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3 years ago

Newtons first law of motion states that the velocity of an object doesn’t change if the forces acting on the object are

1 answer:

4 0

Balenced

Think of it this way if 2 same weight and same speed objects were racing toward each other perfectly straight then their forces will cancel out and the movement afterwords will be 0

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The newest generation of smart phone uses three different batteries connected in a series simultaneously to extend battery life

azamat

Answer:

7.2 V

Explanation:

The three batteries are connected in series to the terminals of the phone: it means that they are connected along the same branch, so the current flowing through them is the same.

This also means that the potential difference across the phone will be equal to the sum of the voltages provided by each battery.

Here, the voltage provided by each battery is

V = 2.4 V

So, the overall voltage will be

V = 2.4 V + 2.4 V + 2.4 V = 7.2 V

8 0

3 years ago

If the diameter of a radar dish is doubled, what happens to its resolving power assuming that all other factors remain unchanged

kirill115 [55]

Answer:

θ’ = θ₀ / 2

we see that the resolution angle is reduced by half

Explanation:

The resolving power of a radar is given by diffraction, for which we will use the Rayleigh criterion for the resolution of two point sources, they are considered resolved if the maximum of diffraction of one coincides with the first minimum of the other.

The first minimum occurs for m = 1, so the diffraction equation of a slit remains

a sin θ = λ

in general, the diffraction patterns occur at very small angles, so

sin θ = θ

θ = λ / a

in the case of radar we have a circular aperture and the equation must be solved in polar coordinates, which introduces a numerical constant.

θ = 1.22 λ /a

In this exercise we are told that the opening changes

a’ = 2 a

we substitute

θ ‘= 1.22 λ / 2a

θ' = (1.22 λ / a) 1/2

θ’ = θ₀ / 2

we see that the resolution angle is reduced by half

8 0

3 years ago

A sharp edged orifice with a 60 mm diameter opening in the vertical side of a large tank discharges under a head of 6 m. If the

Ierofanga [76]

Answer:

The discharge rate is $Q = 0.0192 \ m^3 /s$

Explanation:

From the question we are told that

The diameter is $d = 60 \ mm = 0.06 \ m$

The head is $h = 6 \ m$

The coefficient of contraction is $Cc = 0.68$

The coefficient of velocity is $Cv = 0.92$

The radius is mathematically evaluated as

$r = \frac{d}{2}$

substituting values

$r = \frac{ 0.06 }{2}$

$r = 0.03 \ m$

The area is mathematically represented as

$A = \pi r^2$

substituting values

$A = 3.142 * (0.03)^2$

$A = 0.00283 \ m^2$

The discharge rate is mathematically represented as

$Q = Cv *Cc * A * \sqrt{ 2 * g * h}$

substituting values

$Q = 0.68 * 0.92* 0.00283 * \sqrt{ 2 * 9.8 * 6}$

$Q = 0.0192 \ m^3 /s$

6 0

3 years ago

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz. Find the possible range of wavelengths in ai

taurus [48]

Answer:

The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

Explanation:

Given that,

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.

The speed of sound in air is 343 m/s.

To find,

The wavelength range for the corresponding frequency.

Solution,

The speed of sound is given by the following relation as :

$v=f_1\lambda_1$

Wavelength for f = 45 Hz is,

$\lambda_1=\dfrac{v}{f_1}$

$\lambda_1=\dfrac{343}{45}=7.62\ m$

Wavelength for f = 375 Hz is,

$\lambda_2=\dfrac{v}{f_2}$

$\lambda_2=\dfrac{343}{375}=0.914\ m/s$

So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

6 0

3 years ago

You throw a bouncy rubber ball and a wet lump of clay, both of mass m, at a wall. Both strike the wall at speed v, but while the

lana [24]

Answer:

<em>The fifth option is the correct answer: mv; 2 mv</em>

Explanation:

<u>Change of Momentum</u>

Assume an object has a momentum p1 and after some interaction it now has a momentum p2, the change of momentum is

$\Delta p=p_2-p_1$

The momentum is computed as

$p=mv$

Where m is the mass of the object and v its speed. Now let's analyze the situation of both the ball and the clay.

The clay has an initial speed v and a mass m, thus its initial momentum is

$p_1=mv$

When it hits the wall, it sticks, thus its final speed is 0 and

$p_2=0$

The change of momentum is

$\Delta p=0-mv=-mv$

The absolute change is mv

Now for the ball, the initial condition is the same as it was for the clay, but the ball hits back at the same speed, thus its final momentum is

$p_2=-mv$

The change of momentum is

$\Delta p=-mv-mv=-2mv$

The absolute change is 2mv

The fifth option is the correct answer: mv; 2 mv

3 0

3 years ago