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jekas [21]
3 years ago
11

A car starts from rest and travels for t1 seconds with a uniform acceleration a1. The driver then applies the brakes, causing a

uniform acceleration a2. If the brakes are applied for t2 seconds, determine the following. Answers are in terms of the variables a1, a2, t1, and t2. (a) How fast is the car going just before the beginning of the braking period?
Physics
1 answer:
Triss [41]3 years ago
5 0

Explanation:

It is given that,

Initially the car is at rest and travels for t₁ seconds with a uniform acceleration a₁. The driver then applies the brakes, causing a uniform acceleration a₂, If the brakes are applied for t₂ seconds.

We need to find the speed of the car just before the beginning of the braking period.

Using the formula of acceleration. It is given by :

a_1=\dfrac{v-u}{t_1}

u = 0

v=a_1\times t_1

So, just before the beginning of the braking period the speed of the car is a_1\times t_1. Hence, this is the required solution.

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A spinning wheel having a mass of 20 kg and a diameter of 0.5 m is positioned to rotate about its vertical axis with a constant
AlekseyPX

Answer:

ωf = 8.8 rad/s

v = 2.2 m/s

Explanation:

We will use the third equation of motion to find the maximum angular velocity of the wheel:

2\alpha \theta = \omega_f^2 -\omega_I^2

where,

α = angular acceleration = 6 rad/s²

θ = angular displacemnt = 1 rev = 2π rad

ωf = max. final angular velocity = ?

ωi = initial angular velocity = 1.5 rad/s

Therefore,

2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = \sqrt{77.65\ rad/s^2}

<u>ωf = 8.8 rad/s</u>

Now, for linear velocity:

v = rω = (0.25 m)(8.8 rad/s)

<u>v = 2.2 m/s</u>

7 0
3 years ago
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kogti [31]

violet

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8 0
2 years ago
A stalled car is being pushed up a hill at constant velocity by three people. the net force on the car is
frosja888 [35]
I think you need more information like the force of gravity and the force of the three people. Once you combine the two, however, you should get the net force.
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3 years ago
A uniform rod of mass 1.90 kg and length 2.00 m is capable of rotating about an axis passing through its centre and perpendicula
astraxan [27]

Complete Question:

A uniform rod of mass 1.90 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m1 = 5.40 kgis  attached to one end and a second mass m2 = 2.50 kg is attached to the other end of the rod. Treat the two masses as point particles.

(a) What is the moment of inertia of the system?

(b) If the rod rotates with an angular speed of 2.70 rad/s, how much kinetic energy does the system have?

(c) Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined?

(d) If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.70 rad/s?

Answer:

a) 8.53 kg*m² b) 31.1 J c) 7.9 kg*m² d) 28.8 J

Explanation:

a) If we treat to the two masses as point particles, the rotational inertia of each mass will be the product of the mass times the square of the distance to the axis of rotation, which is exactly the half of the length of the rod.

As the mass has not negligible mass, we need to add the rotational inertia of the rod regarding an axis passing through its centre, and perpendicular to its length.

The total rotational inertia will be as follows:

I = M*L²/12 + m₁*r₁² + m₂*r₂²

⇒ I =( 1.9kg*(2.00)²m²/12) + 5.40 kg*(1.00)²m² + 2.50 kg*(1.00)m²

⇒ I =  8.53 kg*m²

b)  The rotational kinetic energy of the rigid body composed by the rod and  the point masses m₁ and m₂, can be expressed as follows:

Krot = 1/2*I*ω²

if ω= 2.70 rad/sec, and I = 8.53 kg*m², we can calculate Krot as follows:

Krot = 1/2*(8.53 kg*m²)*(2.70)²(rad/sec)²

⇒ Krot = 31.1 J

c) If the mass of the rod is negligible, we can remove its influence of the rotational inertia, as follows:

I = m₁*r₁² + m₂*r₂² = 5.40 kg*(1.00)²m² + 2.50 kg*(1.00)m²

I = 7.90 kg*m²

d) The new rotational kinetic energy will be as follows:

Krot = 1/2*I*ω² = 1/2*(7.9 kg*m²)*(2.70)²(rad/sec)²

Krot= 28.8 J

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3 years ago
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