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3 years ago

Nancy is sailing her boat toward Sam's boat at 5

Physics

2 answers:

Bas_tet [7]3 years ago

8 0

Answer:

Nora's boat is moving towards Sam at 5 km/hr

Explanation:

The question says that Nora is few meters behind Nancy and is still with respect to her that means the relative velocity between Nora and Nancy is zero

Vrel = 5 - Vnora= 0 ⇒ Vnora = 5km/hr

Pictorially we can represent the given condition as:

Nora-------few meters------Nancy ----------------- Sam

5km/hr 5km/hr →

Hence, Nora's boat is moving towards Sam at 5km/hr.

ladessa [460]3 years ago

6 0

Answer:

Nora's boat is moving towards Sam at 5 km/hr

Explanation:

The question says that Nora is few meters behind Nancy and is still with respect to her that means the relative velocity between Nora and Nancy is zero

Vrel = 5 - Vnora= 0 ⇒ Vnora = 5km/hr

Pictorially we can represent the given condition as:

Nora-------few meters------Nancy ----------------- Sam

5km/hr 5km/hr →

Hence, Nora's boat is moving towards Sam at 5km/hr.

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sweet [91]

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5 0

3 years ago

Read 2 more answers

Compute the expected shell-model quadrupole moment of 209Bi () and compare with the experimental value, - 0.37 b

Over [174]

Answer:

0.22 b

Explanation:

Quadrupole moment of the nucleon is,

$Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2}$

And also,

$R^{2}=R^{2} _{0}A^{\frac{2}{3} }$

And, $R _{0}=1.2\times 10^{-15}m$

Now,

$Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2} _{0}A^{\frac{2}{3} }$

For Bismuth $j=\frac{9}{2}$ and A is 209.

$Q=-\frac{2\frac{9}{2} -1}{2(\frac{9}{2} +1)}\frac{3}{5}(1.2\times 10^{-15}) ^{2}(209)^{\frac{2}{3} }\\Q=0.628\times 35.28\times 10^{-30} \\Q=22.15\times 10^{-30} m^{2} \\Q=0.2215\times 10^{-28} m^{2} \\Q=0.22 barn$

Therefore, the expected value of quadrupole is 0.22 b which is quite related with experimental value which is 0.37 b

3 0

3 years ago

a force of 17 acts on a block of mass 4kg the forces of friction opposing the motion is 5n . the acceleration of the block is

Nady [450]

Fk = μK N

N = m a

N = 4 × a

N = 4a

Fk = μK N

17 = 5 × 4a

17 = 20 a

a = ¹⁷/₂₀ = 0.85 m/s²

8 0

2 years ago

If you were to divide the presentday universe into cubes whose sides are 10 million lightyears long, each cube would contain, on

vovikov84 [41]

Answer:

Explanation:

density of galaxies would be $\frac{1}{0.27^3}$ times higher which is equal to 50.81.

It means in a cube that today contains one galaxy the size of the Milky Way, we would instead find 50.81 galaxies this size.

You can round this off to 52

8 0

3 years ago

If the beam carries 1015 electrons per second and is accelerated by a 350 kV source, find the current and power in the beam.

Y_Kistochka [10]

To solve this problem we will apply the concept of current defined as the electron charge flow by the number of electrons per second. That is,

I = q*N

Here q is Flow of electric charge in one second and N the number of electron flow per second.

A the same time the power is described as the applied voltage for the current.

P = VI

We know the charge of electron, $q = 1.602 * 10^{-19}$ Coulombs, then the current is

$I = (1.602*10^{-19})(10^{15})$

$I = 0.1602 mA$

And the power in the Beam is

$P = VI$

$P = (350*10^3)(0.1602)$

$P = 0.05607 Watts$

3 0

3 years ago