Projectile Problems – Type 2:
1. The photograph below shows a basic projectile at several locations on its trajectory.

a) List the location(s) where the vertical component of the velocity would be zero.

b) What is the vertical component of the acceleration at location #3?

c) What is the horizontal acceleration of the projectile?

d) Identify the location where the vertical displacement would be zero.

e) Identify the location with the maximum displacement.

f) Rank each location in terms of the projectile's speed (highest to lowest).

Answer:

Explanation:

a)     Fx = F cos (θ)

           = (200) cos(60)

           = 100 N

b)     FR = ma

       Fx + Ff = ma

      100 + Ff = (50)(1,5)

       Ff     = 75 - 100

               =  -25 N

c)    Fy = F sin θ

           = (200) sin(60)

           = 173,2 N

Answer:

down below

Explanation:

Image 1- wheels of train showing both translatory motion as well as rotatory motion.

Image 2- rotation of ball shows both rotatory motion as well as translatory motion.

Image 3- the earth rotates about its axis, same time it revolves around the sun thus showing both rotatory motion and curvilinear motion in a fixed time. (perodic motion)

Image 4- while cutting wood, the

carpenter's saw has both

translatory motion and oscillatory

motion, as it moves down while

oscillating.

Answer:

the study of the past

Explanation:

dogs and cats

Answer:

the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

Explanation:

Given the data in the question,

near point = 20 cm

far point = 2 m = 200 cm

Now, for an object that is infinitely far away, the image is at is its far point.

so using the following expression, we can determine the focal length

1/f = 1/i + 1/o

where f is the focal length, i is the image distance and o is the object distance.

here, far point i = 2 m = 200 cm  and v is ∞

so we substitute

1/f = 1/(-200 cm)  +  1/∞

f = -200 cm

Also, for object at its closest point, the image appear at near point,

so

1/f = 1/i + 1/o

we make o the subject of formula

o = ( i × f ) / ( i - f )

given that near point i = 20 cm

we substitute

o = ( -20 × -200 ) / ( -20 - (-200) )

o = 4000 / 180

o = 22.2 cm

Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

Answer:

don't know what class are you you are using which mobile or laptop

Answer:

Explanation:

The formula for angular momentum is

L = mvr where L is the angular momentum, m is the mass of the object, v is the velocity of the object, and r is the radius of the object. The problem we have that prevents us from just throwing those numbers in there is that mass has to be in kg and it's not, and radius has to be in meters and it's not.

Changing the mass to kg:

750 g = .750 kg

Changing the radius to m:

35 cm = .35 m

Now we can fill in the variables with their respective values:

L = .750(10.0)(.35) gives us

[tex]L=2.625\frac{kg*m^2}{s}[/tex]

11.54 minutes

Explanation:

The decay rate equation is given by

[tex]N = N_0e^{-\frac{t}{\lambda}}[/tex]

where [tex]\lambda[/tex] is the half-life. We can rewrite this as

[tex]\dfrac{N}{N_0} = e^{-\frac{t}{\lambda}}[/tex]

Taking the natural logarithm of both sides, we get

[tex]\ln \left(\dfrac{N}{N_0}\right) = -\left(\dfrac{t}{\lambda}\right)[/tex]

Solving for [tex]\lambda[/tex],

[tex]\lambda = -\dfrac{t}{\ln \left(\frac{N}{N_0}\right)}[/tex]

[tex]\:\:\:\:= -\dfrac{(24\:\text{minutes})}{\ln \left(\frac{1000\:\text{counts/min}}{8000\:\text{counts/min}}\right)}[/tex]

[tex]\:\:\:\:=11.54\:\text{minutes}[/tex]

Answer:

A. Equals to that of the smaller sphere

B. 3 times less than that of the smaller sphere

Explanation:

(a) Equals to that of the smaller sphere

The potential of an isolated metal sphere, with charge Q and radius R, is kQ=R, so a sphere with charge 3Q and radius 3R has the same potential

b) 3 times less than that of the smaller sphere

However, the electric field at the surface of the smaller sphere is ?=? 0 = kQ=R2 , so tripling Q and R reduces the surface field by a factor of 1/3

Answer

Explanation

:giác mạc

Answer:

Explanation:

its 1000 yards if its going around the track 2 times and that if one whole  around the track is 500 its 500 x 2

Answer:

5.4uC

Explanation:

Answer:

Period is 86811.5 seconds.

Explanation:

[tex]{ \boxed{ \bf{T {}^{2} = (\frac{4 {\pi}^{2} }{GM}) {r}^{3} }}}[/tex]

[tex]{ \tt{T {}^{2} = \frac{4 {(3.14)}^{2} }{(6.6 \times {10}^{ - 11} ) \times (5.98 \times {10}^{24} )} \times {((35880\times {10}^{3}) } + (6370 \times {10}^{3} )) {}^{3} }} \\ \\ { \tt{T {}^{2} = 7.54 \times {10}^{9} }} \\ { \tt{T = \sqrt{7.54 \times {10}^{9} } }} \\ { \tt{T = 86811.5 \: seconds}}[/tex]

Answer:

B

Explanation:

P1/P1 = 40/20

=2

Answer: hello  some of your values are wrongly written hence I will resolve your question using the right values

answer:

stiffness =  1.09 * 10^-6 N/m

Explanation:

Given data:

Length ( l ) = 16 m

radius of wire ( r ) = 3.5 m

mass ( m ) = 5kg

Distance stretched (  Δl ) = 4 * 10^-3 m ( right value )

average bond length ( between atoms ) = 2.3 * 10^-10 m ( right value)

first step : calculate the area

area ( A ) = πr^2 = π * ( 3.5)^2 = 38.48 m^2

        γ          = MgL / A Δl

                    = [ (5 * 9.81 * 16 ) / ( 38.48 * (4.3*10^-3) ) ]

                    = 784.8 / 0.165 = 4756.36 N/m^2

hence : stiffness =   γ  * bond length

                           =  4756.36 * 2.3 * 10^-10  = 1.09 * 10^-6 N/m

Answer:

[tex]I_2=6.8A[/tex]

Explanation:

From the question we are told that:

Turns of inner coil [tex]N_1=120[/tex]

Radius of inner coil [tex]r_1=0.012m[/tex]

Current of  inner coil [tex]I_1=6.0A[/tex]

Turns of Outer coil [tex]N_2=150[/tex]

Radius of Outer coil [tex]r_2=0.017m[/tex]

Generally the equation for Magnetic Field is mathematically given by

[tex]B =\frac{ \mu N I}{2R}[/tex]

Therefore

Condition for the net Magnetic field to be zero

[tex]\frac{N_1* I_1}{( 2 * r_1 )}=\frac{N_2 * I_2}{2 * r_2}[/tex]

[tex]I_2=\frac{(N_1* I_1)*(( 2 * r_2)}{( 2 * r_1)*N_2}[/tex]

[tex]I_2=\frac{(120*6.0)*(( 2 * 0.017)}{( 2 * 0.012)*150}[/tex]

[tex]I_2=6.8A[/tex]

Answer:

a)  [tex]F_g=1.5*10^9Ibf[/tex]

b)  [tex]F_t=12490Ibf/ft^2[/tex]

     [tex]F_b=0[/tex]

Explanation:

From the question we are told that:

Height [tex]h=200ft[/tex]

Width [tex]w=1200ft[/tex]

a)

Generally the equation for Dam's Hydro static force is mathematically given by

[tex]F_g=\rho*g*\frac{h}{2}(w*h)[/tex]

Where

[tex]\rho=Density\ of\ water[/tex]

[tex]\rho=62.4Ibm/ft^3[/tex]

Therefore

[tex]F_g=62.4*32.2*\frac{200}{2}(1200*200)[/tex]

[tex]F_g=1.5*10^9Ibf[/tex]

b)

Generally the equation for Dam's Force per unit area is mathematically given by

[tex]F=\rho*g*h[/tex]

For Top

[tex]F_t=\rho*g*h[/tex]

[tex]F_t=62.4*32.2*200[/tex]

[tex]F_t=12490Ibf/ft^2[/tex]

For bottom

[tex]Here \\H=0 zero[/tex]

Therefore

[tex]F_b=0[/tex]

The hydrostatic force on the dam is [tex]2.995 \times 10^9 \ lbF[/tex].

The force per unit area near the top is 86.74 psi.

The force per unit area near the bottom is zero.

The hydrostatic force on the dam is the force exerted on the dam by the column of the water.

[tex]F = PA\\\\F = (\rho gh) \times (wh)\\\\F = (62.4 \times 32.17 \times 200) \times (1200 \times 200)\\\\F = 9.636 \times 10^{10} \ lb-ft/s^2\\\\1 \ lbF = 32.17\ lb-ft/s^2\\\\F = 2.995 \times 10^9 \ lbF[/tex]

The force per unit area is the pressure exerted near the top of the dam.

[tex]P = \rho gh\\\\P = 0.052 \times \rho h[/tex]

where;

P is pressure in PSI

ρ is density of water in lb/gal

h is the vertical height in ft

[tex]P = 0.052 \times 8.34 \times 200\\\\P = 86.74 \ Psi[/tex]

The pressure near the bottom is zero, become the vertical height is zero.

Learn more about hydrostatic pressure here: https://brainly.com/question/11681616

Answer:

1.25

Explanation:

Answer:

The speed at B is 16.18 ft/s .

Explanation:

Speed at A, u = 11 ft/s

Speed at C, v' = 18 ft/s

Time from A to C = 5 s

Time from B to C = 1.3 s

Let the speed of car at B is v.

Let the acceleration is a.

From A to B

Use first equation of motion

v = u + a t

18 = 11 + a x 5

a = 1.4 ft/s^2

Let the time from A to B is t' .

t' = 5 - 1.3 = 3.7 s

Use first equation of motion from A to B

v = 11 + 1.4 x 3.7 = 16.18 ft/s  

Answer:

a) The proton's speed is 5.75x10⁵ m/s.

b) The kinetic energy of the proton is 1723 eV.  

Explanation:

a) The proton's speed can be calculated with the Lorentz force equation:

[tex] F = qv \times B = qvBsin(\theta) [/tex]     (1)          

Where:

F: is the force = 9.14x10⁻¹⁷ N

q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C

v: is the proton's speed =?

B: is the magnetic field = 3.28 mT

θ: is the angle between the proton's speed and the magnetic field = 17.6°

By solving equation (1) for v we have:

[tex]v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s[/tex]

Hence, the proton's speed is 5.75x10⁵ m/s.

b) Its kinetic energy (K) is given by:

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

Where:

m: is the mass of the proton = 1.67x10⁻²⁷ kg

[tex] K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV [/tex]  

Therefore, the kinetic energy of the proton is 1723 eV.

I hope it helps you!        

Answer:

x = 68.3 m

Explanation:

tan 32.8 = x / 106

Answer:

[tex]R=0.023m[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=1.16*10^{-26}[/tex]

Potential difference [tex]V=523V[/tex]

Magnitude [tex]m=0.370 T[/tex]

Generally the equation for Velocity is mathematically given by

[tex]\frac{1}{2}mv^2=ev[/tex]

[tex]v=\frac{2ev}{m}[/tex]

[tex]v=\frac{2*1.6*10^{-19}*542}{1.16*10^{-26}}[/tex]

[tex]v=12.22*10^4m/s[/tex]

Generally the equation for Force is mathematically given by

[tex]F=qvBsin \theta[/tex]

Where

[tex]qVB=m\frac{v^2}{R}[/tex]

[tex]F=m\frac{v^2}{R}sin\theta[/tex]

Therefore

[tex]R=\frac{mv}{qB sin \theta}[/tex]

[tex]R=\frac{1.6*10^{-26}*12.2*10^{4}}{1.60*10^{-19}*0.394 sin 90}[/tex]

[tex]R=0.023m[/tex]

B will have the greater force

Fc=MV2 /R=Fm

The A particle has less centipetal force and larger radius so larger curve

Answer:

D.

Explanation:

The crest of a wave refers to the highest point of a wave. This is illustrated by D.

Answer:

It's electricity produced from hydropower. It's also a form of energy that controls the power of water motion.

Explanation:

One pro about hydroelectric power is that it's renewable energy. But one con about hydroelectric power is that it can impact the environment in a negative way.

(C)

Explanation:

The capacitance C of a parallel plate capacitor is given by

[tex]C = \epsilon_0 \dfrac{A}{d}[/tex]

Let C' be the new capacitance where the area and the plate separation distance are doubled. This gives us

[tex]C' = \epsilon_0\dfrac{A'}{d'} = \epsilon_0\left(\dfrac{2A}{2d}\right) = \epsilon_0 \dfrac{A}{d} = C[/tex]

Answer:

c) 2Q

Explanation:

From the given information:

The pressure inside a pipe can be expressed by using the formula:

[tex]\Delta P = \dfrac{128 \mu L Q}{\pi D^4}[/tex]

Since the diameter in both pipes is the same, we can say:

[tex]D = D_A = D_B[/tex]

where;

length of the first pipe A [tex]L_A = L[/tex] and the length of the second pipe B [tex]L_B = 2L[/tex]

Since the difference in pressure is equivalent in both pipes:

Then:

[tex]\dfrac{128 \mu L_1Q_1}{\pi D_1^4} = \dfrac{128 \mu L_2Q_2}{\pi D_2^4}[/tex]

[tex]\dfrac{ L_1Q_1}{D_1^4} = \dfrac{ L_2Q_2}{D_2^4}[/tex]

[tex]\dfrac{ LQ_1}{D^4} = \dfrac{ 2LQ}{D^4}[/tex]

[tex]\mathbf{Q_1 = 2Q}[/tex]

The flow rate in pipe B is 2Q of the flow rate of the pipe A

The flow rate is defined as the flow of the fluid across the cross section in per unit time.

From the given information:

The pressure inside a pipe can be expressed by using the formula:

[tex]\Delta p=\dfrac{128\mu LQ}{\pi D^4}[/tex]

Since the diameter in both pipes is the same, we can say:

[tex]D=D_A=D_B[/tex]

where;

length of the first pipe A  [tex]L_A=L[/tex] and the length of the second pipe B  

[tex]L_B=2L[/tex]

Since the difference in pressure is equivalent in both pipes:

Then:

[tex]\dfrac{128\mu L_1Q_1}{\pi D_1^4}=\dfrac{128\mu L_2Q_2}{\pi D_2^4}[/tex]

[tex]\dfrac{L_1Q_1}{D_1^4}=\dfrac{L_2Q_2}{D_2^4}[/tex]

[tex]\dfrac{LQ_1}{D_1^4}=\dfrac{2LQ}{D_2^4}[/tex]

[tex]Q_1=2Q[/tex]

Hence the flow rate in pipe B is 2Q of the flow rate of the pipe A

To know more about Flow rate follow

https://brainly.com/question/26061120

Answer:

(i) When a battery is connected inseries to two long parallel wires, the currents in the two wires will be in opposite directions. Due to which a force of repulsion will be acting between them and they are moving further apart.

(ii) When a battery is connected in parallel to two long parallel wires, the currents in the two wires will be in same direction. Due to it, a force of attraction will be acting between them and they are coming closer to each other.

hope it's help you ....!!!!!

plz mark as brain list and follow me #rishu...!!!!

Explanation:

Hope it will helps you lot!

Answer:

The SI unit of coefficient of linear expansion can be expressed as °C-1 or °K-1. ... The dimension of coefficient of linear expansion will be M0L0T0K−1.