By Newton's second law, we know that force ( $F$ ), measured in newtons, is the product of mass ( $m$ ), measured in kilograms, and net acceleration ( $a$ ), measured in meters per square second. That is:

$F = m\cdot a$ (1)

The initial force applied in the mass is:

$F = (10\,kg)\cdot \left(2.5\,\frac{m}{s^{2}} \right)$

$F = 25\,N$

In addition, we know that force is directly proportional to acceleration. If the smaller force is removed, then the initial force is reduced to $\frac{4}{5}$ of the initial force. The acceleration of the mass is:

$\frac{25\,N}{20\,N} = \frac{2.5\,\frac{m}{s^{2}} }{a}$

$a = 2\,\frac{m}{s^{2}}$

The acceleration of the mass is 2 meters per square second.

4 0

3 years ago

A solid ball is released from rest and slides down a hillside that slopes downward at 65.0" from the horizontal

PilotLPTM [1.2K]

Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
X: mgsin65 - F = mAx 
Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) 
Moment about center of mass: 
Fr = Iα 
Now Ax = rα 
and F = umgcos65 
mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) 
umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) 
ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) 
ugcos65 = 0.4(gsin65 - ugcos65) 
1.4ugcos65 = 0.4gsin65 
1.4ucos65 = 0.4sin65 
u = 0.4sin65/1.4cos65 
u = 0.613

3 0

3 years ago

How far apart would you have to place the poles of a 1. 5 v battery to achieve the same electric field?

Zarrin [17]

To place the poles of a 1. 5 v battery to achieve the same electric field is 1.5×10−2 m

The potential difference is related to the electric field by:

∆V=Ed

where,

∆V is the potential difference

E is the electric field

d is the distance

what is potential difference?

The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.

We want to know the distance the detectors have to be placed in order to achieve an electric field of

E=1v/cm=100v/cm

when connected to a battery with potential difference

∆v=1.5v

Solving the equation,we find

$d = \frac{ \:Δv}{e}$