In a gas-powered lawnmower, the gasoline is converted from chemical energy into which other type of energy

A particle of mass m moves under the influence of a force given by F = (−kx + kx3/α2) where k and α are positive constants. a) F

Romashka-Z-Leto [24]

Answer:

Explanation:

a ) F = (-kx + kx³/a²)

intensity of field

I = F / m

= (-kx + kx³/a²) / m

If U be potential function

- dU / dx = (-kx + kx³/a²) / m

U(x) = ∫ (kx - kx³/a²) / m dx

= k/m ( x²/2 - x⁴/4a²)

b )

For equilibrium points , U is either maximum or minimum .

dU / dx = x - 4x³/4a² = 0

x = ± a.

dU / dx = x - x³/a²

Again differentiating

d²U / dx² = 1 - 3x² / a²

Put the value of x = ± a.

we get

d²U / dx² = -2 ( negative )

So at x = ± a , potential energy U is maximum.

c )

U = k/m ( x²/2 - x⁴/4a²)

When x =0 , U = 0

When x= ± a.

U is maximum

So the shape of the U-x curve is like a bowl centered at x = 0

d ) Maximum potential energy

put x = a or -a in

U(max) = k/m ( x²/2 - x⁴/4a²)

= k/m ( a² / 2 - a⁴/4a²)

= k/m ( a² / 2 - a²/4)

a²k / 4m

This is the maximum total energy where kinetic energy is zero.

4 0

3 years ago

A thin metallic spherical shell of radius 0.357 m has a total charge of 5.03 times 10^-6 C placed on it. At the center of the sh

Artyom0805 [142]

Answer:

The electric field is $5.623\times10^{4}\ N/C$

Explanation:

Given that,

Radius = 0.357 m

Charge $Q=5.03\times10^{-6}\ C$

Point charge $q=4.15\times10^{-6}\ C$

Distance = 0.815 m

We need to calculate the total electric field

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q}{r^2}$

Where, q = point charge

r = distance

Put the value into the formula

$E=\dfrac{9\times10^{9}\times4.15\times10^{-6}}{(0.815)^2}$

$E=5.623\times10^{4}\ N/C$

Hence, The electric field is $5.623\times10^{4}\ N/C$

4 0

3 years ago

3. A man hits a 0.2 kg golf ball with a force of 4

pogonyaev

Answer:

Force(f)= mass x acceleration

Acceleration (a) is the rate of change in velocity.

F=4N

M=0.2kg

a=F/M

a=4/0.2

a=20m/s^2

Explanation:

8 0

2 years ago

Two manned satellites approaching one another at a relative speed of 0.550 m/s intend to dock. The first has a mass of 2.50 ✕ 10

NNADVOKAT [17]

Answer: Their final relative velocity is -0.412 m/s.

Explanation:

According to the law of conservation,

$m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v$

Putting the given values into the above formula as follows.

$m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v$

$2.50 \times 10^{3} kg \times 0 m/s + 7.50 \times 10^{3} kg \times -0.550 m/s = (2.50 \times 10^{3} kg + 7.50 \times 10^{3} kg)v$

$-4.12 \times 10^{3} kg m/s = (10^{4} kg) v$

v = $\frac{-4.12 \times 10^{3} kg m/s}{10^{4} kg}$

= -0.412 m/s

Thus, we can conclude that their final relative velocity is -0.412 m/s.

8 0

3 years ago

HELP PLS and HURRY

tangare [24]

Answer:

C - 50,000 * 77 * 3

Explanation:

At the top of the hill the potential energy is E= mgh= (160 kg)(9.81 m s^-2)(30 m)= 47088

hope it helps ,

<u>help me by marking as brainliest....</u>

5 0

2 years ago