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4 years ago

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A 3-kg block, attached to a spring, executes simple harmonic motion according to x= 2cos(50t) where x is in meters and t is in s

econds. The spring constant of the spring is: (a) 1 N/m (b) 100 N/m (c) 150 N/m (d) 7500 N/m (e) none of the above

1 answer:

saw5 [17]4 years ago

4 0

Explanation:

w=50=√k/m=√k/3

SO

(50²)*3 is 7500N/m

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a man exerts 3000.00N of force to push a car 35.00 meters in 90.00 seconds.... 1. what is the work done 2.what is the power gene

vaieri [72.5K]

1). Work = (force) x (distance)

Work = (3,000 newtons) x (35 meters) = 105,000 newton-meters = <em>105,000 joules</em>

2). Power = (work) / (time)

Power = (105,000 joules) / (90 seconds)
= 1,166-2/3 joules per second =<em> 1,166 and 2/3 watts</em> .

<em>Note:</em>
That's no ordinary man.
1,166 watts is the same as roughly 1.6 horsepower.
Not too many people can sustain 1 horsepower or more for 90 seconds.

4 0

4 years ago

X-rays with an energy of 301 keV undergo Compton scattering with a target. If the scattered X-rays are detected at 77.5^{\circ}

Alex73 [517]

Answer:

$6.03\cdot 10^{-12} m$

Explanation:

First of all, we need to find the initial wavelength of the photon.

We know that its energy is

$E=301 keV = 4.82\cdot 10^{-14}J$

So its wavelength is given by:

$\lambda = \frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.82\cdot 10^{-14} J}=4.13\cdot 10^{-12}m$

The formula for the Compton scattering is:

$\lambda' = \lambda +\frac{h}{mc}(1-cos \theta)$

where

$\lambda$ is the original wavelength

h is the Planck constant

m is the electron mass

c is the speed of light

$\theta=77.5^{\circ}$ is the angle of the scattered photon

Substituting, we find

$\lambda' = 4.13\cdot 10^{-12} m +\frac{6.63\cdot 10^{-34} Js)}{(9.11\cdot 10^{-31}kg)(3\cdot 10^8 m/s)}(1-cos 77.5^{\circ})=6.03\cdot 10^{-12} m$

7 0

3 years ago

A car of mass 1000.0 kg is traveling along a level road at 100.0 km/h when its brakes are applied. Calculate the stopping distan

OleMash [197]

Explanation:

It is given that,

Mass of the car, m = 1000 kg

Speed of the car, v = 100 km/h = 27.77 m/s

The coefficient of kinetic friction of the tires, $\mu_k=0.5$

Let f is the net force acting on the body due to frictional force, such that,

$-f=ma$

$a=\dfrac{-f}{m}$

$a=\dfrac{-\mu _k mg}{m}$

$a=\mu_k g$

$a=-0.5\times 9.8$

$a=-4.9\ m/s^2$

We know that the acceleration of the car in calculus is given by :

$v.dv=a.dx$ , x is the stopping distance

$\int\limits^0_v {v.dv}=\int\limits^x_0 {a.dx}$

$\dfrac{v^2}{2}|_v^0=ax$

$0-(27.77)^2=-2\times 4.9x$

On solving the above equation, we get, x = 78.69 meters

So, the stopping distance for the car is 78.69 meters. Hence, this is the required solution.

6 0

3 years ago

How much charge must pass by a point in a wire in 10 s for the current inb the wire to be 0.50 a?

Ostrovityanka [42]

Charge must pass by a point in a wire is 5 Coulomb

Given:

time = 10 s = t

current = 0.50 A = I

To Find:

charge = q

Solution: Current is defined as amount of charge passing through a region per unit time. It is given by the formula

I = q/t

q = I x t = 0.50 A x 10 s = 5 Coulomb

Hence, charge must pass by a point in a wire is 5 Coulomb

Learn more about Charge here:

brainly.com/question/25922783

#SPJ4

3 0

2 years ago

A lightning flash releases about 1010J of electrical energy. Part A If all this energy is added to 50 kg of water (the amount of

zavuch27 [327]

Answer:

water is in the vapor state,

Explanation:

We must use calorimetry equations to find the final water temperatures. We assume that all energy is transformed into heat

E = Q₁ + $Q_{L}$

Where Q1 is the heat required to bring water from the current temperature to the boiling point

Q₁ = m $c_{e}$ ( $T_{f}$ -T₀)

Q₁ = 50 4180 (100 - 37)

Q₁ = 1.317 10⁷ J

Let's calculate the energy so that all the water changes state

$Q_{L}$ = m L

$Q_{L}$ = 50 2,256 106

$Q_{L}$ = 1,128 10⁸ J

Let's look for the energy needed to convert all the water into steam is

Qt = Q₁ + $Q_{L}$

Qt = 1.317 107 + 11.28 107

Qt = 12,597 10⁷ J

Let's calculate how much energy is left to heat the water vapor

ΔE = E - Qt

ΔE = 10¹⁰ - 12,597 10⁷

ΔE = 1000 107 - 12,597 107

ΔE = 987.4 10⁷ J

With this energy we heat the steam, clear the final temperature

Q = ΔE = m $c_{e}$ ( $T_{f}$ -To)

( $T_{f}$ -T₀) = ΔE / m $c_{e}$

$T_{f}$ = T₀ + ΔE / m $c_{e}$

$T_{f}$ = 100 + 987.4 10⁷ / (50 1970)

$T_{f}$ = 100 + 1,002 10⁵

$T_{f}$ = 1,003 10⁵ ° C

This result indicates that the water is in the vapor state, in realizing at this temperature the water will be dissociated into its hydrogen and oxygen components

4 0

3 years ago