Answer:
A. 91 meters north
Explanation:
Take +y to be north.
Given:
v₀ = 13 m/s
a = 0 m/s²
t = 7 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (13 m/s) (7 s) + ½ (0 m/s²) (7 s)²
Δy = 91 m
The displacement is 91 m north.
25,000 Feet = 7620m
PE = mgh where m is mass, g is gravity accel: 9.8 n h is height
= 90 x 9.8 x 7620
= 6720840J
= 6.72MJ
F = ma where m is mass, a is accel = gravity = 9.8
= 90 x 9.8
= 882N
Accel = gravity = 9.8m/s^2
KE = 1/2mv^2 where m is mass n v is vel
if no wind resistance, PE leaving airplane = KE at net
6720840 = 1/2 x 90 x v^2
v^2 = 149352
v = 386.5m/s
Answer:
His average speed was 10.3199 m/s.
Explanation:
You are given the mass of a sphere that is 26 kg sphere and it is released from rest when θ = 0°. You are also given the force of the spring that is F = 100 N. You are asked to find the tension of the spring. Imagine that the sphere is connected to a spring. The spring exerts a tension and the spring exerts gravitational pull. This will follow the second law of newton.
T - F = ma
T = ma + F
T = 26kg (9.81m/s²) + 100 N
T = 355.06 N
9.8 ms^-2 is acceleration