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1 month ago

What is the torque τa about axis a due to the force f⃗ ? express the torque about axis a at cartesian coordinates (0,0)?

Physics

1 answer:

Yuliya22 [3.3K]1 month ago

5 0

Response:

$\tau_a = F a sin \theta$

Explanation:

Torque is defined as:

$\tau = F d sin \theta$

where

F is the force's magnitude,

d represents the distance from where the force is applied to the center of rotation,

$\theta$ and is the angle between the force's direction and d.

In this question, we have:

$F$ , the force

$a$ , the distance of applying the force from the center (0,0)

$\theta$ , and the angle between the force's direction and a.

Thus, the torque is

$\tau_a = F a sin \theta$

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If an electronic circuit experiences a loss of 3 decibels with an input power of 6 watts, what would its output power be, to the

Yuliya22 [3333]

Answer:

The output power of the circuit is 3 Watts.

Given:

a loss in decibels = 3 dB

Input power = 6 Watts

To find:

What is the output power?

Formula used:

Output power = Input power × loss in ratio

Solution:

3 dB loss corresponds to a ratio of 0.5

Output power can be calculated as follows:

Output power = Input power × loss in ratio

Output power = 6 × 0.5

Output power = 3 Watts

Therefore, the output power of the circuit is 3 Watts.

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1 month ago

Read 2 more answers

-. What is the acceleration of 4 kg trolling bag pulled by a girl with a force of 3 N?

inna [3103]

Answer:

Acceleration(a) = 0.75 m/s²

Explanation:

Given:

Force(F) = 3 N

Mass of object(m) = 4 kg

Find:

Acceleration(a)

Computation:

Force(F) = ma

3 = (4)(a)

Acceleration(a) = 3/4

Acceleration(a) = 0.75 m/s²

3 0

2 months ago

A person who climbs up something (e.g., a hill, a ladder, the stairs) from the ground gains potential energy. a person's weight

kicyunya [3294]

The following values have been provided:

weight w = 240 lb = 1,067.52 N

energy E = 3,000 J

The equation for potential energy is:

E = w h

where h indicates the height that the person needs to ascend, therefore:

h = 3000 / 1067.52

h = 2.81 m

Thus, he must ascend 2.81 meters

3 0

1 month ago

a 75 kg man is standing at rest on ice while holding a 4kg ball. if the man throws the ball at a velocity of 3.50 m/s forward, w

Keith_Richards [3271]

Answer:

The resulting velocity for him will be 0.187 m/s in reverse direction.

Explanation:

Given:

The mass of the man is, $M=75\ kg$

The mass of the ball is, $m=4\ kg$

The initial velocity of the man is, $u_m=0\ m/s(rest)$

The initial velocity of the ball is, $u_b=0\ m/s(rest)$

The final velocity of the ball is, $v_b=3.50\ m/s$

The final velocity of the man is, $v_m=?\ m/s$

To determine this scenario, we employ the principle of momentum conservation.

This principle states that the total initial momentum equals the total final momentum.

Momentum is calculated by multiplying mass by velocity.

Initial momentum = Initial momentum of the man and the ball

Initial momentum = $Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm$

Final momentum = Final momentum of the man and the ball

Final momentum = $Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14$

Hence, the total initial momentum equals the total final momentum

$0=75v_m+14\\\\75v_m=-14\\\\v_m=\frac{-14}{75}\\\\v_m=-0.187\ m/s$

The negative sign indicates that the man moves backward.

Thus, his final velocity ends up being 0.187 m/s backward.

3 0

1 month ago

What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Yuliya22 [3333]

Details that are not provided: the problem figure is included.

We can address the exercise by applying Poiseuille's law. This law indicates that for a fluid flowing in a laminar manner within a confined pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ represents the pressure difference across the two ends

$\mu$ denotes the viscosity of the fluid

L signifies the length of the pipe

$Q=Av$ indicates the volumetric flow rate, where $A=\pi r^2$ is the cross-sectional area of the tube and $v$ refers to the fluid's velocity

r stands for the pipe's radius.

This law can be utilized for the needle, allowing us to compute the pressure difference between point P and the needle's end. In this scenario, we have:

$\mu=0.001 Pa/s$ is the dynamic viscosity of water at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Substituting these values into the formula yields:

$\Delta P = 3200 Pa$

This pressure difference specifies the value between point P and the needle's termination. As the end of the needle is under atmospheric pressure, the gauge pressure at point P, relative to atmospheric pressure, is exactly 3200 Pa.

8 0

2 months ago