Answer:
D(11) = 37.660 m
dD/dt = 2.7260 m/s
Explanation:
given data
two path apart = 17 m
walks east one path = 4 km/h = 1.111 m/s
walks west other path = 7 km/h = 1.944 m/s
pass each other time t = 0
solution
we consider here east is the positive direction and west is the negative direction
so that
the east - west distance between them is = 1.111 + 1.944 = 3.055 m/s
and
the actual distance between them time t is
D(t) = ![\sqrt{(3.055 t)^2 + 17 ^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%283.055%20t%29%5E2%20%2B%2017%20%5E2%7D)
at time 11 s
D(11) = ![\sqrt{(3.055 *11)^2 + 17 ^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%283.055%20%2A11%29%5E2%20%2B%2017%20%5E2%7D)
D(11) = 37.660 m
and
increase rate is dD/dt
dD/dt =
so for 11 sec
dD/dt =
dD/dt = 2.7260 m/s
Answer:
Option B. 28 N
Explanation:
From the question given above, the following data were obtained:
Coefficient of static friction (μ) = 0.35
Normal force (N) = 80 N
Frictional force (F) =?
The coefficient of static friction, frictional force and the normal force are related according to the following equation:
Coefficient of friction = frictional force / normal force
μ = F/N
With the above formula, we can obtain the frictional force as follow:
Coefficient of static friction (μ) = 0.35
Normal force (N) = 80 N
Frictional force (F) =?
μ = F/N
0.35 = F / 80
Cross multiply
F = 0.35 × 80
F = 28 N
Thus the frictional force is 28 N
Answer:
2m/s²
Explanation:
- V=Vstart+at
- rewrite that to find a so a=(V-Vstart)/t
part A of graph
part B of graph
the average between the two is both answers added divided by the number of answers
- (4m/s²–0m/s²)/2
- 4m/s²/2
- 2m/s²
Answer:
h1/h2 = ![\frac{2R^2}{3R^2 + h^2}](https://tex.z-dn.net/?f=%5Cfrac%7B2R%5E2%7D%7B3R%5E2%20%2B%20h%5E2%7D)
Explanation:
Using two rolls of tissue paper : One roll dropped normally while the other drops as some holds onto a sheet of the toilet paper ( I.e. the tissue paper drops rotating about its axis )
<u>Determine the ratio of heights h1/h2</u>
mass of tissues = same
radius of tissues = same
h1 = height of tissue 1
h2 = height of tissue 2
<em>For the first tissue ( Tissue that dropped manually ) </em>
potential energy = kinetic energy
mgh = 1/2 mv^2
therefore the final velocity ( v^2 ) = 2gH ----- ( 1 )
<em>second tissue ( Tissue that dropped while rotating ) </em>
gh =
( 3 +
) ------ ( 2 )
To determine the ratio of heights we will equate equations 1 and 2
hence :
gh =
( 3 +
)
∴ h1/h2 = ![\frac{2R^2}{3R^2 + h^2}](https://tex.z-dn.net/?f=%5Cfrac%7B2R%5E2%7D%7B3R%5E2%20%2B%20h%5E2%7D)
It could measure the person's speed or also known as velocity. The object could be similar to a police officer using a radar gun to measure someone's MPH/speed.