I remember c/d. That's not a problem. But if you want 'c', you'll have to give me 'd'.
Answer:
a)
b)
Explanation:
Given:
mass of bullet, 
compression of the spring, 
force required for the given compression, 
(a)
We know

where:
a= acceleration


we have:
initial velocity,
Using the eq. of motion:

where:
v= final velocity after the separation of spring with the bullet.


(b)
Now, in vertical direction we take the above velocity as the initial velocity "u"
so,

∵At maximum height the final velocity will be zero

Using the equation of motion:

where:
h= height
g= acceleration due to gravity


is the height from the release position of the spring.
So, the height from the latched position be:



Considering that the book is moving with constant speed, the force applied by Anna must be the same that the friction force:

If we clear the previous equation:
Answer:
a=0.212 m/s²
Explanation:
Given that
q= 10⁻⁹ C
m = 5 x 10⁻⁹ kg
Magnetic filed ,B= 0.003 T
Speed ,V= 500 m/s
θ= 45°
Lets take acceleration of the mass is a m/s²
The force on the charge due to magnetic filed B
F= q V B sinθ
Also F= m a ( from Newton's law)
By balancing these above two forces
m a= q V B sinθ



a=0.212 m/s²
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