How many nitrogen atoms are in one molecule of (NH4)2SO4?

1. Two forces F~ 1 and F~ 2 are acting on a block of mass m=1.5 kg. The magnitude of force F~ 1 is 12N and it makes an angle of

VLD [36.1K]

Answer:

Normal force=7.48 N

Explanation:

N+F~1 sinθ-mg=0

=>N=1.5*9.8-12 sin37◦

=>N=14.7-7.22=7.48 N

6 0

2 years ago

A positively charged object is brought near but not in contact with the top of an uncharged gold leaf electroscope. The experime

Olin [163]

Answer:

The leaves of the electroscope move further apart.

Explanation:

This is what happens; when the positive object is brought near the top, negative charges migrating from the gold leaves to the top. This is because the negative charges in the gold are attracted by the positive charge. Thus, it leaves behind a net positive charge on the leaves, though the scope remains neutral overall. To that effect, the leaves repel each other and move apart. If a finger touches the top of the electroscope at the moment when the positive object remains near the top, it basically grounds the electroscope and thus the net positive charge in the leaves flows to the ground through the finger. However, the positive object continues to "hold" negative charges in place at the top. Ar this moment the gold leaves have lost their net positive charge, so they no longer repel, and they move closer together. If the positive object is moved away, the negative charges at the top are no longer attracted to the top, and they redistribute themselves throughout the electroscope, moving into the leaves and charging them negatively.

Thus, the leaves move apart from each other again and we now have a negatively charged electroscope. If a negatively charged object is now brought close to the top, but without touching, the negative charges already in the electroscope will be repelled down toward the leaves, thereby making them more negative, causing them to repel more, and hence move even further apart.

So, the leaves move further apart.

7 0

2 years ago

The "atomic weight" of an atom reflects the average number of

Illusion [34]

E) Protons, neutrons, and electrons

8 0

3 years ago

Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s

alekssr [168]

Answer:

a) θ = 58.3º

b) vfh = 13.7 m/s

c) g = -9.8 m/s2

d) h = 22.2 m

e) vfb = 15.5 m/s

Explanation:

a)

Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
vₓ₀ = v * cos θ (1)
where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:

$x_{f} = v_{ox} * t = v_{o} * cos \theta * t (2)$

Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:

$cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)$

⇒θ = cos⁻¹ (0.526) = 58.3º (4)

b)

At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)

c)

At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)

d)

Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:

$\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)$

Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
Replacing (7) in (6), we get:

$\Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)$

e)

When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
The horizontal component, since it keeps constant, is just v₀x:
v₀ₓ = 13.7 m/s
The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:

$v_{fy} = v_{oy} - g*t (9)$

Replacing by the time t (a given), g, and v₀y from (7), we can solve (9) as follows:

$v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s (10)$

Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:

$v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)$

3 0

2 years ago

The pressure of liquid varies as per<br>its depth

Slav-nsk [51]

Answer:

Yes.

Explanation:

The pressure varies as per the depth of the container

4 0

3 years ago