Based on the chemical equation, use the drop-down menu to choose the coefficients that will balance the chemical

Feliz [49]

a. 46 m/s east

The jet here is moving with a uniform accelerated motion, so we can use the following suvat equation to find its velocity:

$v=u+at$

where

v is the velocity calculated at time t

u is the initial velocity

a is the acceleration

The jet in the problem has, taking east as positive direction:

u = +16 m/s is the initial velocity

$a=+3 m/s^2$ is the acceleration

Substituting t = 10 s, we find the final velocity of the jet:

$v=16 + (3)(10)=46 m/s$

And since the result is positive, the direction is east.

b. 310 m

The displacement of the jet can be found using another suvat equation

$s=ut+\frac{1}{2}at^2$

where

s is the displacement

u is the initial velocity

a is the acceleration

t is the time

For the jet in this problem,

u = +16 m/s is the initial velocity

$a=+3 m/s^2$ is the acceleration

t = 10 s is the time

Substituting into the equation,

$s=(16)(10)+\frac{1}{2}(3)(10)^2=310 m$

4 0

3 years ago

A wall clock has a minute hand with a length of 0.55 m and an hour hand with a length of 0.26 m. Take the center of the clock as

Lemur [1.5K]

Answer:

The magnitude of the acceleration of the tip of the minute hand of the clock $1.675\times10^{-6}\ m/s^2$ .

Explanation:

Given that,

Length of minute hand = 0.55 m

Length of hour hand = 0.26 m

The time taken by the minute hand to complete one revelation is

$T= 3600\ sec$

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=\dfrac{2\pi}{T}$

Put the value into the formula

$\omega=\dfrac{2\pi}{3600}$

$\omega=0.001745\ rad/s$

We need to calculate the magnitude of the acceleration of the tip of the minute hand of the clock

Using formula of acceleration

$a=r\omega^2$

Put the value into the formula

$a=0.55\times(0.001745)^2$

$a=1.675\times10^{-6}\ m/s^2$

Hence, The magnitude of the acceleration of the tip of the minute hand of the clock $1.675\times10^{-6}\ m/s^2$ .

3 0

3 years ago

In an atomic clock there are approximately 9.193 × 109oscillations of the specified light emitted by cesium-133 atoms. The text

Aleks [24]

Explanation:

6000 years = 6000 x 365 x 24 x 60 x 60

= 1.892 x 10¹¹ second

gain is 1 second

1 second is equivalent to 9.193 × 10⁹ oscillations .

In 1.892 x 10¹¹ second , change in oscillation is 9.193 × 10⁹ oscillation

in one second change in oscillation = (9.193 / 1.892 ) x 10⁹⁻¹¹

= 4.859 x 10⁻² oscillations .

5 0

2 years ago

A horizontal clothesline is tied between 2 poles, 16 meters apart. When a mass of 3 kilograms is tied to the middle of the cloth

Alla [95]

Answer:

The magnitude of the tension on the ends of the clothesline is 41.85 N.

Explanation:

Given that,

Poles = 2

Distance = 16 m

Mass = 3 kg

Sags distance = 3 m

We need to calculate the angle made with vertical by mass

Using formula of angle

$\tan\theta=\dfrac{8}{3}$

$\thta=\tan^{-1}\dfrac{8}{3}$

$\theta=69.44^{\circ}$

We need to calculate the magnitude of the tension on the ends of the clothesline

Using formula of tension

$mg=2T\cos\theta$

Put the value into the formula

$3\times9.8=2T\times\cos69.44$

$T=\dfrac{3\times9.8}{2\times\cos69.44}$

$T=41.85\ N$

Hence, The magnitude of the tension on the ends of the clothesline is 41.85 N.

4 0

3 years ago

Read 2 more answers

The gas in a balloon has T=280 K and V=0.0279 m^3. if the temperature increases to 320 K at constant pressure, what is the new v

victus00 [196]

Answer:

0.0319 m³

Explanation:

Use ideal gas law:

PV = nRT

where P is pressure, V is volume, n is amount of gas, R is the gas constant, and T is temperature.

Since P, n, and R are held constant:

n₁ R / P₁ = n₂ R₂ / P₂

Which means:

V₁ / T₁ = V₂ / T₂

Plugging in:

0.0279 m³ / 280 K = V / 320 K

V = 0.0319 m³

8 0

3 years ago