One strategy in a snowball fight is to throw a snowball at a high angle over level ground. Then, while your opponent is watching that snowball, you throw a second one at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of 26.0 m/s. The first one is thrown at an angle of 67.5° with respect to the horizontal. At what angle should you throw the second snowball to make it hit the same point as the first?

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:

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:

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:

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:

171.5Hz,514.5Hz and 857.5Hz

Explanation:

We are given that

Distance between two loudspeaker,d=2.65 m

Distance of listener from one end=19.1 m

Distance of listener from other end=20.1 m

[tex]\Delta L=20.1-19.1=1[/tex]m

Speed of sound,v=343m/s

For destructive interference

[tex]f_{min,n}=\frac{(n-0.5)v}{\Delta L}[/tex]

Using the formula  and substitute n=1

[tex]f_{min,1}=\frac{(1-0.5)\times 343}{1}=171.5Hz[/tex]

For n=2

[tex]f_{min,2}=\frac{(2-0.5)\times 343}{1}=514.5Hz[/tex]

For n=3

[tex]f_{min,3}=(3-0.5)\times 343=857.5Hz[/tex]

Hence, the three lowest frequencies that give minimum signal (destructive interference) at the listener's location is given by

171.5Hz,514.5Hz and 857.5Hz

Answer:

i think its 1-10 ´ 10-15 m

Explanation:

It is found that nuclear radii range from 1-10 ´ 10-15 m. This radius is much smaller than that of the atom, which is typically 10-10 m. Thus, the nucleus occupies an extremely small volume inside the atom. The nuclei of some atoms are spherical, while others are stretched or flattened into deformed shapes.

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.

HEYA!!

Here's your Answer:

Energy is the power we use for transportation, for heat and light in our homes and for the manufacture of all kinds of products. There are two sources of energy: renewable and non-renewable energy

RENEWABLE ENERGY: Renewable energy, often referred to as clean energy, comes from natural sources or processes that are constantly replenished. For example, (sunlight or wind keep shining and blowing), even if their availability depends on time and weather.

NON-RENEWABLE ENERGY:Non-renewable energy comes from sources that will run out or will not be replenished in our lifetimes—or even in many, many lifetimes.

Most non-renewable energy sources are (fossil fuels: coal, petroleum, and natural gas).

HOPE IT HELPS!!!

The so-called "force" of friction always acts exactly opposite to the direction in which an object is moving.  So for this question, since the box is moving to the right, the "force" of friction is acting toward the left.

There's not enough information given in the question to determine its magnitude.

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:

33 m/s

Explanation:

v = √(T / (m/L))

where v is the velocity,

T is the tension,

and m/L is the mass per length.

v = √(200 N / (0.055 kg / 0.30 m))

v = 33 m/s

Answer:

a) h_max = 400ft

b) v = 1024 ft/s

Explanation:

The general equation of a motion is:

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

You have the following equation of motion is given by:

[tex]s(t)=160t-16t^2[/tex]  (2)

You compare both equations (1) and (2) and you obtain:

vo: initial velocity = 160ft/s

g: gravitational acceleration = 32ft/s

a) The maxim height reached by the ball is given by:

[tex]h_{max}=\frac{v_o^2}{2g}=\frac{(160ft/s)^2}{64ft/s^2}=400ft[/tex]

b) The velocity is given by:

[tex]v^2=v_o^2-2gh\\\\v=\sqrt{(160ft/s)^2-2(32ft/s^2)(384ft)}=1024ft/s[/tex]

Answer:

a) Smax = 400 ft

b) v = 32 ft/s

c) v = - 32 ft/s

Explanation:

(a)

The function given for the height of ball is:

s = 160 t - 16 t²

Therefore, in order to find the time to reach maximum height (in-flexion point), we must take the derivative with respect to t and set it equal to zero:

Therefore,

160 - 32 t = 0

32 t = 160

t = 160/32

t  = 5 sec

Therefore, maximum distance is covered at a time interval of 5 sec.

Smax = (160)(5) - (16)(5)²

Smax = 800 - 400

Smax = 400 ft

b)

First we calculate the time at which ball covers 384 ft

384 = 160 t - 16 t²

16 t² - 160 t + 384 = 0

Solving Quadratic Equation:

Either:

t = 6 sec

Or:

t = 4 sec

Since, the time to reach maximum height is 5 sec. Therefore, t < 5 sec

Therefore,

t = 4 sec

Now, we find velocity at 4 sec by taking derivative od s with respect to t at 4 sec:

v = 160 - 32 t

v = 160 - (32)(4)

v = 32 ft/s

c)

Since, t = 6 s > 5 s

The second value of t = 6 sec must correspond to the instant when the ball is 384 ft above the ground while traveling downward.

Hence, velocity at that time will be:

v = 160 - (32)(6)

v = -32 ft/s

Negative sign due to downward motion.

Answer:

T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N

Explanation:

This is a balance exercise where we must apply the expressions for translational balance in the two axes

     ∑  F = 0

Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º

decompose the tension of the two upper cables

          cos 41 = T₁ₓ / T1

          sin 41 = T₁y / T1

          T₁ₓ = T₁  cos 41

          T₁y= T₁  sin 41

for cable gold

           cos 63 = T₂ / T₂

           sin 63 = [tex]T_{2y}[/tex] / T₂

We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.

Let's start by analyzing the point where the traffic light meets the vertical cable

              T₃ - W = 0

              T₃ = W

              T₃ = 200 N

now let's write the equations for the single point of the three wires

X axis

   - T₁ₓ + T₂ₓ = 0

  T₁ₓ = T₂ₓ

   T1 cos 41 = T2 cos 63

   T1 = T2 cos 63 / cos 41                (1)

y Axis

      [tex]T_{1y}[/tex] + T_{2y} - T3 = 0

       T₁ sin 41 + T₂ sin 63 = T₃          (2)

to solve the system we substitute equation 1 in 2

        T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W

         T₂ (cos 63 tan 41 + sin 63) = W

         T₂ = W / (cos 63 tan 41 + sin 63)

We calculate

          T₂ = 200 / (cos 63 tan 41 + sin 63)

          T₂ = 200 / 1,2856

           T₂ = 155.6 N

we substitute in 1

            T₁ = T₂ cos 63 / cos 41

             T₁ = 155.6 cos63 / cos 41

             T₁ = 93.6 N

therefore the tension in each cable is

            T₁ = 93.6 N

             T₂ = 155.6 N

             T₃ = 200 N

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!        

Answer:

B. less

Explanation:

acceleration due to gravity on Earth, g = 9.8 m/s²

acceleration due to gravity on Moon, g = 1.6 m/s²

Given mass of the object as, m = 5 kg

Weight of an object is given as, W = mg

Weight of the object on Earth, W = 5 x 9.8 = 49 N

Weight of the object on Moon, W = 5 x 1.6 = 8 N

Therefore, the object weighs less on the moon compared to its weight on Earth.

The correct option is "B. less"

Answer:

A

Explanation:

Firstly, recall the equation for the circumference of a circle (C=2*pi*r). Then apply the formula speed = distance/time.

Answer:

i see they did already i am just supporting them

Answer:

A

Explanation:

Firstly, recall the equation for the circumference of a circle (C=2*pi*r). Then apply the formula speed = distance/time.

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:

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:

[tex]v = \sqrt{20h} [/tex]

Explanation:

The potential energy (PE) we are looking here is gravitational potential energy (GPE).

GPE= mgh,

where m is the mass of an object,

g is the gravitational field strength

h is the height of the object

KE= ½mv²,

where m is the mass and v is the velocity

loss in GPE= gain in KE

mgh= ½mv²

gh= ½v² (divide by m throughout)

Assuming that the object is on earth, then g= 10N/Kg

½v²= 10h (substitute g=10)

v²= 20h (×2 on both sides)

v= √20h (square root both sides)

Answer: No.

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

Answer:

v_avg = 2.9 cm/s

Explanation:

The average velocity of the object is the sum of the distance of all its trajectories divided the time:

[tex]v_{avg}=\frac{x_{all}}{t}[/tex]

x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm

Then, x_all = 150cm + 140cm = 290cm

The average velocity is, for t = 100s

[tex]v_{avg}=\frac{290cm}{100s}=2.9\frac{cm}{s}[/tex]

hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s

Answer:

The average velocity of the object is 0.1cm/s

Explanation:

Given that the object travels from point 15cm to 165cm and back to 25cm within 100 seconds

The average velocity is calculated as thus.

Average Velocity = ∆D/t

Where ∆D represent the displacement.

The displacement is calculated as follows.

∆D = End point - Start Point.

From the question, the end and start point are 25cm and 15cm respectively.

Hence,

∆D = 25cm - 15cm

∆D = 10cm.

t = 100 seconds

So, Average Velocity = 10cm/100s

Average Velocity = 0.1cm/s

Hence, the average velocity of the object is 0.1cm/s

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

learn more about kinematics here:

https://brainly.com/question/11503629

Answer:

Boat A wins and 50km/hr

Explanation:

To solve this problem we need to calculate the time required for both boats to make the round trip.

For Boat A.

The time required for the first trip =distance/ speed = 50/50 = 1h

For the return trip the time is same since it's the same distance on the same speed.

Hence the time taken for boat A to make the round trip is 2hrs.

For boat B,

The time required for the first trip =distance/ speed = 50/25 = 2h

The time for the return trip is;

=distance/ speed = 50/75 = 2/3h= 2/3×60= 40mins

By comparing both time.

Boat A requires less time hence Boat A wins.

2.The average velocity of the winning boat is the velocity of boat A.

Which is total distance/ total time for the trip.

100km/2hrs = 50km/hr

boat A would win the race and the average velocity of the winning boat would be 50 km/h

The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object.

As given in the problem Two boats start together and race across a 50-km-wide lake and back. Boat A goes across at 50 km/h and returns at 50 km/h. Boat B goes across at 25 km/h, and its crew, realizing how far behind it is getting, returns at 75 km/h.

Turnaround times are negligible, and the boat that completes the round-trip first wins

Average velocity of the boat A = ( 50 + 50 ) / 2

                                                              = 50 km /h

Thus, boat A would win the race and the average velocity of the winning boat would be  50 km/h

Learn more about Velocity from here,

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#SPJ2

Answer:

Explanation:

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Answer:

The normal force is the force that the floor does as a reaction of the gravitational force that an object does against the floor (is the resistance that objects have when other objects want to move trhough them, and the force comes by the 3rd Newton's law, and this is specially used in cases where the first object is fixed, like walls or the floor). With this in mind, the point in where the normal force will be greater is the point that is closer to the center of mass of the object (the point with more mass)

If the wheels are in the extremes of the object, and the center of mass is in the middle of the object, the normal force will be equal. Now if for example, you put a little mass in one end of the object, now the center of weight displaces a little bit and is not centered, and the side is where you put the weight on will receive a bigger normal force from the floor than the other side.

The force on the front and rear wheels can be determined by finding the

force acting under equilibrium conditions.

The normal reaction between the floor and the front wheel is larger. The

correct option is c) The force on the front wheels must be greater than the

force on the rear wheels.

Reasons:

The forces acting on the wheels are;

Vertical forces acting on the cart:

The weight of the cart acting on the floor

The normal reaction of the floor on the wheels

The horizontal acting on the cart:

Pushing force acting horizontally

Friction force acting in the reverse direction

At equilibrium, the cart does not tip over

Sum of moment about the rear wheel = [tex]\mathbf{W \times \dfrac{d}{2} + F \times h - N_{front} \times d} = 0[/tex]

Where;

h = The height of the cart

d = depth of the cart

[tex]N_{front}[/tex]  = The normal reaction at the front wheels

Therefore;

[tex]\displaystyle N_{front} = \frac{W \times \dfrac{d}{2} + F \times h}{d} = \mathbf{ \frac{W}{2} + \frac{F \times h}{d}}[/tex]

Sum of moment about the front wheel = [tex]\mathbf{ F \times h + N_{rear} \times d - W \times \dfrac{d}{2} } = 0[/tex]

Therefore;

[tex]\displaystyle N_{rear} = \frac{W \times \dfrac{d}{2} - F \times h}{d} = \mathbf{ \frac{W}{2} - \frac{F \times h}{d}}[/tex]

Which gives;

[tex]\displaystyle N_{front} = \mathbf{ \displaystyle N_{rear} + 2 \times \frac{F \times h}{d}}[/tex]

[tex]\displaystyle 2 \times \frac{F \times h}{d} > 0[/tex]

Therefore;

[tex]\displaystyle N_{front} > \displaystyle N_{rear}[/tex]

The correct option is c) The force on the front wheels must be greater than

the force on the rear wheels.

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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:

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.

Answer:

The resistance of the aluminium wire is 0.541 ohms

Explanation:

Given:

Cross sectional area of the aluminum wire, A = 4.9 x 10-4 m2

Length of the wire, L = 10 km = 10,000 m

Resistance of the wire = (pL) / A

Where:

p is resistivity of aluminium wire = 2.65 x 10^-8 ohm-m

L is length of the wire

A is cross sectional area of the wire

Substitute these values and solve for resistance of the wire.

Resistance = (2.65 x 10^-8 x 10,000) / (4.9 x 10^-4)

Resistance = 0.541 ohms

Therefore, the resistance of the aluminium wire is 0.541 ohms