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wolverine [178]
4 years ago
5

Proposed Kinematic Exercise I

Physics
1 answer:
valentinak56 [21]4 years ago
3 0

Answer:

(a) 20 m

(b) 6 m/s²

(c) Between t=0 and t=2, the body moves to the left.

Between t=2 and t=4, the body moves to the right.

Explanation:

v = 3t² − 6t

x(0) = 4

(a) Position is the integral of velocity.

x = ∫ v dt

x = ∫ (3t² − 6t) dt

x = t³ − 3t² + C

Use initial condition to find value of C.

4 = 0³ − 3(0)² + C

4 = C

x = t³ − 3t² + 4

Find position at t = 4.

x = 4³ − 3(4)² + 4

x = 20

(b) Acceleration is the derivative of velocity.

a = dv/dt

a = 6t − 6

Find acceleration at t = 2.

a = 6(2) − 6

a = 6

(c) v = 3t² − 6t

v = 3t (t − 2)

The velocity is 0 at t = 0 and t = 2.  Evaluate the intervals.

When 0 < t < 2, v < 0.

When t > 2, v > 0.

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Points A and B lie above an infinite plane of negative charge that produces a uniform electric field of strength E=10 N/C. What
olasank [31]

There is no illustration of the problem provided but I'll attempt to provide an answer.

The relationship between the electric potential difference between two points and the average strength of the electric field between those two points is given by:

║E║ = ΔV/d

║E║ is the magnitude of the average electric field, ΔV is the potential difference between A and B, and d is the distance between A and B.

We are given the following values:

║E║= 10N/C

d = 3m

Plug these values in and solve for ΔV

10 = ΔV/3

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4 0
4 years ago
PLEASE HELP
Gennadij [26K]

Answer:

3.675 m

Explanation:

a_{x} =0 v_{xo}=100 a_{y} =-g  v_{yo}=0

X-direction     | Y-direction

R=x_{o}+ v_{xo} t  | y=y_{o}+v_{yo}t+\frac{1}{2}a_{y}t^2

75=100t         |y=0+0+\frac{1}{2} (9.8)(0.75)

\frac{75}{100} =t             | y=3.675 m

0.75s=t              

Hope it helps

3 0
3 years ago
Name two processes that release nuclear energy?
Kaylis [27]

Answer:  Nuclear energy is released from an atom through one of two processes: nuclear fusion or nuclear fission. In nuclear fusion, energy is released when the nuclei of atoms are combined or fused together. This is how the sun produces energy. In nuclear fission, energy is released when the nuclei of atoms are split apart.

Let me know if it helps!

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Researchers have measured the acceleration of racing greyhounds as a function of their speed; a simplified version of their resu
natita [175]

\Huge\boxed{\boxed{\dfrac{4}{5}\ \text{meters}}}

Let's start by finding the time it takes for the dog to reach a velocity of 4 m/s.

We can use the following equation, where v_i is initial velocity, v_f is final velocity, t is time, and a is acceleration.

v_f-v_i=at

We're trying to solve for t first, so divide both sides by a.

\dfrac{v_f-v_i}{a}=t

Substitute in the known values.

\dfrac{4-0}{10}=t

\dfrac{4}{10}=t

\dfrac{2}{5}=t

Now, we can use the following formula to find the distance.

s=v_it+\dfrac{1}{2}at^2

Substitute in the known values.

s=0*\dfrac{2}{5}+\dfrac{1}{2}*10*(\frac{2}{5})^2

Anything multiplied by 0 is

s=\dfrac{1}{2}*10*(\frac{2}{5})^2

Just simplify from there.

s=\dfrac{1}{2}*10*\dfrac{4}{25}

s=5*\dfrac{4}{25}

s=\dfrac{20}{25}

s=\boxed{\dfrac{4}{5}}

5 0
3 years ago
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In a ballistics test, a 28-g bullet pierces a sand bag that is 30 cm thick. If the initial bullet velocity was 55 m/s and it eme
skelet666 [1.2K]

Answer:

Frictional force will be equal to 126.04 N

Explanation:

We have given mass of the bullet m = 28 gram = 0.028 kg

Initial velocity u = 55 m /sec

Width of sand = 30 cm = 0.3 m

So initial kinetic energy E=\frac{1}{2}mu^2=\frac{1}{2}\times 0.028\times 55^2=42.35j

Final velocity of the bullet v = 18 m /sec

So final kinetic energy E=\frac{1}{2}mv^2=\frac{1}{2}\times 0.028\times 18^2=4.536j

So change in kinetic energy = 42.35 - 4.536 = 37.814 j

From work energy theorem this change in kinetic energy will be equal to work one by frictional force

So f\times 0.3=37.814

f=126.04 N

4 0
4 years ago
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