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tatuchka [14]
2 years ago
15

A block of size 20m x 10 mx 5 m exerts a force of 30N. Calculate the

Physics
1 answer:
Orlov [11]2 years ago
3 0

Answer:

We know that force applied per unit area is called pressure.

Pressure = Force/ Area

When force is constant than pressure is inversely proportional to area.

1- Calculating the area of three face:

A1 = 20m x 10 m =200 Square meter

A2 = 10 mx 5 m = 50 Square meter

A3 = 20m x 5 m = 100 Square meter

Therefore A1 is maximum and A2 is minimum.

2- Calculate pressure:

P = F/ A1 = 30 / 200 = 0.15 Nm⁻²  ( minimum pressure)

P = F / A2 = 30 / 50 = 0.6 Nm⁻²   ( maximum pressure)

Hence greater the area less will be the pressure and vice versa.

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A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is fa
elena55 [62]

Answer:

Answer:

15.67 seconds

Explanation:

Using first equation of Motion

Final Velocity= Initial Velocity + (Acceleration * Time)  

v= u + at

v=3

u=50

a= - 4 (negative acceleration or deceleration)  

3= 50 +( -4 * t)

-47/-4 = t

Time = 15.67 seconds

6 0
2 years ago
Emmy kicks a soccer ball up at an angle of 45° over a level field. She watches the ball's trajectory and notices that it lands,
Elenna [48]

Let u be the initial velocity of the soccer ball at an angle of inclination of \theta_0 with the positive x-axis.

Given that:

\theta_0=45^{\circ}

The horizontal distance covered by the projectile=20 m

Time of flight, t_f=2 seconds

Acceleration due to gravity, g= 10 m/s^2 downward.

As "north" and "up" as the positive x ‑ and y ‑directions, respectively.

So, g= -10 m/s^2

As the acceleration due to gravity is in the vertical direction, so the horizontal component of the initial velocity remains unchanged.

The x-component of the initial velocity, u_x=u\cos\theta_0.

The horizontal distance covered by the projectile = u_x\times t_f

\Rightarrow u_x\times t_f=20

\Rightarrow u_x\times 2=20

\Rightarrow u_x=10 m/s

So, the horizontal component of the velocity is 10 m/s which is constant and the graph has been shown in the figure (i).

Now,  u\cos(45^{\circ})=10 [as u_x=u\cos\theta_0]

\Rightarrow u=10\sqrt{2} m/s.

The vertical component of the initial velocity,

u_y= u\sin\theta_0

\Rightarrow u_y=10\sqrt{2}\sin(45^{\circ})

\Rightarrow u_y=10 m/s

Let v be the vertical component of the velocity at any time instant t.

From the equation of motion,

v=u+at

where u: initial velocity, v: final velocity, a: constant acceleration, and t: time taken to change the velocity from u to v.

In this case, we have u=u_y, a= -10 m/s^2.

So at any time instant, t.

v=u_y+(-10)t

\Rightarrow v=10-10t

The vertical component of the velocity, v, is the function of time and related as v=10-10t.

This is a linear equation.

At 2 second, the vertical component of the velocity

v=10-10x2=-10 m/s.

The graph has been shown in figure (ii).

7 0
2 years ago
In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is
Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

\theta = Angle = 20°

y = -20 cm

Velocity components

u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s

u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s

Acceleration components

a_x=0

a_y=-9.81\ m/s^2

y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0

t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629

Time taken is 0.26088 seconds

x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m

The distance the beetle travels on the ground is 0.3677181864 m

6 0
2 years ago
Determine the number of unpaired electrons in the octahedral coordination complex [fex6]3–, where x = any halide.
juin [17]
There will be four unpaired electrons
The metal complex is [FeX₆]³⁻
X being the halogen ligand 
X = F, CL, Br, and I
The oxidation of metal state is +3
The ground state configuration is
₂₆Fe =Is² 2s²2p⁶ 3s² 3p⁶ 3d⁶ 4s²
Metal, Fe(III) ion electron configures
₂₆Fe³⁺ = Is2 2s² 2p⁶ 3s² 3p⁶ 3d⁵
3 0
2 years ago
A uniform sphere with mass M and radius R is rotating with angular speed ω1 about a frictionless axle along a diameter of the sp
liq [111]

Answer:

W_2=\sqrt{\frac{3}{5} }W_1

Explanation:

For the first ball, the moment of inertia and the kinetic energy is:

I_1 =\frac{2}{5}MR^2

K_1 = \frac{1}{2}IW_1^2

So, replacing, we get that:

K_1 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2

At the same way, the moment of inertia and kinetic energy for second ball is:

I_2 =\frac{2}{3}MR^2

K_2 = \frac{1}{2}IW_2^2

So:

K_2 = \frac{1}{2}(\frac{2}{3}MR^2)W_2^2

Then, K_2 is equal to K_1, so:

K_2 = K_1

\frac{1}{2}(\frac{2}{3}MR^2)W_2^2 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2

\frac{1}{3}MR^2W_2^2 = \frac{1}{5}MR^2W_1^2

\frac{1}{3}W_2^2 = \frac{1}{5}W_1^2

Finally, solving for W_2, we get:

W_2=\sqrt{\frac{3}{5} }W_1

5 0
2 years ago
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