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BARSIC [14]
3 years ago
9

How are the mass and weight an object related? Include a description with words and a equation.

Physics
1 answer:
Basile [38]3 years ago
6 0
The weight is the force experienced, whereas the mass represents the actual quantity of matter inside a body..
weigh on the surface of the earth is equal to mg
mass is m
and at the centre weight is 0 due to 0 acceleration that's 0 g
but mass is always constant and remains m, no matter where you are
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3. According to Hund's rule, what's the expected magnetic behavior of vanadium (V)?
ivanzaharov [21]

Answer:

Diamagnetic

Explanation:

Hunds rule states that electrons occupy each orbital singly first before pairing takes place in degenerate orbitals. This implies that the most stable arrangement of electrons in an orbital is one in which there is the greatest number of parallel spins(unpaired electrons).

For vanadium V ion, there are 18 electrons which will be arranged as follows;

1s2 2s2 2p6 3s2 3p6.

All the electrons present are spin paired hence the ion is expected to be diamagnetic.

6 0
3 years ago
A 400g sample of water absorbs 500j of energy. how did the water temperature change if the specific heat of water is 4.18j/g©. S
Mashutka [201]

In general, the quantity of heat energy, Q, required to raise a mass m kg of a substance with a specific heat capacity of <span>c </span>J/(kg °C), from temperature t1 °C to t2 °C is given by:

<span>Q </span>= <span>mc(t</span><span>2 </span><span>– t</span>1<span>) joules</span>

<span>So:</span>

(t2-t1) =Q / mc

<span>As we know:
Q = 500 J </span>

<span>m = 0.4 kg</span>

<span>c = 4180 J/Kg </span>°c

<span>We can take t1 to be 0</span>°c

t2 - 0 = 500 / ( 0.4 * 4180 )

t2 - 0 = 0.30°c

6 0
3 years ago
Friction is defined as *
den301095 [7]

Answer:

O a force that opposes motion

5 0
3 years ago
A car's bumper is designed to withstand a 5.04 km/h (1.4-m/s) collision with an immovable object without damage to the body of t
emmasim [6.3K]

Answer:

the magnitude of the average force on the bumper is 3189.8 N

Explanation:

Given the data in the question;

In terms of force and displacement, work done is;

W =F^> × x^>

W = Fxcos\theta    ------- let this be equation 1

where F is force applied, x is displacement and θ is angle between force and displacement.

Now, since the displacement of the bumper and force acting on it is in the same direction,

hence, θ = 0°

we substitute into equation 1

W = Fxcos( 0° )

W = Fx ------- let this be equation 2

Now, using work energy theorem,

total work done on the system is equal to the change in kinetic energy of the system.

W_{net = ΔKE

= \frac{1}{2}mv² -  \frac{1}{2}mu² --------- let this be equation 3

where m is mass of object, v is final velocity, u is initial velocity.

from equation 2 and 3

Fx = \frac{1}{2}mv² -  \frac{1}{2}mu²

we make F, the subject of formula

F = \frac{m}{2x}( v² - u² )

given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s

so we substitute

F = \frac{830}{(2)(0.255)}( (0)² - (1.4)² )

F = 1627.45098 ( 0 - 1.96 )

F = 1627.45098 ( - 1.96 )

F = -3189.8 N

The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.

Therefore, the magnitude of the average force on the bumper is 3189.8 N

6 0
2 years ago
Estimate the constant rate of withdrawal (in m3 /s) from a 1375 ha reservoir in a month of 30 days during which the reservoir le
kap26 [50]

Answer:

Explanation:

1 ha = 10⁴ m²

1375 ha = 1375 x 10⁴ m² = 13.75 x 10⁶ m²

In flow in a month = .5 x 10⁶ x 30 m³ = 15 x 10⁶ m³

Net inflow after all loss = 18.5 - 9.5 - 2.5 cm = 6.5 cm = .065 m

Net inflow in volume = 13.75 x 10⁶ x .065 m³= .89375 x 10⁶ m³

Let Q be the withdrawal in m³

Q - 15 x 10⁶ - .89375 x 10⁶ = 13.75 x 10⁶ x .75 = 10.3125 x 10⁶

Q = 26.20 x 10⁶ m³

rate of withdrawal per second

= 26.20 x 10⁶ / 30 x 24 x 60 x 60

= 26.20 x 10⁶ / 2.592 x 10⁶

= 10.11 m³ / s

6 0
3 years ago
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