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kiruha [24]
3 years ago
9

Which element is found in group 17 and period 2

Physics
1 answer:
exis [7]3 years ago
4 0
The answer is fluorine
You might be interested in
A light spring of constant 163 N/m rests vertically on the bottom of a large beaker of water.
LUCKY_DIMON [66]

Answer:

24.71cm

Explanation:

We approach this problem base don Hooke's law which states the elongation produced in an elastic material is proportional to the applied load or force provided that its elastic limit is not exceeded. This is expressed mathematically as follows;

F=ke................(1)

where F is the applied force, k is the force constant and e is the elongation or extension of the material.

In this problem, the applied force F is the weight of the wood which is calculated as follows,

F=mg.............(2)

m = 4.11kg

g=9.8m/s^2

Hence,

F=4.11*9.8\\F=40.278N

Given that k = 163N/m, we make appropriate substitutions into equation (1) to obtain the following;

40.278=163*e\\e=\frac{40.78}{163}\\e=0.2471m

Since it is required in cm, we perform the conversion as follows, knowing that 100cm = 1m

0.2471m=0.2471*100cm=24.71cm

NB: We do not necessarily need the the density of the wood to perform our calculations since other parameters were given from which we were able to obtain its weight.

7 0
4 years ago
A satellite that weighs 4900 N on the launchpad travels around the earth's equator in a circular orbit with a period of 1.667 h.
charle [14.2K]

Answer:

(a)F= 3.83 * 10^3 N

(b)Altitude=8.20 * 10^5 m

Explanation:

On the launchpad weight = gravitational force between earth and satellite.

W = GMm/R²

where R is the earth radius.

Re-arranging:

WR² / GM = m

m = 4900 * (6.3 * 10^6)² / (6.67 * 10^-11 * 5.97 * 10^24) = 488 kg

The centripetal force (Fc) needed to keep the satellite moving in a circular orbit of radius (r) is:

Fc = mω²r

where ω is the angular velocity in radians/second. The satellite completes 1 revolution, which is 2π radians, in 1.667 hours.

ω = 2π / (1.667 * 60 * 60) = 1.05 * 10^-3 rad/s

When the satellite is in orbit at a distance (r) from the CENTRE of the earth, Fc is provided by the gravitational force  between the earth and the satellite:

Fc = GMm/r²

mω²r = GMm / r²

ω²r = GM / r²

r³ = GM/ω² = (6.67 * 10^-11 * 5.97 * 10^24) / (1.05 * 10^-3)²  

r³ = 3.612 * 10^20

r = 7.12 * 10^6 m

(a) F = GMm/r²  

F=(6.67 * 10^-11 * 5.97 * 10^24 * 488) / (7.12 * 10^6 )²

F= 3.83 * 10^3 N

(b) Altitude = r - R = (7.12 * 10^6) - (6.3 * 10^6) = 8.20 * 10^5 m

4 0
3 years ago
Suppose I create a two-slit interference pattern using light with some wavelength lambda and slits with some separation D. If I
Vlad [161]

Answer:

The pattern would spread out.

Explanation:

The fringe  width is given by the relation

fringe width = λ L / D where λ is wavelength of light used, L is distance of wall from the source of light and D is slit separation.

If we increase λ , fringe width will increase and therefore pattern will spread out .

7 0
4 years ago
For an investigation, a teacher dropped a metal bolt into a container of maple syrup and asked students to remove the bolt witho
Aneli [31]

<em>Choice-C</em> is the right one.

-- The bolt didn't react to the magnet, and just laid there.  That tells you that the bolt was made of a non-magnetic material.

-- The bolt didn't float. It went straight to the bottom.  That tells you that it's more dense than the fluid around it.

6 0
3 years ago
The block has a weight of 75 lblb and rests on the floor for which μkμk = 0.4. The motor draws in the cable at a constant rate o
WINSTONCH [101]

The given question is incomplete. The complete question is as follows.

The block has a weight of 75 lb and rests on the floor for which \mu k = 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s. Neglect the mass of the cable and pulleys.

Determine the output of the motor at the instant \theta = 30^{o}.

Explanation:

We will consider that equilibrium condition in vertical direction is as follows.

           \sum F_{y} = 0

         N - W = 0

           N = W

or,      N = 75 lb

Again, equilibrium condition in the vertical direction is  as follows.

        \sum F_{x} = 0

       T_{2} - F_{k} = 0

         T_{2} = \mu_{k} N

                  = 0.4 \times 75 lb

                  = 30 lb

Now, the equilibrium equation in the horizontal direction is as follows.

         \sum F_{x} = 0

       T Cos (30^{o}) + T Cos (30^{o}) = T_{2}

          2T Cos (30^{o}) = T_{2}

    or,             T = \frac{T_{2}}{2 Cos (30^{o})}

                        = \frac{30}{2 Cos (30^{o})}

                        = \frac{30}{1.732}

                        = 17.32 lb

Now, we will calculate the output power of the motor as follows.

             P = Tv

                = 17.32 lb \times 6

                = 103.92 \times \frac{1}{550} \times \frac{hp}{ft/s}

                = 0.189 hp

or,             = 0.2 hp

Thus, we can conclude that output of the given motor is 0.2 hp.

3 0
3 years ago
Read 2 more answers
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