A 1100 kg car is traveling around a flat 82.3 m radius curve. The coefficient of static friction between the car tires and the road is .521. What is the maximum speed in m/s at which the car can take the curve?

Answer:

(A). The rabbit travel the distance 138 m.

(B). The average speed of the hiker for the entire trip is 7.2 km/h.

Explanation:

Given that,

Speed = 12.0 m

Time = 11.5 second

(A). We need to calculate the distance

Using formula of distance

[tex]d=v\times t[/tex]

[tex]d =12.0\times11.5[/tex]

[tex]d=138\ m[/tex]

The rabbit travel the distance 138 m.

(B). Given that,

Distance = 5.10 km

Time = 1 hours

Distance 16.5 km

Time = 2 hours

We need to calculate the average speed of the hiker for the entire trip

Using formula of average speed

[tex]v=\dfrac{D}{T}[/tex]

Where, D = Total distance

T = Total time

Put the value into formula

[tex]v=\dfrac{5.10+16.5}{1+2}[/tex]

[tex]v=7.2\ km/h[/tex]

The average speed of the hiker for the entire trip is 7.2 km/h.

Hence, This is the required solution.

Answer:

X(t) = 9.8 *t - 4.9 * t^2

Explanation:

We set a frame of reference with origin at the hand of the girl the moment she releases the ball. We assume her hand will be in the same position when she catches it again. The positive X axis point upwards.The ball will be subject to a constant gravitational acceleration of -9.81 m/s^2.

We use the equation for position under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a *t^2

X0 = 0 because it is at the origin of the coordinate system.

We know that at t = 2, the position will be zero.

X(2) = 0 = V0 * 2 + 1/2 * -9.81 * 2^2

0 = 2 * V0 - 4.9 * 4

2 * V0 = 19.6

V0 = 9.8 m/s

Then the position of the ball as a function of time is:

X(t) = 9.8 *t - 4.9 * t^2

The largest force constant of the spring can be found by balancing the kinetic energy of the crate at the top of the ramp with the elastic potential energy of the compressed spring at the bottom of the ramp and taking into consideration the static friction.

To calculate the largest force constant of the spring at the bottom of the ramp, we would use Hooke's law which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. This force can be represented by -kx, where k is the force constant and x is the displacement. The displacement in our scenario is the movement of the crate down the ramp. We know that at the base of the ramp, kinetic energy of the crate will be transformed into elastic potential energy as it compresses the spring.

The kinetic energy of the crate at the top of the ramp will be 1/2*m*v^2 (mass * velocity squared) and at the bottom it will be converted into the potential energy of the compressed spring represented by 1/2*k*x^2. Here, given mass of the crate as 1890/9.81 kg, its velocity as 1.8 m/s, and it travels 5m along the ramp.

Setting these two energies equal, we can find the value of spring constant k.

Also, based on the problem, the static friction must be overcome at the bottom of the ramp to avoid the crate rebound. This scenario involves the interplay of several fundamental concepts in physics including kinetic and potential energy, the work-energy principle, and concepts of kinetic and static friction. Different parts of the given information and reference text refer to these aspects of physics.

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Answer:

distance between the charges is 5.12 × 10⁶ m

Explanation:

charges q₁ = -130.0 C                

              q₂ = 180 C                  

force between the charges = 8 N

force between two charge                              

   [tex]F = \dfrac{k q_1q_2}{r^2}[/tex]

value of  K =8.975 × 10⁹ N.m²/C²

    [tex]8 = \dfrac{8.975 \times 10^{9}\times 130 \times 180}{r^2}[/tex]

    [tex]r^2 = \dfrac{8.975 \times 10^{9}\times 130 \times 180}{8}[/tex]

    [tex]r^2 =2.625 \times 10^{13} [/tex]

    r = 5.12 × 10⁶  m                                          

hence, distance between the charges is 5.12 × 10⁶ m.

Answer:

44.1 m

Explanation:

Given:

Assumptions:

We know that the distance traveled by the sound in a particular medium is equal to the product of the speed of sound in that medium and the time taken.

For traveling sound through the rod, we have

[tex]l=V_r t\\\Rightarrow t = \dfrac{l}{V_r}[/tex]..........eqn(1)

For traveling sound through the air to the women ear for traveling the same distance, we have

[tex]l=V_aT\\\Rightarrow l=V_a(t+0.12)\\\Rightarrow l=V_a(\dfrac{l}{V_r}+0.12)\,\,\,\,\,\,(\textrm{From eqn (1)})\\\Rightarrow l=V_a(\dfrac{l}{15V_a}+0.12)\\\Rightarrow l=\dfrac{l}{15}+0.12V_a\\\Rightarrow l-\dfrac{l}{15}=0.12V_a\\\Rightarrow \dfrac{14l}{15}=0.12V_a\\\Rightarrow l = \dfrac{15}{14}\times 0.12V_a\\\Rightarrow l = \dfrac{15}{14}\times 0.12\times 343\\\Rightarrow l = \dfrac{15}{14}\times 0.12\times 343\\\Rightarrow l = 44.1\ m[/tex]

Hence, the length of the rod is 44.1 m.

The length of the rod can be calculated using the difference in hearing times and the speed of sound in the rod and air. Using the formula for distance (speed x time), and given that the speed of sound in the rod is 15 times the speed of sound in air, the length of the rod is found to be approximately 41.16 meters.

In this problem, we know that the speed of sound in the rod is 15 times the speed of sound in the air, and that the woman hears the sound of the strike twice with a 0.12 second gap. The first sound is transmitted through the rod and the second, through the air. Therefore, we can use this information to conclude that the difference in time is the amount of time it takes for the sound to travel the length of the rod in air after it already traveled through the rod.

The speed of sound in the rod is 15 times the speed of sound in air, which is given as 343 m/s. So, the speed of sound in the rod is 15 * 343 = 5145 m/s.

We are looking for the distance travelled, which is the length of the rod. We can find the distance by using the formula distance = speed x time. In this case we are calculating distance as time taken for sound to travel through air minus the time taken to travel through the rod. Therefore, the length of the rod can be calculated to be 343 m/s * 0.12 s = 41.16 meters.

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Answer:

Energy density will be 14.73 [tex]J/m^3[/tex]

Explanation:

We have given capacitance [tex]C=225\mu F=225\times 10^{-6}F[/tex]

Potential difference between the plates = 365 V

Plate separation d = 0.200 mm [tex]0.2\times 10^{-3}m[/tex]

We know that there is relation between electric field and potential

[tex]E=\frac{V}{d}[/tex], here E is electric field, V is potential and d is separation between the plates

So [tex]E=\frac{V}{d}=\frac{365}{0.2\times 10^{-3}}=1825000N/C[/tex]

Energy density is given by [tex]E=\frac{1}{2}\varepsilon _0E^2=\frac{1}{2}\times 8.85\times 10^{-12}\times (1.825\times 10^6)^2=14.73J/m^3[/tex]

Answer:

Decreases.

Explanation:

Electric potential energy is the potential energy which is associated with the configuration of points charge in a system and it is the result of conservative coulomb force.

When the negatively charge ion is at the position of the negative probe than its potential energy is positive when it is move towards the positive probe it's potential energy becomes negative due to the negative ion.

Therefore, potential energy is decreases when negative charge ion moves through the water from negative probe to positive probe.

Answer:

The emission spectrum is always the same and can be used to identify the element and part of the Bohr model proposed that electrons in the hydrogen are located in particular orbits around the nucleos are True.

Explanation:

The Niels Bohr and quantic mecanic theorys are both based on the study of atomics spectrums. The atomic spectrum is a characteristic pattern of a light wavelenght emited wich is unique to each element.

For example, if we put some low pressure hydrogen in a glass tube and in the tp of the glass we apply a voltage big enough to produce a electric current in the hydrogen gas, the tube its going to emit light wich have a color dependig of the gas element in the interior. If we observe this light with a spectrometer we are going to see shining lines and each one of this lines have a wavelenght and diferent colors. This lines are called emission spectrum and the wavelength of that spectrum are unique to eache element.

Summering up, we can identify elements using the emission spectrum because any element produces the same spectrum than other element.

 According to Niels Bhor theory  the electron only can be in especific discret ratios to the nucleus. Where this electron moves himself in circukar orbits under the influence of the Coulomb attraction force.

Answer:

[tex]A. 8.29\times 10^{-18}\ J[/tex]

Explanation:

Given that:

p = magnitude of charge on a proton = [tex]1.6\times 10^{-19}\ C[/tex]

k = Boltzmann constant = [tex]9\times 10^{9}\ Nm^2/C^2[/tex]

r = distance between the two carbon nuclei = 1.00 nm = [tex]1.00\times 10^{-9}\ m[/tex]

Since a carbon nucleus contains 6 protons.

So, charge on a carbon nucleus is [tex]q = 6p=6\times 1.6\times 10^{-19}\ C=9.6\times 10^{-19}\ C[/tex]

We know that the electric potential energy between two charges q and Q separated by a distance r is given by:

[tex]U = \dfrac{kQq}{r}[/tex]

So, the potential energy between the two nuclei of carbon is as below:

[tex]U= \dfrac{kqq}{r}\\\Rightarrow U = \dfrac{kq^2}{r}\\\Rightarrow U = \dfrac{9\times10^9\times (9.6\times 10^{-19})^2}{1.0\times 10^{-9}}\\\Rightarrow U =8.29\times 10^{-18}\ J[/tex]

Hence, the energy stored between two nuclei of carbon is [tex]8.29\times 10^{-18}\ J[/tex].

Answer:

a) 3.56m/s

b) 1.73m

Explanation:

We have to treat this as a parabolic motion problem:

we will use the next formulas:

[tex]V=Vo+a*t\\Vy^2=Vyo^2+2*a*Y\\X=Vox*t[/tex]

we first have to calculate the initial velocity of the basketball player:

[tex]Vy^2=Vyo^2+2*a*Y\\0^2=Vyo^2+2*(-9.8)*(0.650)\\Vyo=\sqrt{2*9.8*0.650} \\Vyo=3.56 m/s[/tex]

the final velocity is zero when he reaches the maximun height.

To answer the second part we need to obtain the time to reach the maximun height, so:

[tex]V=Vo+a*t\\\\0=3.56+(-9.8)*t\\t=0.36 seconds[/tex]

Now having that time, let's find the distance on the X axis, the X axis behaves as constant velocity movement, so:

[tex]X=Vox*t\\X=4.80*0.36\\X=1.73m[/tex]

Answer:

The yearly accumulation for the current year is 130

Explanation:

Accumulation is defined as

input - output = accumulation

The accumulation rate is accumulation per unit of time (in this case a year)

The accumulation rate will then be the amount of births, plus the amount of immigrants minus the amount of deaths.

ar = 500*0.2 + 55 - 0.05*500

ar = 100 + 55 - 25

ar = 130

Answer:

Explanation:

Electric field due to a charge Q at a point d distance away is given by the expression

E = k Q / d , k is a constant equal to 9 x 10⁹

Field due to charge = 3 X 10⁻⁹ C

E = E = [tex]\frac{9\times 10^9\times3\times10^{-9}}{d^2}[/tex]

Field due to charge = 4 X 10⁻⁹ C

[tex]E = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}[/tex]

These two fields will be equal and opposite to make net field zero

[tex]\frac{9\times 10^9\times3\times10^{-9}}{d^2}[/tex] = [tex][tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}[/tex][/tex]

[tex]\frac{3}{d^2} =\frac{4}{(2-d)^2}[/tex]

[tex]\frac{2-d}{d} =\frac{2}{1.732}[/tex]

d = 0.928

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = [tex]71\times 10^{- 9} C[/tex]

Q' = + 42 nC = [tex]42\times 10^{- 9} C[/tex]

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

[tex]\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}[/tex]

[tex]\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}[/tex]

[tex]\vec{E} = 708.03 N/C[/tex]

Now,

Electric field at the mid-point due to charge Q' is given by:

[tex]\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}[/tex]

[tex]\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}[/tex]

[tex]\vec{E'} = 418.84 N/C[/tex]

Now,

The net Electric field is given by:

[tex]\vec{E_{net}} = \vec{E} - \vec{E'}[/tex]

[tex]\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C[/tex]

The basketball's velocity just before hitting the floor is approximately -6.26 m/s (downward), while just after it leaves the floor, it reaches around 5.42 m/s (upward). During the brief 0.02 seconds it's in contact with the floor, its average acceleration is enormous, about 584 m/s², and directed upward.

This question involves the physics concept of velocity and acceleration related to a bouncing basketball. Let's dig into the details one by one:

Using Physics, the velocity of an object just before it hits the ground can be calculated using the equation for motion that involves falling from a height namely v² = u² + 2gh, where u is the initial velocity, g is acceleration due to gravity, v is the final velocity and h is the height. Since the ball was released, the initial velocity (u) is 0, g is approximately 9.8 m/s² (negative because the ball is falling downwards), and h is -2.0 m (negative because it is below the release point). Solving for v, the equation transforms to v = sqrt(u² + 2gh) = sqrt(0 + 2*(-9.8)*(-2)) = sqrt(39.2) which is approximately 6.26 m/s (negative, indicating downward).

Assuming the ball reaches the height of 1.5 m with uniform acceleration, we can use the same equation, but this time treating it as a ground-to-air motion with initial velocity 0, g = 9.8 m/s² (positive because the motion is upward), and h = 1.5 m. Solving for v, we get v = sqrt(u² + 2gh) = sqrt(0 + 2*9.8*1.5) = sqrt(29.4) which is approximately 5.42 m/s (positive, indicating upward).

Acceleration can be calculated using the formula a = (v_final - v_initial) / t, where v_final is the final velocity, v_initial is the initial velocity, and t is the time. The change in velocity here is the difference between the velocity just after the ball leaves the floor and the velocity just before it hits the floor, i.e., (5.42 m/s - -6.26 m/s) = 11.68 m/s. Given that the ball is in contact with the floor for 0.02 seconds, the average acceleration is therefore a = (11.68 m/s) / 0.02 s = 584 m/s². This is considerably higher than g because while in contact with the floor, the ball is being rapidly decelerated and then accelerated in the opposite direction due to the impact force. The direction is upward, same as the final velocity.

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Answer: 1 microgram is equal to 0.001 miligrams

Explanation: The factor micro is equal 10^-3 while the factor mili is equal to 10^-3 so to converte the micro to mile we have to multiply by 0.001.

Answer:[tex]\theta =49.76^{\circ}[/tex] North of east

Explanation:

Given

Research station is 9.6 km away in [tex]42^{\circ}[/tex]North of east

after travelling 3.1 km [tex]25^{\circ}[/tex] north of east

Position vector of safari after 3.1 km is

[tex]r_2=3.1cos25\hat{i}+3.1sin25\hat{j}[/tex]

Position vector if had traveled correctly is

[tex]r_0=9.6cos42\hat{i}+9.6sin42\hat{j}[/tex]

Now applying triangle law  of vector addition we can get the required vector[tex](r_1)[/tex]

[tex]r_1+r_2=r_0[/tex]

[tex]r_1=(9.6cos42-3.1cos25)\hat{i}+(9.6sin42-3.1sin25)\hat{j}[/tex]

[tex]r_1=4.325\hat{i}+5.112\hat{j}[/tex]

Direction is given by

[tex]tan\theta =\frac{y}{x}=\frac{5.112}{4.325}[/tex]

[tex]\theta =49.76^{\circ}[/tex]

Answer:

d=9.462×10^15 meters

Explanation:

Relation between distance, temps and velocity:

d=v*t

t=1year*(365days/1year)*/(24hours/1day)*(3600s/1h)=31536000s

So:

1 light year=d=3*10^8m/s*3.154*10^7s=9.462×10^15 meters

Answer:

The time taken by missile's clock is [tex]4.6\times 10^{6} s[/tex]

Solution:

As per the question:

Speed of the missile, [tex]v_{m = 6.5\times 10^{3}} m/s[/tex]

Now,

If 'T' be the time of the frame at rest then the dilated time as per the question is given as:

T' = T + 1

Now, using the time dilation eqn:

[tex]T' = \frac{T}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]

[tex]\frac{T'}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]

[tex]\frac{T + 1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]

[tex]1 + \frac{1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]

[tex]1 + \frac{1}{T} = (1 + (\frac{v_{m}}{c})^{2})^{- \frac{1}{2}}[/tex]         (1)

Using binomial theorem in the above eqn:

We know that:

[tex](1 + x)^{a} = 1 + ax[/tex]

Thus eqn (1) becomes:

[tex]1 + \frac{1}{T} = 1 - \frac{- 1}{2}.\frac{v_{m}^{2}}{c^{2}}[/tex]

[tex]T = \frac{2c^{2}}{v_{m}^{2}}[/tex]

Now, putting appropriate values in the above eqn:

[tex]T = \frac{2(3\times 10^{8})^{2}}{(6.5\times 10^{3})^{2}}[/tex]

[tex]T = 4.6\times 10^{6} s[/tex]

Answer:

244.64m

Explanation:

First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:

[tex]x = V*t = -8\frac{m}{s} *3s = -24m[/tex]

After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:

1. The velocity will start to decrease untill it reaches 0m/s.

2. Then, the velocity will start to increase at the rate of the acceleration.

The distance that the ball travels in the first phase can be found with the following expression:

[tex]v^2 = v_0^2 + 2a*d[/tex]

Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:

[tex]d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m[/tex]

Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:

[tex]t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s[/tex]

Then, the time of the second phase will be:

[tex]t_2 = 9s - t_1 = 9s - 1.143s = 7.857s[/tex]

Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:

[tex]x = \frac{1}{2}a*t^2 + v_0*t + x_0[/tex]

V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:

[tex]x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m[/tex]

So, the total distance covered by this object in meters will be the sum of all the distances we found:

[tex]x_total = 24m + 4.57m + 216.07m = 244.64m[/tex]

Answer:

Option C

Explanation:

given,

velocity of airplane = 80 m/s

angle with the horizontal = 15°

speed of the ground= ?

when the plane is taking off the horizontal component of the velocity is v cosθ

so,      

        ground speed of the airplane is = [tex]v\times cos\theta[/tex]

                                                              = [tex]80 \times cos 15^0[/tex]

                                                           v  =  77.27 m/s

horizontal velocity of the air plane comes out to be 77.27 m/s ≅ 77 m/s

so, the correct option is Option C

Answer:

(a) 27.1 m/s

(b) 3.9 second

Explanation:

Let the speed is u.

Maximum horizontal range, R = 75 m

The range is maximum when the angle of projection is 45°.

(a) Use the formula for the maximum horizontal range

[tex]R=\frac{u^{2}}{g}[/tex]

[tex]75=\frac{u^{2}}{9.8}[/tex]

u = 27.1 m/s

(b) Let the time of flight is T.

Use the formula for the time of flight

[tex]T=\frac{2uSin\theta}{g}[/tex]

[tex]T=\frac{2\times 27.1 \times Sin45}{9.8}[/tex]

T = 3.9 second

Answer:

A and B

Explanation:

Answer:

The frequency of motion is 5.8 Hz.

Explanation:

frequency of motion of any object is defined as the number of times the object repeats it's motion in 1 second.

mathematically frequency equals [tex]f=\frac{1}{T}[/tex]

where,

'T' is the time it takes for the object to complete one revolution. Since it is given that the ball completes 5.8 revolutions in 1 seconds thus the time it takes for 1 revolution equals [tex]\frac{1}{5.8}s[/tex]

Hence[tex]T=\frac{1}{5.8}s[/tex]

thus the frequency equals

[tex]\frac{1}{\frac{1}{5.8}}s^{-1}\\\\f=5.8s^{-1}=5.8Hz[/tex]

Answer:

5.8 Hz.

Explanation:

Given:

The frequency of the motion of a rotating object is defined the total number of revolutions the object makes in unit time. It is measured in units of Hertz (Hz) or per second.

It is given that the ball is making 5.8 revolutions in one second, therefore, its frequency of motion would be the same, i.e., 5.8 Hz.

The bridge was approximately 11.01 meters high. The swimmers were moving at a speed of around 14.72 m/s when they hit the water. If the bridge were twice as high, the drop time would be about 2.12 seconds.

The subject in question involves principles of physics, specifically gravitational acceleration. When an object is dropped, it falls under the influence of gravity. This is usually around 9.81 m/s2 on Earth. Using the formula for motion d=1/2gt^2, where d is the distance or height, g is the acceleration due to gravity, and t is the time it takes for the object to fall.

a.) For calculation of the bridge's height, plug in the values of time (t=1.5s) and gravity (g=9.81m/s^2) into the equation, we get d=1/2*(9.81)*(1.5)^2 = 11.01 meters. The height of the bridge from which they jumped is approximately 11 meters.

b.) To calculate how fast the swimmers were moving when they hit the water, we can use the equation v=gt, where v is the velocity or speed. Plugging in the values, we get v=(9.81)*(1.5) = 14.72 m/s. So, the swimmers were moving at about 14.7 meters per second when they hit the water.

c.) If the bridge were twice as high, the time of drop would be found using the equation t= sqrt(2d/g), where d is now twice the original distance = 2*11.01 = 22.02 meters. Substituting the given values, we get t= sqrt(2(22.02)/9.81) = 2.12 seconds. The swimmer's drop time would be approximately 2.12 seconds.

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Answer:

0.68 m/s²

Explanation:

Given:

Mass of the dog,  m = 25 kg

Tension in the rope = 10 N

Force applied by the girl on the dog = 7 N

Now,

since the boy is pulling the dog and the girl is pushing the dog

Thus,

The net force on the dog = 10 N + 7 N = 17 N

also,

Net force on the dog = Mass × Acceleration

thus,

25 kg × Acceleration = 17 N

or

Acceleration = [tex]\frac{\textup{17 N}}{\textup{25 kg}}[/tex]

or

Acceleration = 0.68 m/s²

Answer:

The acceleration of the bullet is 212.12 m/s²

Explanation:

Given that,

Force = 1.4 N

Mass = 6.6 g

We need to calculate the acceleration

Using newton's second law

[tex]F=ma[/tex]

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

Where, F = force

m = mass

Put the value into the formula

[tex]a=\dfrac{1.4 }{6.6\times10^{-3}}[/tex]

[tex]a=212.12\ m/s^2[/tex]

Hence, The acceleration of the bullet is 212.12 m/s²

To find the speed at which the shoe hits the ground, we use the equation v = √(u²+2as). To determine the time it takes for the shoe to hit the ground, we use the formula t = √(2h/g). Substituting the given values into these equations, we can find the speed at which the shoe hits the ground and the time it takes to do so.

The question is both a physics problem related to gravity and the free fall of an object. First, we need to find the final velocity of the shoe when it hits the ground. To calculate this, we can use the following equation from physics: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Since the shoe starts from the initial velocity of the parachutist (15.1 m/s), the acceleration a is the acceleration due to gravity (9.81 m/s²), and it falls a distance of 41.2m, we will need to use the equation v = √(u²+2as) to solve for v.

Next, to determine when the shoe hits the ground, we use the equation for time in free fall: t = √(2h/g), where h is the height (41.2 m), and g is acceleration due to gravity (9.81 m/s²). Substituting the given values, we can find the times when the shoe will hit the ground.

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Answer with Explanation:

The person stating that the scientific law cannot be changed is not correct in his claim.

A scientific law can be defined as a statement that is deemed to explain or predict a natural process and is validated by repeated experiments.

In the light of above statement we may conclude that if the law is validated by experiments it should not be subjected to change but this is not the case as an experiment may be done in the future that invalidates the law.

A classical example of this case is the whole of Newtonian law of gravitation. Classically if we analyse this law it is perfect to describe the motion of solar system , planets ,satellites found in nature hence we should find it absolutely correct and since rockets are designed on the basis of this law we can safely assume that it is correct but it is not the case as the gravity is described radically differently by Einstein by his general theory of relativity thus invalidating the newton's law of gravity.

But we still use the Newton's law of gravity as the error's that are involved in the results of the newton's law are not significant to influence any radical departure from the design philosophy of objects such as rocket's or satellites. It always boils down to the accuracy that we need but theoretically we can say that newton's law of gravity was invalidated by Einstein's general theoty of relativity.  

Explanation:

a) Dark matter: a kind of matter we can't directly observe but that we can imply its presence in various observations including gravitational effects which can't be explained by accepting theories of gravity unless more matter is present. This is the reason why dark matter is thought to account for around 80% of the matter in the universe.

b)21cm Radiation: also called hydrogen line, is a spectral line emitted by neutral hydrogen, it has a frequency of 1420megahertz and 21 cm wavelength. In astronomy, this line is used to study the amount and velocity of hydrogen in the galaxy.

Milky way: it's a barred spiral galaxy with more than 200 billion stars, approximately 100000 light-years in diameter and the sun is located about 28000light years from the center.

From the outside its structure has the following characteristics:

Two fundamentals parameters of the milky Way are [tex]R_{0}[/tex] (the radial distance from de sun to the galactic center) and [tex]Θ_{0}[/tex], the galactic rotational velocity at [tex]R_{0}[/tex]

Einstein's Theory of General Relativity: general relativity is a metric theory of gravitation, that defies gravity as a geometric property of space and time, this means there's no gravitational force deflecting objects from their natural straight paths, but a change in properties of space and time that changes this straight path into a curve.

At weak gravitational fields and slow speed, this theory overlaps with Newton's.

Pulsars: they are rotating neutron stars that emit a focused beam of electromagnetic radiation its formation happens when a medium mass star dies and it maintains its angular momentum emitting a powerful blast of radiation along its magnetic field lines. They are useful to search for gravitational waves, and even to find extrasolar planets.

I hope you find this information interesting and useful! Good luck!

Answer:

(a) [tex]v_{1}=0+a_{1}t_{1}   = a_{1}t_{1}[/tex].

(b) [tex]v_{final}=v_{initial}+at\\ v_{2}=v_{1}+a_{2}  t_{2}[/tex]

(c) [tex]219807.5m[/tex]  

Explanation:

To find an expression for the rocket's speed [tex]v_{1}[/tex] at time [tex]t_{1}[/tex], we use the constant acceleration model, which relates these variables with the expression:[tex]v_{final} =v_{intial}+at[/tex]. In this case, the initial velocity is null, because accelerates from rest. So, we take all values of the first interval, and we replace it to find the expression:

[tex]v_{1}=0+a_{1}t_{1}   = a_{1}t_{1}[/tex]. (expression for the first interval).

Then, we do the same process to find the expression of the second interval, we just replace the variables given:

[tex]v_{final}=v_{initial}+at\\ v_{2}=v_{1}+a_{2}  t_{2}[/tex]

In this case, you can notice that the initial velocity used is the one we obtain from the first interval, because the end of the first period is the beginning of the second period.

To calculate the total distance we have to sum the distance covered during the two intervals, that it's translated as: [tex]d_{total} = d_{1} + d_{2}[/tex].

Then, we use this equation to replace in each distance: [tex]v_{final}^{2} = v_{initial}^{2}  +2ad[/tex].

Isolating d we have:

[tex]d=\frac{v_{final}^{2}-v_{initial}^{2}}{2a}[/tex].

Now, we apply the equation to each interval to obtain [tex]d_{1}[/tex] and [tex]d_{2}[/tex]:

[tex]d_{1}=\frac{v_{1}^{2}-0}{2a_{1}}[/tex].

[tex]d_{2}=\frac{v_{2}^{2}-v_{1}^{2}}{2a_{2}}[/tex].

Before calculating the total distance, we need to know the magnitude of each speed.

[tex]v_{1}=a_{1}t_{1}[/tex]

[tex]v_{1}=(67)(39)=[tex]v_{2}=v_{1}+a_{2}  t_{2}= 2613\frac{m}{s^{2}}+34\frac{m}{s^{2}}(49sec)=4279\frac{m}{s}[/tex][/tex]

At last, we use all values known to calculate the total distance:

[tex]d_{total} = d_{1} + d_{2}[/tex]

[tex]d_{total} =\frac{v_{1}^{2}-0}{2a_{1}} + \frac{v_{2}^{2}-v_{1}^{2}}{2a_{2}}[/tex].

[tex]d_{total} =\frac{(2613)^{2} }{2(67)} +\frac{(4279)^{2}-(2613)^{2}  }{2(34)}\\ d_{total}=50953.5+168854=219807.5m[/tex]  

Therefore, the total distance traveled until the ends of the second period is [tex]219807.5m[/tex]. The rocket is in the Thermosphere.