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4 years ago

Would it be better to collide head-on with an identical car traveling at the same speed or collide with a stationary brick wall?

Physics

1 answer:

attashe74 [19]4 years ago

6 0

It makes no difference. The momentum of either car goes to zero in both cases.

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An overnight rainstorm has caused a major roadblock. Three massive rocks of mass m1=584 kg, m2=838 kg, and m3=322 kg have blocke

Elena-2011 [213]

Answer:

Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N

Explanation:

Total force required = Mass x Acceleration,

F = ma

Here we need to consider the system as combine, total mass need to be considered.

Total mass, a = m₁+m₂+m₃ = 584 + 838 + 322 = 1744 kg

We need to accelerate the group of rocks from the road at 0.250 m/s²

That is acceleration, a = 0.250 m/s²

Force required, F = ma = 1744 x 0.25 = 436 N

Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N

8 0

3 years ago

Explain whether Washington, D.C., or

Svetradugi [14.3K]

I would say Orlando since its near more ocean than Washington. Since Ocean can change weather by making it cold or warm depending on the season.

5 0

3 years ago

Read 2 more answers

Vector has a magnitude of 6.0 m and points 30° north of east. Vector has a magnitude of 4.0 m and points 30° east of north. The

Brut [27]

Answer:

The resultant vector is $\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j$ .

Explanation:

First, each vector is determined in terms of absolute coordinates:

6-meter vector with direction: 30º north of east.

$\vec A = (6\,m)\cdot (\cos30^{\circ} \,i + \sin 30^{\circ}\,j)$

$\vec A = 5.196\,i + 3\,j$

4-meter vector with direction: 30º east of north.

$\vec B = (4\,m)\cdot (\cos 60^{\circ}\,i + \sin 60^{\circ}\,j)$

$\vec B = 2\,i + 3.464\,j$

The resultant vector is obtaining by sum of components:

$\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j$

The resultant vector is $\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j$ .

5 0

4 years ago

Difference between uniform velocity and variable velocity .

vichka [17]

Answer:

If a body covers equal displacement in equal interval of time,however small may be the body is said to have uniform velocity. If a body covers in equal displacement in equal interval of time is called variable velocity.

4 0

4 years ago

The electric field between two parallel plates is uniform, with magnitude 646 N/C. A proton is held stationary at the positive p

Ahat [919]

Answer:

The distance from the positive plate at which the two pass each other is 0.0023 cm.

Explanation:

We need to find the acceleration of each particle first. Let's use the electric force equation.

$F=Eq$

$ma=Eq$

<u>For the proton</u>

$m_{p}a_{p}=Eq_{p}$

$a_{p}=\frac{Eq_{p}}{m_{p}}$

$a_{p}=\frac{646*1.6*10^{-19}}{1.67*10^{−27}}$

$a_{p}=6.19*10^{10}\: m/s^{2}$

<u>For the electron</u>

$m_{e}a_{e}=Eq_{e}$

$a_{e}=\frac{Eq_{e}}{m_{e}}$

$a_{e}=\frac{646*1.6*10^{-19}}{9.1*10^{−31}}$

$a_{e}=1.14*10^{14}\: m/s^{2}$

Now we know that the plate separation is 4.26 cm or 0.0426 m. The travel distance of the proton plus the travel distance of the electron is 0.0426 m.

$x_{p}+x_{e}=0.0426$

Both of them have an initial speed equal to zero. So we have:

$\frac{1}{2}a_{p}t^{2}+\frac{1}{2}a_{e}t^{2}=0.0426$

$t^{2}(a_{p}+a_{e})=2*0.0426$

$t^{2}=\frac{2*0.0426}{a_{p}+a_{e}}$

$t=\sqrt{\frac{2*0.0426}{6.19*10^{10}+1.14*10^{14}}}$

$t=2.73*10^{-8}\: s$

With this time we can find the distance from the positive plate (x(p)).

$x_{p}=\frac{1}{2}a_{p}t^{2}$

$x_{p}=\frac{1}{2}6.19*10^{10}*(2.73*10^{-8})^{2}$

$x_{p}=0.0023\: cm$

Therefore, the distance from the positive plate at which the two pass each other is 0.0023 cm.

I hope it helps you!

7 0

3 years ago