All categories

3 years ago

What is the term for the overall change in position?

2 answers:

8 0

Hi!

When we're looking at a basic overview of a change in position, this is known as displacement. Displacement is a vector quantity, compared to distance, which is a scalar quantity.

What this means is that displacement is focusing on magnitude and direction, instead of just the magnitude.

Hopefully, this helps! =)

mart [117]3 years ago

6 0

That's "displacement". It only depends on the beginning and ending locations, and doesn't care about the route between them.

You might be interested in

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago

A 0.2-kg steel ball is dropped straight down onto a hard, horizontal floor and bounces straight up. The ball's speed just before

nadya68 [22]

Answer:

Explanation:

Given

mass of steel ball $m=0.2\ kg$

initial speed of ball $u=10\ m/s$

Final speed of ball $v=-10\ m/s$ (in upward direction)

Impulse imparted is given by change in the momentum of object

therefore impulse J is given by

$J=\Delta P$

$\Delta P=m(v-u)$

$\Delta P=0.2(-10-10)$

$\Delta =-4\ N-s$

so magnitude of Impulse =4 N-s

7 0

3 years ago

Read 2 more answers

What is the final step in the fourth stage of technological design

Ierofanga [76]

Answer:

after a product has been improved and approved? reporting the results finding ways to lower costs selling a prototype determining criteria.

Explanation:

5 0

3 years ago

A 4 kg rock is dropped from 5 m. There is no friction. What kind of energy does is have before? What kind of energy does it have

denpristay [2]

1) The total mechanical energy of the rock is:
$E=U+K$
where U is the gravitational potential energy and K the kinetic energy.

Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
$E_i=U=mgh$
where $m=4 kg$ is the mass, $g=9.81 m/s^2$ is the gravitational acceleration and $h=5 m$ is the height.
Putting the numbers in, we find the potential energy
$U=mgh=(4 kg)(9.81 m/s^2)(5 m)=196.2 J$

2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
$E_f=K= \frac{1}{2}mv^2$
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
$K=U=196.2 J$

3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is:
$W=196.2 J-0 J=196.2 J$

6 0

3 years ago

Calculate the induced electric field (in V/m) in a 40-turn coil with a diameter of 11 cm that is placed in a spatially uniform m

Naddik [55]

Answer:

Explanation:

Given that,

Number of turn N = 40

Diameter of the coil d= 11cm = 0.11m

Then, radius = d/2 = 0.11/2 =0.055m

r = 0.055m

Then, the area is given as

A =πr²

A = π × 0.055²

A = 9.503 × 10^-3 m²

Magnetic Field B = 0.35T

Magnetic field reduce to zero in 0.1s, t = 0.1s

so we want to find induce electric field. To find the electric field,(E) we need to find the electric potential (V).

E.M.F is given as

ε = —N • dΦ/dt

Where magnetic flux is given as

Φ = BA

Then, ε = —N • dΦ/dt

ε = —N • dBA/dt

ε = —NBA/t

Then, its magnitude is

ε = NBA/t

Inserting the values of N, B, A and t

ε = 40×0.35×9.503×10^-3/0.1

ε = 1.33 V

Then, using the relationship between Electric field and electric potential

V = Ed

ε = E•d

E = ε/d

E = 1.33/0.11

E = 12.09 V/m

7 0

3 years ago