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Marta_Voda [28]
3 years ago
6

Compound name: for Al2O3

Physics
1 answer:
Deffense [45]3 years ago
6 0

Answer:

Aluminum oxide

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An 80-kg football player travels to the right at 8 m/s and a 120-kg player on the opposite team travels to the left at 4.0 m/s.
77julia77 [94]

Answer:

See Explanation

Explanation:

m1(v1) + m2(v2)

Opposite turns the plus to subtraction.

80(8) - 120(4.0)

60 - 480 = 160 kg m/s to the right

7 0
2 years ago
A crash test dummy is inside a test car moving at 55 km/h. How fast is the dummy moving relative to the seat he is sitting in?
kvasek [131]

The dummy is moving with a speed 0 km/h relative to the seat in which it is sitting.

If the relative speed was non-zero, the dummy would move away from its seat, which contradicts the problem formulation.

7 0
3 years ago
Read 2 more answers
F F= {mango, apple, banana, orange)​
Stels [109]

Answer:

<h3>n(F) = 4</h3>

Explanation:

Cardinality of a set is the number of elements in that set. Given the set.

F= {mango, apple, banana, orange)​, we are to determine the cardinality of the set i.e the amount of fruit present in the set. Cardinality of the set F is represented as n(F).

Since there are 4 different fruit in the given set F, hence the cardinality of the set F is n(F) = 4

6 0
3 years ago
What is the net force on this object f air = 400n (up) fgrav=600n (down)
n200080 [17]
Fnet=F1+F2 or Fnet=F1-F2
So 400n up - 600n down
Fnet= 400-600= -200N
5 0
3 years ago
Read 2 more answers
If a system has 225 kcal of work done to it, and releases 5.00 × 102 kj of heat into its surroundings, what is the change in int
vovikov84 [41]

We can solve the problem by using the first law of thermodynamics:

\Delta U = Q-W

where

\Delta U is the change in internal energy of the system

Q is the heat absorbed by the system

W is the work done by the system on the surrounding


In this problem, the work done by the system is

W=-225 kcal=-941.4 kJ

with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is

Q=-5 \cdot 10^2 kJ=-500 kJ

with a negative sign as well because it is released by the system.


Therefore, by using the initial equation, we find

\Delta U=Q-W=-500 kJ+941.4 kJ=441.4 kJ

8 0
3 years ago
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