The distance the bird flies is (17,000 + 600 + 14,400) = 32,000 km.
His average speed is (32,000 km)/(122 days) = 262.3 km/day.
The bird ends up (17,000 - 600 + 14,400) = 30,800 km north
from where he started.
His average velocity is (30,800 km)/(122 days) = 252.5 km/day north .
Answer:
Answer:
safe speed for the larger radius track u= √2 v
Explanation:
The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.
Also given that r_1= smaller radius
r_2= larger radius curve
r_2= 2r_1..............i
let u be the speed of larger radius curve
now, \sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}∑F=
r
1
mv
2
=
r
2
mu
2
................ii
form i and ii we can write
v^2= \frac{1}{2} u^2v
2
=
2
1
u
2
⇒u= √2 v
therefore, safe speed for the larger radius track u= √2 v
Answer:
-600 J
Explanation:
F₁ = 8i +29 j + 32k
F₂ = 48 i - 59 j - 22 k
F = F₁ +F₂ = 8i +29 j + 32k +48 i - 59 j - 22 k
F = 56i - 30 j + 10 k
displacement d = ( 0 - 20 )i + ( 0 - 15 )j + ( 7 -0) k
d = - 20 i - 15 j + 7 k
Work Done = F dot product d
F . d = - 56 x 20 - 30 x - 15 + 10 x 7
= - 1120 +450 + 70
= -600 J
Answer:
= 0.417 m/s
Explanation:
Momentum before throwing the rock: m*V = 95.0 kg * 0.460 m/s
= 44.27 N*s
A) man throws the rock forward
mass of rock m1 = 0.310 kg
V1 = 15.5 m/s, in the same direction of the sled with the man
sled and man:
m2 = 95 kg - 0.310 kg = 94.69 kg
v2 = ?
Conservation of momentum:
momentum before throw = momentum after throw
44.27N*s = 0.310kg * 15.5m/s + 94.69kg*v2
⇒ v2 = [44.27 N*s - 0.310 * 15.5N*s ] / 94.69 kg
= 0.417 m/s
