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A truck’s suspension spring each have a spring constant of 769 N/m. If the potential energy of the right front spring is 250 J, how far is the spring compressed?

Answers

Answer 1

Answer:

The energy stored in the body in a rest state is called potential energy.

There are two types of mechanical energy. The mechanical energy is consist of the following:-

Kinetic energy
Potential energy

According to the question, the solution is:-

The formula we used is $PE= (1)/(2)kx^(2)$

After putting the value the equation is stated as follows:-

$250 =(1)/(2) *769*x^(2)$

Hence the $x^(2)$ is equal to:-

$x^(2) = 651\nx= 0.81$ m

The spring compressed in 0.81m

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Answer 2

Answer:

Answer:

x = 0.81 m

Explanation:

given,

spring constant, k = 769 N/m

Potential energy of the spring = 250 J

distance of spring compression = ?

using conservation of energy

potential energy will equal to the spring energy

$PE = (1)/(2)kx^2$

$250= (1)/(2)* 769* x^2$

x² = 0.650

x = 0.81 m

Hence, the spring is compressed to 0.81 m

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A proton with a velocity in the positive x-direction enters a region where there is a uniform magnetic field B in the positive y-direction. You want to balance the magnetic force with an electric force so that the proton will continue along a straight line. The electric field should be in the ______ direction.

Answers

Answer:

Negative z-direction.

Explanation:

We need to determine the direction of the magnetic force. Since the velocity of the proton is in the positive x direction, and the magnetic field is in the positive y direction, we know by the vectorial formula $F=q(v* B)$ (or, alternatively, with the left hand rule) that the magnetic force points in the positive z-direction (also taking into account that the charge is positive), so the electric field should be in the negative z-direction to balance it.

The switch in the circuit has been in the left position for a long time. At t=0 it moves to the right position and stays there.a. Write the expression for the capacitor voltage v(t), fort≥0. b. Write the expression for the current through the 2.4kΩ resistor, i(t), fort≥0+.

Answers

Answer:

Pls refer to attached file

Explanation:

A car traveling at 45 km/h starts to brake, and comes to a stop over a distance of 18 m. Calculate the accelerationof the braking car.

Answers

Answer:

Acceleration, $a=8.68\ m/s^2$

Explanation:

Given that,

Initial speed of a car, u = 45 km/h = 12.5 m/s

Final speed, v = 0 (as they comes to rest)

Distance, d = 18 m

We need to find the acceleration of the breaking car. Using third equation of motion as follows :

$v^2-u^2=2ad\n\n\text{Where a is acceleration of the car}\n\na=(v^2-u^2)/(d)\n\na=((12.5)^2)/(18)\n\na=8.68\ m/s^2$

So, the acceleration of the braking car is $8.68\ m/s^2$ .

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg), who is sitting at the very bottom end, and whom he holds onto when he arrives. Laughing, John & William leave the slide horizontally and land in the muddy ground near the foot of the slide. (A) If John starts out 1.8 m above William, and the slide is essentially frictionless, how fast are they going when they leave the slide? (B) Thanks to the mud he acquired, John will now experience an average frictional force of 105 N as he slides down. How much slower is he going when he reaches the bottom than when friction was absent?

Answers

Answer:

$v=3.564\ m.s^(-1)$

$\Delta v =2.16\ m.s^(-1)$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

Using the equation of motion:

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2* (9.8* (1.8)/(3))* 3$

$v_J=5.94\ m.s^(-1)$

Now using the law of conservation of momentum at the bottom of the slide:

Sum of initial momentum of kids before & after collision must be equal.

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30* 5.94+0=(30+20)v$

$v=3.564\ m.s^(-1)$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

Now we find the force along the slide due to the body weight:

$F=m_J.g.sin\theta$

$F=30* 9.8* (1.8)/(3)$

$F=176.4\ N$

Hence the net force along the slide:

$F_R=71.4\ N$

Now the acceleration of John:

$a_j=(F_R)/(m_J)$

$a_j=(71.4)/(30)$

$a_j=2.38\ m.s^(-2)$

Now the new velocity:

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2* 2.38* 3$

$v_J_n=3.78\ m.s^(-1)$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^(-1)$

An Australian emu is running due north in a straight line at a speed of 13.0 m/s and slows down to a speed of 9.40 m/s in 3.50 s. (a) What is the magnitude and direction of the bird’s acceleration? (b) Assuming that the acceleration remains the same, what is the bird’s velocity after an additional 1.20 s has elapsed?

Answers

(a) The acceleration of the bird is $a = -1.02\ m/s$ . The negative sign indicated the opposite direction of motion. (b) The final speed is $v = 11.76\ m/s$ .

Given:

Initial speed, $u = 13\ m/s$

Final speed, $9.4\ m/s$

Time, $t = 3.5\ s$

The acceleration can be computed from the velocities and time. The standard unit of acceleration is a meter per second square.

(a)

The acceleration is computed as:

$a = v-u/t\na = 9.4-13/3.5\na = -1.02\ m/s$

Hence, the acceleration of the bird is $a = -1.02\ m/s$ . The negative sign indicated the opposite direction of motion.

(b)

The final speed as the given time can be computed from the first equation of motion. The first equation of motion gives the relation between final and initial speed, acceleration, and time.

The final speed at time 1.2 seconds is equal to:

$v = u+at\nv = 13+(-1.02)*1.20\nv = 11.76\ m/s$

Hence, the final speed is $v = 11.76\ m/s$ .

To learn more about Acceleration, here:

brainly.com/question/2303856

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If you were unfortunate enough to be 5.5 mm away from such a lightning bolt, how large a magnetic field would you experience

Answers

The question is incomplete. Here is the complete question.

Lightning bolts can carry currents up to approximately 20kA. We can model such a current as the equivalent of a very long, straight wire.

(a) If you were unfortunate enough to be 5.5m away from such a lightning bolt, how large a magnetic field would you experience?

(b) How does this field compare to one you would experience by being 5.5cm from a long, straight household current of 5A?

Answer: (a) B = 7.27 x 10⁻⁴ T

(b) Approximately 40 times higher than a household one.

Explanation: Using Biot-Savart Law, the magnetic field in a straight, long wire is given by

$B=(\mu_(0)I)/(2.\pi.r)$

where:

$\mu_(0)$ (permeability of free space) = $4.\pi.10^(-7)$ T.m/A

(a) If lightning bolt is compared to a long and straight wire, then magnetic field is

$B=(4.\pi.10^(-7).10.10^(3))/(2.\pi.5.5)$

B = 7.27 x 10⁻⁴ T

The magnitude of magnetic field in a lightning bolt is 7.27 x 10⁻⁴ T

(b) Magnetic field in a household wire will be

$B=(4.\pi.10^(-7).5)/(2.\pi.5.5.10^(-2))$

B = 1.82 x 10⁻⁵ T

Comparing fields:

$(7.27.10^(-4))/(1.82.10^(-5))$ ≈ 40

The filed for a lightning bolt is approximately 40 times higher than for a household wire.

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