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4 years ago

Which of the following best describes force

Physics

1 answer:

Delicious77 [7]4 years ago

4 0

B.a push or pull(which of the following best describes force)

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On which moment of inertia of an object does depend on?

Likurg_2 [28]

Additional Information:

I couldn't get your question very clearly. In order to solve the question, I will define moment of inertia, state the formula and factors that the moment of inertia of a body depends and does not depend on.

Answer:

<u>Moment of inertia depends on;</u>

1. Mass of the body

2. Axis of rotation and

3. Distribution of the body

<u>Moment of inertia does not depend on;</u>

1. Angular velocity of the body.

Explanation:

The moment of inertia is defined as a quantity that determines the torque needed for a desired angular acceleration or a property of a body due to which it resists angular acceleration about a rotational axis.

Moment of Inertia, I = ∑mr²

Where,

I is the moment of Inertia

m is the mass

r is the distance from the axis of the rotation

The moment of inertia of a body depends on distribution of body, axis of rotation and mass of the body. However, the moment of Inertia of a body is not dependent on angular velocity of the body.

4 0

4 years ago

You perform nine (identical) measurements of the acceleration of gravity (units of m/s2): 10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.6

AveGali [126]

Answer:

0.01

Explanation:

Given the data:

10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, 9.90

True value = 9.81

Mean value :

Σx / n

Sample size, n = 9

(10.1 + 9.87 + 9.76 + 9.91 + 9.75 + 9.88 + 9.69 + 9.83 + 9.90) / 9

= 88.69 / 9

= 9.854

Standard deviation (σ) :

Sqrt (Σ(X - m)² / n)

[(10.1 - 9.854)^2 + (9.87 - 9.854)^2 + (9.76 - 9.854)^2 + (9.91 - 9.854)^2 + (9.75 - 9.854)^2 + (9.88 - 9.854)^2 + (9.69 - 9.854)^2 + (9.83 - 9.854)^2 + (9.90 - 9.854)^2] / 9

Sqrt(0.113824 / 9)

Sqrt(0.0126471)

σ = 0.1124593

Standard Error = σ / sqrt(n)

Standard Error = 0.1124593 / 9

Standard Error = 0.0124954

Standard Error = 0.01 ( 1 significant digit)

3 0

3 years ago

a block suspended from a spring. The spring is stretch 0.5 m. If the spring constant is 500 N/m, what is the weight of the block

ANEK [815]

Answer:250N

Explanation:

Spring constant=500N/m

Extension=0.5m

Weight=spring constant x extension

Weight=500 x 0.5

Weight=250

Weight is 250N

8 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

PLS PLS PLS SOMEONE HELP ME LOL

mihalych1998 [28]

Jack Nicklaus once said about golf that "hitting a perfectly straight shot with a big club is a fluke." Science of Golf (SOG): Newton’s Third Law of Motion and Momentum explains what happens when a golfer swings a golf club and applies a big force to the small golf ball. The video features amateur golfer and Stanford University student Patrick Rodgers who gives his opinion on the importance of knowing what causes power and speed in the golf swing. Jim Hubbell, research engineer at the United States Golf Association (USGA) discusses what forces are and how they work in pairs. Hubbell also defines momentum and how it is transferred to the ball.

Hope this helps

8 0

3 years ago