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3 years ago

In Hooke’s law, what does the x represent?

Physics

1 answer:

Mice21 [21]3 years ago

6 0

Answer: C

X = Displacement of the spring

Hooke's law: It states that the applied force F is proportional to the displacement of spring .

F ∝ x

Where, x = displacement of spring in meters

F = force, measured in Newtons

In another words The force F is equal to the constant K times the disparagement.

F = k.x

Where k is constant and it depends on elastic material.

Spring has restorative force.

If the spring moves in opposite direction then,

F = - k.x

A negative sign indicates that the spring resists and force is to the left. The compression of the spring is greater than the restoring force.

Example: A mass 'm' stretches a spring at a displacement x.

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A(n) ___ keeps a reaction from occurring by increasing the activation energy required for a reaction to occur. A. inhibitor B. c

skelet666 [1.2K]

Due to requirements i am typing this but the ans is a

8 0

3 years ago

Read 2 more answers

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

On a Saturday afternoon, you decide to pay a neighborhood kid to mow your lawn. The kid usesa manual push lawn mower with a mass

Mars2501 [29]

Answer with Explanation:

We are given that

A.Mass,m=12 kg

$\theta=53^{\circ}$

$\mu_k=0.16$

Speed,v=1.5m/s

Net force in x direction must be zero

$F_{net}=0$

$Fsin\theta-f=0$

$Fsin\theta=f$

Net force in y direction

$N-mg-Fcos\theta=0$

$N=mg+Fcos\theta$

$f=\mu_kN=\mu_k(mg+Fcos\theta)$

$Fsin\theta=\mu_k(mg+Fcos\theta)$

$Fsin\theta=\mu_kmg+\mu_kFcos\theta$

$Fsin\theta-\mu_kFcos\theta=\mu_kmg$

$F(sin\theta-\mu_kcos\theta)=\mu_kmg$

$F=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}$

Power,P=Fv

$P=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}v$

Where $g=9.8m/s^2$

B.Substitute the values

$P=\frac{0.16\times 12\times 9.8}{sin53-0.16cos53}\times 1.5$

$P=40.17W$

6 0

4 years ago

Can someone please help me

guajiro [1.7K]

ANSWER:
C. Small, minimize

Hope it helps u!

5 0

3 years ago

Two polarizers A and B are aligned so that their transmission axes are vertical and horizontal, respectively. A third polarizer

Reptile [31]

Answer:

Explanation:

Let the angle between the first polariser and the second polariser axis is θ.

By using of law of Malus

(a)

Let the intensity of light coming out from the first polariser is I'

$I' = I_{0}Cos^{2}\theta$ .... (1)

Now the angle between the transmission axis of the second and the third polariser is 90 - θ. Let the intensity of light coming out from the third polariser is I''.

By the law of Malus

$I'' = I'Cos^{2}\left ( 90-\theta \right )$

So,

$I'' = I_{0}Cos^{2}\theta Cos^{2}\left ( 90-\theta \right )$

$I'' = I_{0}Cos^{2}\theta Sin^{2}\theta$

$I'' = \frac{I_{0}}{4}Sin^{2}2\theta$

(b)

Now differentiate with respect to θ.

$I'' = \frac{I_{0}}{4}\times 2 \times 2 \times Sin2\theta \times Cos 2\theta$

$I'' = \frac{I_{0}}{2}\times Sin 4\theta$

7 0

3 years ago