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NISA [10]
3 years ago
5

A crate is pulled by a force (parallel to the incline) up a rough incline. The crate has an initial spped shown in the figure be

low. The crate is pulled a distance of 8.24m on the incline by a force of 150N force. b) What is the speed of the crate after it is pulled the 8.24M? Answer in units of m/s.​
Physics
1 answer:
Phantasy [73]3 years ago
7 0

Answer:

(a) Kinetic energy of the cart is 648.32 J.

(b) Speed of the cart is 11.44 m/s.

Explanation:

From the attached below figure we came to know that,

Mass of the crate is 10 kg (m)

Initial speed of the crate is 1.2 m/s (v)

Coefficient of friction on a rough plane is 0.286 (µ)

Force on the cart is 150 N (F)

The distance which crate is moved (d) is 8.05 m.

Calculation of kinetic energy:

Acceleration of the cart:

We know that,

F = F – m × g × sinθ - µ × m × g × cosθ (F = ma)

m × a =  F – m × g × sinθ - µ × m × g × cosθ

10 × a = 150 – (10 × 9.8 × sin 27) – (0.286 × 10 × 9.8 × cos 27)

10 × a = 150 – 44.49 – 24.97

10 × a = 80.54

Calculation of speed:

Change in kinetic energy = m × a × d

Change in kinetic energy = 10 × 8.054 × 8.05

Change in kinetic energy = 648.32 J

Explanation:

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When a garden hose with an output diameter of 20 mm is directed straight upward, the stream of water rises to a height of 0.13m
SpyIntel [72]

Answer: h = 0.52m

Explanation:

Using the equation of out flow;

A1 × V1 = A2 ×V2

Where A1 = area of the first nozzle

A2 = area of the second nozzle

V1= velocity of flow out from the first nozzle

V2 = velocity of flow out from 2nd nozzle

But AV= area of nozzle × velocity of water = volume of water per second(m³/s).

Now we can set A×V = Area of nozzle × height of rise.

Henceb A1× h1 = A2 × h2 ( note the time cancel on both sides)

D1 = 20mm= 0.02m; h1 = 0.13m

D2 = 10mm = 0.01m; h2= ?

h2 = π(D1/2)²× h1 /π(D2/2)²

h2 = (0.02/2)² × 0.13/(0.01/2)²

= (0.01)² ×0.13 /(0.005)²

= 1.3 × 10^-5/(5 × 10^-3)²

= 1.3 × 10^-5/25 × 10^-6

= (1.3/25) 10^-5 × 10^6

= 0.052× 10

= 0.52m

7 0
3 years ago
Rank the angular speed of the following objects, from highest to lowest:- A bowling ball of radius 12.3 cm rotating at 8.21 radi
dlinn [17]

Answer:

A bowling ball (8.21rad/s) > A tire (7.94rad/s) > A square (7.75rad/s) > A rock (6.98rad/s) > A top spinning (6.54rad/s)

Explanation:

<u>To rank the angular speed (ω) of the objects, we need first calculate its value for every object:</u>

A bowling ball of radius 12.3cm rotating at 8.21 radians per second:

ω = 8.21 rad/s

A tire of radius 0.321m rotating at 75.8 rpm:

\omega = 75.8 \frac{rev}{min}\cdot \frac{2\pi rad}{1rev}\cdot \frac{1min}{60s} = 7.94rad/s

A 6.84cm diameter top spinning at 375 degrees per second:

\omega = 375 \frac{^\circ}{s}\cdot \frac{2\pi rad}{360^ \circ} = 6.54rad/s    

A square with sides (b) 0.458m long, whose corners are moving with tangential speed (v) 2.51 m/s as it rotates about its center:

\omega = \frac{v}{r} = \frac{v}{\frac{b}{2}\sqrt 2} = \frac{2.51 m/s}{\frac{0.458 m}{2} \sqrt 2} = 7.75rad/s

A rock on a string, being swung in a circle of radius 0.521 m with a centripetal acceleration (a) of 25.4 m/s²:

\omega = \sqrt \frac{a}{r} = \sqrt \frac{25.4 m/s^{2}}{0.521m} = 6.98rad/s

<u>Now, the rank of the angular speed of the objects, from highest to lowest is: </u>

A bowling ball (8.21rad/s) > A tire (7.94rad/s) > A square (7.75rad/s) > A rock (6.98rad/s) > A top spinning (6.54rad/s) 

I hope it helps you!      

4 0
3 years ago
What is the potential energy of a 25 kg bicycle resting at the top of a hill 3 m high (formula:pe=mgh)
yKpoI14uk [10]

The potential energy of a 25 kg bicycle resting at the top of a hill 3 m high will be 735.75 J.

<h3>What is potential energy?</h3>

The potential energy is due to the virtue of the position and the height. The unit for the potential energy is the joule.

The potential energy is mainly depending upon the height of the object. when the cyclist is at the highest position, the height is maximum. Therefore, the potential energy is also maximum.

The potential energy is found as;

PE=mgh

PE=25 kg× 9.81 m/s² ×3 m

PE= 735.75 J.

Hence, the potential energy of a 25 kg bicycle resting at the top of a hill 3 m high will be 735.75 J.

To learn more about the potential energy, refer to the link;

brainly.com/question/24284560

#SPJ1

5 0
2 years ago
The earth exerts the necessary centripetal force on an orbiting satellite to keep it moving in a circle at constant speed. Which
jenyasd209 [6]

Answer:

E) The centripetal force is always perpendicular to the velocity.

Explanation:

Due to gravity and inertia, the satellite follows a uniform circular motion. In this movement, the velocity is always tangent to the orbit and the centripetal force is directed towards the center. Therefore, there is no net acceleration in the same direction of velocity, which implies that it remains constant.

8 0
4 years ago
Since the aluminum bar is not an isolated system, the second law of thermodynamics cannot be applied to the bar alone. Rather, i
max2010maxim [7]

Answer:

ΔS total ≥ 0 (ΔS total = 0 if the process is carried out reversibly in the surroundings)

Explanation:

Assuming that the entropy change in the aluminium bar is due to heat exchange with the surroundings ( the lake) , then the entropy change of the aluminium bar is, according to the second law of thermodynamics, :

ΔS al ≥ ∫dQ/T

if the heat transfer is carried out reversibly

ΔS al =∫dQ/T  

in the surroundings

ΔS surr ≥ -∫dQ/T = -ΔS al → ΔS surr ≥ -ΔS al = - (-1238 J/K) = 1238 J/K

the total entropy change will be

ΔS total = ΔS al + ΔS surr

ΔS total ≥ ΔS al + (-ΔS al) =

ΔS total ≥ 0

the total entropy change will be ΔS total = 0 if the process is carried out reversibly in the surroundings

4 0
4 years ago
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