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3 years ago

What rule does human nature play in science?

Physics

2 answers:

nadya68 [22]3 years ago

8 0

Ur mom :)))))))))))))))))

Greeley [361]3 years ago

8 0

Skills, knowledge, behavior, feeding, communication

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A 40kg kid riding their bike down the street at 5m/s reaches the edge of a hill and coast down to the bottom. If the hill was 10

Romashka [77]

Answer

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Explanation:v

6 0

3 years ago

The Heaviside function H is defined by H(t)={0 if t<0, 1 if t≥0 It is used in the study of electric circuits to represent the

Studentka2010 [4]

Answer:

$V(t)= 240V* H(t-5)$

Explanation:

The heaviside function is defined as:

$H(t) =1 \quad t\geq 0\\H(t) =0 \quad t$

so we see that the Heaviside function "switches on" when $t=0$ , and remains switched on when $t>0$

If we want our heaviside function to switch on when $t=5$ , we need the argument to the heaviside function to be 0 when $t=5$

Thus we define a function f:

$f(t) = H(t-5)$

The $-5$ term inside the heaviside function makes sure to displace the function 5 units to the right.

Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. ( $H(t-5) =1$ when $t\geq 5$ , so it becomes just a 1, which we can safely ignore.)

Therefore our final result is:

$V(t)= 240V* H(t-5)$

I have made a sketch for you, and added it as attachment.

5 0

3 years ago

A 950-kg car strikes a huge spring at a speed of 22m/s (fig. 11-54), compressing the spring 5.0m. (a) what is the spring stiffne

alukav5142 [94]

(a) The spring stiffness constant of the spring is 18,392 N/m.

(b) The time the car was in contact with the spring before it bounces off in the opposite direction is 0.23 s.

<h3>Kinetic energy of the car</h3>

The kinetic energy of the car is calculated as follows;

K.E = ¹/₂mv²

K.E = ¹/₂ x 950 x 22²

K.E = 229,900 J

<h3>Stiffness constant of the spring</h3>

The stiffness constant of the spring is calculated as follows;

K.E = U = ¹/₂kx²

k = 2U/x²

k = (2 x 229,900)/(5)²

k = 18,392 N/m

<h3>Force exerted on the spring</h3>

F = kx

F = 18,392 x 5

F = 91,960 N

<h3>Time of impact</h3>

F = mv/t

t = mv/F

t = (950 x 22)/(91960)

t = 0.23 s

Learn more about spring constant here: brainly.com/question/1968517

#SPJ4

3 0

1 year ago

A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi

Serggg [28]

Answer:

The spring force constant is $k=243\ \frac{N}{m}$ .

Explanation:

We are told the mass of the ball is $m=0.0555\ kg$ , the height above the spring where the ball is dropped is $h=0.536\ m$ , the length the ball compresses the spring is $d=0.04897\ m$ and the acceleration of gravity is $9.8\ \frac{m}{s^{2}}$ .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy $E_{mi}$ is equal to the final mechanical energy $E_{mf}$ :

$E_{mf}=E_{mi}$

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

$\frac{1}{2}kd^{2}=mgh$