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Romashka-Z-Leto [24]
3 years ago
12

A car is traveling clockwise around a circular racetrack of radius 1440m. When the car is at the northernmost point of the circl

e, it has a speed of 36 m/s and is slowing down at a rate of 1.2m/s^2. What is the direction of the velocity the car has?
Physics
1 answer:
JulijaS [17]3 years ago
7 0

Answer:

East

Explanation:

In circular motion, the velocity of the object will always be tangential to the radius of the circle. In this case, since the car is at the northernmost point of the circular track, then it must be moving right or towards the east, no matter what the acceleration might be at that very moment or what the radius of the track is. If it's moving clockwise, then it moves right (towards the east).

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The astronomer Kepler develop the three laws a planetary motion what is meant by the love distance. Relationship
Mila [183]

Explanation:

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4 0
2 years ago
Volume of 10 coin is 25ml what is the volume of 1 coin
timama [110]

Answer: 2.5 ML

Explanation:

10 ÷ 10 = 1

25 ÷ 10 = 2.5

The volume of 1 coin is 2.5 ML

5 0
2 years ago
Superman does an exhibition run at a track meet. When he runs the 200 m
Ber [7]

Answer:

6.32s

Explanation:

Given parameters:

Length of track and distance covered  = 200m

Acceleration  = 10m/s²

Unknown:

Time taken to cover the track  = ?

Solution:

To solve this problem, we apply one of the motion equations as shown below:

       S  = ut + \frac{1}{2} at²  

S is the distance covered

t is the time taken

a the acceleration

u is the initial velocity

The initial velocity of Superman is 0;

 So;

     S  =  \frac{1}{2} at²  

        200  =  \frac{1}{2} x 10 x t²  

          200  = 5t²  

            t²  = 40

            t  = 6.32s

5 0
2 years ago
In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal
kramer

Explanation:

The gravitational force equation is the following:

F_G = G * \frac{m_1 m_2}{r^2} \\

Where:

G = Gravitational constant = 6.67408 * 10^{-11} m^3 kg^{-1} s^{-2}

m1 & m2 = the mass of two related objects

r = distance between the two related objects

The problem gives you everything you need to plug into the formula, except for the gravitational constant. Let me know if you need further clarification.

8 0
3 years ago
When is the magnitude of the acceleration of a mass on a spring at its maximum value?
Andreas93 [3]

Answer:

A. when the mass has a speed of zero

Explanation:

In mass-spring system, the velocity and the acceleration are in anti-phase, which means that when one of the two quantities is maximum, the other one is zero, and vice-versa.

In fact:

- When the displacement of the spring is zero (x=0), the velocity is maximum, due to conservation of energy. In fact, as the displacement is zero, the elastic potential energy of the system (given by \frac{1}{2}kx^2) is zero, therefore the kinetic energy (given by \frac{1}{2}mv^2) must be maximum, and so the velocity (v) is also maximum. On the cotnrary, acceleration (a) is directly proportional to the restoring force of the spring, given by

F=-kx

so we see that when x=0, then the force is zero: F=0, and so the acceleration is zero as well.

- When the displacement of the spring is maximum, the velocity is zero, due to conservation of energy. In fact, as the displacement is maximum, the elastic potential energy of the system (given by \frac{1}{2}kx^2) is maximum, therefore the kinetic energy (given by \frac{1}{2}mv^2) must be zero, and so the velocity (v) is also zero. On the cotnrary, since acceleration (a) is directly proportional to the restoring force of the spring, given by

F=-kx

so we see that when x=maximum, then the force is maximum, and so the acceleration is maximum as well.

Based on this, the correct answer is

A. when the mass has a speed of zero


5 0
2 years ago
Read 2 more answers
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