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Morgarella [4.7K]
3 years ago
8

Raising 100 grams of water from 40 to 60 °C (the specific heat capacity of water is 1

Physics
2 answers:
asambeis [7]3 years ago
6 0

Answer:i think it is c

ANTONII [103]3 years ago
5 0
C. 2000 calories.

Explanation/calculation:

Specific heat capacity = calories / mass * (final temperature - initial temperature)

1 = calories / 100 * (60 - 40)
1 = calories / 100 * 20
1 * (100 * 20) = calories
1 * 2000 = calories
2000 = calories
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PLS HELP. WILL GIVE BRAINLIEST. PLS THIS IS WORTH 35 POINTS ON MY TEST
Nata [24]

Answer:

18m/s^2

Explanation:

Vf = Vi + at

t = distance/ average velocity

(120 + 0)/2 = 60 (average velocity)

400m/60m/s = 20/3 s

insert into first equation:

120 = 0 + a(20/3)

360 = 20a

18 = a

HOPE THIS HELPS!!!

5 0
2 years ago
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Where is the antinode of a standing wave?
anastassius [24]
What's the answer choices?
4 0
4 years ago
3.
Alexxx [7]

Answer:

120°

Explanation:

Draw a free body diagram.  There are three forces acting on the traffic light.  Two tension forces acting along the cables, and weight.

The tension forces have an angle θ between them.  That means each tension force forms an angle of θ/2 with respect to the vertical.  So the y component of each tension force is:

Ty = T cos (θ/2)

Sum of the forces in the y direction:

∑F = ma

Ty + Ty − W = 0

2 Ty = W

Substituting:

2 T cos (θ/2) = W

If W = T, then:

2 W cos (θ/2) = W

2 cos (θ/2) = 1

cos (θ/2) = 1/2

θ/2 = 60°

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6 0
4 years ago
What best describes myotibrils
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8 0
3 years ago
Before 1960, people believed that the maximum attainable coefficient of static friction for an automobile tire on a roadway was
Eduardwww [97]

Answer:

4.18

Explanation:

Givens  

The car's initial velocity  v_{i}= 0 and covering a distance Δx = 1/4 mi = 402.336 m in a time interval t = 4.43 s.  

Knowns

We know that the maximum static friction force is given by:

f_{s_max} =μ_s*n                         (1)

Where μ_s is the coefficient of static friction and n is the normal force.  

Calculations  

(a) First, we calculate the acceleration needed to achieve this goal by substituting the given values into a proper kinematic equation as follows:

Δx=v_{i} +\frac{1}{2} at^2

a=41 m/s

This is the acceleration provided by the engine. Applying Newton's second law on the car, so in equilibrium, when the car is about to move, we find that:  

f_{y}=n-mg=0\\ n=mg\\f_{x}=F-f_{s,max} =0\\ f_{s,max}=F=ma\\

Substituting (3) into (1), we get:

f_{s,max}= μ_s*m*g

Equating this equation with (4), we get:

ma=  μ_s*m*g

 μ_s=a/g

      =4.18

3 0
4 years ago
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