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Firlakuza [10]
3 years ago
11

If energy (E),Velocity (v) and force (f) are taken as fundamental quantities ,then what are the dimensions of mass??

Physics
1 answer:
maxonik [38]3 years ago
7 0

Answer: E^{1}V^{-2}  T^{0}

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Another name for Newton’s 2nd law
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According to Newton’s second law of motion, also know as the law of force and Accelerate , a force upon an object causes it to accelerate according to the formula net force = mass x acceleration
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The land between two normal faults moves upward to form a
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3.00Kg toy falls from a height of 1.00m. What is the kinetic energy just before the ground?
ivanzaharov [21]

Answer:K E = 29.4 J

Explanation:

7 0
3 years ago
FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine
Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

\tau=Fd=mg\cdot R

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

\tau=I\alpha

therefore, equating this to the above equation gives

mg\cdot R=I\alpha

solving for alpha gives

\alpha=\frac{mgR}{I}

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}\boxed{\alpha=1.47s^{-2}}

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

\theta=\frac{1}{2}\alpha t^2

Therefore, ar t = 4.2 s, the above gives

\theta=\frac{1}{2}(1.47)(4.2)^2

\boxed{\theta=12.97}

So how many revolutions is this?

To find out we just divide by 2 pi:

\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}\boxed{\#\text{rev}=2.06}

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

\omega^2=w^2_0+2\alpha(\Delta\theta)_{}

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

\omega^2=0+2(1.47)(12.97)w^2=38.12\boxed{\omega=6.17.}

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

3 0
1 year ago
A car moving with an initial speed of 25 m/s slows down to a speed of 5 m/s in 10 seconds Calculate a) the acceleration of the c
stealth61 [152]

Answer :

(a) The acceleration  of the car is, -2m/s^2

(b) The distance covered by the car is, 150 m

Explanation :  

By the 1st equation of motion,

v=u+at ...........(1)

where,

v = final velocity = 5 m/s

u = initial velocity  = 25 m/s

t = time = 10 s

a = acceleration  of the car = ?

Now put all the given values in the above equation 1, we get:

5m/s=25m/s+a\times (10s)

a=-2m/s^2

The acceleration  of the car is, -2m/s^2

By the 2nd equation of motion,

s=ut+\frac{1}{2}at^2 ...........(2)

where,

s = distance covered by the car = ?

u = initial velocity  = 25 m/s

t = time = 10 s

a = acceleration  of the car = -2m/s^2

Now put all the given values in the above equation 2, we get:

s=(25m/s)\times (10s)+\frac{1}{2}\times (-2m/s^2)\times (10s)^2

By solving the term, we get:

s=150m

The distance covered by the car is, 150 m

8 0
3 years ago
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