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3 years ago

ASAP!! Please help me out here

Physics

1 answer:

Ipatiy [6.2K]3 years ago

3 0

Answer:

<h2>Option: D </h2>

Explanation:

As,

In Option, 

A:

There are two same positive ions so they will move away from each other.

B:

There is one negative and positive ion so they will move towards each other.

C:

Again there is one negative and positive ion so they will move towards each other. 

D:

Here, there is neutral ions so they will not move and its the correct option. 

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Which is a chemical reaction? Select one: a. a peanut shell cracking open into two halves b. solid chocolate melting into liquid

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My best educated guess would be d

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4 years ago

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With what minimum speed must you toss a 130 gg ball straight up to just touch the 15-mm-high roof of the gymnasium if you releas

xxTIMURxx [149]

Answer:

The initial velocity is 0.5114 m/s or 511.4 mm/s

Explanation:

Let the initial velocity be 'v'.

Given:

Mass of the ball (m) = 130 g = 0.130 kg [ 1 g = 0.001 kg]

Initial height of the ball (h₁) = 1.4 mm = 0.0014 m [ 1 mm = 0.001 m]

Final height of the ball (h₂) = 15 mm = 0.015 m

Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.

Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.

Change in kinetic energy is given as:

$\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity$

As it just touches the 15 mm high roof, the final velocity will be zero. So,

$v_f=0\ m/s$ .

Now, the change in kinetic energy is equal to:

$\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2$

Change in gravitational potential energy = Final PE - Initial PE

So,

$\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J$ [ g = 9.8 m/s²]

Now, Change in KE = Change in PE

$0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s$

Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s

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3 years ago

An electric motor rotating a workshop grinding wheel at 1.06 102 rev/min is switched off. Assume the wheel has a constant negati

kvasek [131]

Answer:

t = 106π / 30*2.1

Explanation:

$w_{i} = 1.06*10^{2}$ => 106

=> 106 x 2π/60

=> 106/30π

∝ = -2.1 rad/sec²

$w_{f}$ => 0

$w_{f}$ = $w_{i}$ + ∝t

∴ ( $w_{f}$ - $w_{i}$ ) / ∝ = t

t = 106π / 30*2.1

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3 years ago

A 1.0 kg copper rod rests on two horizontal rails 1.0 m apart and carries a current of 50 A from one rail to the other.

vagabundo [1.1K]

Answer

given,

mass of copper rod = 1 kg

horizontal rails = 1 m

Current (I) = 50 A

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magnetic force acting on a current carrying wire is

F = B i L

Rod is not necessarily vertical

$F_x =i L B_d$

$F_y= i L B_w$

the normal reaction N = mg-F y

static friction f = μ_s (mg-F y )

horizontal acceleration is zero

$F_x-f = 0$

$iLBd = \mu_s(mg-F_y )$

B_w = B sinθ

B_d = B cosθ

iLB cosθ= μ_s (mg- iLB sinθ)

$B = \dfrac{\mu_smg}{i(cos\theta +\mu_s sin\theta)}$

$\theta =tan{-1}{\mu_s}$

$\theta =tan{-1}{0.6}$

$\theta = 31^0$

$B = \dfrac{0.6\times 1 \times 9.8}{50(cos31^0 +0.6 sin31^0)}$

B = 0.1 T

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3 years ago

ASAP!! Please help me out here ​

ASAP!! Please help me out here