<span>Ik it has something to do with not always being able to be used. Example: Goes dark at night therefore no sunlight some people say a but i would say d but the person that said it was a was not very trustable so yea i would go with D hope this helped:)</span>
Answer:
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Explanation:
For this problem let's use Newton's second law applied to each body
Body A
X axis
T = m_A a
Axis y
N- W_A = 0
Body B
Vertical axis
W_B - T = m_B a
In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension
We write the equations
T = m_A a
W_B –T = M_B a
We solve this system of equations
m_B g = (m_A + m_B) a
a = m_B / (m_A + m_B) g
In this initial case
m_A = M
m_B = M
a = M / (1 + 1) M g
a = ½ g
Let's find the tension
T = m_A a
T = M ½ g
T = ½ M g
Now we change the mass of the second block
m_B = 2M
a = 2M / (1 + 2) M g
a = 2/3 g
We seek tension for this case
T’= m_A a
T’= M 2/3 g
Let's look for the relationship between the tensions of the two cases
T’/ T = 2/3 M g / (½ M g)
T’/ T = 4/3
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Answer:
B . energy cannot be created or destroyed
Explanation:
first you have to find accelerarion, it is given that the initial velocity(u) is 3 m/s, distance travelled(s) be 2m finall it came to rest so final velocity be 0m/s
now using the 3rd law of motion
v^2=u^2+2as
0=9+2a2
a= -9/4m/s^2
now force=mass×accelration
=2kg×(-9/4)m/s^2
=4.5 N
4.5 newton force applied on the book!
✌️:)
Answer:
a) P = 44850 N
b) 
Explanation:
Given:
Cross-section area of the specimen, A = 130 mm² = 0.00013 m²
stress, σ = 345 MPa = 345 × 10⁶ Pa
Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa
Initial length, L = 76 mm = 0.076 m
a) The stress is given as:
on substituting the values, we get
or
Load, P = 44850 N
Hence<u> the maximum load that can be applied is 44850 N = 44.85 KN</u>
b)The deformation (
) due to an axial load is given as:
on substituting the values, we get
or
