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gladu [14]
3 years ago
12

Two racecars are driving at constant speeds around a circular track. both cars are the same distance away from the center of the

track, but car 2 is driving twice as fast as car 1. the acceleration of car 2 is ___________ the acceleration of car 1.
Physics
1 answer:
emmainna [20.7K]3 years ago
8 0
If the track has circumference c miles, then and the car's speeds are x and y mi/s, then since time = distance/speed, 

<span>In the same direction, the slow car has a "lead" of 1 mile which the faster car has to make up. </span>

<span>1/(y-x) = 120 </span>
<span>1/(x+y) = 30 </span>

<span>120(y-x) = 30(x+y) </span>
<span>120y-120x = 30x+30y </span>
<span>90y = 150x </span>
<span>y = 5/3 x </span>


<span>1/(x + 5/3 x) = 30 </span>
<span>8/3 x * 30 = 1 </span>
<span>80x = 1 </span>
<span>x = 1/80 mi/sec = 45 mph </span>
<span>y = 75 mph</span>
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If an object weighs 80 Newtons, what force must be applied to the rope to raise the object when the mechanical advantage is four
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80 -4f(45.56 - 56) solve and get answer
5 0
3 years ago
An electric field of magnitude 2.35 V/m is oriented at an angle of 25.0° with respect to the positive z-direction. Determine the
zzz [600]

Answer:

The magnitude of the electric flux is 3.53\ N-m^2/C

Explanation:

Given that,

Electric field = 2.35 V/m

Angle = 25.0°

Area A= 1.65 m^2

We need to calculate the flux

Using formula of the magnetic flux

\phi=E\cdot A

\phi = EA\cos\theta

Where,

A = area

E = electric field

Put the value into the formula

\phi=2.35\times1.65\times\cos 25^{\circ}

\phi=2.35\times1.65\times0.91

\phi=3.53\ N-m^2/C

Hence, The magnitude of the electric flux is 3.53\ N-m^2/C

8 0
3 years ago
A rabbit is hopping along at an approximately constant speed of 3.9 m/s. The rabbit passes a crouched cat ready to chase the rab
Firlakuza [10]

Answer:

t  = 7,8 s

Explanation:

From the instant, the rabbit passes the cat. The cat star running acceleration of 0,5 m/s² .

When the cat arrives at the speed of 3,9 m/s the cat catches the rabbit

Then for the cat arrives at 3,9 m/s nedds

v = vo + a*t     vo  = 0  then   v = a*t

3,9 ( m/s) = 0,5 ( m/s² ) * t

t  = 7,8 s

v  =  3,9 m/s =

4 0
2 years ago
Who is juan pablo duarte?
jeka57 [31]

Answer:

Juan Pablo Duarte was a Dominican military leader, writer, activist, and nationalist politician who was the foremost of the founding fathers of the Dominican Republic.

3 0
2 years ago
If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g
Bas_tet [7]

Answer:

  W = 1,307 10⁶ J

Explanation:

Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)

              F = G m₁ M / r²

              W = ∫ F. dr

              W = G m₁ M ∫ dr / r²

we integrate

             W = G m₁ M (-1 / r)

                 

We evaluate between the limits, lower r = R_ Moon and r = ∞

           W = -G m₁ M (1 /∞ - 1 / R_moon)

            W = G m1 M / r_moon

Body weight is

             W = mg

             m = W / g

The mass is constant, so we can find it with the initial data

For the capsule

            m = 1000/32 = 165 / g_moon

            g_moom = 165 32/1000

            .g_moon = 5.28 ft / s²

I think it is easier to follow the exercise in SI system  

           W_capsule = 1000 pound (1 kg / 2.20 pounds)

           W_capsule = 454 N

           W = m_capsule g

           m_capsule = W / g

           m = 454 /9.8

           m_capsule = 46,327 kg

Let's calculate

          W = 6.67 10⁻¹¹ 46,327   7.36 10²² / 1.74 10⁶

          W = 1,307 10⁶ J

7 0
3 years ago
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