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Ganezh [65]
3 years ago
12

A speeding car is travelling at a constant 30.0 m/s when it passes a stationary police car. If the police car delays for 1.00 s

before starting, what must be the magnitude of the constant acceleration of the police car to catch the speeding car after the police car travels a distance of 300 m?
(A) 6.00 m/s2
(B) 3.00 m/s2
(C) 7.41 m/s2
(D) 1.45 m/s2
(E) 3.70 m/s2
Physics
1 answer:
pashok25 [27]3 years ago
8 0

Answer:

option (C)

Explanation:

Speed of car = 30 m/s

Let the time taken by the police car to catch the speeding car is t

The distance traveled by the speeding car in t + 1 second is equal to the distance traveled by the police car in time t

Distance traveled by the police car in time t

s=ut + 0.5 at^{2}    .... (1)

Distance traveled by the speeding car in t + 1 second

s = 30 (t + 1) = 300

t + 1 = 10

t = 9 s

Put the value of t in equation (1), we get

300 = 0 + 0.5 x a x 9 x 9

a = 7.41 m/s^2

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qaws [65]

let the length of the beam be "L"

from the diagram

AD = length of beam = L

AC = CD = AD/2 = L/2

BC = AC - AB = (L/2) - 1.10

BD = AD - AB = L - 1.10

m = mass of beam = 20 kg

m₁ = mass of child on left end = 30 kg

m₂ = mass of child on right end = 40 kg

using equilibrium of torque about B

(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)

30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)

L = 1.98 m

4 0
2 years ago
For a specific volume of 0.2 m3/kg, find the quality of steam if the absolute pressure is (a) 40 kPa and (b) 630 kPa. What is th
ICE Princess25 [194]

Answer:

x=0.0498

x'=0.659

Explanation:

Specific Volume V=0.2m_3/kg

Absolute Pressure (a) P_a= 40kpa

Giving

T_a=75.87

v_f=1.265*10^{-3}m^3/kg

v_g=3.993m^3/kg

                               (b) P_a= 630kpa

Giving

T_b=160.13C

v_f'=1.10282*10^{-3} m^3/kg

v_g'=0.30286 m^3/kg

(a)

Generally the equation for quality of Steam X  is mathematically given by

x=\frac{v-v_f}{v_g-v_f}

x=\frac{0.2-1.0265*10^{-3}}{3.993-1.0265*10^{-3}}

x=0.0498

(b)

Generally the equation for quality of Steam X  is mathematically given by

x'=\frac{v-v_f'}{v_g'-v_f'}

x'=\frac{0.2-1.10*10^{-3}}{3.30-1.1*10^{-3}}

x'=0.659

4 0
3 years ago
A coin released at rest from the top of a tower hits the ground after falling 5.7 s. What is the speed of the coin as it hits th
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Given parameters:

Initial velocity of Coin = 0m/s

Time taken before coin hits ground  = 5.7s

Unknown:

Final velocity of the coin  = ?

Velocity is displacement with time. To solve this problem, we have to apply one of the equations of motion.

The fitting one of them here is shown below;

             V = U + gt

where;

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

t is the time taken

Here we use positive value of acceleration due to gravity because the coin is falling with the effect of acceleration and not against it.

Now input the parameters and solve;

               V  = 0 + 9.81 x 5.7

               V = 55.917m/s

Therefore, the final velocity is 55.917m/s.

8 0
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Nitroglycerin flows through a pipe of diameter 3.0 cm at 2.0 m/s. If the diameter narrows to 0.5 cm, what will the velocity be?
Korolek [52]

Answer:

72 m/s

Explanation:

D1 = 3 cm, v1 = 2 m/s

D2 = 0.5 cm,

Let the velocity at narrow end be v2.

By use of equation of continuity

A1 v1 = A2 v2

3.14 × 3 × 3 × 2 = 3.14 × 0.5 ×0.5 × v2

v2 = 72 m/s

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