(a) The minimum force F he must exert to get the block moving is 38.9 N.
(b) The acceleration of the block is 0.79 m/s².
<h3>
Minimum force to be applied </h3>
The minimum force F he must exert to get the block moving is calculated as follows;
Fcosθ = μ(s)Fₙ
Fcosθ = μ(s)mg
where;
- μ(s) is coefficient of static friction
- m is mass of the block
- g is acceleration due to gravity
F = [0.1(36)(9.8)] / [(cos(25)]
F = 38.9 N
<h3>Acceleration of the block</h3>
F(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32
a = F(net)/m
a = 28.32/36
a = 0.79 m/s²
Thus, the minimum force F he must exert to get the block moving is 38.9 N.
The acceleration of the block is 0.79 m/s².
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Answer:

Explanation:
As the path is straight, so the speed is equivalent to velocity. Now. assuming that the acceleration and deceleration of the train are constant. So, change of velocity with respect to time for acceleration as well as deceleration is constant. Hence, the slope of the speed-time graph is constant for the time of acceleration as well as deceleration. The speed for the time from
to
is constant, so slope for this interval of time is zero. The speed-time graph is shown in the figure.
The total distance covered by the train during the entire journey is the area of the speed-time graph.
Area


As velocity is in
and time is in
so the unit of area is 
Hence, the total distance is
.
According to the given statement Final velocity when they stick together is 8.735i^ + 11.25j^
<h3>What is collision and momentum?</h3>
The unit of momentum is kg ms -1. Momentum is a vector parameter that is influenced by the object's direction. During collisions involving objects, momentum is a relevant concept. The final velocity before a collision between two objects equals the total motion after the impact (in the absence of external forces).
<h3>Briefing:</h3>
From conservation of momentum
Initial momentum = final momentum
m u +M U =(m+M) V
2000×25 i^ +1500×30 j^ =(2000+1500) V
V = 8.735i^ + 11.25j^
Final velocity when they stick together is 8.735i^ + 11.25j^
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The complete question is -
A 2000 kg truck is moving eastward at 25 m/s. it collides inelastically with a 1500 kg truck traveling southward at 30 m/s. they collide at the intersection. Find the direction and magnitude of velocity of the wreckage after the collision, assuming the vehicles stick together after the collision.
Answer:
= 201.53 meters
Explanation:
A car started from rest and accelerated at 9.54 m/s^2 for 6.5 seconds. How much distance was covered by the car?
Use the formula d = 
where d is the distance, t is the time and "a" is the acceleration.

Answer: The radial acceleration of a point on the rim in two ways is 13.20 m/s^2
Explanation: Please see the attachments below