All categories

4 years ago

Which one of the following quantities is measured by meters per-second squared?

Physics

2 answers:

Akimi4 [234]4 years ago

6 0

Answer is acceleration

Ann [662]4 years ago

4 0

(any unit of length) divided by (any unit of time)²

is a unit of acceleration.

You might be interested in

A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s2. the tangential acceleration of a point on its ri

vodka [1.7K]

We know that tangential acceleration is related with radius and angular acceleration according the following equation:
at = r * aa
where at is tangential acceleration (in m/s2), r is radius (in m) aa is angular acceleration (in rad/s2)
So the radius is r = d/2 = 1.2/2 = 0.6 m
Then at = 0.6 * 5 = 3 m/s2
Tangential acceleration of a point on the flywheel rim is 3 m/s2

5 0

3 years ago

A drunken sailor stumbles 580 meters north, 530 meters northeast, then 480 meters northwest. What is the total displacement and

kvv77 [185]

Answer:

(a) 1294.66 m

(b) 88.44°

Explanation:

d1 = 580 m North

d2 = 530 m North east

d3 = 480 m North west

(a) Write the displacements in vector forms

$\overrightarrow{d_{1}}=580\widehat{j}$

$\overrightarrow{d_{2}}=530\left ( Cos45\widehat{i}+Sin45\widehat{j} \right )$

$\overrightarrow{d_{2}}=374.77\widehat{i}+374.77\widehat{j}$

$\overrightarrow{d_{3}}=480\left ( - Cos45\widehat{i}+Sin45\widehat{j} \right )$

$\overrightarrow{d_{3}}=-339.41\widehat{i}+339.41widehat{j}$

The resultant displacement is given by

$\overrightarrow{d}\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}$

$\overrightarrow{d}=\left ( 374.77-339.41 \right )\widehat{i}+\left ( 580+374.77+339.41 \right )\widehat{j}$

$\overrightarrow{d}=35.36\widehat{i}+1294.18\widehat{j}$

magnitude of the displacement

$d ={\sqrt{35.36^{2}+1294.18^{2}}}=1294.66 m$

d = 1294.66 m

(b) Let θ be the angle from + X axis direction in counter clockwise

$tan\theta =\frac{1294.18}{35.36}=36.6$

θ = 88.44°

4 0

3 years ago

An emf is induced in a conducting loop of wire 1.22 m long as its shape is changed from square to circular. Find the average mag

Afina-wow [57]

Answer:

The induced emf in the loop is $7.35\times 10^{-4}\ V$

Explanation:

Given that,

Length of the wire, L = 1.22 m

It changes its shape is changed from square to circular. Then the side of square be its circumference, 4a = L

4a = 1.22

a = 0.305 m

Area of square, $A=a^2=(0.305)^2=0.0930\ m^2$

Circumference of the loop,

$C=2\pi r=L\\\\r=\dfrac{L}{2\pi}\\\\r=\dfrac{1.22}{2\pi}=0.194\ m$

Area of circle,

$A'=\pi r^2\\A'=\pi (0.194)^2\\\\A'=0.118\ m^2$

The induced emf is given by :

$\epsilon=\dfrac{\d\phi}{dt}\\\\\epsilon=\dfrac{\d(BA)}{dt}\\\\\epsilon=B\dfrac{A'-A}{t}\\\\\epsilon=0.125 \times \dfrac{0.118-0.0930}{4.25}\\\\\epsilon=7.35\times 10^{-4}\ V$