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2 years ago

9

A 1.0 x 10^3 kg car traveling at 7.0 x 10^1 m/s takes 3.0 m to stop under full braking. The same car under similar road conditio

ns, traveling at 1.4 x 10^2 m/s, takes how many meters to stop under full brakin?

1 answer:

kumpel [21]2 years ago

8 0

Answer:

So Mass is 2000 kg and acceleration is three m/s square. So this gives you a net force equal to 6000 newton or 6.0 and 210

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A car moves for 10 minutes and travels 5,280 meters .What is the average speed of the car?

xxTIMURxx [149]

Answer:use the formular distance over time i.e distance/time. Make sure to convert the distance from metres to kilometers and time from minutes to hours .

Explanation:

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A teacher asks students to make a model of a transform plate boundary the students use blocks to represent tectonic plates and S

nirvana33 [79]

Answer:

Hello your question is incomplete hence I will give you a general answer as regards to tectonic plates sliding past each other in a sideways direction

answer : The Transform boundary is been demonstrated by the students when sliding tectonic plates past each other in sideways directions

Explanation:

The event that can be demonstrated by the students using blocks to represent tectonic plates and sliding the clocks past each other in sideways direction is Transform Boundary of the tectonic plates

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3 years ago

In the form of radioactive decay known as alpha decay, an unstable nucleus emits a helium-atom nucleus, which is called an alpha

trapecia [35]

Answer:

Velocity will be $v=4.06\times10^7m/sec$

Explanation:

We have given mass of alpha particle m = 4 u $=4\times 1.67\times 10^{-27}kg=6.68\times 10^{-27}kg$

Charge on alpha particle $q=2e=2\times 1.6\times 10^{-19}C=3.2\times 10^{-19}C$

Charge on thorium particle $=90\times 1.6\times 10^{-19}=144\times 10^{-19}C$

Diameter is given as d = 15 fm

So radius $r=\frac{d}{2}=\frac{15}{2}=7.5fm=7.5\times 10^{-15}m$

Potential energy is given by $E=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r}=\frac{Kq_1q_2}{r}=\frac{9\times 10^9\times 144\times 10^{-19}\times 3.2\times 10^{-19}}{7.5\times 10^{-15}}=5.529\times 10^{-12}J$

From energy conservation $\frac{1}{2}mv^2=5.529\times 10^{-12}$

$\frac{1}{2}\times 6.68\times 10^{-27}v^2=5.529\times 10^{-12}$

$v=4.06\times10^7m/sec$

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3 years ago

A sports car and a minivan run out of gas and are pushed to the side of the road. Which is easier to push, and why?

astraxan [27]

Answer:

The correct option is;

The sports car, because it has less mass and therefore less inertia

Explanation:

Newton's first law of motion states that an object will continue in its state of rest or uniform motion in a straight line unless acted on by a force. The property exhibited by the object is known as inertia

Newton's second law states that force is directly proportional to the rate of change of momentum produced

The rate of change of momentum of an object is directly proportional to the resultant force applied and is in the direction of the resultant force. The resultant force is equal to the rate of change of momentum.

Therefore, we have;

F = m·dv/dt = m×a

Given that the force required to move an object is directly proportional to its mass therefore, the inertia or the object resistance that requires a force to bring change is directly proportional to the mass of the object.

The mass of a sports car being considered lesser than the mass of the minivan will require less force to push and therefore has less inertia.

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3 years ago