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2 years ago

À stone is thrown

1 answer:

3 0

A = -9.8

v = -9.8t -8

s = -4.9 t2 -8t +25

So… -5t^2 -8t + 25 =0, we’ll rearrange to 5t^2 + 8t - 25. We get two roots, one is positive and is 1.59 seconds

V = -9.8(1.59) - 8 = -23.6

So… it takes 1.59 seconds to hit the ground at -23.6 m/s.

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Grop 17 of the periodic table contains the ?

Andrei [34K]

Group 17 is the second column from the right in the periodic table and contains six elements: fluorine (F), chlorine (Cl), bromine (Br), iodine (I), astatine (As), and (Ts). Astatine and are radioactive elements with very short half-lives and thus do not occur naturally.

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3 years ago

a stone attached to 1m long string is moving with the speed of 5ms in a circle find the centripetal acceleration of the stone

Dafna1 [17]

Answer:

The centripetal acceleration of the stone is 5 m/s²

Explanation:

The length of the string to which the stone is attached, r = 1 m

The speed with which the string is rotated, v = 5 m/s

The centripetal acceleration, $a_c$ , is given as follows;

$a_c = \dfrac{v^2}{r}$

Therefore, the centripetal acceleration of the stone found as follows;

$a_c = \dfrac{(5 \ m/s)^2}{1 \ m} = 5 \ m/s^2$

The centripetal acceleration of the stone, $a_c$ = 5 m/s².

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2 years ago

How does Lenz's Law illustrate the concept that "you can't get

Crazy boy [7]

Answer:

When there is a change in magnetic flux linkage through a loop of wire, an electromotive force is induced in the loop, according to the Faraday-Newmann-Lenz Law:

$\epsilon=-\frac{N\Delta \Phi}{\Delta t}$

where

N is the number of turns in the loop

$\Delta \Phi$ is the change in magnetic flux through the loop

$\Delta t$ is the time elapsed

The negative sign in the formula represents Lenz's Law, and tells us about the direction of the electromotive force.

In fact, the negative sign means that the direction of the induced emf is such that to oppose to the change in the magnetic flux that originated the induced emf.

This is a consequence of the law of conservation of energy: no energy can be created out of nowhere. In fact, when the emf is induced in the loop, electrical energy appears in the circuit; however, this electric energy cannot come out of nowhere. Instead, it is just "created" from the transformation of some other form of energy (for instance, the mechanical energy that is used to move the loop in the magnetic field, and changing its magnetic flux).

The negative sign in Lenz's Law tells exactly this: the direction of the induced emf is such that it opposes the initial change in magnetic flux that generated the induced emf, so that overall the total energy is conserved.

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2 years ago

You're carrying a 3.2-m-long, 24kg pole to a construction site when you decide to stop for a rest. You place one end of the pole

Aleksandr [31]

Answer:

Tension of 132N

Explanation:

We need to apply Summatory of Force to find the tension in the hand.

We define te tensión in the hand as $F_2$ and the Tension in fence post as $F_1$ , then

$\sum F = 24(9.8)$

$F_1 + F_2= 24(9.8)$

We apply summatory of moments then

$F_2*1.25 = F_1*1.6$

Where the Force 2 is 1.25m from the center of summatory,

We can note that,

$1.6 m - 0.35m=1.25m$

We have two equation and two incognites, then replacing (1) in (2)

$1.6(235.2 -F_2) = 1.25F_2$

$376.32 = F2(1.6+1.25)$

$F_2= \frac{376.32}{2.85}$

$F_2 =132 N$

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3 years ago

I.Solve the following problems and answer the following questions. Show all your work and provide answers rounded off to the app

Ilia_Sergeevich [38]

The sprinter’s average acceleration is 1.98 m/s²

The given parameters;

initial velocity of the sprinter, u = 18 km/h

final velocity of the sprinter, v = 27 km/h

time of motion of the sprinter, t = 3.5 x 10⁻⁴ h

Convert the velocity of the sprinter to m/s;

$initial \ velocity, u = 18 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 5 \ m/s\\\\final \ velocity, v =27 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 7.5 \ m/s\\\\$

The time of motion is seconds;

$t = 3.5 \times 10^{-4} \ h \times \frac{3600 \ s}{1 \ h} = 1.26 \ s$

The sprinter’s average acceleration is calculated as follows;

$a = \frac{v- u}{t} \\\\a = \frac{7.5 \ m/s \ - \ 5 \ m/s}{1.26 \ s} \\\\a = 1.98 \ m/s^2$

Thus, the sprinter’s average acceleration is 1.98 m/s²

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6 0

3 years ago