A 78 N block is supported by a spring whose constant is 12 N/m. Calculate the elongation of the spring under this load.

Technician A says that cabin filters are accessible behind the glove compartment. Technician B says that cabin filters are acces

BartSMP [9]

Answer:

A

but it depends on the car

Explanation:

3 0

3 years ago

Based on the graph of kinetic energy given (gray curve in the graphing window), sketch a graph of the baseball's gravitational p

chubhunter [2.5K]

1) Physical principles:

a) Total mechanical energy = kinetic energy + potential energy.

b) Total mechanical energy is conserved (neglecting external forces, like drag and friction)

2) Notation:

a) Total mechanical energy: ME.

b) Kinetic energy: KE

c) Gravitational potential energy: PE

∴ ME = KE + PE = constant

3) Solution:

a) Since, ME is conserved, it is constant and would be represented in the graph by a horizontal line.

b) At start (t = 0), the ball has only KE, so KE =ME = E and PE = 0

c) As the time goes, the ball gains altitude (PE increases) and loses speed (KE decreases).

d) PE increases from 0 to a maximum value. In the graph that happens at t = 2s.

At that point, KE = 0, and PE = ME.

That is the point of highest altitude and where the speed is zero.

d) From t = 2 seg, the ball starts to lose altitude, then the ball loses PE, and gains KE.

Just before reaching the ground, at t = 4s, the ball has the same initial KE and PE as at t = 0: KE = ME and PE = 0.

The PE may be sketched on the same graph along with the KE and the ME.

The graph is attached. The red line is the ME and the blue line is the PE.

Note that at any point in the graph PE + KE = ME.

5 0

4 years ago

Read 2 more answers

PLZ HELP ME

bonufazy [111]

Answer:

1. The stone will strike the ground 49.46 m from the base of the cliff

2. A) Approximately 0.542 seconds

B) Approximately 3.69 m/s

3. A) The time the ball spends in the air is approximately 4.0775 s

B) The horizontal range is approximately 141.25 m.

Explanation:

1. The time it takes the stone to land is given by the equation, t = √(h/(1/2 × g)

∴ t = √(30/(1/2 × 9.81)) ≈ 2.473 seconds

The horizontal distance covered by the stone in that time = 20 × 2.473 ≈ 49.46 m

The stone will strike the ground 49.46 m from the base of the cliff

2. A) The time the ball spends in the air = t = √(h/(1/2 × g)

∴ t = √(1.44/(1/2 × 9.81)) ≈ 0.542 seconds

B) The initial horizontal velocity, u = Horizontal distance/(Time) = 2/0.542 ≈ 3.69 m/s

The initial horizontal velocity ≈ 3.69 m/s

3. A) The time the ball spends in the air is given by the following equation;

t = 2 × u × sin(θ)/g = 2 × 40 × sin(30)/9.81 ≈ 4.0775 s

t ≈ 4.0775 s

B) The horizontal range, R, of the ball is given by the equation for the range of a projectile as follows;

$Range, R = \dfrac{u^2 \times sin (2 \cdot \theta) }{g}$

Substituting the known values, gives;

$Range, R = \dfrac{40^2 \times sin (2 \times 30^{\circ}) }{9.81} \approx 141.25 \ m$

The horizontal range ≈ 141.25 m.

4 0

3 years ago

In your own words, explain how stalagmites are formed in caves.

romanna [79]

Stalagmite formation occurs only under certain pH conditions within the cave. They form through deposition of calcium carbonate and other minerals

7 0

3 years ago

A flexible loop has a radius of 10.5 cm and it is in a magnetic field of B = 0.117 T. The loop is grasped at points A and B and

MAXImum [283]

Explanation:

Given that,

Radius = 10.5 cm

Magnetic field = 0.117 T

Time = 0.243 s

After stretched, area is zero

(I). We need to calculate the magnetic flux through the loop before stretched

Using formula of magnetic flux

$\phi=B\times A$

$\phi=B\times \pi r^2$

Where, B = magnetic field

r = radius

Put the value into the formula

$\phi=0.117\times3.14\times(10.5\times10^{-2})^2$

$\phi=4.05\times10^{-3}\ Tm^2$

(II). We need to calculate the magnetic flux through the loop after stretched

$\phi=B\times A$

Here, A = 0

$\phi=0$

So, The magnetic flux through the loop after stretched is zero.

(III). We need to calculate the magnitude of the average induced electromotive force

Using formula of the induced electromotive force

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{\phi_{after}-\phi_{before}}{t}$

$\epsilon=-\dfrac{0-4.05\times10^{-3}}{0.243}$

$\epsilon =16.67\times10^{-3}\ V$

Hence, This is the required solution.

3 0

3 years ago