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const2013 [10]
3 years ago
8

4. Deformation of the hood in a head-on collision-

Physics
1 answer:
Rufina [12.5K]3 years ago
6 0

Answer:

D

Explanation:

not sure but the answer is on quizlet . & i got a 100 so yeah

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Where would you find convergent and divergent plate boundaries relative to
Sergio [31]

Answer:

When two tectonic plates meet, we get a “plate boundary.” There are three major types of plate boundaries, each associated with the formation of a variety of geologic features.

Explanation:

8 0
2 years ago
A truck moves 70 m east, then moves 120 m west, and finally moves east again a distance of 90 m. If east is chosen as the positi
asambeis [7]

Answer:

140m east

Explanation:

If East is positive then lets rephrase the problem into integers

A truck moves +70 m, then moves -120m, and finally moves +90m.

So totally Displacement = +70-120+90= +140m

Since east is positive, the trucks resultant displacement is 140 m east of origin

6 0
3 years ago
1. A car starting from rest accelerates uniformly at 3
amid [387]

The car's final speed in  m/s after the acceleration is 30.

<u />

<u>Given the following data:</u>

  • Initial velocity, U = 0 m/s (since the cars starts from rest)
  • Time, t = 10 seconds
  • Acceleration, a = 3 meter per seconds square.

To find the car's final speed in  m/s after the acceleration, we would use the first equation of motion;

Mathematically, the first equation of motion is given by the formula;

V = U + at\\\\V = 0 + 3(10)

<em>Final speed, V </em><em>=</em><em> 30 m/s</em>

Therefore, the car's final speed in  m/s after the acceleration is 30.

Read more: brainly.com/question/8898885

5 0
3 years ago
A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca
svetoff [14.1K]

Answer:

q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Explanation:

Given that L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V, we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000  \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}

#To find the particular solution:

Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Hence the charge at any time, t is q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

6 0
3 years ago
a cart's initial velocity is +3.0 meters per second. What is its final velocity after accelerating at a rate of 1.5m/s2 for 8.0
Andreas93 [3]

The final velocity is +15.0 m/s

Explanation:

The motion of the cart is a uniformly accelerated motion (=at constant acceleration), therefore we can use the following suvat equation:

v=u+at

where

v is the velocity at time t

u is the initial velocity

a is the acceleration

t is the time

For the cart in this problem, we have:

u = +3.0 m/s (initial velocity)

a=1.5 m/s^2 (acceleration)

t = 8.0 s (time)

Substituting, we find the final velocity:

v=3.0+(1.5)(8.0)=+15.0 m/s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

4 0
3 years ago
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