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KatRina [158]
3 years ago
5

If you are asked to modify the robots or drones that are currently used , what kind of modifications you would do and suggest an

y two fields where it can be used.
Physics
1 answer:
Eddi Din [679]3 years ago
4 0

Explanation:

In recent times of pandemic, robots can be use as replacement of labor in the industries. Mundane tasks can be programmed in their system so that they can used readily.

Drones can used delivery for essential goods and services, so that human interference can be least and the spread of virus can be curbed.

In a recent example, Argentina where aerial data has reportedly been used to accelerate the construction of emergency hospitals.

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It's d

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Scientists study dark matter by looking at how it affects visible matter.

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What is lower body strength
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Find the resultant of an easterly force of 100 N and a southeast force of 80 N acting at 65 degrees to the 100 N force
saveliy_v [14]

Answer:

Resultant is 152 N at 28.5 degrees south to the 100 N force

Explanation:

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3 years ago
A black widow spider hangs motionless from a web that extends vertically from the ceiling above. If the spider has a mass of 1.2
leva [86]

Answer:

Tension = 0.012 N

Explanation:

If the black widow spider is hanging vertically motionless from the ceiling above. Then, the weight of the spider must be balancing the tension in the spider web. Therefore,

Tension = Weight

Tension = mg

where,

m = mass of spider = 1.27 g = 0.00127 kg

g = acceleration due to gravity = 9.8 m/s²

Therefore,

Tension = (0.00127 kg)(9.8 m/s²)

<u>Tension = 0.012 N</u>

8 0
3 years ago
In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m
ikadub [295]

(a) -39.4^{\circ}

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

p_y =0

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2

where

m is the mass of each ball

v_1= 4.60 m/s is the velocity of the cue ball after the collision

v_2 = 3.40 m/s is the velocity of the second ball after the collision

\phi_1=28.0^{\circ} is the angle of the cue ball with the x-axis

\phi_2 is the angle of the second ball

Solving for \phi_2, we find the angle between the direction of motion of the second ball and the original direction of motion:

sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m

The initial total momentum along the x-direction as

p_x = m u

where

m is the mass of the cue ball

u is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

mu = 6.69 m\\u = 6.69 m/s

7 0
3 years ago
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