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horrorfan [7]
2 years ago
6

A 208.4 N force is pulling a 197 kg object across a horizontal surface.

Physics
2 answers:
Komok [63]2 years ago
5 0

Work done

  • FscosØ
  • 208.4(161.5)cos32
  • 33656.6cos32
  • 28542.4J

Work done by kinetic friction:-

  • \mu_kN
  • \sf \mu_Kmg
  • 0.455(197)(9.8)
  • 878.423N
LenaWriter [7]2 years ago
3 0

Answer:

The Work done is 28542.4J.

The work done by the kinetic friction is 878.423N

Given:

FscosØ

We put the values together.

208.4(161.5)cos32

33656.6cos32

28542.4J

Work done by kinetic friction:-

Given the formula, we put the values.

0.455(197)(9.8)

878.423N

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The moon Phobos orbits Mars
Soloha48 [4]

Answer:

Explanation:

We are basically needing to solve for the time in the equation d = rt, where d is the distance around Mars (aka the circumference), r is the velocity, and t is time. We need to find the circumference and the velocity. We will begin with the velocity.

Because the gravitational attraction between Phobos and Mars provides the centripetal acceleration necessary to keep Phobos in its (sort of) circular path, the equation we use for this is:

F_g=F_c which says that Force supplied by gravity is equal to the centripetal force. Expanding that:

\frac{Gm_{Phobos}m_{Mars}}{r^2}=\frac{m_{Phobos}v^2}{r}

When we move that around mathematically to solve for the velocity value, what we end up with is:

v=\sqrt{\frac{Gm_{Mars}}{r} and filling in:

v=\sqrt{\frac{(6.67*10^{-11})(6.42*10^{23})}{9.38*10^6} } and we get that

v = 2100 m/s

Now for the circumference:

C = 2πr and

C = 2(3.1415)(9.38 × 10⁶) so

C = 5.9 × 10⁷

Putting that all together in the C = vT equation:

5.9 × 10⁷ = 2100T so

T = 2.8 × 10⁴ sec or 7.8 hours

8 0
3 years ago
car 1 is traveling south at 18 m/s and has a full load, giving it a total mass of 14,650 kg. Car 2 is traveling north at 11 m/s
kvasek [131]

Answer:

v_{4}= 80.92[m/s] (Heading south)

Explanation:

In order to calculate this problem, we must use the linear moment conservation principle, which tells us that the linear moment is conserved before and after the collision. In this way, we can propose an equation for the solution of the unknown.

ΣPbefore = ΣPafter

where:

P = linear momentum [kg*m/s]

Let's take the southward movement as negative and the northward movement as positive.

-(m_{1}*v_{1})+(m_{2}*v_{2})=-(m_{1}*v_{3})+(m_{2}*v_{4})

where:

m₁ = mass of car 1 = 14650 [kg]

v₁ = velocity of car 1 = 18 [m/s]

m₂ = mass of car 2 = 3825 [kg]

v₂ = velocity of car 2 = 11 [m/s]

v₃ = velocity of car 1 after the collison = 6 [m/s]

v₄ = velocity of car 2 after the collision [m/s]

-(14650*18)+(3825*11)=(14650*6)-(3825*v_{4})\\v_{4}=80.92[m/s]

4 0
3 years ago
______________ controls the flow of electric current.<br> Answer!!! <br> Please
stepladder [879]

Answer:

Explanation:

Its resistor

8 0
3 years ago
50 POINTS! A Boy throws a ball horizontally a distance of 22m downrange from the top of a tower that is 20.0m tall. What is his
DerKrebs [107]

The ball's horizontal and vertical velocities at time t are

v_x=v_{xi}

v_y=v_{yi}-gt

but the ball is thrown horizontally, so v_{yi}=0. Its horizontal and vertical positions at time t are

x=v_{xi}t

y=20.0\,\mathrm m-\dfrac g2t^2

The ball travels 22 m horizontally from where it was thrown, so

22\,\mathrm m=v_{xi}t

from which we find the time it takes for the ball to land on the ground is

t=\dfrac{22\,\rm m}{v_{xi}}

When it lands, y=0 and

0=20.0\,\mathrm m-\dfrac{9.8\frac{\rm m}{\mathrm s^2}}2\left(\dfrac{22\,\rm m}{v_{xi}}\right)^2

\implies v_i=v_{xi}=11\dfrac{\rm m}{\rm s}

7 0
3 years ago
Need some help
Keith_Richards [23]

The answer is C) the density of the rock

Density of rock is the dependent variable, because it depends on the temperature. The temperature can’t be the dependent variable because ,the density of a rock does not have magical powers that can change temperature of a room. However changing the temperature of the room ,will change the density of the rock. Hope this helps !

4 0
3 years ago
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