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Anettt [7]
3 years ago
10

The period of a simple pendulum is 3.5 s. The length of the pendulum is doubled. What is the period T of the longer pendulum?

Physics
1 answer:
Rudik [331]3 years ago
6 0

Answer:

Explanation:

T = 2π√(L/g)

If you increase L to 2L, the period is increased by a factor of √2

T = 3.5√2 ≈ 4.9 s

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The momentum of the car is 4.4x10^3 kg•m/sec
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Give three examples of objects in equilibrium found in classroom?​
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Answer:

Book. Bottle. table are some examples of objects in equilibrium found in classroom

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2 years ago
An object has a mass of 60 kg. It decelerates from 50 m/s to 20 m/s when a resultant force of 300 N acts on it. For how long doe
Pavlova-9 [17]
  • Force=300N
  • Mass=60kg

Find Acceleration

\\ \rm\longmapsto F=ma

\\ \rm\longmapsto a=\dfrac{F}{m}

\\ \rm\longmapsto a=\dfrac{-300}{60}

\\ \rm\longmapsto a=-5m/s^2

Now

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  • v=20m/s

Using 1st equation of kinematics

\\ \rm\longmapsto v=u+at

\\ \rm\longmapsto t=\dfrac{v-u}{a}

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\\ \rm\longmapsto t=6

4 0
2 years ago
How can I answer question 19? please help me, I need the answer now
Crank
The spring starts out 22 cm long with nothing hanging on it.

Hanging 35 newtons of weight on it stretches the spring 1 meter.

Ellen is going to hang 250 grams of mass on the spring.
What's the weight of 250 grams of mass ?

   Weight = (mass) x (acceleration of gravity in the place where the mass is) .

On Earth, the acceleration of gravity is 9.8 m/s² .
250 grams is 0.25 of a kilogram.

Weight of 250 grams  =  (0.25 kilogram) x (9.8 m/s²)

                                 =    (0.25 x 9.8)  kg-m/s²

                                 =        2.45 newtons .

2.45 newtons of weight is  (2.45 / 35) of 35 newtons,
so it'll stretch the spring  (2.45 / 35) of a meter.

   2.45/35 = 0.07 of a meter  =  7 centimeters.

The spring was 22 cm long with nothing hanging on it,
and the 250-gm weight stretched it 7 cm.
So with the weight hanging on it, it's (22 + 7) = 29 cm long.

6 0
3 years ago
a 15-nC point charge is at the center of a thin spherical shell of radius 10cm, carrying -22nC of charge distributed uniformly o
Aleks [24]

Answer:

A) E = 278925.62 N/C with direction; radially out.

B) E = 43048.47 N/C with direction radially out.

C) E = -3214.29 N/C with direction radially in.

Explanation:

From Gauss' Law, the Electric field for any spherically symmetric charge or charge distribution is the same as the point charge formula. Thus;

E = kQ/r²

where;

Q is the net charge within the distance r.

We are given the charge Q = 15-nC and

spherical shell of radius 10cm

A) The distance r = 2.2 cm = 0.022 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C

While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²

E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.022)²

E = 278925.62 N/C

This will be radially out ,since the net charge is positive.

B) The distance r = 5.6 cm = 0.056 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C

While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²

E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.056)²

E = 43048.47 N/C

This will be radially out ,since the net charge is positive.

C) The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;

Q = 15 nC - 22 nC

Q = -7 nC = -7 x 10^(-9) C

and;

E = (9 x 10^(9)*(-7 x 10^(-9))/(0.14)²

E = -3214.29 N/C

This will be radially in, since the net charge is negative. You can indicate this with a negative answer.

8 0
3 years ago
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