Which two features are part of a line graph

5.

saul85 [17]

Explanation:

Newton thus the S I unit of force is newton

8 0

3 years ago

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The 20 kg at angle of 53⁰ in an inclined plane is realsed from rest the coefficient of friction bn the block and the inclined pl

Scorpion4ik [409]

Given :-

A 20kg block at an angle 53⁰ in an inclined plane is released from rest .
$\mu_s = 0.3 \ \& \ \mu_k = 0.2$

To Find :-

Would the block move ?
If it moves what is its speed after it has descended a distance of 5m down the plane .

Solution :-

For figure refer to attachment .

So the block will move if the angle of the inclined plane is greater than the angle of repose . We can find it as ,

$\longrightarrow \theta_{repose}= tan^{-1}(\mu_s)$

Substitute ,

$\longrightarrow \theta_{repose}= tan^{-1}( 0.2)$

Solve ,

$\longrightarrow\underline{\underline{\theta_{repose}= 16.6^o }}$

Hence ,

$\longrightarrow\theta_{plane}>\theta_{repose}$

Hence the block will slide down .

Now assuming that block is released from the reset , it's initial velocity will be 0m/s .

And the net force will be ,

$\longrightarrow F_n = mgsin53^o - \mu_k N$

Substitute, N = mgcos53⁰ ( see attachment)

$\longrightarrow ma_n = mgsin53^o - \mu_k mgcos53^o$

Take m as common,

$\longrightarrow\cancel{m }(a_n) = \cancel{m}( gsin53^o - \mu gcos53^o)$

Simplify ,

$\longrightarrow a_n = gsin53^o - \mu_k g cos53^o$

Substitute the values of sin , cos and g ,

$\longrightarrow a_n = 10( 0.79 - 0.2 (0.6))$

Simplify ,

$\longrightarrow a_n = 10 ( 0.79 - 0.12 ) \\\\ \longrightarrow a_n = 10 (0.67)\\\\ \longrightarrow \underline{\underline{a_n = 6.7 m/s^2}}$

Now using the Third equation of motion namely,

$\longrightarrow2as = v^2-u^2$

Substituting the respective values,

$\longrightarrow2(6.7)(5) = v^2-(0)^2$

Simplify and solve for v ,

$\longrightarrow v^2 = 67 m/s\\\\\longrightarrow v =\sqrt{67} m/s \\\\\longrightarrow\underline{\underline{ v = 8.18 m/s }}$

Hence the velocity after covering 5m is 8.18 m/s .

7 0

3 years ago

1. How much heat energy is required to raise the temperature of a 5 kg aluminium bar

Ray Of Light [21]

Answer:

180 kJ

Explanation:

Given that:

Mass (m) = 5 kg

Initial temperature (T1) = 28°C

Final temperature (T2) = 68°C

The change in temperature (ΔT) = T2 - T1 = 68°C - 28°C = 40°C

Specific heat capacity of aluminium (c) = 900 J/kg°C

The quantity of heat energy required (q) is given by:

q = mcΔT

q = 5 kg × 900 J/kg°C × 40°C

q = 180000 Joules

q = 180 kJ

Therefore 180 kJ is required to raise the temperature of aluminium from 28°C to 68°C.

5 0

3 years ago

Efficiency of electrons causes negative charge why?

Paraphin [41]

Answer:

Electron is negatively charged because you don't get work or in simple words charge out of it.

Explanation:

4 0

3 years ago

In the study of Physics, the "Universal Gas Law" is not considered a law at all.true or false

LiRa [457]

Answer: It’s False hope this helps

Explanation:

4 0

3 years ago