Answer:
v = 5.949 m / s
, a = 59.18 m / s²
Explanation:
Linear and angular quantities are related
v = w r
we search the magnitudes to the SI system
w = 95 rev / min (2π rad / 1rev) (1 min / 60 s) = 9.948 rad / s
d = 598 mm = 0.598 m
a) let's calculate the linear velocity
v = 9.948 0.598
v = 5.949 m / s
b) Cenripetal acceleration
a = v² / r
a = 5.94 2 / 0.598
a = 59.18 m / s²
Each Celsius degree is the size of 1.8 Fahrenheit degrees. So you need dip your Fahrenheit thermometer into the sample, see where you're starting, and then warm it up to a temperature that reads (37.1 x 1.8) = 66.8 Fahreheit degrees higher.
Answer:
t=L/
Explanation:
<u>solution:</u>
Let E be an observer, and B a second observer traveling with velocity 
as measured by E. If E measures the velocity of an object A as 
then B will measure A velocity as
= -
Applied here,
the walkway (W) and the man (M) are moving relative to Earth (E}, the velocity of the man relative to the moving walkway is
= 
- 
,
The time required for the woman, traveling at constant speed relative to the ground, to travel distance L relative to the ground is
:
t=L/
Answer:
a. 05cm from x axis
b. 8cm from x axis
Explanation:
If the net magnetic field is zero and the currents are in the same direction then the thanks point is between the currents i1 and i2 as show in the attachment below
a. Given that i1= 5A and i2=3A
Let assume the null point is xcm from current i1, then the null point will be (4-x)cm from current i2 since the total length is 4cm.
Now the magnetic field of the current i1 from the null point= to magnetic field of current i2 from the null point
B1=B2
μi1/2πx=μi2/2π(4-x)
i1/x=i2/(4-x)
5/x=3/(4-x)
20-5x=3x
8x=20
8x=2.5cm
since from the left of x axis is 2cm, then the null point is 2.5-2 which 0.5cm from the origin x axis.
The null point is 0.5cm from the origin x axis
b. If both current are flowing in opposite direction, the null point lies outside of the current.
Then with same analysis let assume the first current i1 is xcm from the null point and since the total length is 4cm the second current i2 will be (x-4)cm from the null point.
Also the magnetic field of the current i1 from the null point = to magnetic field of current i2 from the null point
B1=B2
μi1/2πx=μi2/2π(x-4)
i1/x=i2/(x-4)
5/x=3/(x-4)
5x-20=3x
2x=20
x=10cm.
This shows that the distance of the null point from current i1 is 10cm and the current i1 is 2cm from the x axis, then the null point is 10-2=8cm from the origin x axis.
The null point is 8cm from the x axis.
Check the attachment to see the diagram of the current and the null points
Answer:
6.65m/s
Explanation:
Using the equation of motion
S = ut + 1/2gt²
S is the height of fall
t is the time
u is the horizontal velocity
g is the acceleration due to gravity
Given
S = 300 + 50
S = 350m
t = 7.8seconds
g = 9.8m/s^2
Get S
S = 7.8u + 1/2(9.8)(7.8)²
S = 7.8u + 298.116
350 = 7.8u + 298.116
7.8u = 350 - 298.116
7.8u = 51.884
u = 51.884/7.8
u = 6.65m/s
Hence the rock's horizontal velocity was 6.65m/s
