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Kruka [31]
3 years ago
13

Some coastal regions of the world have cooler summers and warmer winters than inland regions at the same latitude. What accounts

for this difference in climates
Physics
1 answer:
maks197457 [2]3 years ago
8 0
Water holds in heat very well. Keep the temperature more steady and average. The areas around the water will also have a less variant change in temperature as a result. This property of water is known as high specific heat.
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An object with a mass of 78 kg is lifted through a height of 6 meters how much work is done
il63 [147K]
We got the following:
m = 78kg
h = 6m
g = 9.8 m/sec^{2} ≈ 10 m/sec^{2}
----------------------------------------------------------------------------
A = ?

As we know A = Ep = mgh    ->      potential energy
so the answer would be A = 78 * 6 * 10 = 4680 Joule


5 0
3 years ago
The Heat required to raise the temp. of 20 g water from 25 C to 36 C
Dennis_Churaev [7]
What is the question
5 0
3 years ago
How many electrons do alkali metals have in their outer shell?
BARSIC [14]
The outer shell can hold 1 electron
3 0
3 years ago
Read 2 more answers
A light, flexible cable is wrapped around a solid cylinder with mass 3.3 kg and a radius of 0.8 meters. The cylinder rotates on
kari74 [83]

Answer:

9.16rad/s^2

Explanation:

We are given that

Mass,m_1=3.3 kg

Radius,r=0.8 m

m_2=4.9 kg

Height,h=2.9 m

We have to find the angular acceleration of the cylinder.

According to question

4.9g-T=4.9a

Tr=I\alpha

Where

\alpha=\frac{a}{r}

Tr=\frac{1}{2}m_1ra

T=\frac{1}{2}m_1a=\frac{1}{2}(3.3)a

Substitute the value

4.9g-\frac{1}{2}(3.3a)=4.9a

4.9\times 9.8=4.9a+\frac{3.3a}{2}

Where g=9.8 m/s^2

48.02=a(4.9+1.65)=6.55a

a=\frac{48.02}{6.55}=7.33m/s^2

Angular acceleration,\alpha=\frac{a}{r}=\frac{7.33}{0.8}=9.16rad/s^2

7 0
3 years ago
Two blocks A and B have a weight of 11 lb and 6 lb, respectively. They are resting on the incline for which the coefficients of
Andreas93 [3]

Answer:

the block that starts moving first is block A ,    fr = 1.625 N ,  fr = 1.5 N

Explanation:

For this exercise we use Newton's second law, for which we take a reference system with the x axis parallel to the plane and the y axis perpendicular to the plane

X axis

       fr- Wₓ = 0

       fr = Wₓ

Axis y

      N- W_{y} = 0

      N = W_{y}

Let's use trigonometry to find the components of the weight

     sin θ = Wₓ / W

     Cos θ = W_{y} / W

     Wₓ = W sin θ

     W_{y} = W cos θ

     Wₓ = 11 sin θ

     W_{y} = 11 cos θ

The equation for friction force is

      fr = μ N

   

We substitute

      μ (W cos θ) = W sin θ

      μ = tan θ

We can see that the system began to move the angle.

         θ = tan⁻¹ μ

So the angles are

Block A      θ = tan⁻¹ 0.15

           θ = 8.5º

Block B      θ = tan⁻¹ 0.26

             θ = 14.6º

So the block that starts moving first is block A

The friction force is

         

Block A

         fr = Wx = W sin θ

         fr = 11 sin 8.5

         fr = 1.625 N

Block B

         fr = 6 sin 14.6

         fr = 1.5 N

5 0
3 years ago
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