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jolli1 [7]
3 years ago
6

Two automobiles of equal mass approach an intersection. One vehicle is traveling with velocity 14.0 m/s toward the east and the

other is traveling north with speed v2i. Neither driver sees the other. The vehicles collide in the intersection and stick together, leaving parallel skid marks at an angle of 61.5° north of east. Determine the initial speed v2i of the northward-moving vehicle.
Physics
1 answer:
Sergeeva-Olga [200]3 years ago
4 0

Answer:

kqwwwwj

Explanation:

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At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl
Debora [2.8K]

Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

s=ut+0.5at^2 \\

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\

Solving both equations

600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\

Passing of B occurs at 4108.31 height.

6 0
3 years ago
A 19kg block is being pulled with a constant horizontal force of 95 Newton’s while also experiencing a constant friction force o
yuradex [85]

Answer:

A

⋅(19kg)=19Akg

95=19Akg

19=19Ak

Explanation:

5 0
3 years ago
How many legs does an arachnid have??
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An arachnid has eight legs.
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A force of 6.00 N acts in the positive direction on a 3.00 kg object, originally traveling at +15.0 m/s, for 10.0 s. (a) What is
frozen [14]

Answer:

60 kg m/s

Explanation:

Let a\;\; m/s^2 be the acceleration of the object.

As the acceleration of the object is constant, so

a=\frac {v-u}{t}\cdots(i)

Given that applied force, F=6.00 N,

From Newton's second law, we have

F= m\times a,

\Rightarrow F=\frac {m(v-u)}{t} [from equation (i)]

\Rightarrow Ft=m(v-u)

\Rightarrow Ft=mv-mu

\Rightarrow mv-mu=6\times 10 [given that time, t=10 s and F=6 N]

\Rightarrow mv-mu=60 kg \;m/s

Here mv is the final momentum of the object and mu is the initial momentum of the object.

So, the change in the momentum of the object is mv-mu.

Hence, the change in the momentum of the object is 60 kg m/s.

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Zinc can not be changed into a simpler substance by a chemical or physical process. Therefore, zinc is classified as:
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