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Kisachek [45]
2 years ago
14

During a medieval siege of a castle, the attacking army uses a trebuchet to heavy stones at the castle the trebuchet launches th

e stones with a velocity of + 48.5 m/s at an angle of 42.0 degrees how long does take the stone to hit the ground? For those settings, what is the maximum How high will the stones go Show all your work

Physics
1 answer:
vlada-n [284]2 years ago
7 0

Answer:

Explanation:

The question relates to time of flight of a projectile .

Time of flight = 2 u sinθ / g

u is speed of projectile , θ is angle of projectile

= 2 x 48.5 sin42 / 9.8

= 6.6 seconds  .

Maximum height attained

= u² sin²θ / g

= 48.5² sin²42 / 9.8

= 107.47 m .

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When the displacement in SHM is equal to 1/3 of the amplitude xm, what fraction of the total energy is (a) kinetic energy and (b
Nesterboy [21]

Answer:

Explanation:

Given

Displacement is \frac{1}{3} of Amplitude

i.e. x=\frac{A}{3} , where A is maximum amplitude

Potential Energy is given by

U=\frac{1}{2}kx^2

U=\frac{1}{2}k(\frac{A}{3})^2

U=\frac{1}{18}kA^2

Total Energy of SHM is given by

T.E.=\frac{1}{2}kA^2

Total Energy=kinetic Energy+Potential Energy

K.E.=\frac{1}{2}kA^2 -\frac{1}{18}kA^2

K.E.=\frac{8}{18}kA^2

Potential Energy is \frac{1}{8} th of Total Energy

Kinetic Energy is \frac{8}{9} of Total Energy

(c)Kinetic Energy is 0.5\times \frac{1}{2}kA^2

P.E.=\frac{1}{4}kA^2

\frac{1}{2}kx^2=\frac{1}{4}kA^2

x=\frac{A}{\sqrt{2}}                  

7 0
3 years ago
Read 2 more answers
Q1. Tractors are often used on sloping fields, so stability is important in their design. On the diagram, the center of the X ma
Vesna [10]

Answer:

a) The tractor has not toppled over because the the center of mass is acting in between the wheels

b) The features which affect its stability are;

1) The location of the center of gravity of the tractor close to the ground

2) The increased spacing between the left and the right wheels of the tractor

Q2. a) The center of mass wheel act in the region to the right away the wheel span (space in between the wheel) which creates a clockwise effect to, topple the buses over

b) It raises the bus's center of mass higher

c) To raise the center of mass to its highest practical level

Q3. At the bottom middle location within the tractor

Explanation:

a) The tractor has not toppled over because the vertical line from the center of gravity to the ground is still in between the wheel base such that since the tractor will topple over by turning clockwise about the right tire due to its weight, the weight of the tractor is currently acting in the anticlockwise direction to the right end tire of the tractor keeping the tractor's tires in good contact with the ground

b) The features which affect its stability are;

1) The location of the center of gravity of the tractor close to the ground

2) The increased spacing between the left and the right wheels of the tractor

Q2. a) When either of the buses are tilted further, the mass of the buses is acting on the tires and in their tilted condition, the mass is acting mainly on the right tire, whereby as the vertical line from the center of mass to the ground is in between the wheel bases the mass of the buses still serve s the restoring force to keep the bus on ground by providing anticlockwise moment where clockwise lotion is required for the bus to topple over

If the buses are tilted further, the vertical line from the center of mass will cross to the other side of the right tire such that the mass now provides clockwise moment turning the buses clockwise and since there are no opposing anticlockwise moment to balance that of the now clockwise moment of the buses it will continue to turn clockwise with the result that they eventually  topple over

b) When the upper deck is full of passengers, there will be an appreciable proportion of the total mass of the bus in the upper deck such that the level of the center of mass will be raised up higher

c) Sand bags are only put upstairs because by allowing for the disproportionate distribution of load such that the majority of the load of the bus is concentrated upstairs, the center of mass will be at its maximum height allowing the bus to be tested for stability in the worst case scenario

Therefore, the sands are only put upstairs to raise the center of mass

Q3. Being that the tractor is being used on a rough ground, the safest position will be at the lowest possible location at the middle of both the front and rear wheels and the left and right tires.

5 0
3 years ago
Which of the following modifications to a solenoid would be most likely to decrease the strength of its magnetic field?
Goryan [66]

By reading the fine details of the question, carefully and analytically, I have determined that there's no list of modifications to choose from.

The strength of the magnetic field of a solenoid depends on the electric current in its coil windings, the number of wire turns in its coil windings, and the material in its core.

In order to <em>DE</em>crease the strength of its magnetic field, any one or more of these steps could do the job:

-- DEcrease the electric current in its coil windings.  This can be accomplished by decreasing the voltage of the power source that energizes the coil, and/or increasing the resistance of the wire in the coil.

-- DEcrease the number of wire turns in the coil.

-- If the solenoid has anything in its core, change the core to something with a lower magnetic 'permeability'.  An Iron core will produce the greatest magnetic field strength.  Air, vacuum, or NO core will produce the lowest magnetic field strength.

8 0
3 years ago
What is kinematics????<br>explain briefly!!!!! ​
Marysya12 [62]

Answer:

i have only this much information

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3 0
2 years ago
Read 2 more answers
1. A ball is launched horizontally at 40 m/s while 6.0 m above a level field . How far will the ball move horizontally before st
Tresset [83]

Answer:

44 m

Explanation:

Given that,

Horizontal velocity of the ball, u = 40 m/s

It is 6 m above the level field.

We need to find the distance covered by the ball when move horizontally before striking the ground. Let it is d.

Firstly, we will find time taken for the ball to hit the ground. Using second equation of motion as follows :

s=ut+\dfrac{1}{2}at^2

Put u = 0 and a = g

s=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2s}{g}} \\\\t=\sqrt{\dfrac{2\times 6}{9.8}} \\\\t=1.1\ s

No finding the horizontal distance as follows :

d = ut

d = 40 m/s × 1.1 s

d = 44 m

So, the ball will move 44 m horizontally before striking the ground.

5 0
2 years ago
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