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3 years ago

9. Which is the type of force that causes acceleration?a.balancedb.zeroc.unbalanced

Physics

1 answer:

Artist 52 [7]3 years ago

6 0

Explanation:

<h3>Answer:-</h3>

<u>Unbalanced</u><u> </u>Force causes acceleration.

But why?

This is because for accelerating a body we need unbalanced forces to move it.

Hope it helps :)

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If a wave y(x, t) (6.0 mm) sin(kx (600 rad/s)t f) travels along a string, how much time does any given point on the string take

kipiarov [429]

Answer:

t₁ - t₂ = 0.0011 s

Explanation:

given,

y(x, t) = (6.0 mm) sin( kx + (600 rad/s)t + Φ)

now,

y m = 6 mm ω = 600 rad/s

y₁ = + 2.0 mm y₂ = -2 .0 mm

now,

2 = (6.0 mm) sin( kx + (600 rad/s)t + Φ)

-2 = (6.0 mm) sin( kx + (600 rad/s)t + Φ)

so,

kx + (600 rad/s)t₁ + Φ = $\dfrac{\pi}{180}sin^{-1}(\dfrac{1}{3})$ ......(1)

we have multiplied with π/180 to convert angle into radians

kx + (600 rad/s)t₂ + Φ = $\dfrac{\pi}{180}sin^{-1}(-\dfrac{1}{3})$ ......(2)

subtracting both the equation (1)-(2)

600(t₁-t₂) = $\dfrac{2\pi}{180}sin^{-1}(\dfrac{1}{3})$

now,

t₁ - t₂ = 0.0011 s

time does any given point on the string take to move between displacements is equal to 0.0011 s

3 0

4 years ago

After watching a video about submarines, Jamil wants to learn more about the ocean. which question could be answered through sci

SOVA2 [1]

<span>Submarines are equipped with water tanks called ballast tanks that fill up to submerge the vessel. Emptying the tanks and filling them with air causes the submarine to surface.</span>

5 0

3 years ago

If you calculate W, the amount of work it took to assemble this charge configuration if the point charges were initially infinit

sp2606 [1]

Answer:

W = 0×(kq2L)

Explanation:

We know that the work to assemble a charge configuration of two charges a distance r from each other is simply W = kq2/r

If we want to assemble three charges A, B, and C. It's necessary to consider the distances between them

WABC = kq2/(rAB + rAC + rBC)

So, to assemble four charges A, B, C, & D....

WABCD = kq2/(rAB + rAC + rAD + rBC + rBD + rCD)

Considering a square charge configuration with sides L, such as in figure attached A, B, & C are positive & D is negative

rAB = L

rAC = L√2

rAD = L (-)

rBC = L

rBD = L√2 (-)

rCD = L (-)

⇒ W = kq2/(L + L√2 + (-L) + L + (-L√2) + (-L)

⇒ ∴ W = 0 × (kq2/L)

This way, working through each option...

(a)

The positive charges are equidistant from each other at a distance of L.

rAB = L

rAC = L

rAD = ½L⋅sin(60) (-)

rBC = L

rBD = ½L⋅sin(60) (-)

rCD = ½L⋅sin(60) (-)

Wa = kq2/(3L - (3/2)L⋅(0.866))

⇒ ∴ Wa = (1/1.7) × (kq2/L) = (0.5879)× (kq2/L)

(b)

rAB = L

rAC = 2L

rAD = 3L (-)

rBC = L

rBD = 2L (-)

rCD = L (-)

Wb = kq2/(4L - 6L)

⇒ ∴ Wb = (-1/2) × (kq2/L) = (-0.5)× (kq2/L)

(c)

The factor doesn't matter, so Wc = 0 × (kq2/L)

In this case, the greater work is actually the less work. Therefore, the positive work represents the amount of work the system actually exhibits, that we don't have to do. If there is negative work, we have to make up that work in order to place the charges as desired.

This way, charge configuration (a) requires the least amount of work.

5 0

3 years ago

A proton and an alpha particle (q = +2e, m = 4 u) are fired directly toward each other from far away, each with an initial speed

stich3 [128]

Answer:

Distance of closest approach, $r=1.91\times 10^{-14}\ m$

Explanation:

It is given that,

Charge on proton, $q_p=e$

Charge on alpha particle, $q_a=2e$

Mass of proton, $m_p=1.67\times 10^{-27}\ kg$

Mass of alpha particle, $m_a=4m_p=6.68\times 10^{-27}\ kg$

The distance of closest approach for two charged particle is given by :

$r=\dfrac{k2e^2(m_p+m_a)}{2m_am_pv_p^2}$

$r=\dfrac{9\times 10^9\times 2(1.6\times 10^{-19})^2(1.67\times 10^{-27}+6.68\times 10^{-27})}{2\times 6.68\times 10^{-27}\times 1.67\times 10^{-27}(0.01\times 3\times 10^8)^2}$