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Alexxandr [17]
3 years ago
11

What should you do when you cut your palm on a small piece of broken glass that is lying on the lab bench?

Physics
1 answer:
natali 33 [55]3 years ago
8 0

First, report this to your instructor. Small cuts should be cleaned and checked for broken glass. Bandages and dressings are available in the First-Aid Kit found in the laboratory. If bleeding is not stopping, apply pressure to the wound or affected area. Any treatment outside emergency first aid will be referred to the student infirmary. Severe emergencies will be referred to the Hospital emergency room.

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Where might someone with a pharmaceutical manufacturing career work?
klemol [59]

Answer:  People with careers in pharmaceutical manufacturing can expect to work in: research laboratories or manufacturing plants, and to maintain work hours during the day.

Explanation:

8 0
3 years ago
Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude
labwork [276]

Answer:\tau=1.03\times 10^{-4}\ N-m

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

\tau=NIAB\ \sin\theta

Let us assume that, A=0.0008\ m^2

\theta is the angle between normal and the magnetic field, \theta=90^{\circ}-30^{\circ}=60^{\circ}

Torque is given by :

\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m

So, the net torque about the vertical axis is 1.03\times 10^{-4}\ N-m. Hence, this is the required solution.

4 0
3 years ago
Two balls with equal masses, m, and equal speed, v, engage in a head on elastic collision. what is the final velocity of each ba
Allushta [10]
The collision is elastic. This means that both momentum and kinetic energy are conserved after the collision.

- Let's start with conservation of momentum. The initial momentum of the total system is the sum of the momenta of the two balls, but we should put a negative sign in front of the velocity of the second ball, because it travels in the opposite direction of ball 1. So ball 1 has mass m and speed v, while ball 2 has mass m and speed -v:
p_i = p_1-p_2 = mv-mv =0
So, the final momentum must be zero as well:
p_f = 0
Calling v1 and v2 the velocities of the two balls after the collision, the final momentum can be written as
p_f = mv_1 + mv_2 = 0
From which
v_1 = -v_2

- So now let's apply conservation of kinetic energy. The kinetic energy of each ball is \frac{1}{2} mv^2. Therefore, the total kinetic energy before the collision is
K_i = \frac{1}{2} mv^2 +  \frac{1}{2} mv^2 = mv^2
the kinetic energy after the collision must be conserved, and therefore must be equal to this value:
K_f = K_i = mv^2 (1)
But the final kinetic energy, Kf, is also
K_f =  \frac{1}{2} mv_1^2 +  \frac{1}{2}mv_2^2
Substituting v_1 = -v_2 as we found in the conservation of momentum, this becomes
K_f = mv_2 ^2
we also said that Kf must be equal to the initial kinetic energy (1), therefore we can write 
mv_2^2 = mv^2

Therefore, the two final speeds of the balls are
v_2 = v
v_1 = -v_2 = -v

This means that after the collision, the two balls have same velocity v, but they go in the opposite direction with respect to their original direction.

8 0
3 years ago
Which has the greater momentum when moving?
Basile [38]
I assume there are choices to this question that you forgot to include. No matter, I could just lay out the concept so that you can understand the gist.

The best way to approach this is to know the definition of momentum. In physics, momentum is always defined in terms of equation. For momentum, it is the product of the mass and velocity. Therefore, any increase of these two parameters would promote greater momentum. The greater the mass paired with the faster the velocity, the greater the momentum.
7 0
3 years ago
In addition to stand-alone discussion boards, all of these sites include discussion boards as part of their features except: ​
adell [148]

Answer:

A. Linkedln

Explanation:

In addition to stand-alone discussion boards, all of these sites include discussion boards as part of their features except: ​

A. Linkedln

All other application in addition to stand-alone Google, Yahoo, Microsoft, etc. include discussion board.

8 0
3 years ago
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