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aniked [119]
3 years ago
11

A 20.0 kg crate sits at rest at the bottom of a 15.0-m-long ramp that is inclined at 34.0° above horizontal. A constant horizont

al force of 290 N is applied to the crate to push it up the ramp. While the crate is moving, the ramp exerts a constant frictional force on it that has magnitude 65.0 N.
(A) Is the total work done on the crate during its motion from the bottom to the top of the ramp?
Express your answer to three significant figures and include the appropriate units.
(B) How much time does it take the crate to travel to the top of the ramp?
Express your answer to three significant figures and include the appropriate units.

Physics
1 answer:
ankoles [38]3 years ago
7 0

Answer:

987 joules, 3.01s

Explanation:

(A)

from the attached diagram

net force, Fnet, pulling the crate up the ramp is given by

Fnet = FcosФ - WsinФ - Fr

where FcosФ is the component of horizontal force 290N resolved parallel to the plane

WsinФ = mgsinФ = component of the weight of the crate resolved parallel to the plane

Fr = constant opposing frictional force

Fnet = 290cos34⁰ - 20 × 9.8 × sin34° - 65

Fnet = 240.421 - 109.602 - 65

Fnet = 65.82N

Work done on the crate up the ramp, W, is given by

W = Fnet × d (distance up the plane)

W = 65.819 × 15

W = 987.285 joules

W = 987 joules (to 3 significant Figures)

(B)

to calculate the time of travel up the ramp

we use the equation of motion

s = ut + \frac{1}{2}at^{2}

where s = distance up the plane, 15m

u = Initial velocity of the crate, which is 0 for a body that is initially at rest

a = acceleration up the plane, given by

a = \frac{Fnet}{m}

where m = mass of the crate, 20 kg

now, a = \frac{65.819}{20} \\a = 3.291\frac{m^{2} }{s}

from, s = ut + \frac{1}{2}at^{2}

15 = 0*t + \frac{1}{2}* 3.291 * t^{2}

15 = 0 + 1.645t^{2}

15 = 1.645t^{2}

t = \sqrt{\frac{15}{1.645} }

t = 3.019

t = 3.01s (to 3 sig fig)

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A 2.00-kg box is suspended from the end of a light vertical rope. A time-dependent force is applied to the upper end of the rope
Alex

Answer:

T = 27.92 N

Explanation:

For this exercise let's use Newton's second law

      T - W = m a

The weight

      W = mg

The acceleration can be found by derivatives

     a = dv / dt

     v = 2 t + 0.6 t²

     a = 2 + 0.6 t

We replace

      T - mg = m (2 + 0.6t)

      T = m (g + 2 + 0.6 t)               (1)

Let's look for the time for the speed of 15 m / s

       15 = 2 t + 0.6 t²

       0.6 t² + 2 t - 15 = 0

We solve the second degree equation

        t = [-2 ±√(4 - 4 0.6 (-15))] / 2 0.6

        t = [-2 ±√40] / 1.3 = [-2 ± 6.325] / 1.2

We take the positive time

       t = 3.6 s

Let's calculate from equation 1

       T = 2.00 (9.8 + 2 + 0. 6  3.6)

        T = 27.92 N

4 0
3 years ago
An emergency vehicle blowing its siren is moving
Georgia [21]

The frequency produced by the siren is 631.12 Hz

<h3>Doppler effect</h3>

The variation in frequency when a source of sound moves relative to an observer is determined by the doppler effect.

<h3>Frequency of observer</h3>

So, the frequency of the observer  f' = (v ± v')f/(v ± v") where

f' = 590 Hz

f = frequency of source or siren ,

v = speed of sound = 330 m/s,

v' = speed of observer = 0 m/s (since you are stationary) and

v" = speed of source = 23 m/s

Since the source moves away from the detector, the sign in the denominator is positive and v' = 0 m/s

So, f' = (v + 0)f/(v + v")

f' = vf/(v + v")

Since, we require the frequency of the source, make f subject of the formula, we have

<h3>Frequency of siren</h3>

f = (v + v")f'/v

Substituting the values of the variables into the equation, we have

f = (v + v")f'/v

f' = (330 m/s + 23 m/s)  × 590 Hz/330 m/s

f' = 353 m/s × 590 Hz/330 m/s

f' = 208270 m/sHz/330 m/s

f' = 631.12 Hz

The frequency produced by the siren is 631.12 Hz

Learn more about doppler effect here:

brainly.com/question/2169203

5 0
2 years ago
A firefighter with a weight of 707 N slides down a vertical pole with an acceleration of 2.79 m/s2, directed downward. (a) What
DedPeter [7]

Answer:

The vertical force acting on the firefighter = 908.27 N

Explanation:

Force: Force of a body is defined as the product of mass and its acceleration. The S.I unit of force is Newton (N)

The vertical force acting on the firefighter = Force due to the weight of the firefighter + force due to acceleration.

Ft = Fw - Fa

Where Ft = The vertical force acting on the firefighter, Fw = Force due to the weight of the firefighter, Fa = force due to acceleration.

Fw = mg

Making m the subject of formula in the equation above

m = Fw/g................... Equation 1

Where m = mass of the firefighter, g = acceleration due to gravity,

<em>Given: Fw = 707 N, </em>

<em> Constant: g = 9.8 m/s²</em>

Substituting these values into eqaution 1

m = 707/9.8

m = 72.14 kg.

But, Fa = ma

Where a = acceleration of the firefighter.

<em>Given: a = 2.79 m/s², </em>

<em>And m = 72.14 kg</em>

Fa = 72.14 × 2.79

Fa = 201.27 N

Therefore, Ft = 707 + 201.27  = 908.27 N

Ft = 908.27 N

The vertical force acting on the firefighter = 908.27 N

7 0
4 years ago
What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential di
yan [13]
What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 1.50×10^4v ?
3 0
2 years ago
A bullet with a mass m b = 11.5 g is fired into a block of wood at velocity v b = 249 m/s. The block is attached to a spring tha
Ostrovityanka [42]

Answer:

0.358Kg

Explanation:

The potential energy in the spring at full compression = the initial kinetic energy of the bullet/block system

0.5Ke^2 = 0.5Mv^2

0.5(205)(0.35)^2 = 12.56 J = 0.5(M + 0.0115)v^2

Using conservation of momentum between the bullet and the block

0.0115(265) = (M + 0.0115)v

3.0475 = (M + 0.0115)v

v = 3.0475/(M + 0.0115)

plugging into Energy equation

12.56 = 0.5(M + 0.0115)(3.0475)^2/(M + 0.0115)^2

12.56 = 0.5 × 3.0475^2 / ( M + 0.0115 )

12.56 = 0.5 × 9.2872/ M + 0.0115

12.56 = 4.6436/ M + 0.0115

12.56 ( M + 0.0115 ) = 4.6436

12.56M + 0.1444 = 4.6436

12.56M = 4.6436 - 0.1444

12.56 M = 4.4992

M = 4.4992÷12.56

M = 0.358 Kg

4 0
3 years ago
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