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3 years ago

A crate with a mass of 1000 kg is being pulled along greased tracks by a winch. The winch is exerting a force of 2000 N

Physics

1 answer:

Vesnalui [34]3 years ago

3 0

<h3><u>We are given:</u></h3>

Coefficient of Kinetic Friction = 0.2

Mass of the crate = 1000 kg

Force applied = 2000 N

<h2 /><h2><u>Net force Acting on the crate in the Horizontal Direction:</u></h2><h3><u>Normal Force applied on the Crate:</u></h3>

Normal force is the force applied on the object by the surface. It is equal and opposite to the force of gravity

So, we can say that Normal force = | Gravitational Force |

Normal Force = | mg |

Normal Force = 1000 * 9.8

Normal Force = 9800 N

<h3><u>Finding the Frictional Force:</u></h3>

We know that:

coefficient of Kinetic friction = Friction force / Normal force

<em>replacing the known values</em>

0.2 = Friction force / 9800

Friction Force= 0.2 * 9800

Friction Force = 1960 N

<h3><u>Net force acting on the Crate:</u></h3>

We know that a force of 2000 N is being applied on the crate in the Horizontal direction

Frictional force is always opposite to the horizontal force. So, we can say that:

Applied force - Friction Force = Net Force

<em>replacing the variables</em>

2000 - 1960 = Net force

Net Force = 40N

Therefore, a net force of 40N is being applied on the Crate

<h2></h2><h2><u>Acceleration of the Crate:</u></h2>

From newton's second law of motion:

F = ma

<em>replacing the variables</em>

40 = 1000 * a

a = 40/1000

a = 0.04 m/s²

Hence, the crate will have an acceleration of 0.04 m/s²

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The membrane of the axon of a nerve cell is a thin cylindrical shell of radius r = 10-5 m, length L = 0.32 m, and thickness d =

maksim [4K]

Answer:

5.3 x 10⁻⁹ C

Explanation:

r = radius of cylindrical shell = 10⁻⁵ m

L = length = 0.32 m

A = area

Area is given as

A = 2πrL

A = 2 (3.14) (10⁻⁵) (0.32)

A = 20.096 x 10⁻⁶ m²

d = separation = 10⁻⁸ m

$k_{appa}$ = dielectric constant = 4

Capacitance is given as

$Q=\frac{k_{appa}\epsilon _{o}A}{d}$ eq-1

V = Potential difference across the membrane = 74 mV = 0.074 Volts

Q = magnitude of charge on each side

Magnitude of charge on each side is given as

Q = CV

using eq-1

$Q=\frac{k_{appa} \epsilon _{o}AV}{d}$

Inserting the values

$Q=\frac{4 (8.85\times 10^{-12})(20.096\times 10^{-6})(0.074)}{10^{-8}}$

Q = 5.3 x 10⁻⁹ C

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3 years ago

A bowling ball and a soccer ball are both rolling at the same speed. Which has more kinetic energy and why?

MissTica

Ek = 1/2 mv^2

If they are travelling at the same speed the one with the larger mass will have more kinetic energy as kinetic energy and mass are proportional.

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4 years ago

Which of these results from destructive interference?

Mazyrski [523]

Based on the options given, the answer is two waves subtract from each other. Destructive interference happens when two waves meet each other however with different frequencies cancel each other out.

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3 years ago

Which of the following statements about this experiment is FALSE? Before each trial one should reshape the bob into something li

Troyanec [42]

Which of the following statements about this experiment is FALSE? Before each trial one should reshape the bob into something like a ball. You may assume the collision between the bob and the box is completely inelastic. The initial position for the box should be just touching the pendulum bob when it is hanging straight down. To make the box move, the pendulum bob should hit close to the bottom of the box during the collision is given below

Explanation:

So, The bob is held at angle theta from initial position, P. Find the displacement from P.

displacement= L sin Θ. I hope you get it here. The pendulum forms a triangle with length L, the initial position and horizontal displacement.

So,the point is at displacement point the kinetic energy of bob is 0 and when it is released it will have maximum kinetic energy at P. It will hit the box and all he kinetic energy of bob will get transferred into driving force(F.D) of box. Now kinetic energy = Force * displacement.

K.E= 0.5 m V^2= F.D * Lsin Θ.

Find F.D

F.D = (0.5 m V^2) / Lsin Θ.

Now for the box, F.D - Friction = m(box) a.

Friction = F.D + m(box) a.

Find the accelaration of box from an equation of motion. ( u=0, find displacement of box,s,time taken and so on)

U get the friction. Now,

coeff of friction = µ = friction/ reaction.

Note that reaction here = weight of box= m(box) g. g= acc of free fall= 9.81.

So here u go.. U get the coeff of friction.. I hope am right here.. and made no mistake.. Anyway try it with the values to confirm! ;)

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3 years ago

From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum up

NemiM [27]

Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

K.E = (0.5)(mv₀²)

where,

K.E = initial kinetic energy of projectile = 1430 J

m = mass of projectile = 0.41 kg

v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ = (83.5 m/s) Cos(40.5°)

<u>v₀ₓ = 63.5 m/s</u>

y- component = v₀y = v₀Sinθ = (83.5 m/s) Sin(40.5°)

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3 years ago