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marta [7]
3 years ago
12

Pls answerrrrr thisssss

Physics
1 answer:
Jobisdone [24]3 years ago
4 0

The amplitude of wave-c is 1 meter.

The speed of all of the waves is (12meters/2sec)= 6 m/s.

The period of wave-a is 1/2 second.

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Fluid of intelligent​
GarryVolchara [31]

"Fluid intelligence involves being able to think and reason abstractly and solve problems. This ability is considered independent of learning, experience, and education. Examples of the use of fluid intelligence include solving puzzles and coming up with problem-solving strategies."  

- Verywell Mind

8 0
2 years ago
A uniform solid disk of mass 5.00 kg and diameter 47.0 cm starts from rest and rolls without slipping down a 40.0 ∘ incline that
saveliy_v [14]

Answer:

The speed it reaches the bottom is

v=6.51m/s

Explanation:

Given: m=5.0kg, r=47cm\frac{1m}{100cm}=0.47m

Using the conservation of energy theorem

U_i=K_E+K_{ER}

m*g*h=\frac{1}{2}*m*v^2+\frac{1}{2}*I*w^2

v=r*w, I=\frac{1}{2}*m*r^2

m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2

m*g*h=\frac{3}{4}*m*r^2*w^2

g*h=\frac{3}{4}*r^2*w^2

Solve to w'

w^2=\frac{4*g*h}{3*r^2}

h=x*sin(30)=6.5m*sin(30)=3.25m

w=\sqrt{\frac{4*9.8m/s^2*3.25m}{3*(0.235m)^2}}

w=27.74rad/s

v=27.74rad/s*0.235m=6.51m/s

7 0
3 years ago
You shoot an arrow into the air. Two seconds later (2.00 s) the arrow has gone straight upward to a height of 35.0 m above its l
sdas [7]

This question can be solved by using the equations of motion.

a) The initial speed of the arrow is was "9.81 m/s".

b) It took the arrow "1.13 s" to reach a height of 17.5 m.

a)

We will use the second equation of motion to find out the initial speed of the arrow.

h= v_it + \frac{1}{2}gt^2\\

where,

vi = initial speed = ?

h = height = 35 m

t = time interval = 2 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

35\ m = (v_i)(2\ s)+\frac{1}{2}(9.81\ m/s^2)(2\ s)^2\\\\v_i(2\ s)=19.62\ m\\\\v_i = \frac{19.62\ m}{2\ s}

<u>vi =  9.81 m/s</u>

b)

To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

h= v_it + \frac{1}{2}gt^2\\

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 17.5 m

vi = initial speed = 9.81 m/s

t = time = ?

Therefore,

17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0

solving this quadratic equation using the quadratic formula, we get:

t = -3.13 s (OR) t = 1.13 s

Since time can not have a negative value.

Therefore,

<u>t = 1.13 s</u>

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

4 0
2 years ago
A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T
Nady [450]
The first collision because a greater amount of momentum must be taken and used in order to push the cart back, giving it a greater mass and impulse
6 0
3 years ago
At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco
Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

A_1*m=M*A_2

A_i =Area

M,m = Counts per second

Our radios are given by

r_1 = 11cm

R_2 = 20cm

m = 65cps

Therefore replacing we have that,

A_1*m=M*A_2

4\pi r_1^2*m = M * 4\pi R_2^2 M

r^2*m=MR^2

M = \frac{m*r^2}{R^2}

M = \frac{65*11^2}{20^2}

M = 19.6625cps

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

7 0
3 years ago
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