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2 years ago

Through x rays is harmful for human health it is used in medical field why?

1 answer:

6 0

It is strong enough to penetrate through flesh but not bone so we can see if there are fractures or breaks in our skeleton

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the wheel of a car has a radius of .350 m. the engine of the car applies applies a torque of 295 N m to this wheel, which does n

maria [59]

The magnitude of static friction force is f_s = 842.8 N

Explanation:

Write down the values given in the question

The wheel of a car has radius r = 0.350 m

The car applies the torque is τ = 295 N m

It is said that the wheels does not slip against the road surface,

Here we apply a force of static friction,

It can be calculated as

Frictional force f_s = τ / r

= 295 Nm / 0.350 m

f_s = 842.8 N

6 0

3 years ago

A lightning bolt occurs when billions of protons are transferred at the same time. ____________________

RSB [31]

True. You're suppose to answer true or false, right?

3 0

3 years ago

The kinetic energy of a rotating body is generally written as K=12Iω2, where I is the moment of inertia. Find the moment of iner

stira [4]

Answer:

See explanation

Explanation:

We have a mass $m$ revolving around an axis with an angular speed $\omega$ , the distance from the axis is $r$ . We are given:

$\omega = 10 [rad/s]\\r=0.5 [m]\\m=13[Kg]$

and also the formula which states that the kinetic rotational energy of a body is:

$K =\frac{1}{2}I\omega^2$ .

Now we use the kinetic energy formula

$K =\frac{1}{2}mv^2$

where $v$ is the tangential velocity of the particle. Tangential velocity is related to angular velocity by:

$v=\omega r$

After replacing in the previous equation we get:

$K =\frac{1}{2}m(\omega r)^2$

now we have the following:

$K =\frac{1}{2}m(\omega r)^2 =\frac{1}{2}Iw^2$

therefore:

$mr^2=I$

then the moment of inertia will be:

$I = 13*(0.5)^2=3.25 [Kg*m^2]$

3 0

3 years ago

A person exerts a tangential force of 37.7 N on the rim of a disk-shaped merry-go-round of radius 2.75 m and mass 144 kg. If the

Shkiper50 [21]

Answer:

ω = 0.467 rad/s

Explanation:

given,

tangential force exerted by the person = 37.7 N

radius of merry-go-round = 2.75 m

mass of merry-go-round = 144 Kg

angle = 33.2°

moment of inertia

$I = \dfrac{1}{2} m R^2$

$I = \dfrac{1}{2}\times 144 \times 2.75^2$

I = 544.5 kg.m²

torque = force x radius

τ = 37.7 x 2.75

τ = 103.675 N.m

angular acceleration

$\alpha= \dfrac{\tau}{I}$

$\alpha= \dfrac{103.675}{544.5}$

α = 0.190 rad/s²

now ,

distance = $33.2\times \dfrca{2\pi}{360}$

d = 0.579 rad

we know,

using equation of rotational motion

$d = \omega t + \dfrac{1}{2}\alpha t^2$

$0.579 = \dfrac{1}{2}\times 0.190\times t^2$

t = 2.46 s

angular speed

ω = α x t

ω = 0.19 x 2.46

ω = 0.467 rad/s

7 0

2 years ago

A spaceship moves radially away from Earth with acceleration 29.4 m/s 2 (about 3g). How much time does it take for sodium street

Gemiola [76]

Answer:

doppler shift's formula for source and receiver moving away from each other:

λ'=λ°√(1+β/1-β)

Explanation:

acceleration of spaceship=α=29.4m/s²

wavelength of sodium lamp=λ°=589nm

as the spaceship is moving away from earth so wavelength of earth should increase w.r.t increasing speed until it vanishes at λ'=700nm

using doppler shift's formula:

λ'=λ°√(1+β/1-β)

putting the values:

700nm=589nm√(1+β/1-β)

after simplifying:

β=0.17

by this we can say that speed at that time is: v=0.17c

to calculate velocity at an acceleration of a=29.4m/s²

we suppose that spaceship started from rest so,

v=v₀+at

where v₀=0

so v=at

as we want to calculate t so:-

t=v/a v=0.17c ,c=3x10⁸ ,a=29.4m/s²

putting values:

=0.17(3x10⁸m/s)/29.4m/s²

t=1.73x10⁶

7 0

3 years ago