Calculate the extension of a 20cm spring that has a spring contrast of 45000N/m and 1500N of elastic potential energy.

Answer:

112.63km/hr

Explanation:

The given dimension is :

         70mph

We are to convert this to km/hr

         1 mile  = 1.609km

so;

     70mph x 1.609    = 112.63km/hr

So,

  The solution is 112.63km/hr

Answer:

30 degrees

Explanation: reflection, same angle

For a ray of monochromatic light is incident on a plane mirror at and angle of 30°. The angle of reflection for the light is 30°.

Reflection occurs when radiation bounces off from a surface. Light is an electromagnetic wave and it can be reflected. According to the laws of reflection, the angle of incidence is equal to the law of reflection.

Hence, for a ray of monochromatic light is incident on a plane mirror at and angle of 30°. The angle of reflection for the light is 30°.

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Answer:

reducing temperature of the surrounding

Explanation:

combustion reactions are exothermic so they give off heat. reducing the temperature of the surrounding will enable more efficient energy transfer

Answer:

the tack's tangential speed is  5.59 m/s

Explanation:

Given that;

R = 0.331 m

wheel rotates 2.69 times a second which means, the wheel complete 2.69 revolutions in a second, so

ω = 2.69 rev/s × 2π/1s = 16.9 rad/s

using the relation of angular speed with tangential speed

tangential speed v of the tack is expressed as;

v = R × ω

so we substitute

v = 0.331 m × 16.9 rad/s

v = 5.59 m/s

Therefore, the tack's tangential speed is  5.59 m/s

Answer:

C. Y

Explanation:

From Newton's second law of motion, we know that:

       Force = mass x acceleration

So;

      acceleration  = [tex]\frac{Force }{mass}[/tex]  

 Therefore, to have the highest acceleration at a constant force, the mass must be low. Acceleration is inversely proportional to mass.

Y has the least mass and it will have the highest acceleration

Answer:

7.35 m/s

Explanation:

Using y - y' = ut - 1/2gt², we find the time it takes the ball to fall from the 1.2 m table top and hit the floor.

y' = initial position of ball = 1.2 m, y = final position of ball = 0 m, u = initial vertical velocity of ball = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for ball to hit the ground.

So, substituting the values of the variables into the equation, we have

y - y' = ut - 1/2gt²

0 - 1.2 m = (0 m/s)t - 1/2(9.8 m/s²)t²

- 1.2 m = 0 - (4.9 m/s²)t²

- 1.2 m = - (4.9 m/s²)t²

t² = - 1.2 m/- (4.9 m/s²)

t² = 0.245 s²

t = √(0.245 s²)

t = 0.49 s

Since d = vt where d = horizontal distance ball moves = 3.6 m, v = horizontal velocity of ball = unknown and t = time it takes ball to land = 0.49 s.

So, d = vt

v = d/t

= 3.6 m/0.49 s

= 7.35 m/s

Since the initial velocity of the ball is 7.35 m/s since the initial vertical velocity is 0 m/s.

It is shown thus V = √(u² + v²)

= √(0² + v²)

= √(0 + v²)

= √v²

= v

= 7.35 m/s

Answer:

A)5524J,

B) 29.2Nm

Explanation:

This question can be treated using work- energy theorem

Work= change in Kinectic energy

W= Δ KE

Work= difference between the final Kinectic energy and intial Kinectic energy.

We know that

Kinectic energy= 1/2 mv^2 .............eqn(1)

This can be written in term of angular velocity, as

KE= 1/2 I

Answer:

12.31 m/s

Explanation:

If we recall from the previous knowledge we had about speed,

we will know that:

speed = distance/ time.

As such:

The average speed of the rider bicycle is

average speed = total distance/ total time

Mathematically, it can be computed as:

[tex]v_{avg} = \dfrac{d+d}{\dfrac{d}{v_1}+ \dfrac{d}{v_2}}[/tex]

[tex]v_{avg} = \dfrac{2d}{\dfrac{d}{10 \ m/s}+ \dfrac{d}{16 \ m/s}}[/tex]

[tex]v_{avg} = \dfrac{2}{\dfrac{1}{10 \ m/s}+ \dfrac{1}{16 \ m/s}}[/tex]

[tex]v_{avg} = \dfrac{2}{\dfrac{13}{80 \ m/s}}[/tex]

[tex]\mathbf{v_{avg} =12.31 \ m/s}[/tex]

Answer:

15 seconds

Explanation:

If car was moving at 20m/s for 3 sec.

if car traveled 100m = 15 sec total

Answer:

Hello your question is incomplete attached below is the missing part of the question

In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Assume that the pivot is attached tot he center of the bar.

You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m1>m2.I

answer : part A = 2*[(M1 - M2)/(M1 + M2)]*g/L

              part A = attached below

Explanation:

Part A :

Assuming that mass of swing is negligible

α = T/I

where ; T = torque, I = inertia,

hence T =  L/2*9*(M1 - M2)

also;  I = [tex]M1*(L/2)^2 + M2*(L/2)^2[/tex]=  ( M1 + M2) * (L/2)^2

Finally the magnitude of the angular acceleration α

α = 2*[(M1 - M2)/(M1 + M2)]*g/L

Part B attached below

Answer:

Analysis

Explanation:

Because you must Analysis each and every cold too find out which virus caused this.

It’s weird because Interpretation and Analysis have the meaning of examination

Answer:

20 J

Explanation:

Work done is given force by distance .

W= F * d  where F is force given by the product of pressure and area

W= P* Δv  where Δv  is change in volume.

Given that ;

1 mole of an ideal gas is kept at 0°C, the pressure of the gas is : 1 atm.

Δv  is change in volume , 30 l - 10l = 20 l

W= 1 * 20 = 20 J

Answer:

742.3N

Explanation:

Given parameters:

Distance  = 39m

Mass 1  = 9.2 x 10⁷kg

Mass 2  = 1.84 x 10⁸kg

Unknown:

Gravitational force between the ships  = ?

Solution:

To solve this problem, we apply the newton's law of universal gravitation:

            Fg  = [tex]\frac{G x mass 1 x mass 2}{r^{2} }[/tex]  

G is the universal gravitation constant  = 6.67 x 10⁻¹¹

r is the distance or separation

   Fg  = [tex]\frac{6.67 x 10^{-11} x 9.2 x 10^{7} x 1.84 x 10^{8} }{39^{2} }[/tex]   = 742.3N

Answer:

8.34 x 10⁻⁶N

Explanation:

Given parameters:

Mass 1  = 100kg

Mass 2 = 200kg

Distance of separation  = 40cm = 0.4m

Unknown:

Gravitational force of attraction between them  = ?

Solution:

To solve this problem, we use the expression below which is derived from the Newton's law of universal gravitation:

           Fg  = [tex]\frac{G x mass 1 x mass 2}{d^{2} }[/tex]

G is the universal gravitation constant = 6.67 x 10⁻¹¹

d is the separation

 Now;

  Fg = [tex]\frac{6.67 x 10^{-11} x 100 x 200}{0.4^{2} }[/tex]   = 8.34 x 10⁻⁶N

Answer:

Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).

Explanation:

Answer:

[tex] \underline{ \boxed{ yes}}\\[/tex]

Explanation:

[tex]given : initial \: velocity \: (u )= 40 {ms}^{ - 1} \\ given : final \: velocity \: (u )= 0 {ms}^{ - 1} \\ given : - (acceleration) \: (a_r) = 2 {ms}^{ - 2} \\ given : distance \: (s) \: = \: ? : \\ but \: {v}^{2} = {u}^{2} + 2( a)s\\ {0}^{2} = {40}^{2} + 2( - 2)s \\ - {40}^{2} = - 4s \\ s = \frac{ - {40}^{2} }{ - 4} \\ s = \frac{1600}{4} \\s = 400 \: m[/tex]

Answer:

0.25 kg

Explanation:

p = mv

2 = m(8)

2/8 = m(8)/8 *cancels

m = 1/4 OR 0.25 kg

Answer:

2 m/s²

Explanation:

From the given information:

The first mass m_1 = 0.6 kg

The second mass m_2 = 0.3 kg

The magnitude for the acceleration of 0.3 kg is:

a = net force/ effective mass

Mathematically, it can be computed as follows:

[tex]a = \dfrac{F}{m}[/tex]

[tex]a = \dfrac{(m_2 +m_1 -m_1) }{(m_2+m_1+m_1)}(g)[/tex]

[tex]a = \dfrac{0.3 +0.6 -0.6}{(0.3 +0.6+0.6)}(9.8)[/tex]

a ≅ 2 m/s²

Answer:

f" = 40779.61 Hz

Explanation:

From the question, we see that the bat is the source of the sound wave and is initially at rest and the object is in motion as the observer, thus;

from the Doppler effect equation, we can calculate the initial observed frequency as:

f' = f(1 - (v_o/v))

We are given;

f = 46.2 kHz = 46200 Hz

v_o = 21.8 m/s

v is speed of sound = 343 m/s

Thus;

f' = 46200(1 - (21/343))

f' = 43371.4285 Hz

In the second stage, we see that the bat is now a stationary observer while the object is now the moving source;

Thus, from doppler effect again but this time with the source going away from the obsever, the new observed frequency is;

f" = f'/(1 + (v_o/v))

f" = 43371.4285/(1 + (21.8/343))

f" = 40779.61 Hz

Answer:False

Explanation:The Earth rotates faster at the equator than at the poles.

Answer:

α = 36.21 °

β = 143.79°

Explanation:

To do this, we need to know the expression to calculate the angle.

In this case:

α₁ = tan⁻¹ (Fy₁/Fx₁)   (1)

Now, let's analize the given data.

We have a charge q₁ at the origin of the cartesian coordinate system, so, it's at the 0. The charge q₂ is 20 cm above q₁, meaning is on the y-axis. Finally q₃ it's 20 cm to the right, meaning it's on the x-axis.

Knowing this,we can calculate the force that q₂ and q₃ are exerting over q₁. As these forces are in the x and y-axis respectively, we also are calculating the value of the forces in the x and y axis, that are needed to calculate the direction.

The expression to calculate the force would be Coulomb's law so:

F = K q₁q₂ / r²    (2)

The value of K is 9x10⁹ N m² / C². Let's calculate the forces:

F₁₂ = Fy = 9x10⁹ * (7.1x10⁻⁴) * (6.5x10⁻⁴) / (0.020)²

Fy = 1.04x10⁷ N

F₁₃ = Fx = 9x10⁹ * (7.1x10⁻⁴) * (8.9x10⁻⁴) / (0.020)²

Fx = 1.42x10⁷ N

Now that we have both forces, we can calculate the magnitude of the force:

F = √(Fx)² + (Fy)²

F = √(1.04x10⁷)² + (1.42x10⁷)²

F = 1.76x10⁷ N

Finally, the direction would be applying (1):

α = tan⁻¹ (1.04x10⁷/1.42x10⁷)

α = 36.21 °

And counter clockwise it would be:

β = 180 - 36.21 = 143.79°

Hope this helps

Answer:

C) The natural frequency of the building should be very different from typical earthquake frequencies

Explanation:

We shall apply the concept of resonance in this problem .

When a body is applied an external harmonic force ( forced vibration) such that natural frequency of body is equal to frequency of external force or periodicity of external force , the body vibrates under resonance ie its amplitude of vibration becomes very high .

In the present case if natural frequency of building becomes equal to the earthquake's frequency ( external force ) , the building will start vibrating with maximum amplitude , resulting into quick collapse of the whole building . So to avoid this situation , natural frequency of building should be very different from typical earthquake frequencies .

Answer:

C. A heat engine must deposit some energy in a cold reservoir.

Explanation:

The second law of thermodynamics says that "It is impossible to extract an amount of heat Q from a hot reservoir and use it all to do work W. Some amount of heat q must be exhausted to a cold reservoir."

This means that if we extract an amount of heat Q from the hot reservoir, the work W can never be exactly equal to Q, then there is a surplus of heat q that must be deposited in a cold reservoir.

Then we have the equation:

Q = W + q

From this we can conclude that the correct option is:

C. A heat engine must deposit some energy in a cold reservoir.

There will be always some energy that is not transformed into work, and is deposited in a cold reservoir.

C. A heat engine must deposit some energy in a cold reservoir.

Therefore, option C is correct.

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Answer: [tex]5.944\times 10^{23}\ kg[/tex]

Explanation:

Given

Time period [tex]T=7.08\times 10^3\ s[/tex]

Orbital speed [tex]v=3.40\times 10^3\ m/s[/tex]

mass of GS [tex]m_{GS}=930\ kg[/tex]

Radius of Mars [tex]r=3.43\times 10^6\ m[/tex]

Consider the mass of mars is M

Here, Gravitational pull will provide the centripetal force

[tex]F_G=F_c[/tex]

[tex]\dfrac{GMm_{GS}}{r^2}=\dfrac{m_{GS}v^2}{r}\\M=\dfrac{v^2\cdot r}{G}\\M=\dfrac{(3.43\times 10^3)^2\cdot 3.43\times 10^6}{6.67\times 10^{-11}}[/tex]

[tex]M=5.944\times 10^{23}\ kg[/tex]

In March 1999 the Mars Global Surveyor (GS) entered its final orbit on Mars, sending data back to Earth. The mass of Mars is approximately 6.419 × 10²³ kg.

Kepler's Third Law states that the square of the orbital period (T) is proportional to the cube of the semi-major axis (a) of the orbit:

T² = (4π² / GM) × a³

In a circular orbit, the semi-major axis is equal to the radius of the orbit (r).

Given:

Orbital period (T) = 7.08 × 10³ s

Orbital speed (v) = 3.40 × 10³ m/s

Mass of GS (m) = 930 kg

Radius of Mars (r) = 3.43 × 10⁶ m

The orbital speed (v) is related to the radius (r) and the gravitational constant (G) by:

v = √(GM / r)

v² = GM / r

G = (v² × r) / M

T² = (4π² / [(v² × r) / M]) × r³

T² = (4π² × M × r²) / v²

M = (T² × v²) / (4π² × r²)

M = ( (7.08 × 10³)² × (3.40 × 10³)² ) / (4π² × (3.43 × 10⁶)²)

M = 6.419 × 10²³ kg

Therefore, the mass of Mars is approximately 6.419 × 10²³ kg.

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