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3 years ago

9

Se aplican dos fuerzas concurrentes a un objeto de 4N a la derecha y 5N a la izquierda. ¿Hacia donde se movió y con cuanta fuerz

a?

1 answer:

Maksim231197 [3]3 years ago

3 0

The forces move strongly towards the left by 1N

Given the following

Force towards the right = 4N

Force towards the left = 5N

Note that the force acting towards the left is negative, hence the force acting towards the left is -5N

Take the sum of force

Resultant force = -5N + 4N

Resultant force = -1N

This shows that the forces move strongly towards the left by 1N

Learn more here: brainly.com/question/24629099

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wo lacrosse players collide in midair. Jeremy has a mass of 120 kg and is moving at a speed of 3 m/s. Hans has a mass of 140 kg

Julli [10]

2.71 m/s fast Hans is moving after the collision.

<u>Explanation</u>:

Given that,

Mass of Jeremy is 120 kg ( $M_J$ )

Speed of Jeremy is 3 m/s ( $V_J$ )

Speed of Jeremy after collision is ( $V_{JA}$ ) -2.5 m/s

Mass of Hans is 140 kg ( $M_H$ )

Speed of Hans is -2 m/s ( $V_H$ )

Speed of Hans after collision is ( $V_{HA}$ )

Linear momentum is defined as “mass time’s speed of the vehicle”. Linear momentum before the collision of Jeremy and Hans is

= $=\mathrm{M}_{1} \times \mathrm{V}_{\mathrm{J}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{H}}$

Substitute the given values,

= 120 × 3 + 140 × (-2)

= 360 + (-280)

= 80 kg m/s

Linear momentum after the collision of Jeremy and Hans is

= $=\mathrm{M}_{\mathrm{J}} \times \mathrm{V}_{\mathrm{JA}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{HA}}$

= 120 × (-2.5) + 140 × $V_{HA}$

= -300 + 140 × $V_{HA}$

We know that conservation of liner momentum,

Linear momentum before the collision = Linear momentum after the collision

80 = -300 + 140 × $V_{HA}$

80 + 300 = 140 × $V_{HA}$

380 = 140 × $V_{HA}$

380/140= $V_{HA}$

$V_{HA}$ = 2.71 m/s

2.71 m/s fast Hans is moving after the collision.

4 0

3 years ago

Help with this physics task pls

cupoosta [38]

Answer:

Answers can be seen below

Explanation:

First we must explain the essential when we clear equations, and that is that if the term we need to clear is accompanied by other terms that are being added up, then those terms go to the other side of the equation to subtract if those terms are subtracting, then they go to the other side to add, if those terms are found multiplying then they go to the other side of the equation to divide and if those other terms are found dividing then they go to the other side of the equation to multiply.

(Primero debemos explicar lo esencial cuando despejamos ecuaciones, y es que si el término que necesitamos despejar va acompañado de otros términos que se están sumando, entonces esos términos van al otro lado de la ecuación para restar si esos términos están restando, luego van al otro lado para sumar, si esos términos se encuentran multiplicando luego van al otro lado de la ecuación a dividir, y si esos términos se encuentran dividiendo, pasan al otro lado de la ecuación a multiplicar.)

1 )

$t=\frac{v}{a}$ ; $d=s*(t-t_{0} )$

2)

$k=\frac{2*U}{x^{2} }$ ; $T_{2}=\frac{P_{2}*V_{2}*T_{1} }{P_{1}*V_{1} } \\$

3)

$L=\frac{F}{\pi*r*P}$ ; $d=\frac{w}{F*cos(o)}$

4)

$t^{2}=\frac{2*x}{g} ; V_{2}=\frac{A_{1}*V_{1} }{A_{2} } \\$

5)

$h=\frac{V}{\pi *r^{2} } ; r=\frac{t}{F*sin(o)}$

6)

$h=\frac{m}{(1/2)*\pi *r^{2} } ; h_{2}=\frac{F_{2}*(1/2)*b_{1} *h_{1} }{F_{1}*(1/2)*b_{2}*h_{2} }$

7)

$b=\frac{mg-ma}{v}; m=\frac{F+kx}{g*cos(o)}$

8)

$a=\frac{v-v_{o} }{t} ; u=\frac{m_{1}+m_{2} }{M}$

9)

$v_{o}=\frac{x-\frac{1}{2}*a*t^{2} }{t} ; F=\frac{W+uNd}{d*cos(o)}$

10)

$h=\frac{E-\frac{1}{2}*m*v^{2} }{mg} ; v_{2} ^{2} = \frac{Dk-\frac{1}{2} m*v_{1}^{2} }{\frac{1}{2}m }$

11)

$N=\frac{mg*sin(o)-F}{u} ; x^{2}=\frac{W+\frac{1}{2}k*x_{1}^{2} }{\frac{1}{2}*k }$

12)

$x=x_{o} +\frac{v^{2-v_{o}^{2} } }{2a} ; m=\frac{P*A-F_{1}-F_{2} }{g}$

13)

$x_{o} = x-\frac{F}{k} ; u=\frac{cos(o)-\frac{a}{g} }{sin(o)}$

14)

$t=\frac{d}{v} +t_{o}$ ; $t_{o} = t-(\frac{v-v_{o} }{a} )$

15)

$F_{2}=\frac{W-F_{1} *d}{d}+F_{3} ; v_{2}^{2}=v_{1}^{2}+\frac{2*Dk}{m}$

16)

$y_{1}=y-\frac{u}{mg} ; x^{2} = \frac{2W}{k}+x_{o} ^{2}$

7 0

3 years ago

The graphs display velocity data Velocity is on the y-axis (m/s), while time is on the x-axis (s). Based on the graphs, which da

RUDIKE [14]

Answer:

The first graph is showing the constant acceleration (1 m/s)

Explanation:

The second graph showing the flexible velocity therefore a in the graph is different at t1, t2, t3, t4

The last graph is showing constant velocity therefore there is no acceleration (a = 0)

5 0

3 years ago

Read 2 more answers

A spinning coil of wire is what moves the cone in a speaker, producing the sound. True/False

Georgia [21]

False, the spinning coil of wire that moves the cone in a speaker does not produces sound.

<u>Explanation</u>:

The wire coil is an electromagnet that is fixed to speaker cone. A normal magnet attached to the back of the speaker cone.When audio is sent in the form of short bursts of electric current to the speaker cone through the wire.

A magnetic field is induced when electric current allowed to pass through the coil. This magnetic field is repelled by the other magnet. This repulsion cause the cone to move forward. In the absence of electric current in the coil, the cone moves backward.

Thus sound waves are produced due to the resulting rarefaction and compression. So it is not the spinning coil of wire but he permanent magnet that produces the sound.

5 0

4 years ago

Read 2 more answers

In the Bohr model of the hydrogen atom, the speed of the electron is approximately 2.2 106 m/s.

Murrr4er [49]

The central force acting on the electron as it revolves in a circular orbit is $9.52 \times 10^{-8} \ N$ .

The given parameters;

<em>speed of electron, v = 2.2 x 10⁶ m/s</em>
<em>radius of the circle, r = 4.63 x 10⁻¹¹ m</em>

<em />

The central force acting on the electron as it revolves in a circular orbit is calculated as follows;

$F = \frac{M_e v^2}{r} \\\\$

where;

$M_e$ is mass of electron = 9.11 x 10⁻³¹ kg

$F = \frac{(9.11 \times 10^{-31}) \times(2.2\times 10^6)^2 }{4.63 \times 10^{-11}} \\\\F = 9.52 \times 10^{-8} \ N$

Thus, the central force acting on the electron as it revolves in a circular orbit is $9.52 \times 10^{-8} \ N$ .

Learn more about centripetal force here:brainly.com/question/20905151

8 0

3 years ago