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4 years ago

Can some answer 7 a and b please

Physics

1 answer:

pav-90 [236]4 years ago

4 0

Answer:

a.After $15$ second Mr Comer's speed $=$ $1.7 m/s$

b.Distance travelled by Mr.Comer in $15$ seconds $=$ $18.75\ m$

Explanation:

a. Lets recall our first equation of motion $V_{f} =V_{i} + at$

Now we know that $V_{i} = 0.8m/s$ , $a = 0.06\ m/s^2$ and

$t = 15 \ sec$

Plugging the values we have.

$V_{f} =V_{i} + at$

$V_{f} =0.8 + 0.06 \times 15$

$V_{f} =0.8+ 0.9$

$V_{f} =1.7 m/s$

Then Mr.Comer's speed after $15$ sec $=$ $1.7ms^-1$

Lets find the distance and recall our third equation of motion.

$V_{f} ^2-V{i}^2 = 2as$

So $s =$ distance covered.

Dividing both sides with 2a we have.

$\frac{V_{f} ^2-V{i}^2}{2a} = 2as$

Plugging the values.

$\frac{(1.7)^2-(0.8)^2}{2\times0.06} = 2as$

$s= 18.75\ m$

So Mr.Comer will travel a distance of $s= 18.75\ m$ .

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The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold 24 x (mass o

matrenka [14]

Answers:

(a) $0.0073kg$

(b) Volume gold: $3.79(10)^{-7}m^{3}$ , Volume cupper: $7.6(10)^{-8}m^{3}$

(c) $17633.554kg/m^{3}$

Explanation:

We are told the total mass $M$ of the coin, which is an alloy of gold and copper is:

$M=m_{gold}+m_{copper}=7.988g=0.007988kg$ (1)

Where $m_{gold}$ is the mass of gold and $m_{copper}$ is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats $K$ and mass is:

$K=24\frac{m_{gold}}{M}$ (2)

Finding ${m_{gold}$ :

$m_{gold}=\frac{22}{24}M$ (3)

$m_{gold}=\frac{22}{24}(0.007988kg)$ (4)

$m_{gold}=0.0073kg$ (5) This is the mass of gold in the coin

<h2>(b) Volume of gold and cupper</h2><h2 />

The density $\rho$ of an object is given by:

$\rho=\frac{mass}{volume}$

If we want to find the volume, this expression changes to: $volume=\frac{mass}{\rho}$

For gold, its volume $V_{gold}$ will be a relation between its mass $m_{gold}$ (found in (5)) and its density $\rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}$ :

$V_{gold}=\frac{m_{gold}}{\rho_{gold}}$ (6)

$V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}$ (7)

$V_{gold}=3.79(10)^{-7}m^{3}$ (8) Volume of gold in the coin

For copper, its volume $V_{copper}$ will be a relation between its mass $m_{copper}$ and its density $\rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}$ :

$V_{copper}=\frac{m_{copper}}{\rho_{copper}}$ (9)

The mass of copper can be found by isolating $m_{copper}$ from (1):

$M=m_{gold}+m_{copper}$

$m_{copper}=M-m_{gold}$ (10)

Knowing the mass of gold found in (5):

$m_{copper}=0.007988kg-0.0073kg=0.000688kg$ (11)

Now we can find the volume of copper:

$V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}$ (12)

$V_{copper}=7.6(10)^{-8}m^{3}$ (13) Volume of copper in the coin

<h2>(c) Density of the sovereign coin</h2><h2 />

Remembering density is a relation between mass and volume, in the case of the coin the density $\rho_{coin$ will be a relation between its total mass $M$ and its total volume $V$ :

$\rho_{coin}=\frac{M}{V}$ (14)

Knowing the total volume of the coin is:

$V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3}$ (15)

$\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}}$ (16)

Finally:

$\rho_{coin}=17633.554kg/m^{3}}$ (17) This is the total density of the British sovereign coin

6 0

3 years ago

I would like to know the correct answer for this

Mnenie [13.5K]

The last choice on the list is the correct one, for both #2 and #3.

3 0

4 years ago

Viewed through a spectroscope, the spectral profile of a yellow street lamp has a narrow line in the yellow region of the visibl

muminat

Answer:

discrete lines are observed by the spectroscope, the emission of the lamp is of the ATOMIC source

Explanation:

Bulbs can emit light in several ways:

* When the emission is carried out by the heating of its filament, the bulb is called incandescent, in general its spectrum is similar to that of a black body, this is a continuous spectrum with a maximum dependent on the fourth power of the temperature of the filament.

* The emission can be by atomic transitions, in this case there is a discrete spectrum formed by the spectral lines of the material that forms the gas of the lamp, in general for the yellow emission the most used materials are mercury and sodium or a mixture of they.

Consequently, as discrete lines are observed by the spectroscope, the emission of the lamp is of the ATOMIC type

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3 years ago

A plane is flying east at 115 m/s. The wind accelerates it at 2.88 m/s^2 directly northwest. After 25.0s, what is the magnitude

kondaur [170]

Answer:

81.8 m/s

Explanation:

The initial velocity of the plane is:

$v_0=115 m/s$ (toward east)

So, decomposing along the x- and y- directions:

$v_{x0} = 115 m/s\\v_{y0} = 0$

(we took east as positive x-direction and north as positive y-direction)

The acceleration is

$a=2.88 m/s^2$ (northwest, so the angle with the positive x-direction is 135 degrees)

Decomposing it along the two directions:

$a_x = a cos 135^{\circ} = (2.88 m/s^2)(cos 135^{\circ})=-2.04 m/s^2\\a_y = a sin 135^{\circ} = (2.88 m/s^2)(sin 135^{\circ})=2.04 m/s^2$

So the two components of the velocity after a time t = 25.0 s will be

$v_x = v_{x0} + a_x t = 115 m/s + (-2.04 m/s^2)(25.0 s)=64 m/s\\v_y = v_{y0} + a_y t = 0 m/s + (2.04 m/s^2)(25.0 s)=51 m/s$

So, the magnitude of the velocity of the plane will be

$v=\sqrt[v_x^2+v_y^2}=\sqrt{(64 m/s)^2+(51 m/s)^2}=81.8 m/s$

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3 years ago

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2 years ago