$ is the change in height

Also, force exerted on the man is given by $F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}$

Substituting 80 Kg for m, 50 cm which is equivalent to 0.5m for \triangle h and other values as used in part c

$F=ma=m\times \frac {v^{2}-u^{2}}{2\triangle h}=80\times \frac {7.75^{2}-0^{2}}{2\times 0.5}= 4805 N\approx 4.8 kN$

Part E

$P=1.7\times 10^{8}=1.7\times 10^{8}\times (\frac {10^{5} Dyn}{10^{4} cm^{2}}=1.7\times 10^{9} Dyn/cm^{2}$

Part F

Bending his legs increases the time over which the ground applies force, thus decreasing the force applied by the ground

7 0

2 years ago