Which of the following is a positive statement?

A force off 700 newtons is applied to a 600 kg bowling ball. What is the acceleration of the bowling ball once the force is appl

Readme [11.4K]

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{700}{600} = \frac{7}{6} \\ = 1.1666666...$

We have the final answer as

Hope this helps you

6 0

2 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

2 years ago

A student is holding a stone at a certain height. The stone has 50 joules of potential energy and 0 joules of kinetic energy. Th

Hitman42 [59]

Answer:

d) The stone will have about 50 joules of kinetic energy and 0 joules of potential energy .

Explanation:

Given :

Initial Potential energy , $P_i=50\ J$ .

Initial Kinetic energy , $K_i=0\ J$ . ( because ball is in rest )

Now , we know , kinetic energy is maximum when an object reaches ground .

Also , potential energy is zero when an object is in ground .

We know , by conservation of energy :

Initial total energy = Final total energy

$P_i+K_i=P_f+K_f\\\\50+0=0+K_f\\\\K_f=50 \ J$

Therefore , option d) is correct .

6 0

2 years ago

Nick, a 60 kg physics student, wants to go bungee jumping, but doesn't have a bungee cord. He finds a 15 m long, strong spring (

vitfil [10]

Answer:

h = 24.81 m

Explanation:

Given:-

- The mass of the student, m = 60 kg

- The length of the spring, L = 15 m

- The spring constant, k = 60 N/m

Find:-

How far below the bridge is he hanging

Solution:-

- First realize that after the student attempts a bungee jump he oscillates violently ( dynamic motion ). After some time all the kinetic energy has been converted to Elastic and gravitational potential energy student is (stationary) and hanging down on one end of the spring.

- We will apply equilibrium condition on the student. We see that there are two forces acting on the student. The weight (W) of the student acting downward is in combat with the spring restoring force (Fs) acting upwards.

- Apply equilibrium condition in vertical direction:

Fs - W = 0

Fs = W

- The weight and spring force can be expressed as:

k*x = m*g

Where, g : Gravitational acceleration constant = 9.81 m/s^2

x : The extension of the spring from original position

- Solve for the extension (x) of the spring for this condition.

x = m*g / k

- Plug in the values and evaluate:

x = (60 kg)*(9.81 m/s^2) / (60 N/m)

x = 9.81 m

- The spring extends for about 9.81 m from its original length. So the distance (h) from edge of the bridge would be:

h = L + x

h = 15 + 9.81

h = 24.81 m

4 0

2 years ago

The potential energy of a 25 kg bicycle resting at the top of a 3 m high hill is _______ J

frozen [14]

The potential energy:
E p = m · g · h
m = 25 kg, g = 9.81 m/s², h = 3 m
E p = 25 kg · 9.81 m/s² · 3 m = 735.75 J

5 0

2 years ago

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