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3 years ago

Who is known as the "father of modern chemistry" because he first organized all known elements into four different

Physics

2 answers:

iVinArrow [24]3 years ago

5 0

Answer:

D. Henry Moseley

Explanation:

I'm chemistry right now

Zolol [24]3 years ago

4 0

Answer:Antoine lavoisier

Explanation:

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Calculate the minimum frequency of ultrasound (in Hz) that will allow you to see details as small as 0.193 mm in human tissue. (

STatiana [176]

Answer:

f = 7.97 x 10⁶ Hz = 7.97 MHz

Explanation:

The speed of a wave is given by the following formula:

$v = f\lambda$

where,

v = speed of the ultrasound wave through human tissue = 1540 m/s

f = frequency of ultrasound wave required = ?

λ = wavelength of ultrasound waves = smallest detail required = 0.193 mm

λ = 0.193 mm = 1.93 x 10⁻⁴ m

Therefore,

$1540\ m/s = f(1.93\ x\ 10^{-4}\ m)\\f = \frac{1540\ m/s}{1.93\ x\ 10^{-4}\ m}$

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3 years ago

If the N pole of a magnet is brought near to an unmagnetized iron, nail, then the nail becomes

Andreyy89

Answer:

The nail also magnatized .

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2 years ago

When humans started launching rockets into outer space, they needed to make certain the rockets had enough power to escape Earth

GaryK [48]

Tha answer has to be a

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3 years ago

An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

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3 years ago

A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately

Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car $V_{c}$ = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

$V_{f}$ = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

$V_{f}$ = $V_{i}$ + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁ = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁ will be

$d_{c}$ = $V_{c}$ × t₁

we substitute

$d_{c}$ = 22.352 × 7

$d_{c}$ = 156.46 m

now distance between polive car and speeding car

Δd = $d_{c}$ - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( $V_{f}$ - $V_{c}$ )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = $V_{f}$ × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b) how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

6 0

3 years ago