A spring is compressed by 0.02m. Calculate the energy stored in the spring if the force constant is 400Nm-1

For an object starting from rest and accelerating with constant acceleration, distance traveled is proportional to the square of

natali 33 [55]

The problem states that the distance travelled (d) is directly proportional to the square of time (t^2), therefore we can write this in the form of:

d = k t^2

where k is the constant of proportionality in furlongs / s^2

Using the 1st condition where d = 2 furlongs, t = 2 s, we calculate for the value of k:

2 = k (2)^2

k = 2 / 4

k = 0.5 furlongs / s^2

The equation becomes:

d = 0.5 t^2

Now solving for d when t = 4:

d = 0.5 (4)^2

d = 0.5 * 16

d = 8 furlongs

It traveled 8 furlongs for the first 4.0 seconds.

8 0

2 years ago

A baseball player runs 27.4 meters from the

sleet_krkn [62]

6.0 m longer because the player ran 3 and came back 3 at the very end, which looks like he went nowhere but in reality he ran 6.

6 0

2 years ago

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A planar loop consisting of seven turns of wire, each of which encloses 200 cm2, is oriented perpendicularly to a magnetic field

PtichkaEL [24]

Answer:

The induced current is 0.084 A

Explanation:

the area given by the exercise is

A = 200 cm^2 = 200x10^-4 m^2

R = 5 Ω

N = 7 turns

The formula of the emf induced according to Faraday's law is equal to:

ε = (-N * dφ)/dt = (N*(b2-b1)*A)/dt

Replacing values:

ε = (7*(38 - 14) * (200x10^-4))/8x10^-3 = 0.42 V

the induced current is equal to:

I = ε /R = 0.42/5 = 0.084 A

3 0

2 years ago

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A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by

aleksklad [387]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The position of the car as a function of time t is given by

$x(t)=\alpha t^2 - \beta t^3$

where

$\alpha = 1.50 m/s^2$

$\beta = 0.05 m/s^3$

The average velocity is given by the ratio between the displacement and the time taken:

$v=\frac{\Delta x}{\Delta t}$

The position at t = 0 is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

The position at t = 2.00 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

So the displacement is

$\Delta x = x(2)-x(0)=5.6-0=5.6 m$

The time interval is

$\Delta t = 2.0 s - 0 s = 2.0 s$

And so, the average velocity in this interval is

$v=\frac{5.6 m}{2.0 s}=2.8 m/s$

b)

The position at t = 0 is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

While the position at t = 4.00 s is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

So the displacement is

$\Delta x = x(4)-x(0)=20.8-0=20.8 m$

The time interval is

$\Delta t = 4.0 - 0 = 4.0 s$

So the average velocity here is

$v=\frac{20.8}{4.0}=5.2 m/s$

c)

The position at t = 2 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

While the position at t = 4 s is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

So the displacement is

$\Delta x = 20.8 - 5.6 = 15.2 m$

While the time interval is

$\Delta t = 4.0 - 2.0 = 2.0 s$

So the average velocity is

$v=\frac{15.2}{2.0}=7.6 m/s$

Learn more about average velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

6 0

2 years ago

Fields of Point Charges Two point charges are fixed in the x-y plane. At the origin is q1 = -6.00 nC . and at a point on the x-a

My name is Ann [436]

Answer:

Part A) Electric fields at the point due to q₁ and q₂:

E₁ = 33.75*10³ N/C (-j) , E₂= ( 6.48 (-i) + 8.64 (+j) )*10³ N/C

Part B) Net electric field at P (Ep)

Ep= (6.48*10³ (-i)+25.11 10³ (-j) )N/C

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence

1nC= 10⁻⁹C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

q₁ = -6.00 nC = -6 *10⁻⁹C

q₂ = +3.00 nC = +3*10⁻⁹C

d₁ = 4cm = 4 *10⁻²m

$d_{2} =\sqrt{(4*10^{-2})^{2}+((3*10^{-2})^{2} }$

d₂ = 5 *10⁻²m

Part A) Calculation of the electric fields at the point due to q₁ and q₂

Look at the attached graphic:

E₁: Electric Field at point P(0,4) cm due to charge q₁. As the charge q₁ is negative (q₁-), the field enters the charge

E₂: Electric Field at point P(0,4) cm due to charge q₂. As the charge q₂ is positive (q₂+) ,the field leaves the charge

E₁ = k*q₁/d₁² = 9*10⁹ *6 *10⁻⁹/ (4 *10⁻²)² = 33.75*10³ N/C

E₂ = k*q₂/d₂²= 9*10⁹ *3*10⁻⁹/(5 *10⁻²)² = 10.8*10³ N/C

E₁ = 33.75*10³ N/C (-j)

E₂x=E₂cosβ = 10.8*(3/5) = 6.48*10³ N/C

E₂y=E₂sinβ = 10.8*(4/5) = 8.64*10³ N/C

E₂= ( 6.48 (-i) + 8.64 (+j) )*10³ N/C

Part B) Calculation of the net electric field at P (Ep)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Ep=Epx (i) + Epy (j)

Epx= E₂x= 6.48*10³ N/C (-i)

Epy= E₁y+E₂y= (33.75*10³ (-j) + 8.64*10³ (+j) ) N/C=25.11 10³ (-j) N/C

Ep= (6.48*10³ (-i)+25.11 10³ (-j) )N/C

Ep= (6.48*10³ (-i)+25.11 10³ (-j) )N/C

3 0

2 years ago