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Aloiza [94]
3 years ago
14

which of the following is accurate in describing the placement and classification of iodine? a. iodine is located in period 5, g

roup 17 and is classified as a nonmetal. b. iodine is located in period 17, group 5 and is classified as a nonmetal. c. iodine is located in period 6, group 7 and is classified as a metal. d. iodine is located in period 7, group 6 and is classified as a metal.
Physics
1 answer:
Elden [556K]3 years ago
8 0
A. iodine is located in period 5, group 17 and is classified as a nonmetal

Group is the column, period is the raw.

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A train travels 200km/hr. how much distance will the train be from the station in 2.5 hours?
SCORPION-xisa [38]

Answer:

500km

Explanation:

Given parameters:

Speed  = 200km/hr

Time taken  = 2.5hrs

Unknown:

Distance  = ?

Solution:

To solve this problem, we use the speed, time and distance equation.

   Therefore;

  Distance  = Speed x time

So;

  Distance  = 200 x 2.5  = 500km

6 0
2 years ago
If an object were released in space far away from planted or stars and given an initial momentum, describe what would happen to
mamaluj [8]
Strange as it may seem, the object would keep moving, in a straight line and at the same speed, until it came near another object. Its momentum and kinetic energy would never change. It might continue like that for a billion years or more.

Have a look at Newton's first law of motion.
7 0
3 years ago
Read 2 more answers
Two cylindrical resistors are made of the same material and have the same resistance. The resistors, R1 and R2, have different r
Paladinen [302]

Answer:

Option d is correct.

Explanation:

We know , resistance of a body is directly proportional to its length and inversely proportional to its area.

R=\dfrac{\rho\ L}{A}=\dfrac{\rho\ L}{\pi r^2}      ( Here, \rho is constant dependent on object material )

Writing R_1 \ and\ R_2 also :

R_1=\dfrac{\rho\ L_1}{\pi r_1^2}\ , \  R_2=\dfrac{\rho\ L_2}{\pi r_2^2}      ( since they are of same material therefore, \rho is same.)

Now , if r_2=2r_1 \ and \ 4L_1=L_2.

Then R_1=R_2

Therefore, option d. is correct.

Hence, this is the required solution.

7 0
3 years ago
What happens when the elevator is in free-fall, that is, what is the value for FN and what is the sensation experienced by the p
Komok [63]

Answer:

Weightlessness

Explanation:

When the elevator is in free fall, this can only occur when the cord of the elevator breaks.

The acceleration of the elevator will be equal to the acceleration due to gravity. That is:

a = g

Then, the body will experience what we called WEIGHTLESSNESS

Where the normal reaction N of the person will tend to zero. That is

N = 0

The value of FN < 0 because the person inside the lift will experience weightlessness.

7 0
3 years ago
The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g
Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:  

k=\sqrt{\frac{I}{M} }

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration  k=\sqrt{\frac{I}{M} }

So, moment of rotational inertia (I) of a cylinder about it axis = \frac{MR^2}{2}

k=\sqrt{\frac{\frac{MR^2}{2}}{M} }

k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }

k=\sqrt{{\frac{R^2}{2}}

k={\frac{R}{\sqrt{2}}

k={\frac{1.20m}{\sqrt{2}}

k = 0.8455 m

k ≅ 0.85 m

For the spherical shell of radius

(I) = \frac{2}{3}MR^2

k = \sqrt{\frac{\frac{2}{3}MR^2}{M}  }

k = \sqrt{\frac{2}{3} R^2}

k = \sqrt{\frac{2}{3} }*R

k = \sqrt{\frac{2}{3}}  *1.20

k = 0.9797 m

k ≅ 0.98 m

For the solid sphere of  radius

(I) = \frac{2}{5}MR^2

k = \sqrt{\frac{\frac{2}{5}MR^2}{M}  }

k = \sqrt{\frac{2}{5} R^2}

k = \sqrt{\frac{2}{5} }*R

k = \sqrt{\frac{2}{5}}  *1.20

k = 0.7560

k ≅ 0.76 m

6 0
3 years ago
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