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3 years ago

Name the applied force in the free body diagram:

Physics

2 answers:

MAVERICK [17]3 years ago

6 0

The applied force is the force from an external agent that tries to start the motion or stop the motion. Since, the block is being dragged in the horizontal direction, the applied force will be spring balance force because it drags the block to the left. Hence, it is the applied force. The forces in the vertical direction are gravity and normal force and the dotted force is frictional force.

Answer: The spring scale drags the box.

guapka [62]3 years ago

5 0

The spring scale dragging the box, since applied force is a force acting on an object due to a person or another object.

Even though the table is an object, the table pushing up is the normal force, which is perpendicular to the contact force.

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Will give brainliest to right answer!

viva [34]

The correct answer is D.

8 0

4 years ago

How can the IMA of a first- class lever be increased?

Dimas [21]

IMA = Ideal Mechanical Advantage

First class lever = > F1 * x2 = F2 * x1

Where F1 is the force applied to beat F2. The distance from F1 and the pivot is x1 and the distance from F2 and the pivot is x2

=> F1/F2 = x1 /x2

IMA = F1/F2 = x1/x2

Now you can see the effects of changing F1, F2, x1 and x2.

If you decrease the lengt X1 between the applied effort (F1) and the pivot, IMA decreases.

If you increase the length X1 between the applied effort (F1) and the pivot, IMA increases.

If you decrease the applied effort (F1) and increase the distance between it and the pivot (X1) the new IMA may incrase or decrase depending on the ratio of the changes.

If you decrease the applied effort (F1) and decrease the distance between it and the pivot (X1) IMA will decrease.

Answer: Increase the length between the applied effort and the pivot.

4 0

4 years ago

Read 2 more answers

Calculate the average linear momentum of a particle described by the following wavefunctions: (a) eikx, (b) cos kx, (c) e−ax2 ,

Maksim231197 [3]

Answer:

a) p=0, b) p=0, c) p= ∞

Explanation:

In quantum mechanics the moment operator is given by

p = - i h’ d φ / dx

h’= h / 2π

We apply this equation to the given wave functions

a) φ = $e^{ikx}$

.d φ dx = i k $e^{ikx}$

We replace

p = h’ k $e^{ikx}$

i i = -1

The exponential is a sine and cosine function, so its measured value is zero, so the average moment is zero

p = 0

b) φ = cos kx

p = h’ k sen kx

The average sine function is zero,

p = 0

c) φ = $e^{-ax^{2} }$

d φ / dx = -a 2x $e^{-ax^{2} }$

.p = i a g ’2x $e^{-ax^{2} }$

The average moment is

p = (p₂ + p₁) / 2

p = i a h ’(-∞ + ∞)

p = ∞

6 0

4 years ago

Help

malfutka [58]

The answer for this is 1200N

6 0

3 years ago

Se golpea una pelota de golf de manera que su velocidad inicial forma un ángulo de 45° con la horizontal. La pelota alcanza el s

nordsb [41]

Answer:

42m/s

6.06s

Explanation:

To find the initial velocity and time in which the ball is fling over the ground you use the following formulas:

$x_{max}=\frac{v_o^2sin(2\theta)}{g}\\\\x_{max}=vt_{max}$

θ: angle = 45°

vo: initial velocity

g: gravitational constant = 9.8m/s^2

x_max: max distance = 180 m

t_max: max time

by replacing the values of the parameters and do vo the subject of the first formula you obtain:

$v_o=\sqrt{\frac{gx_{max}}{sin(2\theta)}}\\\\v_o=\sqrt{\frac{(9.8m/s^2)(180m)}{sin(2(45\°))}}=42\frac{m}{s}$

with this value of vo you calculate the max time:

$t_{max}=\frac{x_{max}}{v}=\frac{x_{max}}{v_ocos(45\°)}\\\\t_{max}=\frac{180m}{(42m/s)cos(45\°)}=6.06s$

hence, the initial velocity of the ball is 42m/s and the time in which the ball is in the air is 6.06s

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TRANSLATION:

Para encontrar la velocidad inicial y el tiempo en el que la pelota está volando sobre el suelo, use las siguientes fórmulas:

θ: ángulo = 45 °

vo: velocidad inicial

g: constante gravitacional = 9.8m / s ^ 2

x_max: distancia máxima = 180 m

t_max: tiempo máximo

reemplazando los valores de los parámetros y haciendo el tema de la primera fórmula que obtiene:

con este valor de vo usted calcula el tiempo máximo:

por lo tanto, la velocidad inicial de la pelota es de 42 m / sy el tiempo en que la pelota está en el aire es de 6.06 s

4 0

3 years ago