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4 years ago

7

A 10.-newton force is required to hold a stretched spring 0.20 meter from its rest position. What is the potential energy stored

in the stretched spring?

1 answer:

Alex4 years ago

3 0

$F_{r}$ = 10 N
x = 0.2 m

$F_{r}$ = -Kx

-K = $\frac{10 N}{0.2 m}$

K = -50 $\frac{N}{m}$

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The speed limit sign in indiana reads 60 miles per hour. convert these units to meters per second

cluponka [151]

Answer:

26.822 m/s

Explanation:

60 mi/hr * 5280 ft/mile * 1 hr / 3600 sec * 12 in / foot * 1 meter / 39.37 in = <u>26.822 m/s</u>

3 0

2 years ago

Two particles oscillate in simple harmonic motion along a common straight-line segment of length 1.0 m. Each particle has a peri

igor_vitrenko [27]

Answer:

a) the particles are <em>0.217 m </em>apart

b) <em>the particles are moving in the same direction</em>.

Explanation:

a) The amplitude of the oscillations is A/2 and the period of each particle is

T = 1.5 s however, they differ by a phase of π/6 rad. Let the phase of the first particle be zero so that the phase of the second particle is π/6. So we can write the coordinates of each of the particles as,

x₁ = A/2 cos(ωt)

x₂ = A/2 cos(ωt + π/6)

we can write the angular frequency ω, as

ω = 2π / T

so,

x₁ = A/2 cos(2π / T)

x₂ = A/2 cos(2π / T + π/6)

Thus, the coordinates of the particles at t = 0.45 s are,

x₁ = A/2 cos((2π × 0.45) / 1.5)) = -0.155 A

x₂ = A/2 cos((2π × 0.45) / 1.5) + π/6) = -0.372 A

Their separation at that time is, therefore,

Δx = x₁ - x₂

= -0.155 A + 0.372 A

= 0.217 A

since A = 1 m

Thus,

<em>Δx = 0.217 m</em>

<em></em>

<em></em>

b) In order to find their directions, we must take the derivatives at t = 0.45 s.

Therefore,

v₁ = dx₁ / dt

= (-πA / T) sin(2πt / T)

= -(π(1) / 1.5) sin(2π(0.45) / 1.5)

= -1.99

and,

v₂ = dx₂ / dt

= (-πA / T) sin((2πt / T) + π/6)

= -(π(1) / 1.5) sin((2π(0.45) / 1.5) + π/6)

= -1.40

Since both v₁ and v₂ are negative, this shows that <em>the particles are moving in the same direction</em>.

6 0

3 years ago

Please help me with questions 1, 2 and 3. <br> i need a step by step explanation

kifflom [539]

Answer:

1) d

2) 5 m/s

3) 100

Explanation:

The equation of position x for a constant acceleration a and an initial velocity v₀, initial position x₀, time t is:

(i) $x=\frac{1}{2}at^2+v_0t+x_0$

The equation for velocity v and a constant acceleration a is:

(ii) $v=at+v_0$

1) Solve equation (ii) for acceleration a and plug the result in equation (i)

(iii) $a = \frac{v -v_0}{t}$

(iv) $x = \frac{v-v_0}{2t}t^2+v_0t + x_0$

Simplify equation (iv) and use the given values v = 0, x₀ = 0:

(v) $x=-\frac{v_0}{2}t + v_0t= \frac{v_0}{2}t$

2) Given v₀= 3m/s, a=0.2m/s², t=10 s. Using equation (ii) to get the final velocity v: $v=at+v_0=0.2\frac{m}{s^2} * 10s+3\frac{m}{s}=2\frac{m}{s}+3\frac{m}{s}=5\frac{m}{s}$

3) Given v₀=0m/s, t₁=10s, t₂=1s and x₀=0. Looking for factor f = x(t₁)/x(t₂) using equation(i) to calculate x(t₁) and x(t₂):

$f=\frac{x(t_1)}{x(t_2)}=\frac{\frac{1}{2}at_1^2 }{\frac{1}{2}at_2^2}=\frac{t_1^2}{t_2^2}=\frac{10^2}{1^2}=\frac{100}{1}$

5 0

4 years ago

Why are SI units used for scientific works ?

sineoko [7]

Energy can be one answer! There are many, but energy is a main one.

5 0

4 years ago

Read 2 more answers

A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the

olga2289 [7]

Answer:

a) F = 30 N, b) I = 12 N s , c) I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

p₀ = m v

The speed can be found by kinematics

v = v₀ - g t

v = 0 - 10 0.4

v = -4.0 m / s

Final after division

pf = m₁ v₁f + m₂ v₂f

p₀ = pf

M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

F = mg

F = 3 10 = 30 N

b) The expression for momentum is

I = Ft

Before the explosion the only force that acts is the weight

I = mg t

I = 3 10 0.4

I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

M v = m₁ v₁f + m₂ v₂f

v₂f = (M v - m₁ v₁f) / m₂

v₂f = (3 (-4) - 1 6) / 2

v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

I = F t = Dp

F t = pf –po)

F t= [m₁ v₁f + m₂ v₂f]

I = [1 6 + 2 (-9) -0]

I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

ΔI =If – I₀

ΔI = - 12 – (-12)

ΔI = -0 N s

3 0

4 years ago