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Leto [7]
2 years ago
6

A fighter jet lands on the flight deck of an aircraft carrier that has a length of 300.0 m. The jet must reduce its speed from a

bout 153 km/h to exactly 0 km/h in 2.0 s. What is the jet’s acceleration?
Physics
1 answer:
Yakvenalex [24]2 years ago
3 0

Answer:

-76.5m/s²

Explanation:

Vf=Vi+at

When equation is rearranged to solve for acceleration

a=(Vf-Vi)/t

a=(0-153)/2

a=-76.5

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It d bro it’s d bro it’s d
3 0
3 years ago
Find the tension in the two groups that are holding the 2.9 kg object in Pl., One makes an angle of 35.6° with respect to the ve
Stella [2.4K]

Answer:

≈ 20.35 N [newton's of tension]

Explanation:

( (2.9 × 9.8) ÷ cos(35.6°) ) × sin (35.6°) =

( (28.42) ÷ (≈0.813) ) × (≈0.582) =

(≈34.96) × (≈0.582) = 20.3449446.... ≈ 20.35

4 0
2 years ago
The Grand Canyon is 1800 meters deep at its deepest point. A rock is dropped from the rim above this point. Write the height of
snow_tiger [21]

Answer:

Explanation:

Given

height of grand canyon is h=1800\ m

Rock is dropped i.e. initial velocity is zero

u=0

Using Equation of motion

h=ut+\frac{1}{2}at^2

h=height

u=initial velocity

a=acceleration

t=time

here a=acceleration due to gravity

1800=0+\frac{1}{2}\times 9.8\times t^2

t^2=\frac{2\times 1800}{9.8}

t=19.166\ s

3 0
3 years ago
A container is filled with an ideal diatomic gas to a pressure and volume of P1 and V1, respectively. The gas is then warmed in
lilavasa [31]

Answer:

Explanation:

The  change is as follows

P₁ V₁ to 3P₁, V₁ ( constt volume )  --- first process

3P₁,V₁ to 3P₁ , 5V₁ ( constt pressure ) ---- second process

In the first process Temperature must have been increased 3 times . So if initial temperature is T₁ then final temperature will be 3 T₁

P₁V₁ = n R T₁ , n is no of moles of gas enclosed.

nRT₁ = P₁V₁

Heat added at constant volume  = n Cv ( 3T₁ - T₁)

= n x 5/3 R X 2T₁ ( for diatomic gas Cv = 5/3 R)

= 10/3 x nRT₁

= 10/3x P₁V₁

In the second process,  Temperature must have been increased 5 times . So if initial temperature is 3T₁ then final temperature will be 15 T₁

Heat added at constant pressure in second case  

= n Cp ( 15T₁ - 3T₁)

= n x 7/3 R X 12T₁ ( For diatomic gas Cp = 7/3 R)

= 28 x nRT₁

= 28 P₁V₁

6 0
3 years ago
Sketch the resultant field pattern around the following current carrying conductors and
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Cccccccvvvvvcjjjjjjjjjjkllk ki e
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