Answer:
Acceleration a ≤ 3 m/s^2
the greatest acceleration that the truck can have without losing its load is 3 m/s^2
Explanation:
For the truck to accelerate without losing its load.
Acceleration force of truck must be less than or equal to the maximum friction force between the truck bed and the load.
Fa ≤ F(friction)
But;
Fa = mass × acceleration
Fa = ma
ma ≤ F(friction)
a ≤ (F(friction))/m ......1
Given;
Fa = mass × acceleration
Fa = ma
mass m = 800 kg
F(friction) = 2400 N
Substituting the given values into equation 1;
a ≤ F(friction)/m
a ≤ 2400N/800kg
a ≤ 3 m/s^2
the greatest acceleration that the truck can have without losing its load is 3 m/s^2
Answer:
They will not meet
h-hX=1.2*g*t²
hX=v0*t-(1/2*g*t²)
Explanation:
fall h=1/2*g*t²
elevation time if v0=20 m/s te=v0/g=20 m/s /9.81 m/s²=2.0387s
hmax=v0²/(2*g)=(400 m²/s²)/19.62 m/s²2=20.387 m
free fall
t=2.0387s yields hX=1/2*g*t²=20.387 m
h-hX=200m - 20.387 m=179,613 m.
so, the second body has not enough initianoal speed to reach a meeting point
Answer:
0.130
Explanation:
From the given data, the coefficient of static friction for each trial are:
1. 0.053
2. 0.081
3. 0.118
4. 0.149
5. 0.180
6. 0.198
The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198
= 0.779
So that;
the average coefficient of static friction = 
= 
= 0.12983
The average coefficient of static friction is 0.130
You can find your answer here www.rsc.org/education/teachers/resources/jesei/minerals/students.htm
