Physics - Vectors-- Could you help me please?

A trumpet player, using her ear, hears 5 beats per second when she plays a note on her trumpet and simultaneously sounds a 440 H

german

<h2>Frequency of trumpet wire = 445 Hz</h2>

Explanation:

The given frequency of tuning fork is 440 Hz . It produces 5 beats with trumpet wire . That means , the frequency of wire can be 440 ± 5

It can be either 445 or 435

Now the length of wire is increased , by which its frequency decreases . Because frequency is inversely proportional to length of wire .

If we decrease the frequency in 435 , the difference between tuning fork frequency and wire frequency will become greater than 5 even . So it cannot produce 3 beats with it .

If we decrease frequency from 445 , it can become 443 Hz . It gives 3 beat with the tuning fork as given .

Thus the initial frequency of wire is 445 Hz

4 0

3 years ago

The nucleus of a certain type of uranium atom contains

BaLLatris [955]

Answer:

charge = electrons + protons

=92+92

=184

8 0

3 years ago

What is the time feather to fall to the surface of the moon ?

denpristay [2]

Mbccbnb jdkdmmdodbs ndkdkmdnd ndkdkfnndf njdkdmdnxjsbd. Djjdjdndnfk ndkdkfkd

8 0

4 years ago

you will For this problem, you will need to look up physical parameters for objects in space want to keep about 4 significant fi

Tanzania [10]

Answer:

a) r₁ = 3.836 10⁷ m, b) F = - 3,390 10⁸ N , c) R = 120.3 m

Explanation:

a) This is a problem of equilibrium where the force is gravitational, we call F1 the force of the Moon and F2 the force from the earth.

F₁ -F₂ = 0

F₁ = F₂

G m $M_{m}$ / r₁² = G m $M_{e}$ / r₂²

Let's look for the measured distance from a Coordinate System located on the Moon,

r₂ = D - r₁

Where D is the average distance from Terra to the Moon 3.84 10⁸ m

$M_{m}$ / r₁² = $M_{e}$ / (D - r₁)²

(D² - 2 D r₁ + r₁²) = $M_{e}$ / $M_{m}$ r₁²

(1 - $M_{e}$ / $M_{m}$ )r₁² - 2D r₁ + D² = 0

Let's replace and solve the second degree equation

(1 - 5.98 10²⁴ / 7.36 10²²) r₁² - 2 3.84 10⁸ r₁ + (3.84 10⁸)² = 0

-80.25 r₁² - 7.68 10⁸ r₁ + 14.75 10¹⁶ = 0

r₁² + 9.57 10⁶ r₁ - 1.838 10¹⁵ = 0

r₁ = [-9.57 10⁶ ±√ (91.58 10¹² + 7.352 10¹⁵)] / 2

r₁ = [-9.57 10⁶ + - 86.28 10⁶] / 2

The results are:

r₁’= 38.355 10 6 m

r₁ ’’ = -47.915 10 6 m

We take the positive result that a distance between the moon and the Earth, the equalization point is r₁ = 3.836 10⁷ m

b) The force at point R = 2 r₁

R = 2 3.8355 10⁷ = 7.671 10⁷ m

F = F₁ - F₂

F = G m $M_{m}$ / R² - G m $M_{e}$ / (D- R)²

F = G m ( $M_{m}$ / R² - $M_{e}$ / (D-R)²)

F = m 6.67 10⁻¹¹ (7.36 10²² / (7.671 10⁷)² - 5.98 10²⁴ / (3.84 10⁸ - 0.7671 10⁸)²

F = m 6.67 10⁻¹¹ (0.125076 10⁸ - 0.63329 10⁸)

F = m (-3.3897 10⁸) N

The mass m of the rocket must be known, suppose it is worth 1 kg (m = 1 kg)

F = - 3,390 10⁸ N

c) let's use gravitational force from the moon

F = G m $M_{m}$ / R²

R =√ G m $M_{m}$ / F

R = √ (6.67 10⁻¹¹ 1 7.36 10²² / 3.3897 10⁸)

R = √ (1.4482 10⁴)

R = 1.2034 10² m

R = 120.3 m

8 0

3 years ago

Can you go around a curve with zero acceleration? Explain

Serjik [45]

“Yes, it is impossible to go around the curved path with zero acceleration. Acceleration is change in velocity. And velocity( speed+ direction) changes at every point in a curved path because there is change in direction at every point.”

6 0

3 years ago