All categories

3 years ago

In the equation for elastic potential energy below, what is represented by the symbol k? Ee = ½ × k × e²

Physics

1 answer:

yarga [219]3 years ago

5 0

Answer:

K is the spring/elastic constant measured in Newton per metre square (N/m²)

Explanation:

E = ½Ke²

In the equation above,

E is the energy in Joules (J)

K is the elastic/spring constant in Newton per metre square (N/m²)

e is the extension in metre (m)

You might be interested in

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

3 years ago

The two-second rule applies to any speed __________ A. in all weather conditions. B. in ideal weather and road conditions. C. in

kumpel [21]

Answer:

The answer is A

Explanation:

A large risk of tailgating is the collision avoidance time being much less than the driver reaction time. Driving instructors advocate that drivers always use the "two-second rule" regardless of speed or the type of road. During adverse weather, downhill slopes, or hazardous conditions such as black ice, it is important to maintain an even greater distance.

5 0

3 years ago

Read 2 more answers

Laney et al. (false memory):(a) Outline the rating scale used in the Food History Inventory [2 marks]

djverab [1.8K]

Answer: The rating scale that is used in food history inventory is from 1 to 8.

Explanation:

In this case the participants have to rate 24 items on a scale of 1 to 8.There are variety of questions related to food or fooding habits.

The questions like do you go out for food?

Do you like bread ? if yes how will you rate it on a scale of 1 to 8.

The questions like this has to be answered in a way . It is done in order to see the preferences and false memory regarding the food preference that people have.

6 0

3 years ago

A thin coil has 17 rectangular turns of wire. When a current of 4 A runs through the coil, there is a total flux of 5 ✕ 10−3 T ·

Alex73 [517]

Answer:

Inductance, L = 0.0212 Henries

Explanation:

It is given that,

Number of turns, N = 17

Current through the coil, I = 4 A

The total flux enclosed by the one turn of the coil, $\phi=5\times 10^{-3}\ Tm^2$

The relation between the self inductance and the magnetic flux is given by :

$L=\dfrac{N\phi}{I}$

$L=\dfrac{17\times 5\times 10^{-3}}{4}$

L = 0.0212 Henries

So, the inductance of the coil is 0.0212 Henries. Hence, this is the required solution.

7 0

3 years ago

A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3 degrees angle.

chubhunter [2.5K]

Answer:

v_{1fy} = - 0.4549 m / s

Explanation:

This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved

initial. Before the crash

p₀ = m v₁₀

final. After the crash

$p_{f}$ = m $v_{1f}$ + m v_{2f}

Recall that velocities are a vector so it has x and y components

p₀ = p_{f}

we write this equation for each axis

X axis

m v₁₀ = m v_{1fx} + m v_{2fx}

Y Axis

0 = -m v_{1fy} + m v_{2fy}

the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components

sin 23.3 = v_{2fy} / v_{2f}

cos 23.3 = v_{2fx} / v_{2f}

v_{2fy} = v_{2f} sin 23.3

v_{2fx} = v_{2f} cos 23.3

we substitute in the momentum conservation equation

m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3

0 = - m v_{1f} sin θ + m v_{2f} sin 23.3

1.83 = v_{1f} cos θ + 1.15 cos 23.3

0 = - v_{1f} sin θ + 1.15 sin 23.3

1.83 = v_{1f} cos θ + 1.0562

0 = - v_{1f} sin θ + 0.4549

v_{1f} sin θ = 0.4549

v_{1f} cos θ = -0.7738

we divide these two equations

tan θ = - 0.5878

θ = tan-1 (-0.5878)

θ = -30.45º

we substitute in one of the two and find the final velocity of the incident ball

v_{1f} cos (-30.45) = - 0.7738

v_{1f} = -0.7738 / cos 30.45

v_{1f} = -0.8976 m / s

the component and this speed is

v_{1fy} = v1f sin θ

v_{1fy} = 0.8976 sin (30.45)

v_{1fy} = - 0.4549 m / s

8 0

3 years ago