on a aircraft carrier a jet is slowed from 105 mph to a stop in 2 seconds. What is the rate of deceleration?

During the first part of a trip, a canoeist covered 65 km at a certain speed. He then traveled 27 km at a speed that was 4 km/

vitfil [10]

Answer:

13 km/h for 65 km

9 km/h for the next 27 km

Explanation:

Velocity of canoe =x km/h for 65 km

Velocity of canoe =x-4 km/h for 27 km after covering 65 km

Total time taken = 9 hours

So,

$\frac{65}{x}+\frac{27}{x-4}=8\\\Rightarrow \frac{65(x-4)+27x}{x^2-4x}=8\\\Rightarrow \frac{92x-260}{8}=x^2-4x\\\Rightarrow 11.5x-32.5=x^2-4x\\\Rightarrow x^2-15.5x+32.5=0$

Solving this quadratic equation we get,

$x=\frac{15.5\pm \sqrt{240.25+130}}{2}=13\ or\ 2.5$

Speed cannot be 2.5 as the speed will become negative for the 27 km stretch.

So, velocity of canoe is <u>13 km/h</u> for 65 km and <u>9 km/h</u> for the next 27 km

7 0

4 years ago

I really need to know what is on the physics SOL please someone tell me

lawyer [7]

The term sol is used by planetary astronomers to refer to the duration of a solar day on Mars.[7] A mean Martian solar day, or "sol", is 24 hours, 39 minutes, and 35.244 seconds.[6]

“Sol” is often used as a direct replacement for “Day” when concerning Mars. Mission duration for Mars missions is measured in Sols, so saying “Today is Sol xyz” would be normal, but I’m not sure if anyone would say “what a wonderful Sol tomorrow is going to be”.

5 0

4 years ago

A person pushes a 15.7-kg shopping cart at a constant velocity for a distance of 25.9 m on a flat horizontal surface. She pushes

valentinak56 [21]

Answer:

a) F = 35.7 N, b) W = 846.7 J, c) W = - 846.9 J, d) W=0

Explanation:

a) For this exercise let's use Newton's second law, let's set a reference frame with the x-axis horizontally

let's break down the pushing force.

cos (-23,7) = Fₓ / F

sin (-237) = F_y / F

Fₓ = F cos 23.7 = F 0.916

F_y = F sin (-23.7) = - F 0.402

Y axis

N- W - F_y = 0

N = W + F 0.402

X axis

Fₓ - fr = 0

F 0.916 = fr

F = fr / 0.916

F = 32.7 / 0.916

F = 35.7 N

It is asked to calculate several jobs

b) the work of the pushing force

W = fx x

W = 35.7 cos 23.7 25.9

W = 846.7 J

c) friction force work

W = F x cos tea

friction force opposes movement

W = - fr x

W = - 32.7 25.9

W = - 846.9 J

d) The work of the force would gravitate, as the displacement and the force of gravity are at 90º, the work is zero

W = 0

6 0

3 years ago

Jupiter, the largest planet in the solar system, has an equatorial radius of about 7.1 x 10^4km (more than 10 times that of Eart

Zarrin [17]

Answer:

The average speed is $v = 1260 \ km/s$

The average speed is different from the average velocity in this question

Explanation:

From the question we are told that

The equatorial radius of Jupiter is $R_j = 7.1*10^{4} \ km$

The period of oscillation of Jupiter is $T_J = 9 \ hours , 50 \ min = 35400 \ seconds$

Generally the average speed is mathematically represented as

$v = \frac{2 \pi * R_j }{T_J}$

=> $v = \frac{2 *3.142 * 7.1*10^{4} }{35400}$

=> $v = 1260 \ km/s$

Generally in average speed the direction is not considered while in average velocity the direction is considered for the case of this question the movement equitorial point has no direction in that it start from one point and after its periodic motion it still remains at that point

6 0

3 years ago

The engine in a small airplane is specified to have a torque of 500 N⋅m . This engine drives a 2.4- m -long, 40 kg single-blade

Mariulka [41]

Answer:

8.04 second

Explanation:

torque = 500 Nm

Length of blade, L = 2.4 m

mass of blade, m 40 kg

initial angular velocity, ωo = 0 rad/s

frequency, f = 2000 rpm = 2000 / 60 rps

final angular velocity, ωf = 2 x π x 2000 / 60 = 209.33 rad/s

Moment of inertia of the blade,

$I = \frac{1}{12}mL^{2}$

$I = \frac{1}{12}\times 40\times 2.4^{2}$

I = 19.2 kg m^2

Torque = Moment of inertia x angular acceleration

500 = 19.2 x α

where, α be the angular acceleration

α = 26.04 rad/s^2

Use first equation of motion

ω = ωo + αt

where t is the time taken by the propeller to reach 2000 rpm.

209.33 = 0 + 26.04 x t

t = 8.04 second

Thus, the time taken by the propeller is 8.04 second.

5 0

3 years ago