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1 year ago

What is the atomic number z of 73li?

2 answers:

5 0

$^7 _3 {Li}$ means that the element is Lithium (Li), 7 corresponds to its mass number (sum of protons and neutrons), 3 corresponds to its atomic number (number of protons in the nucleus).

The problem asks for the atomic number (Z) of this isotope: based on what we said previously, the answer is 3.

aev [14]1 year ago

3 0

I think thats a trick question on the periodic table there is no Z, theres Zi which is zinc but no Z

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You stand on a bathroom scale in a moving elevator. what happens to the scale reading if the cable holding the elevator suddenly

Viefleur [7K]

A bathroom scales works due to gravity. Under normal conditions, a reading can be obtained when your body is pushing some force on the scale. However in this case, since you and the scale are both moving downwards, so your body is no longer pushing on the scale. Therefore the answer is:

<span>The reading will drop to 0 instantly</span>

7 0

1 year ago

Read 2 more answers

An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its

Alja [10]

Answer:

Terminal velocity of object = 12.58 m/s

Explanation:

We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.

Gravitational force = mg = 80 * 9.8 = 784 N

Drag force = $12.0v+4.00v^2$

Equating both, we have

$784=12.0v+4.00v^2\\ \\ v^2+3v-196=0\\ \\ (v-12.58)(v+15.58)=0$

So v = 12.58 m/s or v = -15.58 m/s ( not possible)

So terminal velocity of object = 12.58 m/s

7 0

1 year ago

Water is to be boiled at sea level (1 atm pressure) in a 30-cm-diameter stainless steel pan placed on top of a 18-kW electric bu

Tamiku [17]

Answer:

Explanation:

18 kW = 18000 J /s

60% of 18kW = 10800 J/s

Latent heat of evaporation of water

= 2260 x 10³ J / kg

kg of water being evaporated per second

= 10800 / 2260 x 10³ kg /s

= 4.7787 x 10⁻³ kg / s

= 4.78 gm / s .

3 0

1 year ago

A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above t

ipn [44]

Answer:

ball clears the net

Explanation:

$v_{o}$ = initial speed of launch of the ball = 20 ms^{-1}

$\theta$ = angle of launch = 5 deg

Consider the motion of the ball along the horizontal direction

$v_{ox}$ = initial velocity = $v_{o} Cos\theta = 20 Cos5 = 19.92 ms^{-1}$

$t$ = time of travel

$X$ = horizontal displacement of the ball to reach the net = 7 m

Since there is no acceleration along the horizontal direction, we have

$X = v_{ox} t \\7 = v_{ox} t\\t = \frac{7}{v_{ox}}$ Eq-1

Consider the motion of the ball along the vertical direction

$v_{oy}$ = initial velocity = $v_{o} Sin\theta = 20 Sin5 = 1.74 ms^{-1}$

$t$ = time of travel

$Y_{o}$ = Initial position of the ball at the time of launch = 2 m

$Y$ = Final position of the ball at time "t"

$a_{y}$ = acceleration in down direction = - 9.8 ms⁻²

Along the vertical direction , position at any time is given as

$Y = Y_{o} + v_{oy} t + (0.5) a_{y} t^{2}\\Y = 2 + (20 Sin5) (\frac{7}{20 Cos5}) + (0.5) (- 9.8) (\frac{7}{20 Cos5})^{2}\\Y = 2.00758 m\\$

Since Y > 1 m

hence the ball clears the net

7 0

1 year ago

A 50 kg rocket generates 990 N of thrust. What will be its acceleration if it is launched straight up?

Ilia_Sergeevich [38]

Answer:

The acceleration of the rocket is 10 m/s².

Explanation:

Let the acceleration of the rocket be $a$ m/s².

Given:

Mass of the rocket is, $m=50\ kg$

Thrust force acting upward is, $F_{th}=990\ N$

Acceleration due to gravity is, $g=9.8\ m/s^2$

Now, force acting in the downward direction is due to the weight of the rocket and is given as:

$W=mg=50\times 9.8=490\ N$

Now, net force acting on the rocket in upward direction is given as:

$F_{net}=F_{th}-W\\F_{net}=990-490=500\ N$

Therefore, from Newton's second law, net force acting on the rocket is equal to the product of mass and acceleration.

$F_{net}=ma\\500=50a\\a=\frac{500}{50}=10\ m/s^2$

Therefore, the acceleration of the rocket is 10 m/s².

4 0

1 year ago