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Murrr4er [49]
2 years ago
7

Dos bolas, de masas mA = 40 g y mB =60 g, esta?n suspendidas como se observa en la figura. La bola ma?s ligera se jala en un a?n

gulo de 60° con respecto a la vertical y se libera. A) ?Cua?l es la velocidad de la bola ma?s ligera antes del impacto? B) ?Cua?l es la velocidad de cada bola despue?s de la colisio?n ela?stica? C) ?Cua?l sera? la altura ma?xima de cada bola despue?s de la colisio?n ela?stica?
Physics
1 answer:
Ymorist [56]2 years ago
3 0
MA=40 or something I really don’t know
You might be interested in
Layers of Earth's Atmosphere
Tamiku [17]

Answer:

going to work my way down

Troposphere

contains weather

contains life forms

stratosphere

contains ozone layer

mesosphere

where asteroids burn up

very low temps

thermosphere

widely varying temps

exosphere

almost no molecules

Explanation:

3 0
3 years ago
Force F = (-6.0 N)ihat + (2.0 N) jhatacts on a particle with position vector r =(4.0 m) ihat+ (4.0 m) jhat.
finlep [7]

Answer:

\tau=(-32k)\ N-m

\theta=116.55^{\circ}

Explanation:

Given that.

Force acting on the particle, F=(-6i + 2.0j)\ N

Position of the particle, r=(4i+4j)\ m

To find,

(a) Torque on the particle about the origin.

(b) The angle between the directions of r and F

Solution,

(a) Torque acting on the particle is a scalar quantity. It is given by the cross product of force and position. It is given by :

\tau=F\times r

\tau=(-6i + 2.0j)\times (4i+4j)

\tau=\begin{pmatrix}0&0&-32\end{pmatrix}

\tau=(-32k)\ N-m

So, the torque on the particle about the origin is (32 N-m).

(b) Magnitude of r, |r|=\sqrt{4^2+4^2}=5.65\ m

Magnitude of F, |F|=\sqrt{(-6)^2+2^2}=6.324\ m

Using dot product formula,

F{\circ}\ r=|F|.|r|\ cos\theta

cos\theta=\dfrac{F{\circ} r}{|F|.|r|}

cos\theta=\dfrac{-24+8}{6.324\times 5.65}

\theta=116.55^{\circ}

Therefore, this is the required solution.

6 0
2 years ago
Read 2 more answers
A ball is thrown vertically upward (assumed to be the positive direction) with a speed of 24.0 m/s from a height of 3.0 m. (a) H
SashulF [63]

Answer:

a) 29.36 m

b) 2.44 s

c) 2.57 s

d) 25.117 m/s

Explanation:

t = Time taken

u = Initial velocity = 24 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

b)

v=u+at\\\Rightarrow 0=24-9.81\times t\\\Rightarrow \frac{-24}{-9.81}=t\\\Rightarrow t=2.44 \s

Time taken by the ball to reach the highest point is 2.44 seconds

a)

s=ut+\frac{1}{2}at^2\\\Rightarrow s=24\times 2.44+\frac{1}{2}\times -9.81\times 2.44^2\\\Rightarrow s=29.35\ m

The highest point reached by the ball above its release point is 29.36 m

c) Total height is 3+29.35 = 32.35 m

s=ut+\frac{1}{2}at^2\\\Rightarrow 32.35=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{32.35\times 2}{9.81}}\\\Rightarrow t=2.57\ s

The ball reaches the ground 2.57 seconds after reaching the highest point

d)

v=u+at\\\Rightarrow v=0+9.81\times 2.57\\\Rightarrow v=25.2117\ m/s

The ball will hit the ground at 25.2117 m/s

8 0
3 years ago
Find τf, the torque about point p due to the force applied by the achilles' tendon.
Luba_88 [7]
The formula for the torque is
<span>τf = p F
where
</span><span>τf is the torque
p is the distance where the force is applied by the tendon
F is force applied by the tendon

If there are given values, substitute in the equation and solve for the torque.</span>
3 0
3 years ago
Which is not a simple harmonic motion (S.H.M.) (a) Simple Pendulum (b) Projectile motion (c) None (d) Spring motion
Scrat [10]

Answer:

b) Projectile MOTION

Explanation:

SHM is periodic motion or to and fro motion of a particle about its mean position in a straight line

In this type of motion particle must be in straight line motion

So here we can say

a) Simple Pendulum : it is a straight line to and fro motion about mean position so it is a SHM

b) Projectile motion : it is a parabolic path in which object do not move to and fro about its mean position So it is not SHM

d) Spring Motion : it is a straight line to and fro motion so it is also a SHM

So correct answer will be

b) Projectile MOTION

7 0
3 years ago
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