TIME REMAINING

A 90 kg painter is standing on a horizontal wooden scaffolding of length 10 m, which is supported on each end by a rope. The pai

RSB [31]

Explanation:

For equilibrium, $\sum M = 0$ .

So, $8 m \times mg - (10 m) T_{1}$ = 0

$T_{1} = \frac{8 \times mg}{10}$

= $\frac{8 \times 90 \times 9.8}{10}$

= 705.6 N

Also, for equilibrium $\sum F_{y}$ = 0

$T_{1} + T_{2} - mg$ = 0

or, $T_{2} = mg - T_{1}$

= $90 \times 9.8 - 705.6$

= 176.4 N

Thus, we can conclude that the tension in the first rope is 176.4 N.

8 0

4 years ago

Why is Darwin criticized for his theory? and who were they

kondaur [170]

A few people criticized Darwin for his theory, such as the left-leaning biologists Stephen Jay Gould and Richard Lewontin, who fear the political implications of Darwinian theory. They fear that evolutionary theory, even when bolstered by modern genetics, and molecular biology, does not make reality probable enough.

4 0

3 years ago

Read 2 more answers

John ( body mass= 65 kg) is taking off for a long jump . Horizontal accerleration ax is 5m/s^2 and vertical acceleration ay is 0

cestrela7 [59]

Answer:

(a) The horizontal ground reaction force $F_{g,x}=325\, N$

(b) The vertical ground reaction force $F_{g,y}=696\, N$

(c) The resultant ground reaction force $F_g=768\, N$

Explanation:

Given

John mass , m = 65 kg

Horizontal acceleration , $a_x= 5.0 \frac{m}{s^{2}}$

Vertical acceleration , $a_y=0.9 \frac{m}{s^{2}}$

(a) Using Newton's 2nd law in horizontal direction

$F_{g,x}=ma_x$

=> $F_{g,x}=65\times 5\, N=325\, N$

Thus the horizontal ground reaction force $F_{g,x}=325\, N$

(b) Using Newton's 2nd law in vertical direction

$F_{g,y}-mg=ma_y$

=> $F_{g,y}=mg+ma_y$

=> $F_{g,y}=65\times (9.81+0.9)\, N=696\, N$

Thus the vertical ground reaction force $F_{g,y}=696\, N$

(c) Resultant ground reaction force is

$F_g=(F_{g,x}^{2}+F_{g,y}^{2})^{\frac{1}{2}}$

=> $F_g=(325^{2}+696^{2})^{\frac{1}{2}}\, N=768\, N$

=> $F_g=768\, N$

Thus the resultant ground reaction force $F_g=768\, N$

5 0

3 years ago

Ask Your Teacher In the air over a particular region at an altitude of 500 m above the ground, the electric field is 120 N/C dir

strojnjashka [21]

Answer:

$1.475\times 10^{-13}\ C/m^3$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

A = Area

h = Altitude = 600 m

Electric flux through the top would be

$-110A$ (negative as the electric field is going into the volume)

At the bottom

$120A$

Total flux through the volume

$\phi=120-110\\\Rightarrow \phi=10A$

Electric flux is given by

$\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow q=\phi\epsilon_0\\\Rightarrow q=10A\epsilon_0$

Charge per volume is given by

$\rho=\dfrac{q}{v}\\\Rightarrow \rho=\dfrac{10A\epsilon_0}{Ah}\\\Rightarrow \rho=dfrac{10\epsilon_0}{h}\\\Rightarrow \rho=\dfrac{10\times 8.85\times 10^{-12}}{600}\\\Rightarrow \rho=1.475\times 10^{-13}\ C/m^3$

The volume charge density is $1.475\times 10^{-13}\ C/m^3$

7 0

3 years ago

A positively charged particle moves through an electric field. As part of a complicated trajectory, the particle passes through

kow [346]

Answer:

(B) The speed is larger at A than at B.

Explanation:

Point B, the final point of the trajectory, has higher electric potential than point A, the initial point of the trajectory, so the electric potential energy of the charged particle increases, which means that its kinetic energy must be decreasing, thus the speed at B must be lower than the speed at A.

8 0

3 years ago