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Alenkasestr [34]
3 years ago
7

0 What is 33 C in absolute temperature? hs

Physics
1 answer:
AysviL [449]3 years ago
7 0

Answer:

33 Celsius is 306.15 in absolute temperature

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HEY CAN ANYONE ANSA DIS!!!!
Nataly [62]

Answer: 7.25 * 10^5

Explanation:

1. The equation simplifies to (7*10 + 2.5) 10000.

2. Simplify in the parenthesis. 72.5 * 10000.

3. In scientific notation, the answer is 7.25* 10^5.

4 0
4 years ago
A person uses a match to light charcoal in a grill. Which statement describes
dolphi86 [110]

Answer:

It is flammable.

Explanation:

6 0
3 years ago
which has a higher acceleration:a 10kg object acted upon with a net force of 20N or an 18kg object acted on by a net force of 20
MA_775_DIABLO [31]
<span>Answer: The acceleration of 10 kg object is greater than that of 18 kg object.

Explanation:
According to Newton's Second law:
F = ma --- (A)

Let's find the acceleration for both 10 kg and 18 kg objects!
The net force on both of these masses = F = 20N

(1) Acceleration of 10 kg object
Mass = m = 10 kg
Plug in the values in equation (A):
20 = 10 * a
Acceleration = a = 2 m/s^2

(2) Acceleration of 18 kg object
Mass = m = 18 kg
Plug in the values in equation (A):

20 = 18 * a
Acceleration = a = 1.11 m/s^2


2 > 1.11; therefore, 10 kg object has the higher acceleration compared to the acceleration of the 18 kg object.</span>
7 0
4 years ago
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Why do the plants at the bottom of a pond grow better than plant on the bottom of a lake
yuradex [85]
Because pond water is the most freshest water than lake water
8 0
4 years ago
A speed skater moving across frictionless ice at 8.4 m/s hits a 5.7 m -wide patch of rough ice. She slows steadily, then continu
malfutka [58]

Answer:

Acceleration, a=-2.48\ m/s^2

Explanation:

Initial speed of the skater, u = 8.4 m/s

Final speed of the skater, v = 6.5 m/s

It hits a 5.7 m wide patch of rough ice, s = 5.7 m

We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :

v^2-u^2=2as

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}

a=-2.48\ m/s^2

So, the acceleration on the rough ice -2.48\ m/s^2 and negative sign shows deceleration.

8 0
3 years ago
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