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melomori [17]
3 years ago
15

Which of the following is not an assumption of stage theories?

Physics
2 answers:
olya-2409 [2.1K]3 years ago
8 0

Answer:

B. Stages are continuous and gradual.

Explanation:

Murrr4er [49]3 years ago
5 0

Answer:

B

Explanation:

You might be interested in
¿Cuál es la aceleración centrípeta de un móvil que recorre una
Lady bird [3.3K]

Answer:

a)  a = 4.57 m/s², b)  a = 6.48 m / s² , c)  a = 1.42 m / s²,d)   r = 82.3 m

 

Explanation:

The centripetal acceleration is the acceleration responsible for the change of direction of the acceleration vector and occurs in circular movements, the expression is

           a = v² / r

let's apply this precaution to our cases

a) let's calculate

          a = 8²/14

         a = 4.57 m/s²

b) an automobile at v = 65 km / h (1000 m / 1km) (1 h / 3600 s) =18,055 m/s

let's reduce feet to meters

          1 ft = 0.3048 m

           r = 165 ft (0.3048 m / 1 ft) = 50.292 m

          a = 18,055 2 / 50,292

           a = 6.48 m / s²

c) we calculate

          a = 1.25²2 / 1.1

          a = 1.42 m / s²

d) we look for the radius

          a = v² / r

          r = v² / a

we reduce

          v = 80 km / h (1000 m / 1km) (1h / 3600s) = 22.22  ms

          r = 22.22²/6

          r = 82.3 m

e) the cenripeta acceleration is used to take the curves on the highway,

    Used in centrifuges to separate compounds

       It is used in the games of the park of atraccio

     Used in CD players and computer hard drives

5 0
3 years ago
The equation of a transverse wave traveling along a string is given by y=0.3sin (0.5x-50t) find the maximum displacement of the
uysha [10]

Answer:

0.3cm

Explanation:

Y = 0.3 sin(0.5x - 50t) compare with,

y = A sin(kx - wt)

A = 0.3m

5 0
3 years ago
A 3.9 g dart is fired into a block of wood with a mass of 24.6 g. The wood block is initially at rest on a 1.5 m tall post. Afte
Galina-37 [17]

Answer:

46.48m/s

Explanation:

The problem is a combination of the principle of conservation of linear momentum and projectile motion.

The principle of conservation of linear momentum states that in a closed system, the total momentum of colliding bodies before impact is equal to the total momentum after impact. The masses stated in the problem experienced an inelastic collision. In an inelastic collision, the bodies involved stick together after the collision and move with a common velocity.

For two bodies of masses m_1 and m_2 moving with velocities u_1 and u_2 before impact, if they experience inelastic collision, the conservation of their momenta is as stated in equation (1);

m_1u_1+m_2u_2=(m_1+m_2)v..................(1)

were v is their common velocity after impact. If the second mass m_2 was at rest before the impact, then its initial velocity u_2=0m/s. therefore m_2u_2=0. Equation (1) then becomes;

m_1u_1=(m_1+m_2)v..............(2)

In the problem stated, the second mass taken as the mass of the wooden block was at rest before the impact and the collision was inelastic since both the wood and the dart stuck together and moved with a common velocity after the impact. Therefore we can use equation (2) for the problem.

Given;

m_1=3.9g=0.0039kg\\u_1=?\\m_2=24.6g=0.0246kg\\v=?

Substituting these values into (2), we get the following;

0.0039*u_1=(0.0039+0.024)v\\0.0039u_1=0.0285v.........(3)

Their common v velocity after impact now makes both the wooden block and the dart (as a single body) to fall vertically through a height h of 1.5m over a range R of 3.5m as stated by the problem; hence by the principle of projectile motion for a body projected horizontally, the following relationship holds;

R= vt............(4)

were t is the time taken to fall through the height h. To obtain t we use the second equation of free fall under gravity;

h=\frac{1}{2}gt^2...........(5)

were g is acceleration due to gravity taken as 9.8m/s^2. Therefore;

1.5=\frac{1}{2}*9.8*t^2\\1.5=4.9t^2\\t^2=\frac{1.5}{4.9}=0.306\\t=\sqrt{0.306} =0.55s

We then substitute R and t into equation (4) to obtain v.

3.5=v*0.55\\v=\frac{3.5}{0.55}\\v=6.36m/s

We now further substitute this value of v into (3) to obtain u_1;

u_1=\frac{0.0285v}{0.0039}\\\\u_1=\frac{0.0285*6.36}{0.0039}\\\\u_1=\frac{0.18126}{0.0039}\\\\u_1=46.48m/s

4 0
4 years ago
The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for 7.00 s, at which time it
Keith_Richards [23]

Answer:

The no. of revolutions does the tub turn while it is in motion is = 52.51 revolutions

Explanation:

Given data

\omega_1 = 0

\omega_2 = 5 \frac{rev}{sec}

Time taken = 7 sec

(1). The angular acceleration is given by

\alpha = \frac{d \omega}{dt}

\alpha  = \frac{5}{7}

\alpha  = 0.714 \frac{rev}{s^{2} }

We know that from the equation of motion

\omega_2^{2} =  \omega_1^{2} + 2 \alpha \theta_1

5^{2} = 0 + 2 (0.714) \theta_1

\theta_1 = 17.5 \ rev  -------- (1)

(2). The angular acceleration is given by

\alpha = \frac{d \omega}{dt}

\alpha = - \frac{5}{14}

\alpha = - 0.357 \frac{rev}{s^{2} }

We know that from the equation of motion

\omega_2^{2} =  \omega_1^{2} + 2 \alpha \theta_2

0^{2} = 5^{2} + 2 (-0.357) \theta_2

\theta_2 = 35.01 rev  -------  (2)

Total no of revolution made by the machine is

\theta = \theta_1 + \theta_2

\theta = 17.5 + 35.01

\theta = 52.51 rev

Therefore the no. of revolutions does the tub turn while it is in motion is = 52.51  rev

8 0
3 years ago
You are determining the mass of an object using a triple beam balance. When the pointer is lined up with the zero mark, the ride
astra-53 [7]
We add the values on each of the riders:
300 + 20 + 8
= 328 grams
3 0
4 years ago
Read 2 more answers
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