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3 years ago

Martes

Physics

1 answer:

sdas [7]3 years ago

5 0

Explanation:

(9) Acceleration due to gravity is defined as the acceleration between two massive bodies when they are moving under free fall.

The relation between g and G is given by :

$g=\dfrac{GM}{r^2}$

Where

G is universal gravitational constant

M is mass of the earth

r is distance

(10) The unwanted product that can toxify the crops are called weeds. To control weeds, a chemical called weedicides are sprayed in the field. Some other methods to control them are Ploughing or tilling of soil etc.

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When a wire loop is connected to a battery, what is produced in the loop?

Vinil7 [7]

Answer:

magnetic field

Explanation:

3 0

3 years ago

Calculate the pressure exerted by a 4000N camel on the sand. The camel’s feet have a

Harrizon [31]

Answer:

The pressure exerted by camel feet is <u>2000 N/m²</u>.

Step-by-step explanation:

<h3><u>Solution</u> :</h3>

Here, we have given that ;

Force applied on camel feet = 4000 N
Total area of camel feet = 2 m²

We need to find the pressure exerted by camel feet.

As we know that :

${\longrightarrow{\pmb{\sf{Pressure= \dfrac{Area}{Force}}}}}$

Substituting all the given values in the formula to find the pressure exerted by camel feet.

$\begin{gathered} \begin{array}{l} {\longrightarrow{\sf{Pressure= \dfrac{Area}{Force}}}} \\ \\ {\longrightarrow{\sf{Pressure= \dfrac{4000}{2}}}} \\ \\ {\longrightarrow{\sf{Pressure= \cancel{\dfrac{4000}{2}}}}} \\ \\ {\longrightarrow{\sf{Pressure= 2000 \: N/{m}^{2}}}} \\ \\\star \: \small\underline{\boxed{\sf{\purple{Pressure= 2000 \: N/{m}^{2}}}}} \end{array}\end{gathered}$

Hence, the pressure exerted by camel feet is 2000 N/m².

$\rule{300}{2.5}$

3 0

2 years ago

where σ(t) and σ(0) represents the time-dependent and initial (i.e., time =0) stresses, respectively, and t and τ denote elapsed

lesya [120]

Answer:

$E_r(6)=4.35614\ MPa$

Explanation:

$\epsilon$ = Strain = 0.49

$\sigma _0$ = 3.1 MPa

At t = Time = 32 s $\sigma$ = 0.41 MPa

$\tau$ = Time-independent constant

Stress relation with time

$\sigma=\sigma _0exp\left(-\frac{t}{\tau}\right)$

at t = 32 s

$0.41=3.1exp\left(-\frac{32}{\tau}\right)\\\Rightarrow exp\left(-\frac{32}{\tau}\right)=\frac{0.41}{3}\\\Rightarrow -\frac{32}{\tau}=ln\frac{0.41}{3}\\\Rightarrow \tau=-\frac{32}{ln\frac{0.41}{3}}\\\Rightarrow \tau=16.0787\ s$