An electric field has a positive test charge of 5.00 C placed in it. The force on the test charge is

Una caja de 5.0kg de masa se acelera desde el reposo a través del piso mediante una fuerza a una tasa de 2.0 /s2 durante 7.0s en

Nady [450]

Responder:

<h2>490 julios </h2>

Explicación:

Se dice que el trabajo se realiza cuando una fuerza aplicada a un objeto hace que el objeto se mueva a través de una distancia. El trabajo realizado por un cuerpo se expresa mediante la fórmula;

Workdone = Fuerza * Distancia

Como Fuerza = masa * aceleración,

Workdone = masa * aceleración * distancia

Masa dada = 5.0kg, aceleración = 2.0m / s² d =?

Para obtener d, usaremos una de las leyes del movimiento,

d = ut + 1 / 2at²

u = 0 (ya que el cuerpo acelera desde el reposo) yt = 7.0s

d = 0 + 1/2 (2) (7) ²

d = 49m

Workdone = 5 * 2 * 49

Workdone = 490 Julios

4 0

3 years ago

I’m so confused. please help. i don’t know what i’m suppose to do

aniked [119]

Answer:

Maybe put them in order ????

Explanation:

4 0

3 years ago

Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface

FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

$\tau=I\alpha$ (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

$\tau=Fd$ (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

$Fd=I\alpha\\\\I=\frac{Fd}{\alpha}$

Finally, you replace the values of all parameters in the previous equation for I:

$I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2$

The moment of inertia of the door around the hinges is 2 kgm^2

3 0

4 years ago

Person is lifting a 250 N dumbbell. The weight is 30 cm from the pivot point of the elbow. What force must be exerted five from

qwelly [4]

Refer to the diagram shown below.

The force, F, is applied at 5 cm from the elbow.

For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N

Answer: 1500 N

4 0

3 years ago

A person stands on the ball of one foot. The normal force due to the ground pushing up on the ball of the foot has magnitude 750

White raven [17]

Answer:

the force exerted on the foot by the tibia would be 2975 N

Explanation:

Given the data in the question;

To maintain equilibrium between the foot and the ball vertically, the addition normal normal force $N^>$ (750 N) and the tension in the Achilles tendon $F^>_{Achilles}$ (2225 N) must be equal to the force exerted on the foot by the tibia;

so

| $N^>$ | + | $F^>_{Achilles}$ | = | $F^>_{Tibia}$ |

so force exerted on the foot by the tibia will be;

| $F^>_{Tibia}$ | = | $N^>$ | + | $F^>_{Achilles}$ |

so we substitute IN OUR VALUES

| $F^>_{Tibia}$ | = 750 N + 2225 N

| $F^>_{Tibia}$ | = 2975 N

Therefore, the force exerted on the foot by the tibia would be 2975 N

3 0

3 years ago