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Soloha48 [4]
3 years ago
5

Tyyyyyyyyyyyyyyyyyyyyyyyy​

Physics
2 answers:
andreyandreev [35.5K]3 years ago
5 0
Tyyyyyyyyyyyyyy for what
Aliun [14]3 years ago
4 0

Answer:

hub9hybygbgybgybgygybsbgydgbydxbgbyxdgbyxdyggdxygbyxdgybzgbydbgyzsbgydgbyzdgxbybgydzs

Explanation:

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Most automobiles have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made o
Colt1911 [192]

Answer:

There is a loss of fluid in the  container of 0.475L

Explanation:

To solve the problem it is necessary to take into account the concepts related to the change of voumen in a substance depending on the temperature.

The formula that describes this thermal expansion process is given by:

\Delta V = \beta V_0 \Delta T

Where,

\Delta V =Change in volume

V_0 =Initial Volume

\Delta T = Change in temperature

\beta = coefficient of volume expansion (Coefficient of copper and of the liquid for this case)

There are two types of materials in the container, liquid and copper, so we have to change the amount of Total Volume that would be subject to,

\Delta V_T = \Delta V_l - \Delta V_c

Where,

\Delta V_l= Change in the volume of liquid

\Delta V_c= Change in the volume of copper

Then replacing with the previous equation we have:

\Delta V = \beta_l V_0 \Delta T- \beta_c V_0 \Delta T

\Delta V = (\beta_l-\beta_c)V_0\Delta T

Our values are given as,

Thermal expansion coefficient for copper and the liquid to 20°C is

\beta_c = 51*10^{-6}/\°C

\beta_l = 400*10^{-6}/\°C

V_0 = 16L

\Delta T = (95\°C-10\°C)

Replacing we have that,

\Delta V = (\beta_l-\beta_c)V_0\Delta T

\Delta V = (400*10^{-6}/\°C-51*10^{-6}/\°C)(16L)(95\°C-10\°C)

\Delta V = 0.475L

Therefore there is a loss of fluid in the container of 0.475L

6 0
3 years ago
Scientific notation for 836
JulijaS [17]

Answer:

8.36e2

Explanation:

Use a scientific calculator

8 0
3 years ago
Read 2 more answers
Which is a standard unit of measurement​
ruslelena [56]
There are some more like Time - second (s)
Amount of substance - mole (mole)
Electric current - ampere (A)
Temperature - kelvin (K)
Luminous intensity - candela (cd)

6 0
3 years ago
Two workers are sliding 350 kgkg crate across the floor. One worker pushes forward on the crate with a force of 390 NN while the
svetoff [14.1K]

Answer:

\mu_k=0.18

Explanation:

First, we write the equations of motion for each axis. Since the crate is sliding with constant speed, its acceleration is zero. Then, we have:

x: T+F-f_k=0\\\\y:N-mg=0

Where T is the tension in the rope, F is the force exerted by the first worker, f_k is the frictional force, N is the normal force and mg is the weight of the crate.

Since f_k=\mu_k N and N=mg, we can rewrite the first equation as:

T+F-\mu_k mg=0

Now, we solve for \mu_k and calculate it:

\mu_k=\frac{T+F}{mg}\\ \\\mu_k =\frac{220N+390N}{(350kg)(9.8m/s^{2})} =0.18

This means that the crate's coefficient of kinetic friction on the floor is 0.18.

6 0
3 years ago
Find the surface area of the prism 14yd 6yd 10yd
castortr0y [4]

the answer is 568 yd^2

.......

5 0
3 years ago
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