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3 years ago

The object moves with BLANK from A to B. it BLANK from B to C. it moves with BLANK from C to D.

Physics

1 answer:

Ganezh [65]3 years ago

4 0

Note that this is a position vs time graph.

From A to B, the graph is a straight line with a nonzero slope. This indicates a constant velocity.

From B to C, the graph is a straight line with 0 slope. This indicates a constant position, i.e. the object remains stationary.

From C to D, the graph is a straight line with a nonzero slope. This indicates a constant velocity.

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Look at the image to answer the question correctly.

r-ruslan [8.4K]

Answer:

1- b: 2- a : 3- c : 4- d

Explanation:

it starts 2 move away from strting point, then no motion, then moves toward the start, the slows up.

3 0

3 years ago

The mass of the uniform cantilever is 1100 kg. Determine the force on the beam at A. Determine the force on the beam at B. Use C

Mashcka [7]

Answer:

Force A=-−2,697.75 N

Force B=13, 488.75 N

Explanation:

Taking moments at point A, the sum of clockwise and anticlockwise moments equal to zero.

25 mg-20Fb=0

25*1100g=20Fb

Fb=25*1100g/20=1375g

Taking g as 9.81 then Fb=1375*9.81=13,488.75 N

The sum of upward and downward forces are same hence Fa=1100g-1375g=-275g

-275*9.81=−2,697.75. Therefore, force A pulls downwards

Note that the centre of gravity is taken to be half the whole length hence half of 50 is 25 m because center of gravity is always at the middle

8 0

3 years ago

a block with length 1.5m width 1m height 0.5m and mass 300kg lays on the table.what is the pressure at the bottom surface of the

Nesterboy [21]

Answer:

your answer will be 320kg that would be the pressure at the bottom surface of the block

6 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

If you wish to take a picture of a bullet traveling at 500 m/s, then a very brief flash of light produced by an RC discharge thr

Agata [3.3K]

Answer:

Resistance in the flash tube, $R=3.97\times 10^{-3}\ \Omega$

Explanation:

It is given that,

Speed of the bullet, v = 500 m/s

Distance between one RC constant, d = 1 mm = 0.001 m

Capacitance, $C=503\ \mu F=503\times 10^{-6}\ F$

The time constant of RC circuit is given by :

$\tau=RC$

R is the resistance in the flash tube

$R=\dfrac{\tau}{C}$ ..........(1)

Speed of the bullet is given by total distance divided by total time taken as :

$v=\dfrac{d}{\tau}$

$\tau=\dfrac{d}{v}$

$\tau=\dfrac{0.001}{500}$

$\tau=0.000002\ s$

Equation (1) becomes :

$R=\dfrac{0.000002}{503\times 10^{-6}}$

$R=3.97\times 10^{-3}\ \Omega$

So, the resistance in the flash tube is $3.97\times 10^{-3}\ \Omega$ . Hence, this is the required solution.

8 0

3 years ago

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