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Bas_tet [7]
3 years ago
7

Can some one help me with number 2 urgent?!

Physics
1 answer:
babunello [35]3 years ago
6 0
A) I = 0
b) to the right side
c) to the left side
d) to the left side
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X = v xo t. x = (10.0 m/s)(3.53 s) x = ????
LenaWriter [7]

x = V<em>x</em> * t

given V<em>x</em> = 10m/s n t = 3.53s

x = 10 * 3.53

= 35.3m


7 0
3 years ago
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A cart with a mass of 0.5 kg is at the top of the ramp. The height is 0.40m .
Tju [1.3M]

A=0.05.0M.

B=68.9244GPE.34

C=0

D it would be 79%HIGHER

3 0
2 years ago
Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k
natima [27]

Answer:

part (a) a_1\ =\ 2.9\ kg

Part (b) a_2\ =\ 6.25\ kg

Explanation:

Given,

  • mass of the smaller disk = M_1\ =\ 0.900\ kg
  • Radius of the smaller disk = R_1\ =\ 2.45\ cm\ =\ 0.0245\ m
  • mass of the larger disk = M_2\ =\ 1.6\ kg
  • Radius of the larger disk =R_2\ =\ 5.0\ cm\ =\ 0.05\ m
  • mass of the hanging block = m = 1.60 kg

Let I be the moment of inertia of the both disk after the welding,\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}(M_1R_1^2\ +\ M_2R_2^2)\\\Rightarrow I\ =\ 0.5\times (0.9\times 0.0245^2\ +\ 1.6\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2

part (a)

A block of mass m is hanging on the smaller disk,

From the f.b.d. of the block,

Let 'a' be the acceleration of the block and 'T' be the tension in the string.

mg\ -\ T\ =\ mg\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)

Net torque on the smaller disk,

\therefore \tau\ =\ I\alpha\\\Rightarrow TR_1\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{Ia}{R_1^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,enq (2)

From eqn (1) and (2), we get,

mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.027^2}\ +\ 1.60}\\\Rightarrow a\ =\ 2.91\ m/s^2

part (b)

In this case the mass is rapped on the larger disk,

From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,

Let 'a_2' be the acceleration of the block in the second case,

From the above expression,

\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.05^2}\ +\ 1.60}\\\Rightarrow a\ =\ 6.25\ m/s^2

5 0
3 years ago
Give three examples of objects in equilibrium found in classroom?​
Charra [1.4K]

Answer:

Book. Bottle. table are some examples of objects in equilibrium found in classroom

7 0
1 year ago
A car with a momentum of 3,200 kg-m/s moves foward at a rate of 2 m/s. What is the mass of the vehicle?
natulia [17]

Answer:

1600 kg

Explanation:

use the formula p=mv. p=3200, v=2. Plug in and rearrange.

3200=(m)(2)

m= 3200/2

m=1600

5 0
2 years ago
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