Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour in 2.22 s starting from rest. Assuming that they have constant acceleration throughout that time, find their acceleration in meters per second squared.

Answer:

It takes 325 seconds for the signal to reach Earth.

Explanation:

First, you must make a unit change from m/s to km/s in order to make a comparison with the distance of the radio signal sent to Earth. For that, you know that 1 m is 0.001 km. So:

[tex]3*10^{8} \frac{m}{s} =3*10^{8}\frac{0.001 km}{s}=300,000\frac{km}{s}[/tex]

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three. Remember that proportionality is a constant relationship or ratio between different magnitudes.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

[tex]x=\frac{c*b}{a}[/tex]

In this case, the rule of three is applied as follows: if by definition of speed, 300,000 km of light are traveled in 1 second, 9.75 * 10⁷ km in how long are they traveled?

[tex]time=\frac{9.75*10^{7}km*1second }{300,000 km}[/tex]

time=325 seconds

It takes 325 seconds for the signal to reach Earth.

The number of seconds does it take for the signal to reach the Earth is  325 seconds.

[tex]= 9.75 \times 10^7 \div 300,000 km[/tex]

= 325 seconds

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Answer:

The wheels turn 101.27 radians in that time.

Explanation:

First, we need to find the angular velocity of the tires. We use the following formula for this purpose:

Δv = rω

ω = Δv/r

where,

ω = Angular Velocity of Tires = ?

Δv = change in linear velocity = 5.33 m/s - 0 m/s = 5.33 m/s

r = radius of tire = 0.33 m

Therefore,

ω = (5.33 m/s)/(0.33 m)

ω = 16.15 rad/s

Now, the angular displacement covered by tires can be found out by using the general formula of angular velocity:

ω = θ/t

θ = ωt

where,

θ = angular displacement = ?

t = time = 6.27 s

Therefore,

θ = (16.15 rad/s)(6.27 s)

θ = 101.27 radians

Answer:

Answer in explanation

Explanation:

The path of flow of circuit is called electric circuit. In an electric circuit, resistances are connected in two different ways, which are:

1- Series Combination of Resistances

2- Parallel Combination of Resistances  

Series combination of resistances

In series combination resistors are connected end to end, and there is only one path for the flow of current.  

Characteristics Of Series Circuit:

1. In series circuit there is only one path for the flow of current.

2. Same amount of current flows through each resistor.

I = I₁ = I₂ = I₃ = In

3. Total voltage of the battery is equal to the sum of voltages across each resistor.

V = V₁ +V₂ + V₃ + ... + Vn

4. In series combination the combined resistance of the resistors can be obtained by adding the value of each resistor.

R  =  R₁ + R₂ + R₃ + … + Rn

Parallel combination of resistances

Resistances are set to be connected in parallel, when each of them is connected directly from the terminals of electric source.  

Characteristics Of Series Circuit:

1.  There are more than one path for the flow of current.

2. The value of potential difference remains constant on each resistor.

V = V₁ = V₂ = V₃ = Vn

3. Total current is equal to the sum of current passing through each resistor.

I = I₁ + I₂ + I₃ +...+ In

4. Reciprocal of equivalent resistance is equal to the sum of reciprocals of resistances connected in parallel.  

1/R   =  1/R₁  + 1/R₂  + 1/R₃  + … + 1/Rn  

The circuit diagrams are in attachment.

Answer:

A.) The temperature of the junction between copper and sliver when they have come to equilibrium is 35 degree Celsius

B.) ice (in grams) melts per second = 0.078 kg/s

Explanation:

A.) Given that the two material are of the same area.

The Silver rod is 15.0 cm in length and the copper rod is 25.0 cm in length.

Silver temperature = 0 degree Celsius

Copper temperature = 100 degree Celsius

Thermal conductivity k of silver = 429 W/m•K

Thermal conductivity k of copper = 385W/m.k

Rate of energy transferred P in the two materials can be expressed as

P = k.A.dT/L

dT = change in temperature

Since the rate and the area are the same

429 ( T -0 )/0.15 = 385( 100 - T )/0.25

2860T = 1540(100 - T)

Open the bracket

2860T = 154000 - 1540T

Collect the like terms

2860T + 1540T = 154000

4400T = 154000

T = 154000/4400

T = 35 degree Celsius

The temperature of the junction between copper and sliver when they have come to equilibrium is 35 degree Celsius

B.) Rate of energy transferred P will be

P = 2860 × 35 = 100100

P = Q/t ..... (1)

Where Q = energy transferred

But Q = mcØ .....(2)

And specific heat capacity c of water = 4182J/k.kg

Substitutes Q into formula 1.

P = mcØ/t

Make m/t the subject of formula

m/t = P/cØ

m/t = 100100/ 4182( 35 + 273 )

m/t = 100100/1288056

m/t = 0.078 kg/s

The linear momentum of the bullet, given the data from the question is 1.225 Kg.m/s

Momentum is defined as the product of mass and velocity. It is expressed as

Momentum = mass × velocity

With the above formula, we can obtain the momentum of the bullet. Details below.

The following data were obtained from the question:

Momentum = mass × velocity

Momentum = 0.035 Kg × 35 m/s

Momentum = 1.225 Kg.m/s

From the calculation made above, we can conclude that the linear momentum of the bullet is 1.225 Kg.m/s

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Answer:

A)Kopya

B)YASAK

Explanation:

kopya yasak dostum adın da belli. Başın belaya girmesin

Answer:

Availability of land for Agricultural activities- The rural areas are known for a lesser degree of development which means lesser factories and other work buildings. The area has undeveloped lands which are usually used for a commercial type of agricultural activities.

Bad road networks: Bad road networks are mainly associated with rural settlements. This hinders to an extent the agricultural activities of planting and harvesting of crops due to difficulties in moving of the crops.

Answer:

h = 0.288m

Explanation:

Assume

[tex]v_1[/tex] = Speed of Sue

[tex]v_2[/tex] = Speed of Chris immediately after the push

Sue's KE = [tex]\frac{1}{2} mv_1\ ^2 = 26 v_1 \ ^2[/tex]

now she swings this is converted into gravitation at PE of

mg n = 52 × 9.8 × 0.65  

= 331.24

[tex]26v_{1}\ ^2[/tex] = 331.24

So, [tex]v_1 = 3.569[/tex]

They started at rest by conservation of momentum in case of push off the magnitude of sue momentum and it is equal to the magnitude of Chris momentum in the opposite or inverse direction

[tex]m_1v_1 = m_2v_2[/tex]

[tex]52 \times 3.569 = 78 \times v_2[/tex]

[tex]v_2 = 2.380[/tex]

Chris kt = [tex]\frac{1}{2} \times 78 \times 2.380^2[/tex]

= 220.827

220.827 = mgh

So, h = 0.288m

Answer:

349 J

Explanation:

Length L of baton = 1.0 m

Mass m of baton = 1.2 kg

Weight W of baton = 1.2 kg x 9.81 m/[tex]s^{2}[/tex] = 11.772 N

Height h reached = 5.0 m

Angular speed ω = 3.5 rev/s = 2π x 3.5 (rad/s) = 21.99 rad/s

Total work done on baton will be the work done in taking it to a height of 5.0 m and the kinetic energy with which the baton rolls.

Work done to bringing it to the height of 5.0 m = weight x height above ground

W x h = 11.772 x 5 = 58.86 J

Velocity v of spinning baton = ω x L = 21.99 x 1 = 21.99 m/s

Kinetic energy = [tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex] =

Total work done on baton = 58.86 + 290.14 = 349 J

Answer: the magnetic wave will travel out of the screen.

Explanation:

Electric field direction is perpendicular to the magnetic field direction. Both are also perpendicular to the direction of the particles.

Using right hand rule to solve this problem,

This pointed finger depicts the electric field direction which the curly fingers depict the direction of the magnetic field. The pointed thumb will depict the direction in which the wave travel. Which is out of the screen.

Answer:

A) U = - GMm/r

B) K = 0.5 mGM/r

Explanation:

A) The potential energy U of the satellite

U = - GMm/r

G = universal gravitational constant which is ( 6.67e-11 Nm^2/c^2 )

M = mass of the planet

m = mass

r = distance ( radius )

B) Kinetic energy

kinetic energy expressed as K = 0.5 m Vo^2

NOTE : Vo^2 = GM / r

hence kinetic energy will be expressed as

K = 0.5 mGM/r

Answer:

m = 4.87 kg

Explanation:

In order to find the required mass you first calculate the spring constant of the spring. When the system reaches the equilibrium you obtain the following equation:

[tex]Mg=kx[/tex]      (1)

That is, the weight of the object is equal to the restoring force of the spring.

M: mass of the object = 3.36 kg

g: gravitational constant = 9.8m/s^2

k: spring constant = ?

x: elongation of the spring = 0.0190m

You solve the equation (1) for k:

[tex]k=\frac{Mg}{x}=\frac{(3.36kg)(9.8m/s^2)}{0.0190m}=1733.05\frac{N}{m}[/tex]

Next, to obtain a frequency of 3.0Hz you can use the following formula, in order to calculate the required mass:

[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]     (2)

You solve the equation (2) for m:

[tex]m=\frac{1}{4\pi^2}\frac{k}{f^2}\\\\m=\frac{1}{4\pi^2}\frac{1733.05N/m}{(3.0Hz)^2}=4.87kg[/tex]

The required mass to obtain a frequency of 3.0Hz is 4.87 kg

Answer:

157.36 N

Explanation:

Contact force is the force which is created due to contact and it is applied on the contact point . The force applied by body on the surface is its weight .

If R be the reaction force of the ground

R = mg + F son30

= 12 x 9.8 + 75 sin 30

= 117.6 + 37.5

= 155.10 N .

friction force = f

Net force in forward direction = F cos 30 - f  = ma

75cos 30 - f = 12 x 3.2

f = 65 - 38.4

= 26.6 N  

Total force on the surface =√( f² + R² )

√ (26.6² + 155.1²)

= √707.56 + 24056²

=√ 24763.57

= 157.36 N.

contact force = 157.36 N .

Answer:

1. 39068.07 N

2. 19534.036 N

Explanation:

depth of water h = 10 m

atmospheric pressure Patm = 101325 Pa

density of water p = 1000 kg/m^3

acceleration due to gravity g = 9.81 m/s^2

pressure due to depth of water = pgh

P = 1000 x 9.81 x 10 = 98100 Pa

total pressure on the bottom of the tank is Patm + p = 101325 + 98100 = 199425 Pa

Left plug has diameter = 50 cm = 0.5 m

radius = 0.5/2 = 0.25 m

height = 1 cm = 0.01 m

height below tank surface = 10 - 0.01 = 9.99

pressure at this depth =  1000 x 9.81 x 9.99 = 98001.9 Pa

total pressure = Patm + P = 101325 + 98001.9 = 199326.9 Pa

surface area of plug = π[tex]r^{2}[/tex] = 3.142 x [tex]0.25^{2}[/tex] = 0.196 [tex]m^{2}[/tex]

force required to lift left plug = pressure x area

F =  199326.9 x 0.196 = 39068.07 N

The right side is a hemisphere with the same diameter, therefore surface area is half of the left plug

A = 0.196/2 = 0.098 [tex]m^{2}[/tex]

force F required to lift right plug =  199326.9  x 0.098 = 19534.036 N

Answer:

t = 47 years

Explanation:

To find the number of years in which the electrons cross the complete transmission, you first calculate the drift velocity of the electrons in the transmission line, by using the following formula:

[tex]v_d=\frac{I}{nAq}[/tex]         (1)

I: current = 1,010A

A: cross sectional area of the transmission line = π(d/2)^2

d: diameter of the transmission line = 2.00cm = 0.02 m

n: free charge density = 8.50*10^28 electrons/m^3

q: electron's charge = 1.6*10^-19 C

You replace the values of all parameters in the equation (1):

[tex]v_d=\frac{1010A}{(8.50*10^{28}m^{-3})(\pi(0.02m/2)^2)(1.6*10^{-19}C)}\\\\v_d=2.36*10^{-4}\frac{m}{s}[/tex]

with this value of the drift velocity you can calculate the time that electrons take in crossing the complete transmission line:

[tex]t=\frac{d}{v_d}=\frac{350km}{2.36*10^{-4}m/s}=\frac{350000m}{2.36*10^{-4}m/s}\\\\t=1,483,050,847\ s[/tex]

Finally, you convert this value of the time to years:

[tex]t=1,483,050,847s*\frac{1\ year}{3.154*10^7s}=47\ years[/tex]

hence, the electrons take around 47 years to cross the complete transmission line.

The ball's position in the air at time t is given by the vector,

p(t) = (60 i + 64 k) t + 1/2 (-6 j - 32 k) t ²

and its velocity is given by

v(t) = (60 i + 64 k) + (-6 j - 32 k) t

The ball is in the air for as long as it takes for the vertical (k) component of the position vector to reach 0, so we solve,

64 t - 32/2 t ² = 0  ==>  t = 0 OR t = 4

and so the ball is in the air for 4 s.

After this time, the ball has position vector

p(4) = (60 i + 64 k) t + 1/2 (-6 j - 32 k) t ² = 240 i - 48 j

which has magnitude

||p(4)|| = √(240² + (-48)²) = 48 √26 ≈ 244.8 ft

in a direction θ in the x,y plane from the positive x axis such that

tanθ = -48/240 = -1/5  ==>  θ = -arctan(1/5) ≈ -11.3º

or 11.3º South of East.

The ball hits the ground with speed

||v(4)|| = ||60 i - 24 j - 64 k|| = √(60² + (-24)² + (-64)²) = 4 √517 ≈ 91.0 ft/s

kinematic relationships can be used in all dimensions and finding the position and velocity when reaching the ground is

        r = 244.8 ft, tea = 21.8º from East to South

        v = 91.0 ft / s

given parameters

to find

Kinematics allows finding the position, velocity and acceleration of the body, in this case we have a problem in three dimensions, where they establish a Cartesian coordinate system, a method to solve this exercise is to solve each component independently

a)  The acceleration of gravity acts on the z axis, so let's find the time it takes to reach the ground, if the initial vertical velocity is v_{oz} = 64 ft/s and the acceleration is a_z = g = -32 ft / s², we assume that the ball leaves the ground (z₀ = 0)

         z = z₀ + v_{oz} t + ½ a_z t²

when reaching the ground its height of zero and

        0 = 0 + v_{oz} t + ½ a_z t²

        t (v_{oz) + ½ a_z t) = 0

        t (64 - 16 t) = 0

the solusion of this squadron is

       y = 0

       t = 4 s

the first time is when it leaves and the second time is for when it reaches the ground, therefore the flight time is t = 4s

with this time we find the displacement is each exercise

X axis

    in this axis there is no acceleration, so we use the uniform motion relationships

        vₓ = x / t

        x = vₓ t

        x = 60 4

        x = 240 ft

Y Axis

on this axis there is an acceleration of a_y = -6 ft/s and an initial velocity v_{oy} = 0

we use the kinematic relation

       y = v_{oy} t + ½ a_y t²

       y = 0 - ½ 6 4²

       y = - 48 ft

let's use the Pythagoras theorem to find the position

       r² = (x -x₀) ² + (y -y₀) ² + (z -z₀) ²

       r² = (240-0) ² + (-48-0) ² + (0-0) ²

       r = 244.8 ft

We use trigonometry to find the direction

      tan θ = y / x

       θ = tan⁻¹ [tex]\frac{y}{x}[/tex]

       θ = tan⁻¹  [tex]\frac{96}{240}[/tex]

       θ = -21.8º

This angle is measured clockwise from the x axis, it can also be read

        θ = 21.8º from East to South

b)  Let's look for the speed when we hit the ground

X axis

       vₓ = v_{ox} + aₓ t

       vₓ = 60 - 0

       vₓ = 60 ft / s

Y Axis

       v_y = v_{oy} + a_y t

       v_y = 0 - 6 4

        v_y = -24 ft / s²

Z axis

       v_z = v_{oz} + a_z t

       v_z = 64 -32 4

       v_z = -64 ft / s

With the Pytagoras  theorem find the modulus of this speed is

        v² = vx² + vy² + vz²

        v² = 60² + 24 ² + 64²

        v = 91.0 ft / s

In conclusion, the kinematic relationships can be used in all dimensions and finding the position and velocity when reaching the ground is

       a) r = 244.8 ft, θ = 21.8º from East to South

       b) v = 91.0 ft / s

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Answer: No.

Explanation: Light exists in 3+1 dimensional space (3 space, 1 time).

Answer:

Explanation:

The magnetic force acting horizontally will deflect the wire by angle φ from the vertical

Let T be the tension

T cosφ = mg

Tsinφ = Magnetic force

Tsinφ = BiL  , where B is magnetic field , i is current and L is length of wire

Dividing

Tanφ = BiL / mg

= .055 x 29 x .11 / .010 x 9.8

= 1.79

φ = 61° .

Tension T = mg / cosφ

= .01 x 9.8 / cos61

= .2 N .

Answer:

a

an apple falling from a tree

Answer an apple falling from a tree

Explanation:

Answer:

emf = 0.02525 V

induced current with a counterclockwise direction

Explanation:

The emf is given by the following formula:

[tex]emf=-\frac{\Delta \Phi_B}{\Delta t}=-B\frac{\Delta A}{\Delta t}[/tex][tex]\ \ =-B\frac{A_2-A_1}{t_2-t_1}[/tex]   (1)

ФB: magnetic flux =  BA

B: magnitude of the magnetic field = 1.00T

A2: final area of the loop; A1: initial area

t2: final time, t1: initial time

You first calculate the final A2, by taking into account that the circumference of loop decreases at 11.0cm/s.

In t = 4 s the final circumference will be:

[tex]c_2=c_1-(11.0cm/s)t=164cm-(11.0cm/s)(4s)=120cm[/tex]

To find the areas A1 and A2 you calculate the radius:

[tex]r_1=\frac{164cm}{2\pi}=26.101cm\\\\r_2=\frac{120cm}{2\pi}=19.098cm[/tex]

r1 = 0.261 m

r2 = 0.190 m

Then, the areas A1 and A2 are:

[tex]A_1=\pi r_1^2=\pi (0.261m)^2=0.214m^2\\\\A_2=\pi r_2^2=\pi (0.190m)^2=0.113m^2[/tex]

Finally, the emf induced, by using the equation (1), is:

[tex]emf=-(1.00T)\frac{(0.113m^2)-(0.214m^2)}{4s-0s}=0.0252V=25.25mV[/tex]

The induced current has counterclockwise direction, because the induced magneitc field generated by the induced current must be opposite to the constant magnetic field B.

Answer:

The velocity is  [tex]v = 5838 \ m/s[/tex]

Explanation:

From the question we are told that

   The electric field is [tex]E = 2.09 kV/m = 2.09 *10^{3} \ V/m[/tex]

    The magnetic field is  [tex]B = 0.358 \ T[/tex]

Generally the force experienced by the electron due to the magnetic field is

         [tex]F_m = qvB[/tex]

Generally the force experienced by the electron due to the electric  field is

       [tex]F_e = qE[/tex]

Since from the question the net force is zero  then

     [tex]F_e = F_m[/tex]

=>    [tex]v = \frac{E}{B}[/tex]

Substituting values

      [tex]v = \frac{2.09*10^{3}}{0.358 }[/tex]

    [tex]v = 5838 \ m/s[/tex]

Answer:

R_A = 2.67 ohms

R_B = 1.14 ohms

Explanation:

When the resistors are connected in parallel, the voltage will be the same across both resistors A and B.

Thus, we now have the current and the voltage across B and so we can use Ohm's Law to find the resistance.

V/I = R

Thus, resistance of B; R_B = 8/3

R_B = 2.67 ohms

Now, when the resistors are connected in series, the voltage drop across B is;

V = 8V - 2.4V = 5.6V

Since we now have the resistance of B , we can find the current using Ohm's Law. Thus;

I = V/R

I = 5.6/2.67

I = 2.1 A

Now, current is the same for all resistances in a series circuit because this is the same current through resistors A and B. So, we can use Ohm's law again to find the resistance across A.

So, R = V/I

R_A = 2.4/2.1

R_A = 1.14 ohms

Answer:

gₓ = 1.36 x 10¹³ g

Explanation:

The value of acceleration due to gravity at a certain place is given by the following formula:

gₓ = GM/R²

where,

gₓ = acceleration due to gravity on the surface of neutron star

G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of the star = 10 * Mass of sun = (10)(2 x 10³⁰ kg) = 2 x 10³¹ kg

R = 10 km = 10⁴ m

Therefore,

gₓ = (6.67 x 10⁻¹¹ N.m²/kg²)(2 x 10³¹)/(10⁴)²

gₓ = 1.334 x 10¹⁴ m/s²

Hence, comparing it with the free-fall acceleration at Earth's Surface:

gₓ/g = (1.334 x 10¹⁴)/9.8

gₓ = 1.36 x 10¹³ g

The acceleration due to gravity on the surface of this star in terms of the free-fall acceleration at Earth's surface is [tex]1.35 \times 10^5 \ g_E[/tex].

The given parameters:

The acceleration due to gravity on the surface of this star in terms of the free-fall acceleration at Earth's surface is calculated as follows;

[tex]F = mg = \frac{GM_sm}{R^2} \\\\(1.5 M_s)g = \frac{GM_s(1.5 M_s)}{R^2} \\\\g = \frac{GM_s}{R^2} \\\\[/tex]

where;

[tex]M_s[/tex] is the mass of the Sun = 1.989 x 10³⁰ kg.

[tex]g = \frac{6.67 \times 10^{-11} \times 1.989 \times 10^{30} }{(10,000,000)^2} \\\\g = 1.326 \times 10^{6} \ m/s^2[/tex]

In terms of gravity of Earth [tex](g_E)[/tex];

[tex]= \frac{1.326 \times 10^6}{9.81} = 1.35 \times 10^5 \\\\= 1.35 \times 10^5 \ g_E[/tex]

Thus, the acceleration due to gravity on the surface of this star in terms of the free-fall acceleration at Earth's surface is [tex]1.35 \times 10^5 \ g_E[/tex].

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Answer:

+1.33 × [tex]10^{-7}[/tex] C and -1.33 × [tex]10^{-7}[/tex] C respectively.

Explanation:

Electric potential (V) is the work done in moving a unit positive charge from infinity to a reference point within an electric field. It is measured in volts.

     V = [tex]\frac{kq}{r}[/tex] ............. 1

where: k is a constant = 9 × [tex]10^{9}[/tex] N[tex]m^{2} C^{-2}[/tex], q is the charge and r is the distance between the charges.

From equation 1,

   q = [tex]\frac{Vr}{k}[/tex] ............... 2

The charge on each ball can be determined as;

given that; V = 1.2 × [tex]10^{3}[/tex], k = 9 × [tex]10^{9}[/tex] N[tex]m^{2} C^{-2}[/tex] and r = 1.00 m.

From equation 2,

  q = [tex]\frac{1.2*10^{3} * 1.0}{9*10^{9} }[/tex]

     = 1.33 × [tex]10^{-7}[/tex] C

Thus, the charge on the first ball is +1.33 × [tex]10^{-7}[/tex] C, while the charge on the second ball is -1.33 × [tex]10^{-7}[/tex] C.

Answer:

C. 10⁻⁷ cm

Explanation:

One nanometer = 10⁻⁹ meter

1 meter = 10² cm

one nanometer = 10⁻⁹ x 10² cm

= 10⁻⁹⁺² cm

= 10⁻⁷ cm .

Answer:

The answer is C

Explanation:

I got it right on my quiz

Answer:

10 s

Explanation:

We are given that

[tex]g=10.0m/s^2[/tex]

Initially

[tex]v_x=86.6m/s,y=50.0m/s[/tex]

We have to find the time after firing taken  by projectile before it hits the level ground.

v=[tex]\sqrt{v^2_x+v^2_y}[/tex]

[tex]v=\sqrt{(86.6)^2+(50)^2}=99.99 m/s[/tex]

[tex]\theta=tan^{-1}(\frac{v_x}{v_y})[/tex]

[tex]\theta=tan^{-1}(\frac{50}{86.6})=30^{\circ}[/tex]

Now,

[tex]t=\frac{vsin\theta}{g}[/tex]

Using the formula

[tex]t=\frac{99.99sin30}{10}[/tex]

[tex]t=4.99\approx 5 s[/tex]

Now, total time,T=2t=[tex]2\times 5=10s[/tex]

Hence, after firing it takes 10 s before the projectile hits the level ground.

Answer:

Explanation:

Increase in gravitational potential energy = m x g x h

where m is mass , g is gravitational acceleration and h is height

In the first case when man climbs

increase in potential = 85 x g x h = 1.85 x 10³ J

gh = 21.7647

when dog climbs

increase in potential = 14.5 x g x h  J

= 14.5 x 21.7647

= 315.6 J

Answer:

The acceleration of the ball is  [tex]a_y = - 0.3672 \ m/s^2[/tex]

Explanation:

From the question we are told that

       The maximum height the ball reachs is [tex]H_{max} = 42.24 \ m[/tex]

       The horizontal component of the initial velocity of the ball is [tex]v_{ix} = 5.57 \ m/s[/tex]

       The vertical component of the initial velocity of the ball is [tex]v_{iy} = = 16.18 m/s[/tex]

The vertically motion of the ball can be mathematically represented as

       [tex]v_{fy}^2 = v_{iy} ^2 + 2 a_{y} H_{max}[/tex]

Here the final velocity at the maximum height is zero so [tex]v_{fy} = 0 \ m/s[/tex]

Making the acceleration [tex]a_y[/tex] the subject we have

        [tex]a_y = \frac{v_{iy} ^2}{2H_{max}}[/tex]

substituting values

      [tex]a_y = - \frac{5.57^2}{2* 42.24}[/tex]

      [tex]a_y = - 0.3672 \ m/s^2[/tex]

The negative sign shows that the direction of the acceleration is in the negative y-axis

Answer:

1. v = 6.67 m/s

2. d = 9.54 m

Explanation:

1. To find the horizontal velocity of the rock we need to use the following equation:

[tex] d = v*t \rightarrow v = \frac{d}{t} [/tex]    

Where:

d: is the distance traveled by the rock

t: is the time

The time can be calculated as follows:

[tex] t = \sqrt{\frac{2d}{g}} [/tex]

Where:

g: is gravity = 9.8 m/s²

[tex] t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*7 m}{9.8 m/s^{2}}} = 1.20 s [/tex]

Now, the horizontal velocity of the rock is:

[tex] v = \frac{d}{t} = \frac{8 m}{1.20 s} = 6.67 m/s [/tex]      

Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.          

2. To calculate the distance at which the projectile will land, first, we need to find the time:

[tex] t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*(7 m + 3 m)}{9.8 m/s^{2}}} = 1.43 s [/tex]

So, the distance is:

[tex] d = v*t = 6.67 m/s*1.43 s = 9.54 m [/tex]    

Therefore, the projectile will land at 9.54 m of the second cliff.

I hope it helps you!