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kvasek [131]
3 years ago
13

Make 10 sentences using prefect continous and present perfect​

Physics
1 answer:
gtnhenbr [62]3 years ago
6 0

Answer:

5 present perfect continous sentences given below:

Explanation:

1,she has been writting a novel.2,I jave been watching tv for five hours.3, she has been riding a car.4,they have been working in this factory since 2001.5, kiran has been working in this shop since march.

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juan aims his boat directly across a river flowing at 8mi/hr. juan's boat travels at 6 miles/hr in still water. how was does jua
son4ous [18]
6mph:

Suppose the boat is traveling on a y axis. The river flow acts on the x axis. Motion on each axes are independent. The linear speed of the boat is not changed. Furthermore the projectile motion is changing, but you're specifically asking about the linear speed of the boat which is unchanged.
4 0
3 years ago
If time is tripled and work remains the same the power will.​
VMariaS [17]

Answer:

power will three times decrease (one-third)

Explanation:

P = W / t

thus when work is constant, power is inversely proportional to time

so, when time is tripled, power is decreased to one-third

7 0
3 years ago
Suppose 310. grams of ethanol (ethyl alcohol) is in an aluminum cup of 90.0 grams. Both of these are at 30.0C. A mass m of ice a
Ira Lisetskai [31]

Answer:

Explanation:

Given

mass of ethanol m_e=310\ gm

mass of aluminium cup m_{al}=90\ gm

both are at an initial temperature of T_i=30^{\circ}C

specific heat of ethanol c_e=2.46\ J/g-K

specific heat of aluminium c_{al}=0.9\ J/g-K

specific heat of ice c_i=2.108\ J/g-K

specific heat of water c_w=4.184\ J/g-K

Latent heat of fusion L=334\ J/gm

suppose m is the mass of ice added

Heat loss by Al cup and ethanol after 18^{\circ}C is reached

Q_1=(310\times 2.46+90\times 0.9)\cdot (30-18)

Heat gained by ice such that ice is melted and reached a temperature of 18^{\circ}C

Q_2=m\times 2.108\times (8.5)+m\times 334

Comparing 1 and 2 we get

m=23.65\ gm

Thus 23.65 gm of ice is added

                 

8 0
3 years ago
A large, 68.0-kg cubical block of wood with uniform density is floating in a freshwater lake with 20.0% of its volume above the
LenaWriter [7]

Answer:

a) V = 0.085 m^3

b) m = 17 kg

Explanation:

1) Data given

mb = 68 kg (mass for the block)

20% of the block volume is floating

100-20= 80% of the block volume is submerged

2) Notation

mb= mass of the block

Vw= volume submerged

mw = mass water displaced

V= total volume for the block

3) Forces involved (part a)

For this case we have two forces the buoyant force (B), defined as the weight of water displaced acting upward and the weight acting downward (W)

Since we have an equilibrium system we can set the forces equal. By definition the buoyant force is given by :

B = (mass water displaced) g = (mw) g   (1)

The definition of density is :

\rho_w = \frac{m_w}{V_w}

If we solve for mw we got m_w = \rho_w V_w  (2)

Replacing equation (2) into equation (1) we got:

B = \rho_w V_w g (3)

On this case Vw represent the volume of water displaced = 0.8 V

If we replace the values into equation (3) we have

0.8 ρ_w V g = mg  (4)

And solving for V we have

 V =  (mg)/(0.8 ρ_w g )

We cancel the g in the numerator and the denominator we got

V = (m)/(0.8 ρ_w)

V = 68kg /(0.8 x 1000 kg/m^3) = 0.085 m^3

4) Forces involved (part b)

For this case we have bricks above the block, and we want the maximum mass for the bricks without causing  it to sink below the water surface.

We can begin finding the weight of the water displaced when the block is just about to sink (W1)

W1 = ρ_w V g

W1 = 1000 kg/m^3 x 0.085 m^3 x 9.8 m/s^2 = 833 N

After this we can calculate the weight of water displaced before putting the bricks above (W2)

W2 = 0.8 x 833 N = 666.4 N

So the difference between W1 and W2 would represent the weight that can be added with the bricks (W3)

W3 = W1 -W2 = 833-666.4 N = 166.6 N

And finding the mass fro the definition of weight we have

m3 = (166.6 N)/(9.8 m/s^2) = 17 Kg

8 0
4 years ago
Covalent bonds tend to make atoms more stable by helping them
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Achieve a full outer shell
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