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Jlenok [28]
3 years ago
6

If an irregularly shaped object (such as a wrench) is dropped from rest in a classroom and feels no air resistance, it will:

Physics
1 answer:
Maslowich3 years ago
7 0

Answer:

D) accelerate but will not spin.

Explanation:

On the off chance that there is no air resistance the object will accelerate yet won't turn, this is on the grounds that without air resistance same force is applied on each bit of the object. Force on each segment is coordinated descending i.e parallel force. So there is no force to deliver spin movement in it.  

It will accelerate because of gravity

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A car is traveling at 15 m/sm/s . Part A How fast would the car need to go to double its kinetic energy
GREYUIT [131]

Answer:

21.21 m/s

Explanation:

Let KE₁ represent the initial kinetic energy.

Let v₁ represent the initial velocity.

Let KE₂ represent the final kinetic energy.

Let v₂ represent the final velocity.

Next, the data obtained from the question:

Initial velocity (v₁) = 15 m/s

Initial kinetic Energy (KE₁) = E

Final final energy (KE₂) = double the initial kinetic energy = 2E

Final velocity (v₂) =?

Thus, the velocity (v₂) with which the car we travel in order to double it's kinetic energy can be obtained as follow:

KE = ½mv²

NOTE: Mass (m) = constant (since we are considering the same car)

KE₁/v₁² = KE₂/v₂²

E /15² = 2E/v₂²

E/225 = 2E/v₂²

Cross multiply

E × v₂² = 225 × 2E

E × v₂² = 450E

Divide both side by E

v₂² = 450E /E

v₂² = 450

Take the square root of both side.

v₂ = √450

v₂ = 21.21 m/s

Therefore, the car will travel at 21.21 m/s in order to double it's kinetic energy.

8 0
2 years ago
A uniform disk, a thin hoop, and a uniform sphere, all with the same mass and same outer radius, are each free to rotate about a
olga nikolaevna [1]

Answer:

4 hoop, disk, sphere

Explanation:

Because

We are given data that

Hoop, disk, sphere have Same mass and radius

So let

And Initial angular velocity, = 0

The Force on each be F

And Time = t

Also let

Radius of each = r

So let's find the inertia shall we!!

I1 = m r² /2

= 0.5 mr² the his is for dis

I2 = m r² for hoop

And

Moment of inertia of sphere wiil be

I3 = (2/5) mr²

= 0.4 mr²

So

ωf = ωi + α t

= 0 + ( τ / I ) t

= ( F r / I ) t

So we can see that

ωf is inversely proportional to moment of inertia.

And so we take the

Order of I ( least to greatest ) :

I3 (sphere) , I1 (disk) , I2 (hoop) , ,

Order of ωf: ( least to greatest)

That of omega xf is the reverse of inertial so

hoop, disk, sphere

Option - 4

5 0
3 years ago
Find the velocity of the student using KE= 1/2 mv^2. A 50-kilogram student is running and has 225 joules of kinetic energy. The
defon
225 = 1/2 (50) (v2)
225 = 25 (v2)
225/25 = v2
9 = v2
√9 = v
v = 3 m/s
3 0
3 years ago
Two vehicles A and B accelerate uniformly from rest.
spayn [35]

Answer:

(i) Please find attached the required velocity time graphs plotted with MS Excel

(ii) The velocity of vehicle A at the 18th second = 20 m/s

The velocity of vehicle B at the 18th second = 0 m/s

(iii) The distance between the two vehicles at the moment in (ii) above is 60 meters

Explanation:

The given parameters of the motion of vehicles A and B are;

The acceleration of vehicles A and B = Uniform acceleration starting from rest

The maximum velocity attained by vehicle A = 30 m/s

The time it takes vehicle A to attain maximum velocity = 10 s

The maximum velocity attained by vehicle B = 30 m/s

The time it takes vehicle B to attain maximum velocity = The time it takes vehicle A to attain maximum velocity = 10 s

The time duration vehicle A maintains its maximum velocity = 6 s

The time duration vehicle B maintains its maximum velocity = 4 s

(i) From the question, we get the following table;

\begin{array}{ccc}Time &V_A&V_B\\0&0&0\\10&30&40\\14&30&40\\16&30&20\\18&20&0\\22&0&\end{array}

From the above table the velocity time graphs of vehicles A and B is created with MS Excel and can included here

(ii) The velocity of vehicle A at the start = 0 m/s

After accelerating for 10 seconds, the velocity of vehicle A = The maximum velocity of vehicle A = 30 m/s

The maximum velocity is maintained for 6 seconds which gives;

At 10 s + 6 s = 16 s, the velocity of vehicle A = 30 m/s

The time it takes vehicle A to decelerate to rest = 6 s

The deceleration of vehicle A, a_A = (30 m/s - 0 m/s)/(6 s) = 5 m/s²

Therefore, we get;

v = u - a_A·t

At the 18th second, the deceleration time, t = 18 s - 16 s = 2 s

u = 30 m/s

∴ v₁₈ = 30 - 5 × 2 = 20

The velocity of vehicle A at the 18th second, V_{18A} = 20 m/s

For vehicle B, we have;

At the 14th second, the velocity of vehicle B = 40 m/s

Vehicle B decelerates to rest in, t = 4 s

The deceleration of vehicle B, a_B = (40 m/s - 0 m/s)/(4 s) = 10 m/s²

For vehicle B, at the 18th second, t = 18 s - 14 s = 4 s

∴ v_{18B} = 40 m/s - 10 m/s² × 4 s = 0 m/s

The velocity of the vehicle B at 18th second, v_{18B} = 0 m/s

(iii) The distance covered by vehicle A up to the 18th second is given by the area under the velocity-time graph as follows;

The area triangle A₁ = (1/2) × 10 × 30 = 150

Area of rectangle, A₂ = 6 × 30 = 180

Area of trapezoid, A₃ = (1/2) × (30 + 20) × 2 = 50

The distance covered in the 18th second by vehicle S_A = A₁ + A₂ + A₃

∴ S_A = 150 + 180 + 50 = 380

The distance covered in the 18th second by vehicle S_A = 380 m

The distance covered by the vehicle B in the 18th second is given by the area under the velocity time graph of vehicle B as follows;

Area of trapezoid, A₅ = (1/2) × (18 + 4) × 40 = 440

The distance covered by the trapezoid, S_B = 440 m

The distance of the two vehicles apart at the 18t second, S_{AB} = S_B - S_A

∴ S_{AB} = 440 m - 380 m = 60 m

The distance of the two vehicles from one another at the 18th second, S_{AB} = 60 m.

5 0
3 years ago
The loudness l of a sound, measured in decibels, is given by l=10log10r, where r is the sound's relative intensity. suppose one
marishachu [46]

Answer

given,                              

I is the loudness of sound

I = 10 Log₁₀ r                  

r is relative intensity                    

at when relative intensity is 10⁶        

I = 60 dB                                                  

how much louder when 100 people would be talking together

I = 10 Log₁₀ r                

I = 10 Log₁₀ (10⁶ x 100)  

I = 10 Log₁₀ (10⁸)                

I = 80 dB                      

hence, the intensity will be increased by (80 dB -60 dB) 20 dB when 100 people start talking together.

5 0
3 years ago
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