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GarryVolchara [31]

3 years ago

11

Is Particle B charged or uncharged? If charged, indicate whether it is charged positively or negatively. (n = neutron, p = proto

n, e = electron) Patice B 11p

1 answer:

diamong [38]3 years ago

7 0

Answer:

community College program and maddie d

Explanation:

work business work and maddie

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Lance arrives early at the airport (with flowers and balloons in hand) to welcome a friend. Her plane is delayed. While waiting,

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Answer:

A

Explanation:

Lance was never a bright young fella so he rolled down a hill and lost his left boot

4 0

2 years ago

The periodic wave in the diagram below has a frequency of 80. hertz.

ASHA 777 [7]

Answer:

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6 0

2 years ago

Light rays from an object after passing through a convex lens form an image at the focal point behind the lens (opposite side of

Delvig [45]

The actual position of the object is <span>at a great distance, effectively infinite. The other options given in the question are not at all correct. The correct option among all the options that are given in the question is the last option or option "D". I hope that this answer has actually come to your great help.</span>

7 0

3 years ago

In our first example we will consider a very simple application of Newton’s second law. A worker with spikes on his shoes pulls

sweet-ann [11.9K]

Answer:

Acceleration= $0.5m/s^2$

Speed=0.67 m/s

Explanation:

We are given that

Horizontal force=F=20 N

Mass of box=m=40 kg

We know that

Acceleration= $a=\frac{F}{m}$

Using the formula

Acceleration of box= $\frac{20}{40}=0.5m/s^2$

The acceleration of the box= $0.5m/s^2$

Initial velocity=u=0

Force=F=30 N

Distance=s=0.3 m

$a=\frac{30}{40}=\frac{3}{4} ms^{-2}$

$v^2-u^2=2as$

Substitute the values

$v^2-0=2\times \frac{3}{4}\times 0.3=0.45$

$v^2=0.45$

$v=\sqrt{0.45}=0.67m/s$

Hence, the speed of the box after it has been pulled a distance of 0.3 m=0.67 m/s

4 0

3 years ago

When the valve between the 2.00-L bulb, in which the gas pressure is 2.00 atm, and the 3.00-L bulb, in which the gas pressure is

padilas [110]

Answer:

$P_{C} = 3.2\, atm$

Explanation:

Let assume that gases inside bulbs behave as an ideal gas and have the same temperature. Then, conditions of gases before and after valve opened are now modelled:

Bulb A (2 L, 2 atm) - Before opening:

$P_{A} \cdot V_{A} = n_{A} \cdot R_{u} \cdot T$

Bulb B (3 L, 4 atm) - Before opening:

$P_{B} \cdot V_{B} = n_{B} \cdot R_{u} \cdot T$

Bulbs A & B (5 L) - After opening:

$P_{C} \cdot (V_{A} + V_{B}) = (n_{A} + n_{B})\cdot R_{u} \cdot T$

After some algebraic manipulation, a formula for final pressure is derived:

$P_{C} = \frac{P_{A}\cdot V_{A} + P_{B}\cdot V_{B}}{V_{A}+V_{B}}$

And final pressure is obtained:

$P_{C} = \frac{(2\,atm)\cdot (2\,L)+(4\,atm)\cdot(3\,L)}{5\,L}$

$P_{C} = 3.2\, atm$

5 0

3 years ago