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True [87]
3 years ago
8

The Ancient Roman economy did not make use of

Physics
2 answers:
denis-greek [22]3 years ago
5 0

Answer:

D paper money. The ancient Romans did not use any paper money instead their currency was coins.

ser-zykov [4K]3 years ago
3 0
D. Paper Money


All of the other ones mentioned were used in Ancient Rome.
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Joanne drives her car at a speed of 20 m/s. when she applied her breaks, a frictional force of 2000 N brought her car to a compl
Papessa [141]

Answer:

A) 1000 kg

Explanation:

vf = vi + at

0 = 20 + (a)(10)

a = -2.0 m/s^2

F = ma

2000 = (m)(2)

m = 1000 kg

8 0
1 year ago
Which statement best describes the relationship between an object's density and its index of refraction? as the optical density
babunello [35]
For future references, the answer is B.
4 0
3 years ago
Read 2 more answers
-. A 2kg cart moving to the right at 5m/s collides with an 8kg cart at rest. As a
bulgar [2K]

Answer:

<em>The velocity of the carts after the event is 1 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2+...+m_nv_n

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2+...+m_nv'_n

Since the total momentum is conserved, then:

P = P'

In a system of two masses, the equation simplifies to:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

The m1=2 kg cart is moving to the right at v1=5 m/s. It collides with an m2= 8 kg cart at rest (v2=0). Knowing they stick together after the collision, the common speed is:

\displaystyle v'=\frac{2*5+8*0}{2+8}=\frac{10}{10}=1

The velocity of the carts after the event is 1 m/s

3 0
2 years ago
A proton is observed to have an instantaneous acceleration 11*10^11. what is the magnitude of e of the electric field at the pro
jek_recluse [69]

The magnitude of the electric field at the proton's location is 10,437.5 N/C.

<h3>What the magnitude of the electric field?</h3>

The size of the electric field is basically characterized as the power per charge on the test charge. On the off chance that the electric field strength is meant by the image E. Very much like gravity, electric fields work the same way. In any case, while gravity generally draws in, an electric field, then again, can either rebuff or draw in. By and large, the Electric Field submits to the super-position guideline. the all out Electric Field from various charges is equivalent to the amount of the electric fields from each charge separately. An electric field is the actual field that encompasses electrically charged particles and applies force on any remaining charged particles in the field, either drawing in or repulsing them.

Learn more about the magnitude of the electric field, visit

brainly.com/question/26898699

#SPJ4

6 0
2 years ago
(4.56 x 10^-13)-(1.17 x 10^-13)
avanturin [10]
3.39 x 10^-13

Please mark brainliest!
3 0
3 years ago
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