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2 years ago

7

A balloon can be charged by rubbing it with the sleeve of your sweater while holding it in your hand. You can conclude from this

that a balloon is

1 answer:

zzz [600]2 years ago

5 0

It is possible to conclude from this that the balloon is an INSULATOR. It is because the balloon tends to stay any excess of charges.

<h3>What is an insulator?</h3>

An insulator is any material and/or substance capable of staying any excess of charges immobile in a place.

The insulator acts to immobilize any excess of charge added to the insulating substance and/or material, thereby remaining indefinitely in a given place.

In consequence, if a given charge is transferred to an insulator, it will remain at its initial charging location.

Learn more about insulators here:

brainly.com/question/454728

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A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

3 years ago

Batman (mass = 96.1 kg) jumps straight down from a bridge into a boat (mass = 458 kg) in which a criminal is fleeing. The veloci

MariettaO [177]

Answer:

The velocity of the boat after the batman lands in it is +9.26 m/s

Explanation:

Applying the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

Note: The collision between the Batman and the boat is an inelastic collision.

m'u'+mu = V(m+m').................... Equation 1

Where m' = mass of the Batman, u' = initial velcoity of the batman, m = mass of the boat, u = initial velocity of the boat, V = common velocity.

make V the subject of equation 1

V = (m'u'+mu)/(m+m')............... Equation 2

Given: m' = 96.1 kg, u' = 0 m/s, m = 458 kg, u = +11.3 m/s.

Substitute these values into equation 2

V = [(96.1×0)+(458×11.2)]/(96.1+458)

V = 5129.6/554.1

V = +9.26 m/s

8 0

3 years ago

The thermal conductivity of a cement is 0.168 British thermal units per foot hour degree Fahrenheit [BTU divided by (ft h Super

Sliva [168]

Answer:

thermal conductivity = 0.2906 W/m.K

Explanation:

thermal conductivity = 0.168 BTU/h.ft⁰F

1 BTU/h.ft⁰F = 1.7295772056 W/m.k

0.168 BTU/h.ft⁰F = 0.168 * 1.7295772056 W/m.k

0.168 BTU/h.ft⁰F = 0.2906 W/m.k

5 0

4 years ago

Oxygen and hydrogen are kept under the same temperature and pressure. Oxygen molecule is 16 times more massive than hydrogen one

olya-2409 [2.1K]

Answer:

The average speed of molecule of oxygen is one-fourth of the speed of the hydrogen molecule.

Explanation:

The kinetic energy of gas molecule is given by following equation:

$K.E = \frac{3}{2} KT$

Since the temperature (T) is constant for both molecules. Therefore, there kinetic energies will also be the same. Therefore,

$K.E_{O_2} = K.E_{H_2}\\\frac{1}{2}m_{O_2}v_{O_2}^2 = \frac{1}{2}m_{H_2}v_{H_2}^2 \\\\but,\\\\m_{O_2} = 16m_{H_2}\\\\therefore,\\16 m_{H_2}v_{O_2}^2 = m_{H_2}v_{H_2}^2\\\\\\taking square root on both sides, we get:\\4v_{O_2} = v_{H_2}\\\\v_{O_2} = \frac{v_{H_2}}{4}$

<u>The average speed of molecule of oxygen is one-fourth of the speed of the hydrogen molecule.</u>

7 0

3 years ago

An airplane flying parallel to the ground undergoes two consecutive displacements. The first is 76 km at 39.8◦ west of north, an

sweet [91]

Answer:

162 km

Explanation:

A diagram can be helpful.

Using the law of cosines, we can find the magnitude of the distance (c) to satisfy ...

c^2 = a^2 +b^2 -2ab·cos(C)

where C is the internal angle of the triangle of vectors and resultant. Its value is ...

180° -39.8° -59.9° = 80.3°

Filling in a=76 and b=156, we get ...

c^2 = 76^2 +156^2 -2·76·156·cos(80.3°) ≈ 26116.78

c ≈ √26116.78 ≈ 161.607

The magnitude of the total displacement is about 162 km.

_____

Please note that in the attached diagram North is to the right and East is up. That alteration of directions does not change the angles or the magnitude of the result.

4 0

4 years ago

Read 2 more answers