1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Ivahew [28]
3 years ago
8

Se lanza verticalmente hacia arriba un cuerpo que para por un punto A con una rapidez de 54 m/s y por otro punto B situado mas a

rriba con 24 m/s.calcular :a) tiempo en ir desde A hasta B;b )la altura vertical entre dichos puntos.
​
Physics
1 answer:
kvv77 [185]3 years ago
5 0

Answer:

a) 3,06 seg

b) 119,39 m

Explanation:

<u>Lanzamiento Vertical </u>

Cuando un cuerpo se lanza verticalmente hacia arriba en el vacío, la única fuerza actuante es el peso. Si asumimos la dirección negativa hacia abajo, las fórmulas necesarias son

v_f=v_o+gt

y=y_o+\frac{gt^2}{2}

v_f^2=v_o^2+2gy

Siendo v_f la velocidad final, y la altura del objeto, g=-9.8\ m/seg^2 , v_o la velocidad inicial y t el tiempo

a)

Sabemos que el cuerpo pasa por un punto A a v_o=54\ m/s, y por otro punto B más arriba, a v_f=24\ m/s. El cuerpo está subiendo, pues pierde velocidad. Sabiendo las dos velocidades, podemos calcular el tiempo que toma en ir de A a B

\displaystyle t=\frac{v_f-v_o}{g}

\displaystyle t=\frac{24-54}{-9.8}

\displaystyle t=\frac{-30}{-9.8}=3,06\ seg

b)

Conociendo las velocidades de los extremos, se encuentra la distancia vertical que recorre durante ese intervalo

v_f^2=v_o^2+2gy

\displaystyle 24^2=54^2+2(-9.8)y

\displaystyle 24^2-54^2=-19.6y

Despejando y

\displaystyle y=\frac{576-2916}{-19.6}

y=119,39\ m

You might be interested in
Irregular galaxies are very large. True or False?
xenn [34]
It's false i hope this helps :)

5 0
3 years ago
Read 2 more answers
the refractive index of cooking oil is 1.47, and the refractive index of water is 1.33. A thick layer of cooking oil is floating
VARVARA [1.3K]
When light travels from a medium with higher refractive index to a medium with lower refractive index, there is a critical angle after which all the light is reflected (so, there is no refraction).

The value of this critical angle can be derived by Snell's law, and it is equal to
\theta_C = \arcsin ( \frac{n_2}{n_1} )
where n2 is the refractive index of the second medium and n1 is the refractive index of the first medium.

In our problem, n1=1.47 and n2=1.33, so the critical angle is
\theta_C = \arcsin( \frac{1.33}{1.47} )=\arcsin (0.91)=65^{\circ}
4 0
3 years ago
The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff
stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet  m = 45 kg

coefficient of static friction μ =  0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

\sum f_y =0

The upward force and the downward force is :

N_A+N_B = mg

where;

\mathbf { N_A  \ and  \ N_B} are the normal contact force at center point A and B respectively .

N_A+N_B = 45*9.8

N_A+N_B = 441    ------- equation (1)

Considering the forces on the horizontal axis:

\sum f_x = 0

F_A +F_B  = P

where ;

\mathbf{ F_A \ and \ F_B } are the static friction at center point A and B respectively.

which can be written also as:

\mu_s N_A + \mu_s N_B  = P

\mu_s( N_A +  N_B)  = P

replacing our value from equation (1)

P = 0.30 ( 441)    

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for  the cabinet is not to tip over is calculated by determining the limiting condition  of the unbalanced torque whose effect is canceled by the normal reaction at N_A and it is shifted to N_B:  

Then:

\sum M _B = 0

P*h = mg*0.24

h =\frac{45*9.8*0.24}{132.3}

h = 0.8 m

Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

6 0
3 years ago
An imaginary line perpendicular to a reflecting surface is called ____refrac_____.
umka2103 [35]
Not totally sure but i would say a normal? its not refraction or incidence if its perpendicular and i dont think its a mirror if its an imaginary line so yeah normal (normals are always perpendicular to their surface too i think so)
4 0
4 years ago
Titus drives his jetski a distance of 1000 meters in 7.045 seconds. How fast was he moving in meters per second? How fast was he
Ksenya-84 [330]

Answer:

a) v = 141.9 m/s

b) v = 317.4 miles/h

Explanation:

a) How fast was he moving in meters per second?

v = \frac{d}{t} = \frac{1000 m}{7.045 s} = 141.9 m/s

Hence, the jet ski is moving at 141.9 meters per second.

b) How fast was he moving in miles per hour?

v = \frac{d}{t} = \frac{1000 m}{7.045 s} = 141.9 m/s*\frac{3600 s}{1 h}*\frac{1 mile}{1609.34 m} = 317.4 miles/h      

Therefore, the jet ski is moving at 317.4 miles per hour.

I hope it helps you!

5 0
3 years ago
Other questions:
  • At a college homecoming 20 students lifted a sports car. Each student exerted 450N of upward force on the car. a) What is the ma
    10·1 answer
  • A student drew this diagram of Woese’s modern system of classification.
    12·2 answers
  • If the absolute value of the price elasticity of demand for DVD movies is 0.8 then the elasticity of demand of the DVD for the m
    8·1 answer
  • If a car has an initial velocity of 20 m/s and accelerates at 2.0 m/s2 for 100 m, what is its final velocity?
    12·1 answer
  • An object is dropped from a height H. During the final second of its fall, it traverses a distance of 53.2 m. What was H? An obj
    11·1 answer
  • A driver moves with initial velocity of 40m/s accelerates uniformly at a rate of 12m/s2.  It attains a velocity 52m/s. calculate
    12·1 answer
  • Which way will the people on the right move? Why? Plz answer
    5·1 answer
  • A body of mass m1 = 1.5 kg moving along a directed axis in the positive sense with a velocity
    7·1 answer
  • If the ball has a positive electric charge, what should the charge of the coilgun be to push the ball away?
    12·1 answer
  • What method could I use to test this hypothesis? If the mass and the volume of and object are known, then its density can be cal
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!