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3 years ago

8

amanda placed a small television on a metal rolling cart. on her way to the library, her friend carlos stepped in front of the c

art. although amanda stopped suddenly, the television stayed in place on the cart. what kept the television on the cart when she stopped suddenly?

1 answer:

kramer3 years ago

3 0

Answer:

Friction. Have a great day! Think positive <3

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Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co

aivan3 [116]

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

$r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m$

The radius is 1.195 m

Time period would be given by

$T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s$

Time period of the motion is 2.8375 s

Angular speed is given by

$\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s$

The angular speed of the motion is 2.21433 rad/s

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4 years ago

A spring with spring-constant of 100 n/m is compressed 0.10 m. what is its maximum stored elastic potential energy

Ganezh [65]

Elastic potential energy stored in a spring is

(1/2) · (spring constant) · (stretch or compress)² .

PE = (1/2) · (100 N/m) · (0.1 m)²

PE = (50 N/m) · (0.01 m²)

PE = (50 · 0.01) (N · m / m²)

PE = 0.5 N · m

PE = 0.5 Joule

5 0

3 years ago

g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p

kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × $10^{-5}$ m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

r = $\frac{mv}{qB}$

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × $10^{-5}$ T, v = 121 m/s, Θ = $90^{0}$ (since it enters perpendicularly to the field), q = e = 1.6 × $10^{-19}$ C and m = 9.11 × $10^{-31}$ Kg.

Thus,

r = $\frac{mv}{qB}$ ÷ sinΘ

But, sinΘ = sin $90^{0}$ = 1.

So that;

r = $\frac{mv}{qB}$

= (9.11 × $10^{-31}$ × 121) ÷ (1.6 × $10^{-19}$ × 1.63 × $10^{-5}$ )

= 1.10231 × $10^{-28}$ ÷ 2.608 × $10^{-24}$

= 4.2266 × $10^{-5}$

= 4.23 × $10^{-5}$ m

The radius 'r' of the electron's path is 4.23 × $10^{-5}$ m.

B. The frequency 'f' of the motion is called cyclotron frequency;

f = $\frac{qB}{2\pi m}$

= (1.6 × $10^{-19}$ × 1.63 × $10^{-5}$ ) ÷ (2 × $\frac{22}{7}$ × 9.11 × $10^{-31}$ )

= 2.608 × $10^{-24}$ ÷ 5.7263 × $10^{-30}$

= 455442.4323

f = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.

3 0

4 years ago

Read 2 more answers

What force must the deltoid muscle provide to keep the arm in this position?

ruslelena [56]

Answer:

Deltoid Force, $F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}$

Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

1. The torque about the point in the shoulder for the deltoid muscle, $T_{Deltoid}$

2. The torque of the arm, $T_{arm}$

Assuming the arm is just being stretched and there is no rotation going on,

$T_{Deltoid}$ = 0

$T_{arm}$ = 0

⇒ $T_{Deltoid} = T_{arm}$

$r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}$

Where,

$r_{d}$ is radius of the deltoid

$F_{d}$ is the force of the deltiod

$\alpha_{d}$ is the angle of the deltiod

$r_{a}$ is the radius of the arm

$F_{a}$ is the force of the arm , $F_{a} = mg$ which is the mass of the arm and acceleration due to gravity

$\alpha_{a}$ is the angle of the arm

The force of the deltoid muscle is,

$F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}$

but $F_{a} = mg$ ,

∴ $F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}$

7 0

3 years ago

True/False: Paraphrasing Einstein’s Theory of General Relativity, gravity "handprint" makes an "indentation" in the fabric of sp

Y_Kistochka [10]

B.False

Einstein's vision of GR is NOT that somehow Gravity comes along and alters (indents?) some existing structure.
It is that Gravity (with its four possible sources) actually determines the entire global structure of Space-Time in which such sources are extant.

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3 years ago