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3 years ago

MathPhys i need your help please helpppo

Physics

1 answer:

Lena [83]3 years ago

7 0

Answer:

6.77 m/s

Explanation:

First, in the x direction:

Given:

Δx = 3.17 m

v₀ = v cos 30.8° = 0.859 v

a = 0 m/s²

Δx = v₀ t + ½ at²

(3.17 m) = (0.859 v) t + ½ (0 m/s²) t²

3.17 = 0.859 v t

3.69 = v t

Next, in the y direction:

Given:

Δy = 0.432 m

v₀ = v sin 30.8° = 0.512 v

a = -9.81 m/s²

Δy = v₀ t + ½ at²

(0.432 m) = (0.512 v) t + ½ (-9.81 m/s²) t²

0.432 = 0.512 v t − 4.905 t²

Two equations, two variables. Solve for t in the first equation and substitute into the second equation:

t = 3.69 / v

0.432 = 0.512 v (3.69 / v) − 4.905 (3.69 / v)²

0.432 = 1.89 − 66.8 / v²

66.8 / v² = 1.458

v² = 45.8

v = 6.77

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Two foam peanuts become equally charged as they fall out of a cardboard box. Each foam peanut carries a charge of 8 x 10^-4 C. I

jok3333 [9.3K]

Answer:

Answer labeled B : 2,304,000 N

Explanation:

Recall the formula for the electric force:

$F_e=9\,\,10^9\, \, \frac{q_1*q_2}{d^2} \\F_e=9\,\,10^9\, \, \frac{8\,10^{-4}*8\,10^{-4}}{0.05^2} \\F_e=2,304,000\,\,N$

which agrees with answer B provided.

7 0

3 years ago

Joey, whose mass is m = 36 kg, stands at rest at the outer edge of a frictionless merry-go-round with the mass M = 300 kg and th

Evgen [1.6K]

Answer:

$\omega=0.24\ rad.s^{-1}$

Explanation:

Given:

mass of person, $m=36\ kg$

mass of merry go-round, $M=300\ kg$

radius of merry go-round, $R=2\ m$

velocity of the person running, $v=4\ m.s^{-1}$

We consider merry go-round as a ring:

Now the moment of inertial of the ring is given as,

$I=M.R^2$

$I=300\times 2^2$

$I=1200\ kg.m^{-2}$

Moment of inertia of the person considering as a point mass:

$I_p=m.R^2$

$I_p=36\times 2^2$

$I_p=144\ kg.m^2$

Now according to the conservation of angular momentum:

$I.\omega=I_p.\omega_p$

where:

$\omega =$ angular velocity of the merry-go-round

$\omega_p=$ angular velocity of the person running

$1200\times \omega=144\times \frac{v}{R}$

$\omega=\frac{144}{1200} \times \frac{4}{2}$

$\omega=0.24\ rad.s^{-1}$

4 0

4 years ago

Read 2 more answers

A dock worker applies a constant horizontal force of 81.0 N to a block of ice on a smooth horizontal floor. The frictional force

AnnZ [28]

Answer:

The correct answer is:

(a) 84.240 kg

(b) 24.038 m

Explanation:

The given values are:

Force,

F = 81.0 N

Distance,

S = 13.0 m

Time,

t = 5.20 s

As we know,

The acceleration of mass will be:

⇒ $a=\frac{2S}{t^2}$

On substituting the given values, we get

⇒ $=\frac{2\times 13.0}{(5.20)^2}$

⇒ $=\frac{26}{27.04}$

⇒ $=0.961538 \ m/s^2$

(a)

The mass of the block will be:

⇒ $m=\frac{F}{a}$

On substituting the given values, we get

⇒ $=\frac{81.0}{0.961538}$

⇒ $=84.240 \ kg$

(b)

The final velocity after a given time i.e.,

t = 5.00 s

⇒ $v=at$

On substituting the values, we get

⇒ $=0.961538\times 5.00$

⇒ $=4.8076 \ m/s$

In time, t = 5.00 s

The distance moved by the block will be:

⇒ $d=vt$

On putting the values, we get

⇒ $=4.8076\times 5.00$

⇒ $=24.038 \ m$

5 0

3 years ago

Calculate the energy absorbed when 13 kg of liquid water raises from 18°C to 100°C and then boils at 100°C.

Cloud [144]

Answer:

the energy absorbed is 4.477 x 10⁶ J

Explanation:

mass of the liquid, m = 13 kg

initial temperature of the liquid, t₁ = 18 ⁰C

final temperature of the liquid, t₂ = 100 ⁰C

specific heat capacity of water, c = 4,200 J/kg⁰C

The energy absorbed is calculated as;

H = mcΔt

H = mc(t₂ - t₁)

H = 13 x 4,200(100 - 18)

H = 4.477 x 10⁶ J

Therefore, the energy absorbed is 4.477 x 10⁶ J

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3 years ago

Syncopation is when we "shift" the beat off of the strong or down beat.

BabaBlast [244]

Given what we know, the statement in this question can be considered as true, since Syncopation does in fact shift the beat off of regular rhythms.

<h3>What is Syncopation?</h3>

This is when regular beats are offset in order to disrupt the melody.
This is done purposefully.
It causes the listener a need to continue in order to hear the beat return to normal.

Therefore, given that Syncopation is defined as a shift or displacement in the rhythms or beats from something strong and normal to something that causes the listener to want the beat to go back to normal.

To learn more about Syncopation visit:

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3 0

3 years ago