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vladimir2022 [97]
3 years ago
11

What term describes an atoms tendencies to hold onto electrons?

Physics
1 answer:
Marat540 [252]3 years ago
5 0
Electronegativity? Not sure
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A man standing in front of a mountain beats a drum at regular intervals. The rate of drumming is generally increased and he find
Likurg_2 [28]
I think it is d I hope this help you if not let me know if it is not right
3 0
3 years ago
Initilal velocity of projectile at an angle of 22.5° is 10m/s. Then what is the magnitude of range of projectile?
crimeas [40]

Answer:

0.707m

Explanation:

from formula of range i.e R=Usin2Q/g

3 0
3 years ago
Two identical cars are parked in the sun on a warm day. One car is black and the other is white. Both cars are parked for three
cestrela7 [59]

Answer:

black will be hotter because black paint absorbs light and white reflects it.

4 0
2 years ago
Read 2 more answers
What ocean depth would the volume of an aluminium sphere be reduced by 0.10%
yKpoI14uk [10]

Answer:

6400 m

Explanation:

You need to use the bulk modulus, K:

K = ρ dP/dρ

where ρ is density and P is pressure

Since ρ is changing by very little, we can say:

K ≈ ρ ΔP/Δρ

Therefore, solving for ΔP:

ΔP = K Δρ / ρ

We can calculate K from Young's modulus (E) and Poisson's ratio (ν):

K = E / (3 (1 - 2ν))

Substituting:

ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)

Before compression:

ρ = m / V

After compression:

ρ+Δρ = m / (V - 0.001 V)

ρ+Δρ = m / (0.999 V)

ρ+Δρ = ρ / 0.999

1 + (Δρ/ρ) = 1 / 0.999

Δρ/ρ = (1 / 0.999) - 1

Δρ/ρ = 0.001 / 0.999

Given:

E = 69 GPa = 69×10⁹ Pa

ν = 0.32

ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)

ΔP = 64.0×10⁶ Pa

If we assume seawater density is constant at 1027 kg/m³, then:

ρgh = P

(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa

h = 6350 m

Rounded to two sig-figs, the ocean depth at which the sphere's volume is reduced by 0.10% is approximately 6400 m.

6 0
3 years ago
A 5 kg block is being pushed horizontally across a level surface at a constant velocity. What is the magnitude of the Normal for
luda_lava [24]

Answer:

50 N.

Explanation:

On top of a horizontal surface, the normal force acting on an object is equivalent to the force of gravity acting on the object. That is:

\displaystyle \begin{aligned} F_N = F_g & = ma \\ & = mg\end{aligned}

The mass of the block is 5 kg and the given force due to gravity is 10 N/kg. Substitute and evaluate:

\displaystyle F_N = F_g = (5\text{ kg})(10 \text{ N/kg}) = 50 \text{ N}

In conclusion, the normal force acting on the block is 50 N.

5 0
2 years ago
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