There is no equation here
You use acceleration due to gravity
and 1/2 atsqr=d
therefore 1/2 * 9.8 * tsqr= d
Answer:
- 3 cm
Explanation:
From the mirror formula;
1/f = 1/v + 1/u ; where f is the focal length, v is the image distance, and u is the object distance.
1/-4.5 = 1/9 + 1/v
1/v = -1/4.5 - 1/9
= -1/3
Therefore;
v = -3 cm
Hence;
Image distance is - 3cm
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:
Primero, definimos el desplazamiento como la distancia entre la posición final y la posición inicial.
Así, si comenzamos abajo, luego subimos la escalera, y luego bajamos, la posición final y la posición inicial serán la misma
por lo que el desplazamiento es igual a cero.
La medida recorrida es el espacio total recorrido.
Es decir, si entre el principio y el final de la escalera hay una distancia D.
La persona que sube y baja, recorre esta distancia dos veces.
Entonces cuando una persona sube y baja la escalera, la medida de su trayectoria será 2*D.
