Answer:
Option C
Explanation:
Given that
Motor force is 250 N
Force of friction is 750 N
Weight is 8500 N
And, the normal force is 8500 N
Now based on the above information
Here length of the rector shows the relative magnitude forward force i.e. 250 N i..e lower than the frictional force i.e. backward and weight i.e. 8500 would be equivalent to the normal force
Queremos crear un diagrama general para calcular el área de un triangulo.
Este será algo como:
- Definir variables
- Pedirle al usuario que introduzca los valores deseados (de las variables).
- Leer los valores deseados y asignarlo a la variable correspondiente.
- Realizar la operación para calcular el área.
- Mostrar en pantalla el resultado.
Como naturalmente habra algunas variaciones segun el programa que utilicemos, lo voy a escribir de forma bastante general.
Primero definamos nuestras variables:
Por ejemple, en fortran usariamos algo como:
real:: B, H, A
Donde B será la variable que usaremos para la base, H para la altura, y A para el área.
Luego tenemos que escribir en pantalla algo que le diga al usario que debe introducir la base y el area.
Luego el programa debe ser capaz de leer ese input.
con algo de la forma:
B = read*input 1
H = read*input 2
Una vez tenemos definidas las variables, simplemente calculamos el área del triangulo:
A = H*B/2
Finalmente la podemos mostrar en pantalla con algo como:
print(A).
Lo que nos mostraría el valor del área.
Concluyendo, el diagrama en general sería:
- Definir variables
- Pedirle al usuario que introduzca los valores deseados (de las variables).
- Leer los valores deseados y asignarlo a la variable correspondiente.
- Realizar la operación para calcular el área.
- Mostrar en pantalla el resultado.
Si quieres aprender más, puedes leer:
brainly.com/question/21949109
Answer:
Explanation:
electric field at the location of electron
= 9 x 10⁹ x 7.2 / .03²
= 72 x 10¹² N/C
force on electron = electric field x charge on electron
= 72 x 10¹² x 1.6 x 10⁻¹⁹
= 115.2 x 10⁻⁷ N .
C )
work done = charge on electron x potential difference at two points
potential at .03 m
= 9 x 10⁹ x 7.2 / .03
= 2.16 x 10¹² V
potential at .001 m
= 9 x 10⁹ x 7.2 / .001
= 64.8 x 10¹² V
potential difference = (64.8 - 2.16 )x 10¹² V
= 62.64 x 10¹² V .
work done = 62.64 x 10¹² x 1.6 x 10⁻¹⁹
= 100.224 x 10⁻⁷ J .
D )
There will be no change in the magnitude of force on positron except that the direction of force will be reversed . In case of electron , there will be repulsion and in case of positron , there will be attraction .
Work done in case of electron will be positive and work done in case of positron will be negative .
electric field due to charge will be same in both the cases .
Answer:
13.6 N
Explanation:
Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.
Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.
We assume the motion is downwards in the direction of the hanging block, M.
We now write equations of motion for each block.
So
Mg - T = Ma (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.
Now, since both blocks do not move, a = 0.
So, Mg - T = M(0) = 0 and T - mgcos57 - F = m(0) = 0
Mg - T = 0 (3) and T - mgcos57 - F = 0 (4)
From (3), T = Mg
Substituting T into (4), we have
T - mgcos57 - F = 0
Mg - mgcos57 - F = 0
So, Mg - mgcos57 = F
F = Mg - mgcos57
F = (M - mcos57)g
Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.
We find F, thus
F = (2.94 kg - 2.85 kgcos57)9.8 m/s²
F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²
F = (2.94 kg - 1.552 kg)9.8 m/s²
F = (1.388 kg)9.8 m/s²
F = 13.6024 kgm/s²
F ≅ 13.6 N
The answer is refracts parallel to the axis of the lens
