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3 years ago

How can this system be brought to equilibrium?

Physics

1 answer:

dezoksy [38]3 years ago

5 0

Hello! The answer is B and B.

When a system is in equilibrium, this means that the net force/total force = 0N. Hence you need to add 10N to the 10N force acting towards the right so that the 20N acting in opposite directions will ‘cancel’ out to get a net force of 0N.

Average velocity is calculated by distance travelled divided by time. Hence, velocity = 50m/5s = 10ms^1

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What is the equivalent resistance of a circuit that contains four 75.02

viktelen [127]

Answer:

option A is the correct answer

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2 years ago

The wattage of a bulb is 24 W when it is connected to a 12 V battery. Calculate its effective wattage if it operates on a 6 V ba

tresset_1 [31]

It would be 12W because: 6v is half of 12v so half of 24w would be 12w

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3 years ago

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If a t-shirt gun can fire t-shirts with an initial speed of 15 m/s, what is the maximum distance (along horizontal, flat ground)

kotegsom [21]

Answer:

h = 11.47 m

Explanation:

Initial speed pf the t-shirt gun is 15 m/s

We need to find the maximum distance covered by the t-shirt. It is based on the conservation of energy. The maximum distance covered is given by :

$h=\dfrac{u^2}{2g}\\\\h=\dfrac{(15)^2}{2\times 9.8}\\\\h=11.47\ m$

So, it will cover a distance of 11.47 m.

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3 years ago

Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

Fantom [35]

a) Angular acceleration: $17.0 rad/s^2$

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

$W=mg$

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

$W_p =mg sin \theta$

where $\theta$ is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

$\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta$

where L is the length of the pencil.

The relationship between torque and angular acceleration $\alpha$ is

$\tau = I \alpha$ (1)

where

$I=\frac{1}{3}mL^2$

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for $\alpha$ , we find:

$\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}$

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when $\theta=10^{\circ}$ is

$\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2$

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude $mg$

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

$\tau = Fd$

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

$\tau =mg\frac{L}{2} cos \theta$

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

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3 years ago

Which if the following doesn't cause an object to accelrate it's q 4

Slav-nsk [51]

B. Constant velocity

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3 years ago

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