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3 years ago

How do atoms of copper differ from atoms of aluminum? *

Physics

1 answer:

Degger [83]3 years ago

8 0

Answer:

The copper atoms are heavier than the aluminum atoms. The copper atoms are smaller than the aluminum atoms so more copper atoms fit in the same volume. 2. Copper is more dense than aluminum.

HOPE THIS HELPS !!!!!

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Why do deer depend on plants to survive

jarptica [38.1K]

Deer depend on plants to survive because they are herbivores. Herbivores are animals that only eats plants and fruits in order to survive. Deer's meal includes grass and evergreen plants. If grass is unavailable, they eat whatever food like fallen leaves, twigs, bushes and other woody plants.

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3 years ago

Models of four atoms are shown as figures A, B, C, and D. Which two atoms represent isotopes of the same element?

timurjin [86]

C and d have the same amount of protons and electrons

3 0

3 years ago

A weightlifter curls a 30 kg bar, raising it each time a distance of 0.60 m. How many times must he repeat this exercise to burn

Sholpan [36]

Answe

given,

mass of the bar, m = 30 Kg

distance of rise, h = 0.60 m

Assuming the efficiency = 25 %

energy from the pizza slice = 300 C = 1260 kJ

To consume Energy from the pizza bar is to be pulled several number of time.

( energy from pizza ) x (efficiency) = n m g h

n is the number of lift

( 1260 x 10³) x (0.25) = n x 30 x 9.8 x 0.6

$n = \dfrac{315\times 10^3}{176.4}$

n = 1786 times.

Weightlifter should lift bar 1786 times to burn off the energy.

8 0

3 years ago

Two rams run toward each other. One ram has a mass of 44 kg and runs south with a speed of 6 m/s, while the other has a mass of

gladu [14]

Answer = D. 114 kg-m/s north

3 0

3 years ago

An object carries a +15.5 µC charge. It is 0.525 m from a -7.25 µC charge. What is the magnitude of the electric force on the ob

lorasvet [3.4K]

Answer:

the force of attraction between the two charges is 3.55 N.

Explanation:

Given;

first charge carried by the object, q₁ = 15.5 µC

second charge carried by the q₂ = -7.25 µC

distance between the two charges, r = 0.525 m

The force of attraction between the two charges is calculated as;

$F = \frac{kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(15\times 10^{-6})(7.25\times 10^{-6})}{(0.525)^2} \\\\F = 3.55 \ N$

Therefore, the force of attraction between the two charges is 3.55 N.

3 0

2 years ago