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topjm [15]
3 years ago
7

A 5.0-kg object is pulled along a horizontal surface at a constant speed by a 15-n force acting 20° above the horizontal. How mu

ch work is done by this force as the object moves 6.0 m?
Physics
1 answer:
puteri [66]3 years ago
3 0

Answer:

84.6 J

Explanation:

The work done by the force is given by

W=Fd cos \theta

where

W is the work done

F = 15 N is the force applied

d = 6.0 m is the displacement

\theta=20^{\circ} is the angle between the force's direction and the displacement

Substituting the numbers into the equation, we find

W=(15 N)(6.0 m)cos 20^{\circ} =84.6 J

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irakobra [83]

Answer:

<u><em>1000 units for breakeven</em></u>

Explanation:

Let x be the number of units sold at breakeven.

The total sales at the point would be $2x.

Variable costs would be $1x and fixed costs are $1000.

Total costs are = $1x + $1000

At breakeven: Sales = Costs

Sales =m Costs

$2x = $1x + $1000

$1x = $1000

x = 1000 units.

At 1000 units the sales are equal to the costs ("breakeven").

4 0
1 year ago
What will occur when the waves peak at the same place at the same time?
dedylja [7]
Hello!

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3 years ago
) a 1.0 kilogram laboratory cart moving with a velocity of 0.50 meter per second due east collides with and sticks to a similar
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Momentum would be the same before and after the collision
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 0.25 * 2 =
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 Now new momentum will be 0.5
 answer
 the same before and after the collision
4 0
3 years ago
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The distance of Saturn from the sun is:<br><br> &lt; 1 A.U.<br> &gt; 1 A.U.<br> = 1 A.U.
sertanlavr [38]

Answer:

1 astronomical unit is the average distance from the Earth to the Sun; approximately 150 million km. At its closest point, Saturn is 9 AU, and then at its most distant point, it's 10.1 AU. Saturn's average distance from the Sun is 9.6 AU. We have written many articles about Saturn for Universe Today.

Explanation:

6 0
3 years ago
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An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an
kirill [66]

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁=  m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

v_{f1} = \sqrt{2*a_{1}*d_{1}  }  Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁  :Newton's second law

-Ff₁+Wx₁ = ma₁   , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁      Equation (2)

∑Fy₁=0   : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁  =  m*g*cos30°

We replace   N₁  =  m*g*cos30 and  Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁   :  We divide by m

-μ₁*g*cos30°+g*sin30° = a₁  

g*(-μ₁*cos30°+sin30°) = a₁  

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

v_{f1} = \sqrt{2*3.2*3}  }

v_{f1} =\sqrt{2*3.2*3}

v_{f1} = 4.38 m/s

Rough surface  kinematics

vf₂²=v₀₂²+2*a₂*d₂   v₀₂=vf₁=4.38 m/s

0   =4.38²+2*a₂*d₂  Equation (3)

Rough surface  kinetics (μ= 0.3)

∑Fx₂=ma₂  :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂   Equation (4)

∑Fy₂= 0  :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂   We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0   =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= -  6.79*3 = 20.4 N*m

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

6 0
3 years ago
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