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LenaWriter [7]
3 years ago
8

Suppose a cart is being moved by a certain net force. If a load is added to the cart so its mass is doubled, what is the ratio o

f the new acceleration to the old acceleration? 1. 4 2. 2 3. 0.625 4. 3 5. 1 6. 0.333 7. 0.25 8. 0.2 9. 0.5 10. 1.5
Physics
1 answer:
Irina-Kira [14]3 years ago
4 0

Answer:

The correct answer is 9

Explanation:

Newton's second law states that force is proportional to the product of mass by acceleration

F = ma

With the first condition

F = m a1

We double the mass

F = 2m a2

.ma1 = 2 m a2

.a2 / ai = 0.5

The correct answer is 9

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2 years ago
The current that flows from and back to the power supply in a parallel circuit is called __________ current.
blagie [28]

Answer:

mainline current

Explanation:

Current that flows from and back to the power supply in a parallel circuit. Fuse. A type of circuit protection device.

6 0
2 years ago
PLS HELP!! WILL GIVE BRAINIEIST!!!!!
8_murik_8 [283]
1, 2 and 4 apply. im not sure that 2 ALWAYS applies though.
4 0
3 years ago
An object of mass 10.0kg is released at point A, slidesto the bottom of the 30° incline, then collides with ahorizontal massless
Elenna [48]

Answer:

Explanation:

Potential energy lost by mass = mgh

= 10 x 9.8 x 2 = 196 J

a ) If v be velocity of mass at the bottom , its kinetic energy will be stored in spring as elastic energy

= 1/2 m v² = 1/2 k x² , k is spring constant , x is compression , m is mass falling down

.5 x 10 v² = .5 x 500 x .75²

v = 5.3 m /s

b ) kinetic energy of mass at the bottom

= /2 m v²

= .5 x 10 x 5.3²

= 140.45 J

energy lost by mass while coming down

=potential energy at the top - kinetic energy at bottom

=  196 - 140.45

= 55.55 J .

This is equal to negative work done by friction

work done by friction = - 55.55 J

c ) Since there will be no loss of energy in compression and extension of spring so , no loss of kinetic energy will take place of mass . So it wil have same velocity that is 5.3 m /s while on its return journey.

d ) kinetic energy at the bottom = 140.45

loss of energy by friction again

= 140.45  - 55.55

= 84.9 J

If h be the height attained

mgh = 84.9

10 x 9.8 x h = 84.9

h = .866 m

( We have assumed that loss of energy in return journey will be same due to friction . )

6 0
3 years ago
A rope of negligible mass is wrapped around a 225-kg solid cylinder of radius 0.400 m. The cylinder is suspended several meters
vodomira [7]

Answer:

a)6.67 m/s2

b)16.7 rad/s2

c)increasing angular acceleration

Explanation:

a) It's because the system is not just mass of the man, it consists of the man holding a rope wrapped around a cylinder, not just a man free falling. So you would have to consider the rotating cylinder under the torque created by the man gravity force.

Let g = 10m/s2

T = mgd =75*10*0.4 = 300 N.m

The from the mass moments inertial of the solid cylinder:

I = \frac{Mr^2}{2} = \frac{225*0.4^2}{2} = 18 kgm^2

we can calculate the angular acceleration of the cylinder:

\alpha = \frac{T}{I} = \frac{300}{18} = 16.7 rad/s^2

then translate that to acceleration:

a = \alpha * r = 16.7*0.4 = 6.67 m/s^2

c) if the mass of the rope is not neglected, that means the force of gravity increases as the rope unwrapping around the cylinder, so the torque increases. Also the moment of inertial of the rope-cylinder system decreases due to rope unwrapping. In the end, the angular acceleration is no longer constant, but increasing.

4 0
3 years ago
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