Which of the following diagrams accurately applies the Law of

A car runs into a flying bug. Why does the bug accelerate more in the collision? *

aleksandr82 [10.1K]

Answer:

It has less mass than the car

Explanation:

Both car and bug have the same amount of force acting on them. As the bug mass is much smaller, it sees a greater acceleration

F = ma

5 0

3 years ago

A spring with a spring constant of 120 J/m2 is fixed to a wall, free to oscillate. On the other end, a ball with a mass of 1500

Neporo4naja [7]

Answer:

A. 4.47 m/s

Explanation:

As the ball oscillates, it mechanical energy, aka the total kinetic and elastics energy stays the same. For the ball to be at maximum speed, its elastic energy i 0 and vice versa. When the ball is at rest, its kinetic energy is 0 and its elastic energy is at maximum at 50 cm, or 0.5 m

1500 g = 1.5 kg

$E_e = E_k$

$kx^2/2 = mv^2/2$

$120*0.5^2/2 = 1.5*v^2/2$

$15 = 0.75v^2$

$v^2 = 15 / 0.75 = 20$

$v = \sqrt{20} = 4.47 m/s$

5 0

3 years ago

An electron (mass 9 × 10-31 kg) is traveling at a speed of 0.92 in an electron accelerator. An electric force of 1.4 × 10-13 N i

pogonyaev

Given Information:

Mass of electron = m = 9x10⁻³¹ kg

initial speed of electron = v₁ = 0.92c

Force = F = 1.4x10⁻¹³ J

Distance = d = 3 m

Required Information:

Final speed of electron = v₂ = ?

Answer:

Final speed of electron = v₂ = 2.974x10⁸ m/s

Explanation:

As we know from the conservation of energy,

E₂ - E₁ = W

E₂ = E₁ + W

Where E₂ is the final energy of electron and E₁ is the initial energy of electron

The above equation can be written in the form of particle energy

γ₂mc² = γ₁mc² + W

where γ₁ and γ₂ are given by

γ₁ = 1/√1 - (v₁/c)²

γ₂ = 1/√1 - (v₂/c)²

First calculate γ₁

γ₁ = 1/√1 - (0.92c/c)²

γ₁ = 2.55 m

Now calculate γ₂

γ₂ = (γ₁mc² + W)/mc²

First we need to find the work done

W = F*d

W = 1.4x10⁻¹³*3

W = 4.2x10⁻¹³ J

so γ₂ is

γ₂ = (2.55*9x10⁻³¹*(3x10⁸)² + 4.2x10⁻¹³)/9x10⁻³¹*(3x10⁸)²

γ₂ = 7.73

Now we can find the new speed of the electron

γ₂ = 1/√1 - (v₂/c)²

Re-arranging the above equation results in

v₂ = c*√(1 - 1/γ₂²)

v₂ = 3x10⁸*√(1 - 1/7.73²)

v₂ = 2.974x10⁸ m/s

4 0

3 years ago

The masses of the earth and moon are 5.98 x 1024 and 7.35 x 1022 kg, respectively. Identical amounts of charge are placed on eac

dybincka [34]

Answer:

The magnitude of charge on each is $5.707\times 10^{13} C$

Solution:

As per the question:

Mass of Earth, $M_{E} = 5.98\times 10^{24} kg$

Mass of Moon, $M_{M} = 7.35\times 10^{22} kg$

Now,

The gravitational force of attraction between the earth and the moon, if 'd' be the separation distance between them is:

$F_{G} = \frac{GM_{E}M_{M}}{d^{2}}$ (1)

Now,

If an identical charge 'Q' be placed on each, then the Electro static repulsive force is given by:

$F_{E} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}$ (2)

Now, when the net gravitational force is zero, the both the gravitational force and electro static force mut be equal:

Equating eqn (1) and (2):

$\frac{GM_{E}M_{M}}{d^{2}} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}$

$(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22}) = (9\times 10^{9}){Q^{2}}$

$\sqrt{\farc{(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22})}{9\times 10^{9}}} = Q$

Q = $\pm 5.707\times 10^{13} C$

7 0

3 years ago

The hydraulic lift in a car repair shop has an output diameter of 30 cm and is to lift cars up to 2000 kg. determine the fluid g

leonid [27]

Given Data: Diameter 'd' = 30 cm = 0.3 m Lifting Weight 'W' = mg = 2000*9.81 N = 19,620 N Calculations: Area of the lift 'A' = <span>pi\over4*d^2=pi\over4*0.3^2=0.07 m^2

Thank you for posting your question here at brainly. I hope the answer helps. </span>

4 0

3 years ago

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