1.
Atennis ball is shot straight up with an initial velocity of 34 m/s. What is its velocity two seconds after launch?

Answer:

weight = 900 N

acceleration  = 11.25 m/s²

Explanation:

given data

mass m1 = 75 kg

mass m2 = 80 kg

gravitational acceleration = 12 m/s²

solution

As we know weight of a mass that is

weight of mass = Mass × Acceleration due to gravity .................1

so Astro 1 weight is

weight = 75kg  × 12 m/s²

weight = 900 N

and

so, when Astro 2 needs this much weight the planet on which he is will have the acceleration

acceleration = Weight ÷ Mass of Astro 2        .....................2

acceleration  = 900 ÷ 80 m/s²

acceleration  = 11.25 m/s²

The volume of snow on the lawn, in cubic m, is 44.63 or 44.63 m³ if the lawn outside your neighbor's house has an approximate area of 175 m².

It is defined as a three-dimensional space enclosed by an object or thing.

It is given that:

The lawn outside your neighbor's house has an approximate area of 175 m². One night it snows so that the snow on the lawn has a uniform depth of 25.5 cm.

As we know, unit conversion can be defined as the conversion from one quantity unit to another quantity unit followed by the process of division, and multiplication by a conversion factor.

25.5 cm =  0.255 m

Volume = area×depth

Volume = 175×0.255

Volume = 44.63 cubic meters

Thus, the volume of snow on the lawn, in cubic m is 44.63 or 44.63 m³ if the lawn outside your neighbor's house has an approximate area of 175 m².

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Answer:

Explanation:

Most environments have many niches. If a niche is "empty" (no organisms are occupying it), new species are likely to evolve to occupy it. This happens by the process of natural selection. By natural selection, the nature of the species gradually changes to become adapted to the niche.

Answer:

Divide then multiply or multiply then divide

Explanation:

to get the answer of a and b

Answer:

the length of an antenna one-half wavelength long for sending radio at 300 Hz is 500 km

Explanation:

Given the data in the question;

we know that wave length is;

λ = c/f

c is the speed of voice ( 3 × 10⁸  m/s )

frequency f = 300 Hz

so we substitute

λ = 3 × 10⁸ / 300

λ = 1000000 m

we know that; 1 km = 1000 m

so

λ = 1000000 m / 1000

λ = 1000 km

hence, an antenna one-half wavelength will be;

λ /2

=  1000 km / 2

= 500 km

Therefore, the length of an antenna one-half wavelength long for sending radio at 300 Hz is 500 km

Answer:

Explanation:

Satellite A and satellite B are approaching the earth from opposite directions , that means they are approaching each other . The velocity of satellite A and B are .648c and .795c respectively . Their velocities are comparable to velocity of light so they will follow relativistic laws .

Their relative velocity will be given by the following relation .

[tex]V_r=\frac{u+v}{1+\frac{uv}{c^2} } }[/tex]

where u and v are velocities of vehicles coming from opposite direction and c is velocity of light .

[tex]V_r=\frac{.795c+.648c}{1+\frac{.648c\times .795c}{c^2} } }[/tex]

[tex]V_r=\frac{1.443c}{1+.515 } }[/tex]

= .952c

The gravitational force exerted by the moon on the satellite is such that

F = G M m / R ² = m a   →   a = G M / R ²

where a is the satellite's centripetal acceleration, given by

a = v ² / R

The satellite travels a distance of 2πR about the moon in complete revolution in time T, so that its tangential speed is such that

v = 2πR / T   →   a = 4π ² R / T ²

Substitute this into the first equation and solve for T :

4π ² R / T ² = G M / R ²

4π ² R ³ = G M T ²

T ² = 4π ² R ³ / (G M )

T = √(4π ² R ³ / (G M ))

T = 2πR √(R / (G M ))

The correct expression for the time T it takes the satellite to make one complete revolution around the moon is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].

We can find the period T (the time it takes the satellite to make one complete revolution around the moon) from the gravitational force:

[tex] F = \frac{GmM}{R^{2}} [/tex]    (1)

Where:

G: is the gravitational constant = 6.67x10⁻¹¹ Nm²/kg²

R: is the distance between the satellite and the center of the moon

m: is the satellite's mass

M: is the moon's mass

The gravitational force is also equal to the centripetal force:

[tex] F = ma_{c} [/tex]   (2)

The centripetal acceleration ([tex]a_{c}[/tex]) is equal to the tangential velocity (v):

[tex] a_{c} = \frac{v^{2}}{R} [/tex]   (3)

And from the tangential velocity we can find the period:

[tex] v = \omega R = \frac{2\pi R}{T} [/tex]   (4)

Where:

ω: is the angular speed = 2π/T

By entering equations (4) and (3) into (2), we have:

[tex] F = m\frac{v^{2}}{R} = m\frac{(\frac{2\pi R}{T})^{2}}{R} = \frac{mR(2\pi)^{2}}{T^{2}} [/tex]   (5)

By equating (5) and (1), we get:

[tex] \frac{mR(2\pi)^{2}}{T^{2}} = \frac{GmM}{R^{2}} [/tex]

[tex] T^{2} = \frac{R^{3}(2\pi)^{2})}{GM} [/tex]  

[tex] T = \sqrt{\frac{R^{3}(2\pi)^{2})}{GM}} [/tex]

[tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex]

Therefore, the expression for the time T is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].

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Answer:

The direction could change

Answer:

So increasing the voltage increases the charge in direct proportion to the voltage. If the voltage exceeds the capacitors rated voltage, the capacitor may fail due to breakdown of the dielectric between the two plates that make up the capacitor.

Explanation:

A option.

Answer:

a. P.E = 3430Joules.

b. Workdone = 3430Nm

Explanation:

Given the following data;

Mass = 70kg

Distance = 5m

We know that acceleration due to gravity is equal to 9.8m/s²

To find the potential energy;

Potential energy = mgh

P.E = 70*9.8*5

P.E = 3430J

b. To find the workdone;

Workdone = force * distance

But force = mass * acceleration

Force = 70*9.8

Force = 686 Newton.

Workdone = 686 * 5

Workdone = 3430Nm

Answer:

4m/s^2

Explanation:

Answer:

64.2 m/s

Explanation:

We are given that

Speed ,v=38 m/s

We have to find the maximum speed when his car reach on flat ground.

Using dimensional analysis

[tex]F_{res}\propto v^2[/tex]

If 35% acceleration reduced by F(res) at 38 m/s

Then, 100% acceleration  can be reduced  by F(res) at v' m/s

[tex]\frac{F_1}{F_2}=\frac{v^2}{v'^2}[/tex]

[tex]v'^2=\frac{F_2}{F_1}v^2[/tex]

[tex]v'=v\sqrt{\frac{F_2}{F_1}}[/tex]

Substitute the values

[tex]v'=38\times \sqrt{\frac{100}{35}}[/tex]

[tex]v'=64.2 m/s[/tex]

Hence, the maximum speed when his car can reach on flat ground=64.2 m/s

The maximum velocity of the car at constant acceleration on the road will be 62.2 m/s.

The air resistance is directly proportional to the square of the velocity of an object.

R ∝ v²

The speed of the car was reduced by 35 % at 38 m/s.

So, the speed of the car was reduced by 100% at v' m/s.

The relationship can be given by,

[tex]\dfrac {R_1}{R_2} = \dfrac {v^2}{v'^2}[/tex]

Put the values in the formula and calculate for [tex]v'[/tex],

[tex]v' = \sqrt {\dfrac {R_2 v^2}{R_1 }}[/tex]

[tex]v' = \sqrt {\dfrac { 100\times 38 ^2}{35 }}\\\\v' = 62.2 \rm \ m/s[/tex]

Therefore, the maximum velocity of the car at constant acceleration on the road will be 62.2 m/s.

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Answer: The mean value = 9.85m/s².

Explanation:

Mean = [tex]\dfrac{\text{Sum of n observations}}{n}[/tex]

The given measurements the acceleration of gravity (units of m/s²): 10.1, 9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90.

Number of measurements =9

Sum of measurements =  88.69

Mean = [tex]\dfrac{88.69}{9}=9.85444444\approx9.85[/tex]

Hence, the mean value = 9.85m/s².

Answer:

 the statement is False       [tex]\frac{x_{ball} }{ x_{rec} }[/tex] = cos θ

Explanation:

Let's analyze this problem, the ball and the receiver leave the same point and we want to know if at the same moment they reach the same point, for this we must have both the ball and the receiver travel the same distance.

Let's start by finding the time it takes for the ball to reach the ground

        y = [tex]v_{oy}[/tex] t - ½ g t²

when it reaches the ground its height is y = 0

       0 = vo sin θ  - ½ g t²

       0 = t (vo sin θ - ½ g t)

The results are

       t = 0                             exit point

       t = 2 v₀ sin θ/g            arrival point

at this point the ball traveled

       [tex]x_{ball}[/tex]= v₀ₓ t

       x_{ball} = v₀ cos θ  2v₀ sin θ / g

       x_{ball}= 2 v₀² cos θ   sin θ/ g

Now let's find that distantica traveled the receiver in time

        [tex]x_{rec}[/tex] = v₀ t

        x_{rec} = v₀ (2 v₀ sin θ / g)

        x_{rec} = 2 v₀² sin θ / g

without dividing this into two distances

          [tex]\frac{x_{ball} }{ x_{rec} }[/tex] = cos θ

therefore the distances are not equal to the ball as long as behind the receiver

Therefore the statement is False

Answer:

Speed of Agnes during her journey was 0.848c

Explanation:

Given that;

Age of Agnes t₀ = 20 years

distance d = 2 × distance of star from Earth = 2 × 16 light-years

= 32 light-years

so get her speed speed; we use the following expression

Yvt₀ = d

( v / √(1 - ([tex]\frac{v}{c}[/tex])²) )² = ( 32 light-years / 20 yrs )²

v² / (1 - ( v²/c²)) = ( 32 × c  / 20 )²

v² / (1 - ( v²/c²)) = 2.56 × c²  

v² / c²-v²/c² = 2.56 × c²  

v²c²  / c² - v²  = 2.56 × c²  

v²  / c² - v²  = 2.56

v²  = 2.56 (c² - v²)

v²  = 2.56 (c² - v²)

v²  = 2.56c² - 2.56v²

v² + 2.56v² = 2.56c²

3.56v² = 2.56c²

v² = (2.56/3.56)c²

v = √((2.56/3.56)c²)

so v = 0.848c

Therefore, Speed of Agnes during her journey was 0.848c

Answer:

Explanation:

From the given information;

If we assume that charge 1 is located at the origin;

Then, using the visual overview for identification, we will realize that the known quantities are:

[tex]\mathbf{= q_1, \ q_2 , \ q_3, \ x_1 \ and \ x_2}[/tex]

However, provided that we do not know the exact location of [tex]x_3[/tex],

Then, the unknown quantity is [tex]\mathbf{ x_3}[/tex]

Answer:

1.19m

Explanation:

Given parameters:

Speed of the trumpet sound  = 351m/s

Frequency of the note = 294Hz

Unknown:

Wavelength of the sound  = ?

Solution:

To solve this problem, we use the wave - velocity equation.

    V  = F x ∧  

V is the velocity of the body

F is the frequency

∧ is the wavelength

 So;

          351   = 294 x ∧  

       ∧   = [tex]\frac{351}{294}[/tex]   = 1.19m

Answer: B>A=D>C

Explanation:

Kinetic Energy is the product of mass and square of the velocity

For Jogger A

[tex]K.E._a=\frac{1}{2}mv^2[/tex]

For Jogger B

[tex]K.E._b=\frac{1}{2}\times\frac{m}{2}\times(3v)^2=\frac{9}{4}mv^2\\K.E._b=2.25mv^2[/tex]

For Jogger C

[tex]K.E._c=\frac{1}{2}\times3m\times(\frac{v}{2})^2=\frac{3}{8}mv^2\\K.E._c=0.375mv^2[/tex]

For Jogger D

[tex]K.E._d=\frac{1}{2}\times4m\times(\frac{v}{2})^2=\frac{1}{2}mv^2[/tex]

Kinetic Energy of Joggers in increasing order

B>A=D>C

Answer:

50

Explanation:

Answer:

10.88kJ

Explanation:

Given data

mass= 10kg

heigth= 111m

Applying

PE= mgh

assume g= 9.81m/s^2

substitute

PE= 10*9.81*111

PE=10889.1 Joules

PE=10.881kJ

Hence the potential energy is 10.88kJ

Answer:

The material that Emily invented can be easily repaired by shining ultraviolet light on it.

Explanation:

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Answer:

The material that Emily invented can be easily repaired by shining ultraviolet light on it.

Explanation:

Explanation:

In physics, a force is any interaction that, when unopposed, will change the motion of an object. A force can cause an object with mass to change its velocity, i.e., to accelerate. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity.

Formula

Newton's Second Law

F = m * a

F = force

m = mass of an object

a = acceleration

Answer:

20 C to F

Ans: 68F

-40 C to F

Ans:-40F

40C to F

Ans:104F

Answer:

a)   F = 1.70 10⁻⁹N,   F = 1.47 10⁻⁸ N,

b) * the electronegative repulsion, from the repulsion by quantum effects

Explanation:

a) The atraicione force comes from the electric force given by Coulomb's law,

           F = [tex]k \frac{ q_1 q_2}{r^2}[/tex]

divalent atoms

In this case q = 2q₀ where qo is the charge of the electron -1,6 10⁻¹⁹ C and the separation is given

           F = k q² / r²

           F = [tex]2 \ 10^9 \ \frac{2 (1.6 \ 10^{-19} )^2}{ (0.52 10^{-9} )^2 }[/tex]

           F = 1.70 10⁻⁹N

monovalent atoms

in this case the load is q = q₀

           F = 2 \ 10^9 \  \frac{ (1.6 \  10^{-19} )^2}{ (0.125 10^{-9} )^2 }

           F = 1.47 10⁻⁸ N

b) repulsive forces come from various sources

* the electronegative repulsion of positive nuclei

* the electrostatic repulsion of the electrons when it comes to bringing the electron clouds closer together

* from the repulsion of electron clouds, by quantum effects