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3 years ago

An endothermic reaction is a chemical reaction where heat is

Physics

2 answers:

Elenna [48]3 years ago

7 0

Answer:

absorbed/taken in

Explanation:

endo means internal; within.

exo means external; from outside.

MrRissso [65]3 years ago

7 0

Absorbed/taken in h the definition of endo is internal or within

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When the Moon is in the position shown, how would the Moon look from Earth?

Vlad [161]

Answer:When the Moon is in the position shown below, how would the Moon look to an observer on the North Pole? The moon's rotational period is equal to its revolutional period around Earth. This statement is false because at night we see one side and during the day we see the opposite.

Explanation:

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3 years ago

An object is placed at a known distance in front of a mirror whose focal length is also known. You apply the mirror equation and

Licemer1 [7]

Answer:

the image is behind the mirror, so the image is virtual and upright.

Explanation:

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4 years ago

A 0.25 kg baseball is in contact with a bat for 0.0075 s and exerts a force of 18,500 N. What impulse is delivered to the ball?

Alecsey [184]

The answer is 138.75 kgm/s

Impulse (I) is the multiplication of force (F) and time interval (Δt): I = F · Δt
It is given:
m = 0.25
F = 18,500 N = 18,500 kgm/s²
Δt = 0.0075 s
I = ?

I = F · Δt = 18,500 5kgm/s² * 0.0075 s = 18,500 * 0.0075 kgm/s² * s = 138.75 kgm/s

3 0

3 years ago

Read 2 more answers

Describe the valence electrons in nitrogen and how it could bond to other atoms?

Gekata [30.6K]

Answer:

Each nitrogen molecule consists of two atoms of nitrogen that are bonded by a triple covalent bond. This is because each nitrogen atom has 5 valence electrons. Each atom can complete its octet by sharing three electrons.

I hope this helps!

4 0

3 years ago

A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C

mario62 [17]

Answer:

3.62 m and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}$

$\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}$

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}$

$\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}$

x = - 1.4 m

6 0

3 years ago