Ask question

Ask question

All categories

3 years ago

15

F = 50 N m = 72 kg m/s2

1 answer:

lyudmila [28]3 years ago

3 0

Answer:

Explanation:

F=ma

a=F/m

a=50/72=

a=0.694

You might be interested in

Define power and discuss how to determine it.

serious [3.7K]

there are different types of power so ill show you all the examples of power.

4 0

3 years ago

Referring to the graph below of data describing a cart pulled across a level surface. When a 4 N force pulls on the cart, the ca

LenKa [72]

Answer:gjv

Explanation:

4 0

3 years ago

Read 2 more answers

The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr

MA_775_DIABLO [31]

Answer:

$q=49Q/64$

and

$x =16R/15$

Explanation:

See attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$ (1)

according to the law of gauss and superposition Law:

$E_{Q}=E_{1}+E_{2}=E_{2}$ ; electric field due to the small sphere with r1=R/4

$E_{Q}=kq_{2}/(r_{1}^{2})=$

$q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}$

then: $E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})$ (2)

on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

$r_{p}=2R-R/4=7R/4$ ⇒ $E_{q}=16kq/(49R^{2})$ (3)

We replace (2) y (3) in (1):

$E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})$

$q=49Q/64$

--------------------

if R<x<2R AND $E_{total}=E_{Q}-E_{q}=0$

$E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})$

remember that $q=49Q/64$

then:

$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R

so : $x =16R/15$

7 0

3 years ago

A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.460 Hz. The pendulum ha

zlopas [31]

Answer:

The moment of inertia is $I =1.0697 \ kg m^2$

Explanation:

From the question we are told that

The frequency is $f = 0.460 \ Hz$

The mass of the pendulum is $m = 2.40 \ kg$

The location of the pivot from the center is $d = 0.380 \ m$

Generally the period of the simple harmonic motion is mathematically represented as

$T = 2 \pi * \sqrt{ \frac{I}{ m * g * d } }$

Where I is the moment of inertia about the pivot point , so making I the subject of the formula it

=> $I = [ \frac{T}{2 \pi } ]^2 * m* g * d$

But the period of this simple harmonic motion can also be represented mathematically as

$T = \frac{1}{f}$

substituting values

$T = \frac{1}{0.460}$

$T = 2.174 \ s$

So

$I = [ \frac{2.174}{2 * 3.142 } ]^2 * 2.40* 9.8 * 0.380$

$I =1.0697 \ kg m^2$

4 0

3 years ago

The third floor of a house is 8m above street level. How much work is needed to move a 136kg refrigerator to the third floor?

jonny [76]

m = Mass of the refrigerator to be moved to third floor = 136 kg

g = Acceleration due to gravity by earth on the refrigerator being moved = 9.8 m/s²

h = Height to which the refrigerator is moved = 8 m

W = Work done in lifting the object

Work done in lifting the object is same as the gravitational potential energy gained by the refrigerator. hence

Work done = Gravitation potential energy of refrigerator

W = m g h

inserting the values

W = (136) (9.8) (8)

W = 10662.4 J

8 0

3 years ago