Answer:
1/7 kg
Explanation:
Refer to the attached diagram for enhanced clarity regarding the question.
One of the blocks weighs 1.0 kg and accelerates downward at 3/4g.
g denotes the acceleration due to gravity.
Let M represent the block with known mass, while 'm' signifies the mass of the other block and 'a' refers to the acceleration of body M.
Given M = 1.0 kg and a = 3/4g.
By applying Newton's second law; 
For the body with mass m;
T - mg = ma... (1)
For the body with mass M;
Mg - T = Ma... (2)
Combining equations 1 and 2 gives;
+Mg -mg = ma + Ma
Ma-Mg = -mg-ma
M(a-g) = -m(a+g)
Substituting M = 1.0 kg and a = 3/4g into this equation leads to;
3/4 g-g = -m(3/4 g+g)
3/4 g-g = -m(7/4 g)
-g/4 = -m(7/4 g)
1/4 = 7m/4
Multiplying gives: 28m = 4
m = 1/7 kg
Hence, the mass of the other box is 1/7 kg
Answer:
The air exiting from the hairdryer is moving at a speed of 10 m/s.
Explanation:
The thrust generated by the hairdryer enables it to maintain an elevation angled at 5° from vertical; thus, we derive from the force diagram
by substituting 
, 
into the equation and resolving for 
we find:
This thrust is linked to the speed of air ejection 
through the equation
where 
signifies the rate of air ejection, which is known to be
and since 
,
by inserting these values into equation (2), we obtain the value of as:
resulting in
which indicates the air velocity discharged from the hairdryer.
Explanation:
The term 'collision' refers to the interaction between two objects. There are two distinct types of collisions: elastic and inelastic.
In this scenario, two identical carts are heading towards each other at the same speed, resulting in a collision. In an inelastic collision, the momentum is conserved before and after the incident, but kinetic energy is lost.
After the event, both objects combine and move together at a single velocity.
The graph representing a perfectly inelastic collision is attached, illustrating that both carts move together at the same speed afterward.
A hiker proceeds 200 m west and subsequently another 100 m north, resulting in a displacement of 223 m. The direction can be determined using the trigonometric function where sin(angle) = opposite/hypotenuse, yielding an angle of 26.6 degrees. Therefore, the total displacement is 223 m at an angle of 26.6 degrees north of west.
Answer:
F=126339.5N
Explanation:
To compute the force required to escape, a free-body diagram for the hatch must be drawn. We will equate the downward and upward forces, thus applying the following equation:
Fw=W+Fi+F
where
Fw= force or weight exerted by the water column above the submarine.
To calculate Fw, we can use:
Fw=h. γ. A
h=height
γ=
specific weight of seawater = 10074N / m ^ 3
A=Area
Fw=28x10074x0.7=197467N
w represents the hatch weight = 200N
Fi denotes the internal pressure force in the submarine, which is 1 atm = 101325Pa. We can calculate this force using:
Fi=PA=101325x0.7=70927.5N
Finally, the force needed to open the hatch is determined by the original equation:
Fw=W+Fi+F
F=Fw-W+Fi
F=197467N-200N-70927.5N
F=126339.5N
