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maxonik [38]
2 years ago
8

A fighter jet is catapulted off an aircraft carrier from rest to 75 m/s. If the aircraft carrier deck is 100 m long, what is the

acceleration of the jet
Physics
1 answer:
egoroff_w [7]2 years ago
4 0

The acceleration of the jet is 28.1 m/s^2

Explanation:

Since the motion of the jet is a uniformly accelerated motion, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the jet in this problem, we have

u = 0

v = 75 m/s

s = 100 m

Solving for a, we find the acceleration:

a=\frac{v^2-u^2}{2s}=\frac{75^2-0}{2(100)}=28.1 m/s^2

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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A system delivers 1275 j of heat while the surroundings perform 855 j of work on it. calculate ∆esys in j.
kakasveta [241]
The first law of thermodynamics says that the variation of internal energy of a system is given by:
\Delta U = Q + W
where Q is the heat delivered by the system, while W is the work done on the system.

We must be careful with the signs here. The sign convention generally used is:
Q positive = Q absorbed by the system
Q negative = Q delivered by the system
W positive = W done on the system
W negative = W done by the system

So, in our problem, the heat is negative because it is releaed by the system: 
Q=-1275 J
while the work is positive because it is performed by the surrounding on the system:
W=+855 J

So, the variation of internal energy of the system is
\Delta U = -1275 J+855 J=-420 J
6 0
1 year ago
A block of size 20m x 10 mx 5 m exerts a force of 30N. Calculate the
Orlov [11]

Answer:

We know that force applied per unit area is called pressure.

Pressure = Force/ Area

When force is constant than pressure is inversely proportional to area.

1- Calculating the area of three face:

A1 = 20m x 10 m =200 Square meter

A2 = 10 mx 5 m = 50 Square meter

A3 = 20m x 5 m = 100 Square meter

Therefore A1 is maximum and A2 is minimum.

2- Calculate pressure:

P = F/ A1 = 30 / 200 = 0.15 Nm⁻²  ( minimum pressure)

P = F / A2 = 30 / 50 = 0.6 Nm⁻²   ( maximum pressure)

Hence greater the area less will be the pressure and vice versa.

3 0
1 year ago
A uniform disk has a mass of 3.7 kg and a radius of 0.40 m. The disk is mounted on frictionless bearings and is used as a turnta
yuradex [85]

Answer:

1.25 kgm²/sec

Explanation:

Disk inertia, Jd =

Jd = 1/2 * 3.7 * 0.40² = 0.2960 kgm²

Disk angular speed =

ωd = 0.1047 * 30 = 3.1416 rad/sec

Hollow cylinder inertia =

Jc = 3.7 * 0.40² = 0.592 kgm²

Initial Kinetic Energy of the disk

Ekd = 1/2 * Jd * ωd²

Ekd = 0.148 * 9.87

Ekd = 1.4607 joule

Ekd = (Jc + 1/2*Jd) * ω²

Final angular speed =

ω² = Ekd/(Jc+1/2*Jd)

ω² = 1.4607/(0.592+0.148)

ω² = 1.4607/0.74

ω² = 1.974

ω = √1.974

ω = 1.405 rad/sec

Final angular momentum =

L = (Jd+Jc) * ω

L = 0.888 * 1.405

L = 1.25 kgm²/sec

5 0
1 year ago
What is the temperature when a solid begins to liquefy
MrRa [10]

Answer:

Explanation:

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6 0
1 year ago
A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then
Elan Coil [88]

Answer:

A) W_{ff} =-744.12J

B) F_f=-W_{ff}*sin\theta /hy = 112.75N

C) F_{f2}=207.58N

Explanation:

This question is incomplete. The full question was:

<em>A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.  </em>

<em>Part (a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement.  </em>

<em>Part (b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder.  </em>

<em>Part (c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.</em>

For part A), we make a balance of energy to calculate the work done by the friction force:

W_{ff}=\Delta E

W_{ff}=1/2*m*vf^2-m*g*hy

W_{ff}=-744.12J

For part B), we use our previous value for the work:

W_{ff}=-F_f*(hy/sin\theta)   Solving for friction force:

F_f=-W_{ff}*sin\theta /hy

F_f=112.75N

For part C), we first calculate the acceleration by kinematics and then calculate the module of friction force by dynamics:

Vf^2=Vo^2+2*a*d

Solving for a:

a=-3.844m/s^2

Now, by dynamics:

|F_f|=|m*a|

|F_f|=207.58N

8 0
1 year ago
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