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djyliett [7]
3 years ago
14

The force needed to accelerate a stationary 2000 racing car to a speed of 144 km/h in a distance of 80 meters

Physics
1 answer:
CaHeK987 [17]3 years ago
7 0

Answer:

518400 Newton

Explanation:

Force F = mass m * Acceleration a

a = velocity v / time t

v = distance d / t

Therefore;

F = m * v² / d

with m = 2000 (unit not defined but say g)

v = 144 Km/h

d = 80 meters =0.08Km

F = 2Kg * 144Km/h * 144Km/h / 0.08Km

F = 518400 Newton

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If the force applied to an object is not greater than the starting friction, what will happen to the object?
bonufazy [111]

Answer:

Explanation:

the object will not move as the force exerted is not sufficient enough to overcome its force of friction

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3 years ago
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An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X
Verizon [17]

Answer:

0.625 c

Explanation:

Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.

In the context,

The relative speed of body 2 with respect to body 1 can be expressed as :

$u'=\frac{u-v}{1-\frac{uv}{c^2}}$

Speed of rocket 1 with respect to rocket 2 :

$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$

$u' = \frac{0.7 c}{1.12}$

u'=0.625 c

Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c

5 0
3 years ago
Which formula can be used to solve problems related to the first law of thermodynamics?
docker41 [41]

Answer: The formula used to solve the problems related to first law of thermodynamics is \Delta U=Q+W

Explanation:

First law of thermodynamics states that the total energy of the system remains conserved. Energy can neither be destroyed, nor be created but it can only be transformed into one form to another.

Its implication is any change in the internal energy will be either due to heat energy or work energy.

Mathematically,

\Delta U=Q+W

where, Q = heat energy

W = work energy

\Delta U = Change in internal energy

Sign convention for these energies:

For Q: Heat absorbed will be positive and heat released will be negative.

For W: Work done by the system is negative and work done on the system is positive.

For \Delta U: When negative, internal energy is decreasing and when positive, internal energy is increasing.

Hence, the formula used to solve the problems related to first law of thermodynamics is \Delta U=Q+W

8 0
3 years ago
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A slender rod of length 80.0 cm and 0.600 kg has its center of gravity at its geometrical center. But its density is not uniform
Vlad [161]

Answer:

Icm = 0.0701 kgm^2

Explanation:

7 0
2 years ago
A traffic light is weighing 200N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper
Levart [38]

Answer:

T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N

Explanation:

This is a balance exercise where we must apply the expressions for translational balance in the two axes

     ∑  F = 0

Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º

decompose the tension of the two upper cables

          cos 41 = T₁ₓ / T1

          sin 41 = T₁y / T1

          T₁ₓ = T₁  cos 41

          T₁y= T₁  sin 41

for cable gold

           cos 63 = T₂ / T₂

           sin 63 = T_{2y} / T₂

We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.

Let's start by analyzing the point where the traffic light meets the vertical cable

              T₃ - W = 0

              T₃ = W

              T₃ = 200 N

now let's write the equations for the single point of the three wires

X axis

   - T₁ₓ + T₂ₓ = 0

  T₁ₓ = T₂ₓ

   T1 cos 41 = T2 cos 63

   T1 = T2 cos 63 / cos 41                (1)

y Axis

      T_{1y} + T_{2y} - T3 = 0

       T₁ sin 41 + T₂ sin 63 = T₃          (2)

to solve the system we substitute equation 1 in 2

        T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W

         T₂ (cos 63 tan 41 + sin 63) = W

         T₂ = W / (cos 63 tan 41 + sin 63)

We calculate

          T₂ = 200 / (cos 63 tan 41 + sin 63)

          T₂ = 200 / 1,2856

           T₂ = 155.6 N

we substitute in 1

            T₁ = T₂ cos 63 / cos 41

             T₁ = 155.6 cos63 / cos 41

             T₁ = 93.6 N

therefore the tension in each cable is

            T₁ = 93.6 N

             T₂ = 155.6 N

             T₃ = 200 N

6 0
2 years ago
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