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jenyasd209 [6]
2 years ago
13

A projectile fired into the air, explodes, and splits into two halves of equal mass that hit the ground at the same time. If the

projectile had not exploded, it would have landed at point X, which is a distance R to the right of the launch point. After the explosion, one of the halves lands at point Y, which is a distance 2R to the right of the launch point. If air resistance is negligible, where did the other half land
Physics
1 answer:
steposvetlana [31]2 years ago
4 0

Answer:

At the Launch point

Explanation:

We are told that after the explosion, one of the halves landed at the point Y, which is a distance of 2R to the right side of the launch point.

This means that the only way one half of the object would have gone double the original distance of R is if it's velocity was two times that of the initial velocity. Furthermore, the only way that will make sense is if the other half fell straight down. So In conclusion, if air resistance is neglected, the other half will land at the initial velocity point which is at launch point.

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30. A box with mass 20kg is on a cement floor. The coefficient of static friction between the box and floor is 0.25. A man is pu
Mnenie [13.5K]

Answer:

84.05

Explanation:

  • F=mg×0.25
  • F=20x9.81×0.25
  • f=49.05N
  • F=35N

F=f+F

F=49.05+35

=84.05

7 0
3 years ago
Which group number below represents the Oxygen Group?
worty [1.4K]

Oxygen has an atomic number 8, because it has 8 protons and 8 electrons.

The first shell of an atom can hold up to 2 electrons but oxygen has 8 electrons, in that eight electrons 2 are in the first shell, so it has 6 more electrons left. The second shell can hold up to 8 electrons, oxygen has only 6 more electrons after the first shell is full, so it will have 6 electrons in the second shell

From this we know that oxygen has 2 shells so it is in period 2, and by counting from left to right, the sixth box in period 2 lies on group 16

Therefore Oxygen lies on group 16 and period 2

Happy to help :)

If you need my help for any other question, feel free to ask

8 0
3 years ago
Suggest values of the following in degree Celsius
ollegr [7]

Answer:

freezing water is 0°C

boiling water is 100°C

refrigerator 4°C

room is 20-22°C

3 0
2 years ago
A box is sliding down an incline tilted at a 12° angle above horizontal. The box is initially sliding down the incline at a spee
OLga [1]

Answer:

The box will cover a distance of 0.9199m before coming to rest

Explanation:

We are given;

Angle of tilt; θ = 12°

Speed of sliding down; u = 1.5 m/s

Coefficient of kinetic friction; μ = 0.34

We are told that the box is sliding down an incline tilted at a 12° angle above horizontal.

Thus,

The components of the weight of the block would be;

Fx = mg sinθ = mg sin 12

Fy = mg cosθ = mg cos 12

For, the normal force on the block, it will be counter balanced by the Y component of weight of block and so we have;

Normal force; Fn = mg cos 12

Now formula for the frictional force would be given by;

Ff = μmg cos 12

So, Ff = 0.34mg cos 12

So, the net force along the inclined plane is;

Fnet = Fx - Ff

Fnet = mg sin 12 - 0.34mg cos 12

Where Fnet = mass x acceleration.

Thus;

ma = mg sin 12 - 0.34mg cos 12

m will cancel out to give;

a = g sin 12 - 0.34g cos 12

a = 9.81(0.2079) - 0.34(9.81 × 0.9781)

a = -1.223 m/s²

According to Newton's equation of motion, we have;

(v² - u²) = 2as

s = (v² - u²)/2a

Final velocity is zero. Thus;

s = (0² - 1.5²)/(2 × -1.223)

s = -2.25/-2.446

s = 0.9199 m

Thus, the box will cover 0.9199m before coming to rest

4 0
3 years ago
Suppose the Sun appeared to you 900 times dimmer than it does now. How far away from the Sun would you be? (A) 1/9 AU (B) 3 AU
Harrizon [31]

Answer:

The correct answer is option 'c': 30 AUs

Explanation:

For a spherical wave front emitted by sun with total energy 'E' the energy density over the surface when it is at a distance 'r' from the sun is given by

e=\frac{E}{4\pi r^{2}}

This energy per unit area is sensed by observer as intensity of the sun.

Let the initial intensity of sun at a distance r_{1} be e_{1}

Thus if the sun becomes 900 times dimmer we have

e'=\frac{e_{1}}{900}\\\\\frac{E}{4\pi r_{2}^{2}}=\frac{1}{900}\times \frac{E}{4\pi r_{1}^{2}}\\\\\Rightarrow r_{2}^{2}={r_{1}^{2}}\times 900\\\\\therefore r_{2}={r_{1}}\times {30}

Thus the distance increases 30 times.

5 0
3 years ago
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