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UNO [17]
3 years ago
9

Jet fighter planes are launched from aircraft carriers with the aid of their own engines and a catapult. If in the process of be

ing launched from rest, the thrust of a jet's engines is 2.50 x 10^5 N and after moving through a distance of 90 m the plane lifts off with a kinetic energy of 5.40 x 10^7 J, what is the work done on the jet by the catapult?
Physics
1 answer:
ANTONII [103]3 years ago
5 0

Answer:

W=315 x 10⁵ J

Explanation:

Given that

F= 2.5 x 10⁵ N

d= 90 m

K.E.=5.4 x 10⁷ J

We know that work done by all force is equal to the change in kinetic energy

Lets take work done by catapult is W

W + F.d= K.E.

W= 5.4 x 10⁷ -  2.5 x 10⁵  x 90 J

W= (540 - 25 x 9) 10⁵ J

W=315 x 10⁵ J

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3. A pendulum with a 1.0-kg weight is set in motion from a position 0.04 m above the lowest point on the path of the weight.
gavmur [86]

Answer: K.E = 0.4 J

Explanation:

Given that:

M = 1.0 kg

h = 0.04 m

K.E = ?

According to conservative of energy

K.E = P.E

K.E = mgh

K.E = 1 × 9.81 × 0.04

K.E = 0.3924 Joule

The kinetic energy of the pendulum at the lowest point is 0.39 Joule

6 0
3 years ago
A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arr
Paha777 [63]

Answer:

The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.

Determine Fx."

F_{x}=-1N.m

Explanation:

We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.

torque=cross product of force and position . mathematically this can be express as

T=r*F

Where

F=F_{x}i+(7N)j-(5N)k  and the position vector

r=(2m)i-(3m)j+(2m)k

using the determinant method to expand the cross product in order to determine the torque we have

\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\

by expanding we arrive at

T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\

since we have determine the vector value of the toque, we now compare with the torque value given in the question

(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\

if we directly compare the j coordinate we have

10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m

8 0
3 years ago
5. Gravitational and magnetic forces are examples of _______forces.
Anvisha [2.4K]

Answer:

Explanation:

5.non contact force

6.balanced force

7.unbalanced force

8.net force

9.zero

10.is the difference between two forces

mark me as brainliest please

8 0
3 years ago
The diagram below shows a 5.00-kilogram block
bixtya [17]

The name and strength of the force holding the block up is 50 N upward - Normal force.

The given parameters:

  • <em>Mass of the block, m = 5 kg</em>

The weight of the block acting downwards due to gravity is calculated as follows;

W = mg

where;

  • <em>g is acceleration due to gravity = 10 m/s²</em>

W = 5 x 10

W = 50 N <em>(</em><em>downwards</em><em>)</em>

Since the block is at rest, an a force equal to the weight of the block must be acting upwards. This force is known as normal reaction.

Fₙ = 50 N <em>(</em><em>upwards</em><em>)</em>

Thus, the name and strength of the force holding the block up is 50 N upward - Normal force.

Learn more about Normal force here: brainly.com/question/14486416

4 0
2 years ago
Read 2 more answers
An ant carries a morsel of food 4.26 meters along a straight path to his nest. He then turns around and follows the path back to
Basile [38]
The displacement is zero. The most important concept to understand is the difference between displacement and total distance traveled. Total distance traveled would be tracking the length of the entire path the ant walked for the whole time (4.26m x 2). Displacement is how far from a designated origin (here, the food source) the ant ended up at the end of the time. Mathematically, the ant walked 4.26m from food source to nest (+4.26m) and then walked from the nest to food source (-4.26m), so the net displacement is [+4.26] + [-4.26] = 0m.
6 0
3 years ago
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