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ELEN [110]
3 years ago
12

The strong crystal structure of an ionic compound is one of the main reasons why ionic compounds have a __________ melting point

.
Physics
1 answer:
Alisiya [41]3 years ago
6 0

Answer:

high

Explanation:

The strong crystal structure of an ionic compound is one of the main reasons why ionic compounds have a high melting point.

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Leslie incorrectly balances an equation as 2C4H10 + 12O2 → 8CO2 + 10H2O.
tatiyna

Answer:

13 behind o2

Explanation:

answer is in photo above

4 0
3 years ago
A force did 80 j of work on an object in 4 m how big was the force
yanalaym [24]

Work = (force) x (distance)

80 J = (force) x (4 m)

Force = (80 J) / (4 m)  =  20 N

That's IF the force was in the same direction as the 4m of motion.
If the force was kind of slanted, then it had to be stronger, and
it had a component of 20N in the direction of the motion.

3 0
3 years ago
The equation of a transverse wave traveling along a very long string is y 6.0 sin(0.020px 4.0pt), where x and y are expressed in
zhannawk [14.2K]

Answer:

given

y=6.0sin(0.020px + 4.0pt)

the general wave equation moving in the positive directionis

y(x,t) = ymsin(kx -?t)

a) the amplitude is

ym = 6.0cm

b)

we have the angular wave number as

k = 2p /?

or

? = 2p / 0.020p

=1.0*102cm

c)

the frequency is

f = ?/2p

= 4p/2p

= 2.0 Hz

d)

the wave speed is

v = f?

= (100cm)(2.0Hz)

= 2.0*102cm/s

e)

since the trignometric function is (kx -?t) , sothe wave propagates in th -x direction

f)

the maximum transverse speed is

umax =2pfym

= 2p(2.0Hz)(6.0cm)

= 75cm/s

g)

we have

y(3.5cm ,0.26s) = 6.0cmsin[0.020p(3.5) +4.0p(0.26)]

= -2.0cm

6 0
3 years ago
Use the diagram to answer each question.
valentina_108 [34]

Figure A shows cross section of a land form or rock. In Figure B, compression stress is applied on it. When compression stresses are applied on a rock, it squeezes the rock cause fold or fracture. The fault formed by compression stress is called thrust fault. If the compression stresses/ force continue to act on a rock it will converge and form thrust fault. In Figure C, tension stresses is applied on the rock. When a tension stress applied on a rock it deforms/ lengthen. There are three type of deformations occur due to tension stresses. One is elastic deformation, in which, rock retains it original shape when force/stresses are removed. Second is plastic deformation, in which rock lengthen and change occur permanently. Third type of deformation is result into fracture or breaking of rock. In Figure C, shear stresses are applied on rock. Shear stresses are applied with equal magnitude but in opposite direction. It cause breaking of rock.

3 0
4 years ago
Read 2 more answers
A projectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same
Softa [21]

Answer:

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

-  175 m/s · sin 36.7° /  - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t

-  175 m/s ·  sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

7 0
3 years ago
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