F=ma so a=F/m
ax=180/270=0.67m/s^2
ay=390/270=1.44m/s^2
Magnitude = sqrt((0.67^2)+(1.44^2))=1.59m/s^2
Direction- Tan(x)=0.67/1.44=0.47 Tan^-1(x)=25 degrees
Answer:
The value is
Explanation:
From the question we are told that
The speed of the rope with hook is
The angle is
The speed at which it hits top of the wall is
Generally from kinematic equation we have that
Here h is the height of the wall so
=>
I do not have a calculator, but let’s think...
Imagine a ballon with a certain amount of gas. When you decrease the volume by squeezing the balloon, you increase the pressure.
Which means volume is inversely proportional to the pressure:
PV=PV
Now to speed things up because class is about to start :/
PV/T=PV/T, I think
Tip: check your units and any questions let me know
Answer:
r = 1.63×10^5 mi
Explanation:
Let r = distance of object from earth
Rs = distance between earth and sun
Ms = mass of the sun
= 3.24×10^5 Me (Me = mass of earth)
At a distance R from earth, the force Fs exerted by the sun on the object is equal to the force Fe exerted by the earth on the object. Using Newton's universal law of gravitation,
Fs = Fe
GmMs/(Rs - r)^2 = GmMe/r^2
This simplifies to
Ms/(Rs - r)^2 = Me/r^2
(3.24×10^5 Me)/(Rs - r)^2 = Me/r^2
Taking the reciprocal and then its square root, this simplifies further to
Rs - r = (569.2)r ----> Rs = 570.2r
or
r = Rs/570.2 = (9.3×10^7 mi)/570.2
= 1.63×10^5 mi
Answer:
Explanation:
It is given that,
Speed of one quoll around a curve, v = 3.2 m/s (maximum speed)
Radius of the curve, r = 1.4 m
On the curve, the centripetal force is balanced by the frictional force such that the coefficient of frictional is given by :
So, the coefficient of static friction between the quoll's feet and the ground in this trial is 0.74. Hence, this is the required solution.