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2 years ago

A 15kg dog jumps out of a 40kg canoe. If the dogs velocity is 1.2m/s, what is the velocity of the canoe?

Physics

1 answer:

vivado [14]2 years ago

6 0

Answer:

v = -0.45 m/s

Explanation:

Assuming the canoe was initially at rest with momentum L = 0

and that the dog's velocity is in the positive direction

conservation of momentum

0 = 15(1.2) + 40v

v = -0.45 m/s

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Radio waves of frequency 1.667 GHzGHz arrive at two telescopes that are connected by a computer to perform interferometry. One p

SVETLANKA909090 [29]

Explanation:

The frequency of radio waves is 1.667 GHz

One portion of the same wave front travels 1.260 mm farther than the other before the two signals are combined.

There are two conditions for interference either constructive or destructive.

For constructive interference , the path difference is n times of wavelength and for destructive interference, the path difference is (n+1/2) times of wavelength

We can find wavelength in this case as follows :

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{1.667\times 10^9}\\\\\lambda=0.1799\ m$

If we divide path difference by wavelength,

$\dfrac{\delta}{\lambda} =\dfrac{1.26}{0.1799}\\\\\dfrac{\delta}{\lambda} =7\\\\\delta =7\lambda$

It means that the path difference is 7 times of the wavelength. it means the two waves combine constructively and the value of m for the path difference between the two signals is 7.

5 0

3 years ago

PLEASE HELP ME ASAP!!!!!!!!!

NNADVOKAT [17]

Answer:

letter C. velocity hope this helps

7 0

3 years ago

A voltage V is applied to the primary coil of a step-up transformer with a 3:1 ratio of turns between its primary and secondary

Snezhnost [94]

Explanation:

Let $N_p\ and\ N_s$ are the number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p}{N_s}=\dfrac{1}{3}$

A resistor R connected to the secondary dissipates a power $P_s=100\ W$

For a transformer, $\dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}$

$V_s=(\dfrac{N_s}{N_p})V_p$

$V_s=3V_p$ ...............(1)

The power dissipated through the secondary coil is :

$P_s=\dfrac{V_s^2}{R}$

$100\ W=\dfrac{V_s^2}{R}$

$V_p^2=\dfrac{100R}{9}$ .............(2)

Let $N_p'\ and\ N_s'$ are the new number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p'}{N_s'}=\dfrac{1}{24}$

New voltage is :

$V_s'=(\dfrac{N_s'}{N_p'})V_p'$

$V_s'=24V_p$ ...............(3)

So, new power dissipated is $P_s'$

$P_s'=\dfrac{V_s'^2}{R}$

$P_s'=\dfrac{(24V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}$

$P_s'=6400\ Watts$

So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.

3 0

3 years ago

Pls help

xenn [34]

Answer:

A. It must be zero

Explanation:

A spacecraft leaves the solar system at a velocity of 1,500 m/s. The net force on this spacecraft is zero. What can we say about the spacecraft's acceleration?

According to Newton's second law

Force = Mass × acceleration

If the net force is zero

0 = mass × acceleration

0 = ma

a = 0/m

a = 0m/s²

this shows that the acceleration will be zero If the net force is zero

4 0

2 years ago

A motorboat traveling on a straight course slows

never [62]

Answer:

2.572 m/s²

Explanation:

Convert the given initial velocity and final velocity rates to m/s:

65 km/h → 18.0556 m/s
35 km/h → 9.72222 m/s

The motorboat's displacement is 45 m during this time.

We are trying to find the acceleration of the boat.

We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.

v² = v₀² + 2aΔx

Substitute the known values into the equation.

(9.72222)² = (18.0556)² + 2a(45)
94.52156173 = 326.0046914 + 90a
-231.4831296 = 90a
a = -2.572

The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².

3 0

3 years ago