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11111nata11111 [884]
3 years ago
12

A 2.0 kg hanging mass stretches a coiled spring by 0.15 m. The spring constant, k, is: (A) 0.075 N/m, (B) 2.9 N/m (C) 131 N/m, (

D) 1,742 N/m, (E) none of the above.
Physics
1 answer:
Alla [95]3 years ago
3 0

Answer:

C

Explanation:

Givens

m = 2 kg

F = 2 * 9.81

F =  19.62 N

x = 0.15 m

Formula

F = k*x

Solution

19.62 = k*0.15

k = 19.62/0.15

k = 130.8 which rounded to the nearest given answer is C

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Amiraneli [1.4K]

Answer:

b. -1.5 m/s²

Explanation:

Given the following data;

Initial velocity = 15m/s

Time = 10 seconds.

Since the car came to rest, final velocity = 0m/s

To find acceleration;

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of final speed from the initial speed all over time.

Mathematically, acceleration is given by the equation;

Acceleration (a) = \frac{initial \; speed  -  final \; speed}{time}

a = \frac{v  -  u}{t}

Where,

a is acceleration measured in ms^{-2}

v and u is initial and final speed respectively, measured in ms^{-1}

t is time measured in seconds.

Substituting into the equation, we have;

Acceleration, a = \frac{0 - 15}{10}

Acceleration, a = \frac{-15}{10}

Acceleration = -1.5 m/s²

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3 years ago
A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s
Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8^{-3}

Explanation:

m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg

v_{bullet}=341\frac{m}{s}

Momentum of the motion the first part of the motion have a momentum that is:

P_{1}=m_{bullet}*v_{bullet}

P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529

The final momentum is the motion before the action so:

a).

P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}

P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}

P_{1}=P_{2}

2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}

b).

kinetic energy

K=\frac{1}{2}*m*(v)^{2}

Kinetic energy after

Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J

Kinetic energy before

Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J

Ratio =\frac{Ka}{Kb}

R=\frac{1.14}{406.4}\\R=2.8x10^{-3}

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Nine tree lights are connected inparallel across 120-V potential difference. The cord to the wall socket carries a current of 0.
jok3333 [9.3K]

Answer:

a)3000ohm

b)4.44mA

Explanation:

a) we were given a Nine tree lights connected inparallel across 120-V potential difference, since the resistor are in parallel we use the expresion below

1/R(total)= 1/R₁ + 1/R₂ + 1/R₃ + 1/R₃ +.... 1/R₉

But according to ohm'law which can be expressed below

V=IR

R=V/I

R(total)= 120/0.36

= 333.33ohm

1/R(total)= 1/R₁ + 1/R₂ + 1/R₃ + 1/R₃ +.... 1/R₉

R₁=R₂ =R₃ =R₄= R₅=R₆=R₇=R₈=R₉

1/R(total)=9/R

1/333.33= 9/R

R= 3000ohm

Therefore, the resistance is 3000ohm

b)the bulbs were connected in series here, then for series connection we use below expression

R₁=R₂ =R₃ =R₄= R₅=R₆=R₇=R₈=R₉

R(total)=9R

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=27000ohm

I=VR

I=V/R

I= 120/27000

= 4.44*10⁻³A

4.44mA

Therefore, the current is 4.4mA

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