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3 years ago

14

Qué fuerza neta actúa sobre una caja que se desliza cuando ejerces sobre ella una fuerza de 110 N y la fricción entre la caja y

el suelo es 100 N?

1 answer:

neonofarm [45]3 years ago

8 0

Answer:

if this is translated ill answer it

Explanation:

sorry i cant reed this language

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Throughout the problem, take the speed of sound in air to be 343 .

anyanavicka [17]

Answer:

A pipe has a length of 1.15 m.

a) Determine the frequency of the first harmonic if the pipe is open at each end. The velocity of sound in air is 343 m/s. Answer in units of Hz

b) What is the frequency of the first harmonic if the pipe is closed at one end?

Answer in units of Hz

Explanation:

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3 years ago

Verizon [17]

Q 24: Newton's first law of motion states that anything that stands still will stay still unless unbalanced force is acted upon that object. Newton's second law of motion states that the velocity of an object changes when subjected to external Force. Newton's third law of motion says that for every action there is an equal and opposite reaction. when bumper cars crash into each other one car goes one way and the other goes the other way.

Q 25: if the bumper cars doubled their Mass the motion would be halfed but if the net force is doubled the motion will double.

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3 years ago

Two balls, each with a mass of 0.5 kg, collide on a pool table. Is the law of conservation of momentum satisfied in the collisio

Dmitriy789 [7]

During the collision between two balls on the pool table there is no external force along the line of collision between them

Since there is no external force on it so here we can say

$F = 0 = \frac{\Delta P}{\Delta t}$

here we have

$\Delta P = 0$

so we can say

$P_i = P_f$

since there is no external force so we can say during the collision the momentum of two balls will remain conserved

7 0

3 years ago

Read 2 more answers

A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the

irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

$x=V_{o} cos \theta t$ (1)

$y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2}$ (2)

Where:

$x$ is the horizontal distance between the cannon and the ball

$V_{o}=23 m/s$ is the cannonball initial velocity

$\theta=0\°$ since the cannonball was shoot horizontally

$t$ is the time

$y=0$ is the final height of the cannonball

$y_{o}=1.96 m$ is the initial height of the cannonball

$g=9.8 m/s^{2}$ is the acceleration due gravity

Isolating $t$ from (2):

$t=\sqrt{-\frac{2(y-y_{o})}{g}}$ (3)

$t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}}$ (4)

$t=0.63 s$ (5)

Substituting (5) in (1):

$x=(23 m/s) cos(0\°) 0.63 s$ (6)

Finally:

$x=14.49 m$

5 0

3 years ago

Vector A⃗ points in the negative y direction and has a magnitude of 5 km. Vector B⃗ has a magnitude of 15 km and points in the p

Alexxandr [17]

Answer:

magnitude of A − B = 15.81 km

Explanation:

Vector A points in the negative y-direction and has a magnitude of 5 km. Vector B points in the positive x-direction and has a magnitude of 15 km.

According to Cartesian coordinate system, the resultant will start either from tail of A and ends at head of B and vice-versa.

A(0,-5)

B(15,0)

A - B = (-15 i - 5 j )

Magnitude of the vector is given by

|A - B| = $\sqrt{(-15)^{2}+(-5)^{2}}$

|A - B| = $\sqrt{250}$

|A - B| = 15.81 km

7 0

3 years ago