A spring is 20cm long is stretched to 25cm by a load of 50N. What will be its length when stretched by 100N. assuming that the elastic limit is not reached

Answer:

A.) 78.4 J for both

B.) 78.4 J for both

C.) 8.85 m/s for both

D.) 17.7 kgm/s

Explanation:

Given information:

Mass m = 2 kg

Distance d = 20 m

High h = 4 m

A.) Gravitational potential energy can be calculated by using the formula

P.E = mgh

P.E = 2 × 9.8 × 4

P.E = 78.4 J

Since the two objects are identical, the gravitational potential energy of the block for both a and b will be 78.4 J

B.) According to conservative energy,

Maximum P.E = Maximum K.E.

Therefore, the kinetic energy of the two blocks will be 78.4 J

C.) Since K.E = 1/2mv^2 = mgh

V = √(2gh)

Solve for velocity V by substituting g and h into the formula

V = √(2 × 9.8 × 4)

V = √78.4

V = 8.85 m/s

The velocities of both block will be 8.85 m/s

D.) Momentum is the product of mass and velocity. That is,

Momentum = MV

Substitute for m and V into the formula

Momentum = 2 × 8.85 = 17.7 kgm/s

Both block will have the same value since the ramp Is frictionless.

Answer:

10 m/s²

Explanation:

Acceleration: This the rate of change of velocity. The unit of acceleration is m/s²

From the question,

a = (v-u)/t.................... Equation 1

Where a = acceleration of the cheetah, v = final velocity of the cheetah, u = initial velocity of the cheetah, t = time.

Given: u = 0 m/s, v = 25 m/s, t = 2.5 s.

Substitute these values into equation 1

a = (25-0)/2.5

a = 25/2.5

a = 10 m/s²

Hence the acceleration of the cheetah = 10 m/s²

Answer:

t = 17.68s

Explanation:

In order to calculate the total elapsed time that skydiver takes to reache the ground, you first calculate the distance traveled by the skydiver in the first 5.00s. You use the following formula:

[tex]y=y_o-v_ot-\frac{1}{2}gt^2[/tex]            (1)

y: height for a time t

yo: initial height = 1000m

vo: initial velocity = 0m/s

g: gravitational acceleration = 9.8m/s^2

t: time = 5.00 s

You replace the values of the parameters to get the values of the new height of the skydiver:

[tex]y=1000m-\frac{1}{2}(9.8m/s^2)(5.00s)^2\\\\y=877.5m[/tex]

Next, you take this value of 877.5m as the initial height of the second part of the trajectory of the skydiver. Furthermore, use the value of 7.00m/s as the initial velocity.

You use the same equation (1) with the values of the initial velocity and new height. We are interested in the time for which the skydiver arrives to the ground, then y = 0

[tex]0=877.5-7.00t-4.9t^2[/tex]       (2)

The equation (2) is a quadratic equation, you solve it for t with the quadratic formula:

[tex]t_{1,2}=\frac{-(-7.00)\pm \sqrt{(-7.00)^2-4(-4.9)(877.5)}}{2(-4.9)}\\\\t_{1,2}=\frac{7.00\pm 131.33}{-9.8}\\\\t_1=12.68s\\\\t_2=-14.11s[/tex]

You use the positive value of t1 because it has physical meaning.

Finally, you sum the times of both parts of the trajectory:

total time = 5.00s + 12.68s = 17.68s

The total elapsed time taken by the skydiver to arrive to the ground from the airplane is 17.68s

Answer:

(a) T = 1.35 s

(b) vmax = 0.17 m/s

(c) v = 0.056 m/s

Explanation:

(a) In order to calculate the period of oscillation you use the following formula for the period in a simple harmonic motion:

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

m: mass of the cart = 350 g = 0.350kg

k: spring constant = 7.5 N/m

[tex]T=2\pi \sqrt{\frac{0.350kg}{7.5N/m}}=1.35s[/tex]

The period of oscillation of the car is 1.35s

(b) The maximum speed of the car is given by the following formula:

[tex]v_{max}=\omega A[/tex]       (2)

w: angular frequency

A: amplitude of the motion = 3.8 cm = 0.038m

You calculate the angular frequency:

[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{1.35s}=4.65\frac{rad}{s}[/tex]              

Then, you use the result of w in the equation (2):

[tex]v_{max}=(4.65rad/s)(0.038m)=0.17\frac{m}{s}[/tex]

The maximum speed if 0.17m/s

(c) To find the speed when the car is at x=2.0cm you first calculate the time t by using the following formula:

[tex]x=Acos(\omega t)\\\\t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\t=\frac{1}{4.65rad/s}cos^{-1}(\frac{0.02}{0.038})=0.069s[/tex]

The speed is the value of the following function for t = 0.069s

[tex]|v|=|\omega A sin(\omega t)|\\\\|v|=(4.65rad/s)(0.038m)sin(4.65rad/s (0.069s))=0.056\frac{m}{s}[/tex]

The speed of the car is 0.056m/s

Answer:

[tex]\mathbf{a_A = 10.267 \ ft/s^2}[/tex]

[tex]\mathbf{V_B = (a_t)_B =-7.26 \ ft/s^2}[/tex]

Explanation:

Firstly, there is supposed to be a diagram attached in order  to complete this question;

I have attached  the diagram below in order to solve this question.

From the data given;

The radius of the car R = 600 ft

Velocity of the car  B, [tex]V_B = 45 mi / hr[/tex]

We are to determine  the magnitude of the acceleration of A and the rate at which the speed of B is changing.

To start with the magnitude of acceleration A;

We all know that

1 mile = 5280 ft and an hour =  3600 seconds

Thus for ; 1 mile/hr ; we have :

5280 ft/ 3600 seconds

= 22/15 ft/sec

However;

for the velocity of the car B = 45 mi/hr; to ft/sec, we have:

= (45 × 22/15) ft/sec

= 66 ft/sec

A free body diagram is attached in the second diagram showing how we resolve the vector form

Now; to determine the magnitude of the acceleration of A; we have:

[tex]^ \to {a_A} = a_A sin 45^0 ^{\to} + a_A cos 45^0 \ j ^{\to} \\ \\ ^\to {a_B} = -(a_t)_B \ i ^ \to + (a_c )_B cos 45 ^0 \ j ^{\to}[/tex]

Where;

[tex](a_c)_B[/tex] = radial acceleration of B

[tex](a_t)_B[/tex] = tangential acceleration of B

From observation in the diagram; The acceleration of B is 0 from A

So;

[tex]a_B ^\to - a_A ^\to = a_{B/A} ^ \to[/tex]

[tex](-(a_t)_B - a_A sin 45^0 ) ^\to i+ ((a_t)_B-a_A \ cos \ 45^0) ^ \to j = 0[/tex]

[tex](a_c)_B = \dfrac{V_B^2}{R}[/tex]

[tex](a_c)_B = \dfrac{(66)^2}{600}[/tex]

[tex](a_c)_B = 7.26 ft/s^2[/tex]

Equating the coefficient of i and j now; we have :

[tex](a_t)_B = -a_A \ sin 45^0 --- (1)\\ \\ (a_c)_B = a_A cos \ 45^0 --- (2)\\ \\[/tex]

From equation (2)

replace [tex](a_c)_B[/tex] with 7.26 ft/s^2; we have

[tex]7.26 \ ft/s^2 = a_A cos \ 45^0 \\ \\ a_A = \dfrac{7.26 \ ft/s^2}{co s \ 45^0}[/tex]

[tex]\mathbf{a_A = 10.267 \ ft/s^2}[/tex]

Similarly;

From equation (1)

[tex](a_t)_B = -a_A \ sin 45^0[/tex]

replace [tex]a_A[/tex] with 10.267 ft/s^2

[tex](a_t)_B = -10.267 \ ft/s^2 * \ sin 45^0[/tex]

[tex]\mathbf{V_B = (a_t)_B =-7.26 \ ft/s^2}[/tex]

Answer:

a_total = 14.022 m/s²

Explanation:

The total acceleration of a uniform circular motion is given by the following formula:

[tex]a=\sqrt{a_c^2+a_T^2}[/tex]         (1)

ac: centripetal acceleration

aT: tangential acceleration

Then, you first calculate the centripetal acceleration by using the following formula:    

[tex]a_c=r\omega^2[/tex]

r: radius of the circular trajectory = 2.0m

w: final angular velocity  of the ball = 7.0 rad/s

[tex]a_c=(2.0m)(7.0rad/s)^2=14.0\frac{m}{s^2}[/tex]        

Next, you calculate the tangential acceleration. aT is calculate by using:

[tex]a_T=r\alpha[/tex]    (2)

α: angular acceleration

The angular acceleration is:

[tex]\alpha=\frac{\omega_o-\omega}{t}[/tex]

wo: initial angular velocity = 13 rad/s

t: time = 15 s

Then, you use the expression for the angular acceleration in the equation (1) and solve for aT:

[tex]a_T=r(\frac{\omega_o-\omega}{t})=(2.0m)(\frac{7.0rad/s-13.0rad/s}{15s})=-0.8\frac{m}{s^2}[/tex]

Finally, you replace the values of aT and ac in the equation (1), in order to calculate the total acceleration:

[tex]a=\sqrt{(14.0m/s^2)^2+(-0.8m/^2)^2}=14.022\frac{m}{s^2}[/tex]

The total acceleration of the ball is 14.022 m/s²

Answer:

the particle arrangement in liquid are close together with no regular arrangement

Answer:

Explanation:

Let the velocity of projectile be v and angle of throw be θ.

The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m

considering its vertical displacement

h = - ut +1/2 g t²

100 = - vsinθ x 5 + .5 x 9.8 x 5²

5vsinθ =  222.5

vsinθ = 44.5

It covers 160 horizontally in 5 s

vcosθ x 5 = 160

v cosθ = 32

squaring and adding

v²sin²θ +v² cos²θ = 44.4² + 32²

v² = 1971.36 + 1024

v = 54.73 m /s

Answer:

55.42 m/s

Explanation:

Along the horizontal direction, the rock travels at constant speed: this means that its horizontal velocity is constant, and it is given by

u_x = d/t

Where

d = 160 m is the distance covered

t = 5.0 s is the time taken

Substituting, we get

u_x =160/5 = 32 m/s.

Along the vertical direction, the rock is in free-fall - so its motion is a uniform accelerated motion with constant acceleration g = -9.8 m/s^2 (downward). Therefore, the vertical distance covered is given by the

[tex]S=u_yt+\frac{1}{2}at^2[/tex]

where

S = -100 m is the vertical displacement

u_y is the initial vertical velocity

Replacing t = 5.0 s and solving the equation for u_y, we find

-100 = u_y(5) + (-9.81)(5)^2/2

u_y = 45.25 m/s

Therefore, the speed with which the rock was thrown u

[tex]u= \sqrt{u_x^2+u_y^2} \\=\sqrt{32^2+45.25^2}\\ = 55.42 m/s[/tex]

Answer:

17.656 m

Explanation:

Initial speed u = 20 m/s

angle of projection α = 60°

at the top of the trajectory, one fragment has a speed of zero and drops to the ground.

we should note that the top of the trajectory will coincide with halfway the horizontal range of the the projectile travel. This is because the projectile follows an upward arc up till it reaches its maximum height from the ground, before descending down by following a similar arc downwards.

To find the range of the projectile, we use the equation

R = [tex]\frac{u^{2}sin2\alpha }{g}[/tex]

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

Sin 2α = 2 x (sin α) x (cos α)

when α = 60°,

Sin 2α  = 2 x sin 60° x cos 60° = 2 x 0.866 x 0.5

Sin 2α  = 0.866

therefore,

R = [tex]\frac{20^{2}*0.866 }{9.81}[/tex] = 35.31 m

since the other fraction with zero velocity drops a top of trajectory, distance between the two fragments assuming level ground and zero air drag, will be 35.31/2 = 17.656 m

Answer:

Work is the energy required to move an object from one point to another. while power is the energy transferred per unit time.

Energy can also be defined as the ability to do work.

4.Moving with constant speed

Answer:

Explanation:

The speed of the water in the large section of the pipe is not stated

so i will assume 36m/s

(if its not the said speed, input the figure of your speed and you get it right)

Continuity equation is applicable for ideal, incompressible liquids

Q the flux of water that is  Av with A the cross section area and v the velocity,

so,

[tex]A_1V_1=A_2V_2[/tex]

[tex]A_{1}=\frac{\pi}{4}d_{1}^{2} \\\\ A_{2}=\frac{\pi}{4}d_{2}^{2}[/tex]

the diameter decreases 86% so

[tex]d_2 = 0.86d_1[/tex]

[tex]v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}\\\\=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\\\\\approx1.35v_{1} \\\\v_{2}\approx(1.35)(38)\\\\\approx48.6\,\frac{m}{s}[/tex]

Thus, speed in smaller section is 48.6 m/s

Answer : The chemical equation for the beta decay process of [tex]_{38}^{90}\textrm{Sr}[/tex] follows:

[tex]_{38}^{90}\textrm{Sr}\rightarrow _{39}^{90}\textrm{Y}+_{-1}^0\beta[/tex]

Explanation :

Beta decay : It is defined as the process in which beta particle is emitted. In this process, a neutron gets converted to a proton and an electron.

The released beta particle is also known as electron.

The beta decay reaction is:

[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta[/tex]

The chemical equation for the beta decay process of [tex]_{38}^{90}\textrm{Sr}[/tex] follows:

[tex]_{38}^{90}\textrm{Sr}\rightarrow _{39}^{90}\textrm{Y}+_{-1}^0\beta[/tex]

Answer:

the blank is 39

Explanation: a p e x

Answer:

F = 2123.33N

Explanation:

In order to calculate the torque applied by the left support, you take into account that the system is at equilibrium. Then, the resultant of the implied torques are zero.

[tex]\Sigma \tau=0[/tex]

Next, you calculate the resultant of the torques around the right support, by taking into account that the torques are generated by the center of mass of the wooden, the person and the left support. Furthermore, you take into account that torques in a clockwise direction are negative and in counterclockwise are positive.

Then, you obtain the following formula:

[tex]-\tau_l+\tau_p+\tau_{cm}=0[/tex]          (1)

τl: torque produced by the left support

τp: torque produced by the person

τcm: torque produced by the center of mass of the wooden

The torque is given by:

[tex]\tau=Fd[/tex]           (2)

F: force applied

d: distance to the pivot of the torque, in this case, distance to the right support.

You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:

[tex]-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N[/tex]

Where d1, d2 and d3 are distance to the right support.

You solve the equation for F and replace the values of the other parameters:

[tex]F=\frac{W_pd_2+W_d_3}{d_1}=\frac{(1960N)(2.0m)+(2940N)(3.0m)}{6.0m}\\\\F=2123.33N[/tex]

The force applied by the left support is 2123.33 N

Answer:

I’m saying kinetic gravitational and electromagnetic and I will comment on this if I got it right

Explanation:.

Explanation:

(a) The period of a wave is the time required for one complete cycle. In this case, we have the time of five cycles. So:

[tex]T=\frac{t}{n}\\\\T=\frac{50.2s}{5}\\T=10.04s[/tex]

(b) The frequency of a wave is inversely proportional to its period:

[tex]f=\frac{1}{T}\\f=\frac{1}{10.04s}\\f=0.01Hz[/tex]

(c) The wavelength is the distance between two successive crests, so:

[tex]\lambda=30.2m[/tex]

(d) The speed of a wave is defined as:

[tex]v=f\lambda\\v=(0.1Hz)(30.2m)\\v=3.02\frac{m}{s}[/tex]

Answer:

% increase in resistance of tungsten = 9.27%

Explanation:

We are given:

Co-efficient of resistivity for the metal gold; α_g = 0.0034 /°C

Co-efficient of resistivity for tungsten;α_t = 0.0045 /°C

% Resistance change of gold wire due to temperature change = 7%

Now, let R1 and R2 be the resistance before and after the temperature change respectively.

Thus;

(R2 - R1)/R1) x 100 = 7

So,

(R2 - R1) = 0.07R1

R2 = R1 + 0.07R1

R2 = 1.07R1

The equation to get the change in temperature is given as;

R2 = R1(1 + αΔt)

So, for gold,

1.07R1 = R1(1 + 0.0034*Δt)

R1 will cancel out to give;

1.07 = 1 + 0.0034Δt

(1.07 - 1)/0.0034 = Δt

Δt = 20.59°C

For this same temperature for tungsten, let Rt1 and Rt2 be the resistance before and after the temperature change respectively and we have;

Rt2 = Rt1(1 + α_t*Δt)

So, Rt2/Rt1 = 1 + 0.0045*20.59

Rt2/Rt1 = 1.0927

From earlier, we saw that;

(R2 - R1)/R1) x 100 = change in resistance

Similarly,

(Rt2 - Rt1)/Rt1) x 100 = change in resistance

Simplifying it, we have;

[(Rt2/Rt1) - 1] × 100 = %change in resistance

Plugging in the value of 1.0927 for Rt2/Rt1, we have;

(1.0927 - 1) × 100 = %change in resistance

%change in resistance = 9.27%

Answer:

h = 1094.69m

The maximum height above the ground the rocket will achieve is 1094.69m.

Explanation:

The maximum height h is;

h = height covered during acceleration plus height covered when the motor stops.

h = h1 + h2 .......1

height covered during acceleration h1 can be derived using the equation of motion;

h1 = ut + 0.5at^2

Initial speed u = 0

h1 = 0.5at^2

acceleration a = 20 m/s^2

Time t = 6.0 s

h1 = 0.5×(20 × 6^2)

h1 = 0.5(20×36)

h1 = 360 m

height covered when the motor stops h2 can be derived using equation of motion;

h2 = ut + 0.5at^2 .......2

Where;

a = g = acceleration due to gravity = -9.8 m/s^2

The speed when the motor stops u;

u = at = 20 m/s^2 × 6.0 s = 120 m/s

Time t2 can be derived from;

v = u - gt

v = 0 (at maximum height velocity is zero)

u = gt

t = u/g

t = 120m/s / 9.8m/s^2

t = 12.24 seconds.

Substituting the values into equation 2;

h2 = 120(12.24) - 0.5(9.8×12.24^2)

h2 = 734.69376 m

h2 = 734.69 m

From equation 1;

h = h1 + h2 . substituting the values;

h = 360m + 734.69m

h = 1094.69m

The maximum height above the ground the rocket will achieve is 1094.69m.

Answer:

Final magnification = -375

Explanation:

The total magnification of a compound microscope is expressed mathematically as the product of the magnifying power of each of the lenses that are combined in the compound microscope.

Final magnification = (magnifying power of the objective lens) × (magnifying power of the eyepiece lens) = m₁ × m₂

Magnifying power of a lens is defined as the ratio of the least distance of distinct vision to the focal length of the lens.

For the eyepiece, the least distance of distinct vision is the distance from the object to the near point = 25 + 1 + 0.15 = 26.15 cm

Focal length = 1 cm

Magnifying power of the eyepiece length = (26.15/1) = 26.15

For the objective lens,

The focal length = 0.14 cm

Least distance of distinct vision = -(1 + 1.00 + 0.15 - 0.14) = -2.01 cm (the negative sign means the image seen is upside down)

Magnifying power of the objective lens = (-2.01/0.14) = -14.357

Final magnification = -14.357 × 26.15 = -375

Hope this Helps!!!

Answer:

Explanation:

Using the equation of motion S = ut + 1/2 gt²

S = the vertical displacement (in m)

u = initial velocity of the object (in m/s)

g = acceleration due to gravity (in m/s²)

t = time taken (in secs)

Given u = 0m/s, g = 10m/s² and t = 10s, substituting this value into the equation to get the vertical displacement w have;

S = 0+1/2 (10)(10)²

S = 1000/2

S = 500m

The vertical displacement after 10seconds is 500m

Answer:

Density depends on the temperature and the gap between particles of the liquid. In most of cases temperature is inversely proportional to density means if the temperature increases then the density decreases and the space between particles of that liquid is also inversely proportional to the density means if the intraparticle space increases then the density decreases.

Answer:

Friction

Explanation:

There are tiny bumps and grooves on every object, which make them rough and more difficult to rub against each other. Even though the crate remains at rest at first, the frictional force causes it to stay in place and accelerate with the truck. Hope this helps!

Answer:

The inner diameter is 57.3 cm

Explanation:

The inner diameter of the hollow spherical iron shell can be found using the weight of the sphere ([tex]W_{s}[/tex]) and the weight of the water displaced ([tex]W_{w}[/tex]):

[tex] W_{s} = W_{w} [/tex]

[tex] m_{s}*g = m_{w}*g [/tex]            

[tex] D_{s}*V_{s} = D_{w}*V_{w} [/tex]    

Where D is the density and V is the volume

[tex] D_{s}*\frac{4}{3}\pi*(\frac{d_{o}^{3} - d_{i}^{3}}{2^{3}}) = \frac{4}{3}\pi*(\frac{d_{o}}{2})^{3} [/tex]    

Where [tex]d_{o}[/tex] is the outer diameter and [tex]d_{i}[/tex] is the inner diameter    

[tex] D_{s}*(d_{o}^{3} - d_{i}^{3}) = d_{o}^{3} [/tex]                    

[tex] D_{s}*d_{i}^{3} = d_{o}^{3}(D_{s} - 1) [/tex]          

[tex] 7.87*d_{i}^{3} = 60.0^{3}(7.87 - 1) [/tex]  

[tex] d_{i} = 57.3 cm [/tex]                  

Therefore, the inner diameter is 57.3 cm.    

I hope it helps you!    

Answer:

0.303s

Explanation:

horizontal distance travel = 2.050 m, vertical distance travel = 0.45 m

Using equation of linear motion

Sy = Uy t + 1/2 gt² Uy is the inital vertical component of the velocity, t is the time taken for the vertical motion in seconds, and S is the vertical distance traveled, taken downward vertical motion as negative

-0.45 = 0 - 0.5 × 9.81×t²

0.45 / (0.5 × 9.81) = t²

t = √0.0917 = 0.303 s

Answer:

Change in momentum = 2.197 kgm/s

Explanation:

Momentum = MV

Initial momentum = MU

Final momentum = MV

Computation:

⇒ Change in momentum = MV - MU

⇒ Change in momentum = M (V - U)

⇒ Change in momentum = 0.13(-10.3 - 6.6)

⇒ Change in momentum = 0.13(16.9)

⇒ Change in momentum = 2.197 kgm/s

Answer:

10.15m/s

Explanation:

The change in the frequency of sound (or any other wave) when the source of the sound and the receiver or observer of the sound move towards (or away from) each other is explained by the Doppler effect which is given by the following equation:

f₁ = [(v ± v₁) / (v ± v₂)] f            ----------------------(i)

Where;

f₁ = frequency received by the observer or receiver

v = speed of sound in air

v₁ = velocity of the observer

v₂ = velocity of the source

f = original frequency of the sound

From the question, the observer is the bicyclist and the source is the car driver. Therefore;

f₁ = frequency received by the observer (bicyclist) = 357Hz

v = speed of sound in air = 330m/s

v₁ = velocity of the observer(bicyclist) = (1 / 3) v₂ = 0.33v₂

v₂ = velocity of the source (driver)

f = original frequency of the sound = 365Hz

Note: The speed of the observer is positive if he moves towards the source and negative if he moves away from the source. Also, the speed of the source is positive if it moves away from the listener and negative otherwise.

From the question, the cyclist and the driver are moving in the same direction. But then, we do not know which one is at the front. Therefore, two scenarios are possible.

i. The bicyclist is at the front. In this case, v₁ and v₂ are negative.

Substitute these values into equation (i) as follows;

357 = [(330 - 0.33v₂) / (330 - v₂)] * 365

(357 / 365) = [(330 - 0.33v₂) / (330 - v₂)]

0.98 =  [(330 - 0.33v₂) / (330 - v₂)]

0.98 (330 - v₂) =  (330 - 0.33v₂)

323.4 - 0.98v₂ = 330 - 0.33v₂

323.4 - 330 = (0.98 - 0.33)v₂

-6.6 = 0.65v₂

v₂ = -10.15

The value of v₂ is not supposed to be negative since we already plugged in the right value polarity into the equation.

iI. The bicyclist is behind. In this case, v₁ and v₂ are positive.

Substitute these values into equation (i) as follows;

357 = [(330 + 0.33v₂) / (330 + v₂)] * 365

(357 / 365) = [(330 + 0.33v₂) / (330 + v₂)]

0.98 =  [(330 + 0.33v₂) / (330 + v₂)]

0.98 (330 + v₂) =  (330 + 0.33v₂)

323.4 + 0.98v₂ = 330 + 0.33v₂

323.4 - 330 = (0.33 - 0.98)v₂

-6.6 = -0.65v₂

v₂ = 10.15

The value of v₂ is positive and that is a valid solution.

Therefore, the speed of the car is 10.15m/s

Answer:

The centripetal acceleration of the car will be 12.32 m/s² .

Explanation:

Given that

radius ,R= 57 m

Velocity , V=26.5 m/s

We know that centripetal acceleration given as follows

[tex]a_c=\dfrac{V^2}{R}[/tex]

Now by putting the values in the above equation we get

[tex]a_c=\dfrac{26.5^2}{57}=12.32\ m/s^2[/tex]

Therefore the centripetal acceleration of the car will be 12.32 m/s² .

Answer:

a) 8.33 x [tex]10^{-6}[/tex] Pa  or  8.22 x [tex]10^{-11}[/tex] atm

b) 1.66 x [tex]10^{-5}[/tex] Pa  or  1.63 x [tex]10^{-10}[/tex] atm

c) 2.77 x [tex]10^{-14}[/tex] kg/m^2-s

Explanation:

Intensity of light = 2500 W/m^2

area = 25 ft^2

a) average radiation pressure on a totally absorbing section of the floor[tex]Pav = \frac{I}{c}[/tex]

where I is the intensity of the light

c is the speed of light = [tex]3*10^{8} m/s[/tex]

[tex]Pav = \frac{2500}{3*10^{8} }[/tex] = 8.33 x [tex]10^{-6}[/tex] Pa

1 pa = [tex]9.87*10^{-6}[/tex]

8.33 x [tex]10^{-6}[/tex] Pa = 8.22 x [tex]10^{-11}[/tex] atm

b) average radiation for a totally radiating section of the floor

[tex]Pav = \frac{2I}{c}[/tex]

this means that the pressure for a totally radiating section is twice the average pressure of the totally absorbing section

therefore,

Pav = 2 x 8.33 x [tex]10^{-6}[/tex]  = 1.66 x [tex]10^{-5}[/tex] Pa

or

Pav in atm = 2 x 8.22 x [tex]10^{-11}[/tex] = 1.63 x [tex]10^{-10}[/tex] atm

c) average momentum per unit volume is

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

[tex]m = \frac{2500}{(3*10^{8}) ^{2} }[/tex] = 2.77 x [tex]10^{-14}[/tex] kg/m^2-s

Answer;

The pond's water level will fall.

Explanation;

Archimedes principle explains that a floating body will displace the amount of water that weighs the same as it, whereas a body resting on the bottom of the water displaces the amount of water that is equal to the body's volume.

When the anchor is in the boat it is in the category of floating body and when it is on the bottom of the pond it is in the second category.

Since anchors are naturally heavy and denser than water, the amount of water displaced when the anchor is in the boat is greater than the amount of water displaced when the anchor is on the bottom of the pond since the way anchors are doesn't make for them to have considerable volume.

When the anchor is dropped to the bottom of the pond, the water level will therefore fall. If the anchor doesn't reach the bottom it is still in the floating object category and there will be no difference to the water level, but once it touches the bottom of the pond, the water level of the pond drops.

Hope this Helps!!!

Buoyancy is an upward force exerted by a fluid on a body partially or completely immersed in it

The pond water level will lower slightly

According to Archimedes principle, the up thrust on the boat by the water is given by the volume of the water displaced

Learn more here;

https://brainly.com/question/24529607