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Artyom0805 [142]

3 years ago

9

A car whose total mass is 800kg travelling with a uniform velocity of 20m/s suddenly observes a stationary dog on it's path 50m

ahead. If the total force applied on the breaking system on the car by the driver is 2000N what would most probably happen

1 answer:

Gelneren [198K]3 years ago

4 0

Answer:

Driver hits the dog

Explanation:

From kinematics equation

$v_f^{2}=v_i^{2}+2as$ and making a the subject

$a=\frac {v_f^{2}-v_i^{2}}{2s}=\frac {0-20^{2}}{2*50}=-4m/s^{2}$

F=ma=800*-4=-3200 N

The driver is required to apply a force of 3200 N to effectively not hit the dog. However, when the driver applies 2000 N, this is less the required force by 1200 N hence the driver hits the dog.

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A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material

My name is Ann [436]

Answer:

a) C = 4,012 10⁻¹⁴ F, b) Q = 1.6 10⁻¹¹ C , c) U = 3.21 10⁻¹¹ J

Explanation:

a) The capacitance of a capacitor is

C = k e₀ A / d

Let's calculate

C = 4 8.85 10⁻¹² 17 10⁻⁴ / 0.150 10⁻²

C = 4,012 10⁻¹⁴ F

b) let's look the charge

C = Q / ΔV

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c) The stored energy

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3 years ago

You will be colliding two carts in this lab. Select all the systems for which you predict the total momentum of the system will

baherus [9]

From the momentum conservation we know that the initial momentum is equal to the final momentum. The momentum in a singular way can be defined as the product between the mass and the velocity of an object. In the presented system, however, there are two objects, therefore the mass of both and the speed of both, before and after the collision must be taken into account. Mathematically we could describe this as

$m_1u_1+m_2u_2 = m_1v_1+m_2v_2$

Here,

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The correct answer is C.

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3 years ago

Define orbital velocity

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the Orbital Velocity is the velocity sufficient to cause a natural or artificial satellite to remain in orbit. Inertia of the moving body tends to make it move on in a straight line, while gravitational force tends to pull it down. The orbital path, elliptical or circular, representing a balance between gravity and inertia, and it follows a rue that states that the more massive the body at the centre of attraction is, the higher is the orbital velocity for a particular altitude or distance.

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3 years ago

If we decrease the amount of force applied to an object, and all other factors remain the same, the amount of work completed wil

Nat2105 [25]

A ) decrease.
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D ) not change.

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( Mind marking me for branliest? ; ) )

4 0

3 years ago

What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex +

Mumz [18]

Complete Question:

Given $\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$ at a point. What is the force per unit area at this point acting normal to the surface with $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$ ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, $\sigma_n = 28 MPa$

There are shear stresses acting on the surface since $\tau \neq 0$

Explanation:

$\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$

equation of the normal, $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$

$\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

Traction vector on n, $T_n = \sigma \b n$

$T_n = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

$T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]$

$T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z$

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

$\sigma_n = T_n . \b n$

$\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa$

If the shear stress, $\tau$ , is calculated and it is not equal to zero, this means there are shear stresses.

$\tau = T_n - \sigma_n \b n$

$\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau = \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z$

$\tau = \sqrt{(-5/\sqrt{2})^2 + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa$

Since $\tau \neq 0$ , there are shear stresses acting on the surface.

3 0

3 years ago