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3 years ago

20. Using the picture, how many neutrons does lithium have?

Physics

1 answer:

Igoryamba3 years ago

8 0

Answer:

No. of Neutrons = 3

Explanation:

The atomic number of Lithium is given as 3 in the symbol while the mass number is given as 5.941 which is approximately equal to 6.

Mass Number = No. of Protons + No. of Neutrons = 6

Atomic Number = Number of Electrons = No. of Protons = 3

Therefore,

Mass Number - Atomic Number = (No. of Protons + No. of Neutrons) - No. of Protons

Mass Number - Atomic Number = No. of Neutrons

No. of Neutrons = 6 - 3

<u>No. of Neutrons = 3 </u>

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What is the % error in using g = 10.0 m/s^2? (Take the value ofg as 9.8 m/s^2)

Alenkasestr [34]

Answer:

So percentage error will be 2 %

Explanation:

We have given initial value of acceleration due to gravity $g=10m/sec^2$

And final value of acceleration due to gravity $g=9.8m/sec^2$

We have to find the percentage error

We know that percentage error is given by $percentage\ error=\frac{initial\ value-final\ value}{initial\ value}\times 100$

So $percentage\ error=\frac{10-9.8}{10}\times 100=2$ %

4 0

3 years ago

If light travels at 3.00

Juliette [100K]

Yes light travels at 3.00

5 0

3 years ago

Read 2 more answers

An aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 30mm in diameter and i

ASHA 777 [7]

Answer:

$\delta = 0.385\,m$ (Compression)

Explanation:

The aluminium bar is experimenting a compression due to an axial force, that is, a force exerted on the bar in its axial direction. (See attachment for further details) Under the assumption of small strain, the deformation experimented by the bar is equal to:

$\delta = \frac{P\cdot L}{A \cdot E}$

Where:

$P$ - Load experimented by the bar, measured in newtons.

$L$ - Length of the bar, measured in meters.

$A$ - Cross section area of the bar, measured in square meters.

$E$ - Elasticity module, also known as Young's Module, measured in pascals, that is, newtons per square meter.

The cross section area of the bar is now computed: ( $D_{o} = 0.04\,m$ , $D_{i} = 0.03\,m$ )

$A = \frac{\pi}{4}\cdot (D_{o}^{2}-D_{i}^{2})$

Where:

$D_{o}$ - Outer diameter, measured in meters.

$D_{i}$ - Inner diameter, measured in meters.

$A = \frac{\pi}{4}\cdot [(0.04\,m)^{2}-(0.03\,m)^{2}]$

$A = 5.498 \times 10^{-4}\,m^{2}$

The total contraction of the bar due to compresive load is: ( $P = -180\times 10^{3}\,N$ , $L = 0.1\,m$ , $E = 85\times 10^{9}\,Pa$ , $A = 5.498 \times 10^{-4}\,m^{2}$ ) (Note: The negative sign in the load input means the existence of compressive load)

$\delta = \frac{(-180\times 10^{3}\,N)\cdot (0.1\,m)}{(5.498\times 10^{-4}\,m^{2})\cdot (85\times 10^{9}\,Pa)}$

$\delta = -3.852\times 10^{-4}\,m$

$\delta = -0.385\,mm$

$\delta = 0.385\,m$ (Compression)

7 0

3 years ago

A 73 kg swimmer dives horizontally off a 462 kg raft initially at rest. If the diver's speed immediately after leaving the raft

murzikaleks [220]

Answer:

Corresponding raft speed = -0.875 m/s (the minus sign indicates that the raft moves in the direction opposite to the diver)

Explanation:

Law of conservation of momentum gives that the momentum of the diver and the raft before the dive is equal to the momentum of the diver and the raft after the dive.

And since the raft and the diver are initially at rest, the momentum of the diver after the dive is equal and opposite to the momentum experienced by the raft after the dive.

(Final momentum of the diver) + (Final momentum of the raft) = 0

Final Momentum of the diver = (mass of the diver) × (diving velocity of the diver)

Mass of the diver = 73 kg

Diving velocity of the diver = 5.54 m/s

Momentum of the diver = 73 × 5.54 = 404.42 kgm/s

Momentum of the raft = (mass of the raft) × (velocity of the raft)

Mass of the raft = 462 kg

Velocity of the raft = v

Momentum of the raft = 462 × v = (462v) kgm/s

404.42 + 462v = 0

462v = -404.42

v = (-404.42/462) = -0.875 m/s (the minus sign indicates that the raft moves in the direction opposite to the diver)

Hope this Helps!!!

5 0

3 years ago

You apply a 56.3N force to a 4.2Kg cart towards the right. The cart accelerates towards the right at a rate of 0.8m/s2. What is

vovangra [49]

Answer:

52,94N

Explanation:

Frictional force is the opposing force that is created between two surfaces that try to move in the same direction or that try to move in opposite directions. The main purpose of a frictional force is to create resistance to the motion of one surface over the other surface.

Calculation:

Le the value of frictional force be Fk, therefore acceleration a = (F- Fk)/m

where :

F is the applied force

m the mass of the cart

a = (F- Fk)/m = 0.8

F - Fk = 0.8m

Fk = F - 0.8m

Fk = 56.3 - (0.8×4.2)

Fk = 52.94 N

The magnitude, in Newtons, of the net frictional force is therefore 52.94N

6 0

3 years ago