These two images show pairs of oppositely charged

To test the resilience of its bumper during low-speed collisions, a 4 060-kg automobile is driven into a brick wall. the car's b

Simora [160]

Mass of car, m = 4060 kg
Spring constant, k = 8.00 x 10⁶ N/m
Spring compression, x = 3.72 cm = 3.72 x 10⁻² m

Let the car strike the wall with speed v m/s.

The kinetic energy of the car is released into the stored energy of the spring (if losses are ignored), so that
(1/2)mv² = (1/2)kx²
(4060 kg)*(v m/s)² = (8 x 10⁶ N/m)(3.72 x 10⁻² m)²
4060 v² = 1.1701 x 10⁴
v = 1.6513 m/s

Answer: 1.65 m/s (nearest hundredth)

4 0

3 years ago

An ideal solenoid 20 cm long is wound with 5000 turns of very thin wire. What strength magnetic field is produced at the center

KIM [24]

<h2>Answer:</h2>

0.31425 Tesla

<h2>Explanation:</h2>

The magnetic field strength of a solenoid can be found by using the Ampere's law as follows;

BL = μ₀ x N x I -------------------(i)

Where;

B = magnetic field strength

L = length of the solenoid

μ₀ = magnetic constant = 1.257 x 10⁻⁶H/m

N = number of turns in the coil of the solenoid

I = current flowing through the coil of the solenoid.

<em>From the question, </em>

L = 20cm = 0.2m

N = 5000 turns

I = 10A

<em>Substitute these values into equation (i) as follows;</em>

B x 0.2 = 1.257 x 10⁻⁶ x 5000 x 10

0.2B = 6.285 x 10⁻²

<em>Solve for B;</em>

B = 6.285 x 10⁻² / 0.2

B = 31.425 x 10⁻²

B = 0.31425 T

Therefore, the magnetic field strength is 0.31425 Tesla

7 0

3 years ago

A spherical object (with non-uniform density) of mass 16 kg and radius 0.22 m rolls along a horizontal surface at a constant lin

Lapatulllka [165]

Answer:

Kr = 0.7618K

Explanation:

Suppose that the object's velocity is V, then his kinetic energy is:

K = $\frac{mv^{2} }{2}$

K = $\frac{(16)v^{2} }{2}$

K = 8 $v^{2}$

The rotational kinetic energy is

Kr = $\frac{Iw^{2} }{2}$

where I: The moment of inertia

ω: angular velocity

Kr = $\frac{(0.59)w^{2} }{2}$

Kr = 0.295 $w^{2}$

How the movement is without slipping, then

ω = $\frac{v}{r}$

ω = $\frac{v}{0.22}$

Thus

Kr = $\frac{0.295v^{2} }{0.22^{2} }$

Kr = 6.095 $v^{2}$

8 $v^{2}$ ----> 1

6.095 $v^{2}$ ----->?

Kr = 0.7618K

4 0

4 years ago

Day to night to morning, keep with me in the moment

krok68 [10]

ANSWER: What.....is this lyrics or smth WHY DONT YOU SAY SO

6 0

3 years ago

You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching e

Degger [83]

Answer:

i) C decreases

ii) Q remains constant

iii) E remains constant

iv) ΔV increases

Explanation:

i)

We know, capacitance is given by:

$C=\frac{\epsilon_0.A}{d}$

$\therefore C\propto \frac{1}{d}$

<em>In this case as the distance between the plates increases the capacitance decreases while area and permittivity of free space remains constant.</em>

ii)

As the amount of charge has nothing to do with the plate separation in case of an open circuit hence the charge Q remains constant.

iii)

Electric field between the plates is given as:

$E=\frac{\sigma}{\epsilon_0}$

where:

charge density, $\sigma=\frac{Q}{A}$

<em>As we know that distance of plate separation cannot affect area of the plate. Charge Q and permittivity are also not affected by it, so E remains constant.</em>

iv)

From the basic definition of voltage we know that it is the work done per unit charge to move it through a distance.

Here we increase the distance so the work done per unit charge increases.

6 0

3 years ago