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3 years ago

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What is the value of the normal force if the coefficient of kinetic friction is 0.22 and the kinetic frictional force is 40 newt

ons?

2 answers:

Salsk061 [2.6K]3 years ago

5 0

1.8x10 sqaured newtons for all plato users

kifflom [539]3 years ago

5 0

The answer is 1.8x10^2 newtons

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g A large tank, at 400 kPa and 450 K, supplies air to a converging-diverging nozzle of throat area 4 cm2 and exit area 5 cm2. Fo

mariarad [96]

Answer:

A) ≥ 325Kpa

B) ( 265 < Pe < 325 ) Kpa

C) (94 < Pe < 265 )Kpa

D) Pe < 94 Kpa

Explanation:

Given data :

A large Tank : Pressures are at 400kPa and 450 K

Throat area = 4cm^2 , exit area = 5cm^2

<u>a) Determine the range of back pressures that the flow will be entirely subsonic</u>

The range of flow of back pressures that will make the flow entirely subsonic

will be ≥ 325Kpa

attached below is the detailed solution

<u>B) Have a shock wave</u>

The range of back pressures for there to be shock wave inside the nozzle

= ( 265 < Pe < 325 ) Kpa

attached below is a detailed solution

C) Have oblique shocks outside the exit

= (94 < Pe < 265 )Kpa

D) Have supersonic expansion waves outside the exit

= Pe < 94 Kpa

7 0

3 years ago

A 4.4 kg mess kit sliding on a frictionless surface explodes into two 2.2 kg parts, one moving at 2.9 m/s, due north, and the ot

sp2606 [1]

Answer:

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

Explanation:

Let north represent positive y axis and east represent positive x axis.

Here momentum is conserved.

Let the initial velocity be v.

Initial momentum = 4.4 x v = 4.4v

Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s

Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s

Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s

We have

Initial momentum = Final momentum

4.4v = 12.364 i + 15.092 j

v =2.81 i + 3.43 j

Magnitude

$v=\sqrt{2.81^2+3.43^2}=4.43m/s$

Direction

$\theta =tan^{-1}\left ( \frac{3.43}{2.81}\right )=50.67^0$

50.67° north of east.

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

6 0

3 years ago

A car accelerates uniformly from rest at a speed of 1.67 ft s^2 over a distance of 5 yards.What is the acceleration of a car?

Nina [5.8K]

Answer:

a= 17.69 m/s^2

Explanation:

Step one:

given data

A car accelerates uniformly from rest to 23 m/s

u= 0m/s

v= 23m/s

distance= 30m

Step two:

We know that

acceleration= velocity/time

also,

velocity= distance/time

23= 30/t

t= 30/23

t= 1.30 seconds

hence

acceleration= 23/1.30

accelaration= 17.69 m/s^2

5 0

2 years ago

Read 2 more answers

g An astronaut must journey to a distant planet, which is 189 light-years from Earth. What speed will be necessary if the astron

Butoxors [25]

Answer:

The value is $v = 2.999 *10^{8} \ m/s$

Explanation:

From the question we are told that

The time taken to travel to the planet from earth is $t = 189 \ light-years$

The time to be spent on the ship is $t_{s} = 12 \ years$

Generally speed can be obtained using the mathematical relation represented below

$t_s = 2 * t * \sqrt{1 - \frac{v^2}{c^2 } }$

The 2 in the equation show that the trip is a round trip i.e going and coming back

=> $12 = 2 * 189 * \sqrt{1 - \frac{v^2}{(3.0*10^{8})^2 } }$

=> $v = 2.999 *10^{8} \ m/s$

5 0

3 years ago

The acceleration due to gravity on the moon is about 1/6 of the acceleration due to gravity on the earth. A net force F acts hor

uysha [10]

Answer:

c. $a_m$

Explanation:

We are given that

Acceleration due to gravity on the moon= $a_m$

Acceleration due to gravity on the earth= $a_e$

$g_m=\frac{1}{6}g_e$

Net force due to am on an object on moon= $F_{net}=ma_m$

There is no friction and no drag force and there is no gravity involved

Then, the force acting on an object on earth= $F=ma_e$

$F=F_{net}$ (given)

$ma_m=ma_e$

$a_e=a_m$

Hence, option c is true.

3 0

3 years ago