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pychu [463]
3 years ago
5

PLEASE ANSWER FAST I WILL GIVE BRAINLIEST

Physics
2 answers:
IrinaVladis [17]3 years ago
5 0

Answer:

B

Explanation:

got right on edg

TiliK225 [7]3 years ago
3 0

Explanation:

The answer is:

A squirrel runs up the trunk of a tree.

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Describe the climate and name the climate zone at alaska; portland, oregon; and key west, florida.
Tpy6a [65]
Alaska-                   Subartic Climate 
Portland, Oregon-  Marine West Coast Climate
Key West, Florida- Tropical Savannah Climate
3 0
3 years ago
Given the following specific heat capacities, which material was have the largest change in temperature if 10 grams of each subs
Ilya [14]

Answer:

Explanation:

Comment

You could calculate it out by assuming the same starting temperature for each substance. (You have to assume that the substances do start at the same temperature anyway).

That's like shooting 12 with 2 dice. It can be done, but aiming for a more common number is a better idea.

Same with this question.

You should just develop a rule.  The rule will look like this

The greater the heat capacity the (higher or lower) the change in temperature.

The greater the heat capacity the lower the change in temperature

That's not your question. You want to know which substance will have the greatest temperature change given their heat capacities.

Answer

lead. It has the smallest heat capacity and therefore it's temperature change will be the greatest.

5 0
1 year ago
Read 2 more answers
How do I solve the following problem?
lukranit [14]
Use the impulse-momentum theorem.

Ft = mv

Substitute your known values:

(120kg)(20m/s) = F(1.5)

F= 1600N

Hope this helps!
7 0
3 years ago
What is the potential energy of a 45 kg object resting on the ground?
yKpoI14uk [10]

Answer:The potential energy is zero

Explanation:

3 0
3 years ago
A tank having a volume of 0.85 m 3 initially contains water as a two-phase liquid-vapor mixture at 260 o C and a quality of 0.7.
Keith_Richards [23]

Answer:=14,160 kJ

Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below

Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase

C MPa m^3/kg kJ/kg kJ/kg

1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture

2 260 4.689 0.0422 2599 2797 1 Saturated Vapor

The mass initially contained in the tank is m1 = V/v1

m1 =0.85 m^3 /0.02993 m^3 /kg

= 28.4 kg

The mass finally contained in the tank is

m2 =V2/v

= 0.85 m^3 /0.0422 m^3 /kg

= 20.14 kg

The heat transfer is then

Qcv = m2u2 − m1u1 − he(m2 − m1)

Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ

4 0
3 years ago
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