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elena-14-01-66 [18.8K]
3 years ago
12

A car accelerates from rest to a velocity of smeters second in 4 seconds What is average acceleration over this period of time

Physics
1 answer:
ipn [44]3 years ago
3 0

Answer:

acceleration=s/4

Explanation:

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A 50-cm-long spring is suspended from the ceiling. A 330 g mass is connected to the end and held at rest with the spring unstret
Nataly [62]

Answer:

a)32.34 N/m

b)10cm

c)1.6 Hz

Explanation:

Let 'k' represent spring constant

'm' mass of the object= 330g =>0.33kg

a) in order to find spring constant 'k', we apply Newton's second law to the equilibrium position 10cm below the release point.

ΣF=kx-mg=0

k=mg / x

k= (0.33 x 9.8)/ 0.1

k= 32.34 N/m

b) The amplitude, A, is the distance from the equilibrium (or center) point of motion to either its lowest or highest point (end points). The amplitude, therefore, is half of the total distance covered by the oscillating object.

Therefore, amplitude of the oscillation is 10cm

c)frequency of the oscillation can be determined by,

f= 1/2π \sqrt{\frac{k}{m} }

f= 1/2π \sqrt{\frac{32.34}{0.33} }

f= 1.57

f≈ 1.6 Hz

Therefore,  the frequency of the oscillation is 1.6 Hz

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3 years ago
How much does Nestle pay the local municipality for access to their spring water?
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3 years ago
The graph represents the reaction 2H2 + 02 32H20 as it reaches
Alex

Answer:

C and D

Explanation:

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As an electron moves, what does it make or cause ?
jolli1 [7]
It causes or makes a magnetic field.
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3 years ago
The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
puteri [66]

Answer:

t=5.3687\ s  is the time taken by the car to accelerate the desired range of the speed from zero at full power.

Explanation:

Given:

Range of speed during which constant power is supplied to the wheels by the car is 0\ mph\ to\ 70\ mph.

  • Initial velocity of the car, v_i=0\ mph
  • final velocity of the car during the test, v_f=32\ mph=14.3052\ m.s^{-1}
  • Time taken to accelerate form zero to 32 mph at full power, t=1.2\ s
  • initial velocity of the car, u_i=0\ mph
  • final desired velocity of the car, u_f=64\ mph=28.6105\ m.s^{-1}

Now the acceleration of the car:

a=\frac{v_f-v_i}{t}

a=\frac{14.3052-0}{1.2}

a=11.921\ m.s^{-1}

Now using the equation of motion:

u_f=u_i+a.t

64=0+11.921\times t

t=5.3687\ s is the time taken by the car to accelerate the desired range of the speed from zero at full power.

8 0
3 years ago
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