Explanation:
Q= mc∆T
∆T= 5-24=- 19
Q= 0.5*4186*-19
Q= -39767 J
negative sign show heat releases
Let the angle be Θ (theta)
Let the mass of the crate be m.
a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.
Normal force (N) = mg CosΘ
μ (coefficient of static friction) = 0.29
Static friction = μN = μmg CosΘ
Now, along the ramp, the equation of net force will be:
mg SinΘ - μmg CosΘ = 0
mg SinΘ = μmg CosΘ
tan Θ = μ
tan Θ = 0.29
Θ = 16.17°
b) Let the acceleration be a.
Coefficient of kinetic friction = μ = 0.26
Now, the equation of net force will be:
mg sinΘ - μ mg CosΘ = ma
a = g SinΘ - μg CosΘ
Plugging the values
a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96
a = 2.7244 - 2.44608
a = 0.278 m/s^2
Hence, the acceleration is 0.278 m/s^2
Answer:
might collapse
Explanation:
y'all r too heavy and tht shi will collapse and all y'all would die
The words "... as shown ..." tell us that there's a picture that goes along
with this question, and you decided not to share it. That's sad and
disappointing, but I think the question can be answered without seeing
the picture.
The net force on the crate is zero. Evidence for this is that fact that
the crate is just sitting there. If the net force on an object is not zero,
then the object is accelerating ... it's either speeding up, slowing down,
or its the direction of its motion is changing. If none of these things is
happening, then the net force on the object must be zero.
