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3 years ago

. A 1.00 kg rock is thrown up into the air from ground level at a speed of 8.00 m/s. The ball travels up

Physics

1 answer:

Liula [17]3 years ago

5 0

Answer:

p = -8 kg-m/s

Explanation:

Given that,

Initial speed of the rock, u = 8 m/s

Mass of the rock, m = 1 kg

The ball travels up to a maximum height, then returns to the ground.

We need to find the rock's momentum as it strikes the ground. Let v be the final speed of the rock. Its final speed is as same as initial speed i.e. 8 m/s but in negative direction. So

p = mv

p = 1 kg × (-8 m/s)

= -8 kg-m/s

So, the rock's momentum as it strikes the ground is (-8 kg-m/s).

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Consider two antennas separated by 9.00 m that radiate in phase at 120 MHz, as described in Exercise 35.3. A receiver placed 150

alexgriva [62]

Answer:

$\phi=4.52 rad$

Explanation:

From the question we are told that

Distance b/e antenna's $d=9.00m$

Frequency of antenna Radiation $F_r=120 MHz \approx 120*10^6Hz$

Distance from receiver $d_r=150m$

Intensity of Receiver $i= 10$

Distance difference of the receiver b/w antenna's $(r^2-r^1)=1.8m$

Generally the equation for Phase difference $\phi$ is mathematically given by

$\phi=\frac{2\pi}{\frac{c}{f_r}} *(r^2-r^1)$

$\phi=\frac{2*\pi}{\frac{3*10^{8}}{120*10^6}} *1.8$

$\phi=\frac{4\pi}{5} *1.8$

Therefore phase difference f between the two radio waves produced by this path difference is given as

$\phi=4.52 rad$

7 0

3 years ago

Concept Simulation 2.3 offers a useful review of the concepts central to this problem. An astronaut on a distant planet wants to

Alexandra [31]

Answer:

1.40 m/s^2

Explanation:

Given data

Velocity= 17.4 m/s

time= 12.4 seconds

We want to find the acceleration of the rock

We know that

acceleration = velocity/time

Substitute

acceleration= 17.4/12.4

acceleration=1.40 m/s^2

Hence the acceleration is 1.40 m/s^2

7 0

3 years ago

A bird lands on a bare copper wire carrying a current of 51

nirvana33 [79]

The resistance of the piece of wire is
$R= \frac{\rho L}{A}$
where
$\rho = 1.68 \cdot 10^{-8}\Omega m$ is the resistivity of the copper
$L=5.1 cm=0.051 m$ is the length of the piece of wire
$A=0.13 cm^2 = 0.13 \cdot 10^{-4} m^2$ is the cross sectional area of the wire
By substituting these values, we find the value of R:
$R= \frac{\rho L}{A}=6.6 \cdot 10^{-5} \Omega$

Then, by using Ohm's law, we find the potential difference between the two points of the wire:
$V=IR=(51 A)(6.6 \cdot 10^{-5} \Omega )=3.4 \cdot 10^{-3} V$

7 0

3 years ago

When a sacred item or symbol is removed from its special place or is duplicated in mass quantities, then it becomes profane as a

Galina-37 [17]

Answer:

It becomes profane as a result of desacralization

Explanation:

Because desacralization means when a dedicated religious structure is no longer used for its intended purpose both rather used for another purpose other than the original purpose

7 0

3 years ago

If the radius of a blood vessel drops to 84.0% of its original radius because of the buildup of plaque, and the body responds by

erma4kov [3.2K]

To develop this problem it is necessary to apply the equations concerning Bernoulli's law of conservation of flow.

From Bernoulli it is possible to express the change in pressure as

$\Delta P = \frac{1}{2}\rho (v_1^2-v_2^2)+ \rho g (h_1h_2)$