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3 years ago

The order of elements in the periodic table is based on?

Physics

2 answers:

frez [133]3 years ago

6 0

Answer:The number of protons in the nucleus.

hope this works :)

ahrayia [7]3 years ago

3 0

It is based on atomic number, which is the number of protons in the nuclei of the atoms of an element.

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A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s west. Find the ve

amm1812

<u>Answer</u>:

The velocity of the tennis racket after the collision 14.966 m/s.

<u>Step-by-step explanation:</u>

let the following:

m₁ = mass of tennis racket = 0.311 kg

m₂ = mass of the ball = 0.057 kg

u₁ = velocity of tennis racket before collision = 30.3 m/s

u₂ = velocity of the ball before collision = -19.2 m/s

v₁ = velocity of tennis racket after collision

v₂ = velocity of the ball after collision

Right (+) , Left (-)

An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies remains the same.

So, the total kinetic energy before collision = the total kinetic energy after collision.

So, 0.5 m₁ u₁² + 0.5 m₂ u₂² = 0.5 m₁ v₁² + 0.5 m₂ v₂² ⇒ (1)

Also, the total momentum before collision = the total momentum after collision.

So, m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂ ⇒ (2)

Solving (1) and (2):

∴ v₁ = [ u₁ * (m₁ - m₂) + u₂ * 2m₂ ]/ (m₁ + m₂)

= ( 30.3 * (0.311 - 0.057) - 19.2 * 2 * 0.057 ) / ( 0.311 + 0.057)

= 14.966 m/s.

So, the velocity of the tennis racket after the collision 14.966 m/s.

7 0

3 years ago

Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu

lora16 [44]

The initial force between the two charges is given by:

$F=k \frac{q_1 q_2}{d^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

$q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F$

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

$q_1' = q_1\\q_2' = q_2\\d' = 2d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}$

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

$q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F$

So the force has increased by a factor 6.

8 0

3 years ago

Read 2 more answers

Which is an advantage of storing data digitally?

ra1l [238]

Answer:

Explanation:

Because I said so

5 0

3 years ago

Read 2 more answers

Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th

aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0

3 years ago

PLEASE ANSWER ASAP bbb

bazaltina [42]

Answer:

if one bulb burns out the other bulbs will also turn off because they are connected to each other.

8 0

3 years ago