Two large insulating parallel plates carry charge of equal magnitude, one positive and the other negative, that is distributed uniformly over their inner surfaces. Rank the points 1 through 5 according to the magnitude of the electric field at the points, least to greatest.
A. 1, 2, 3, 4, 5
B. 2, then 1, 3, and 4 tied, then 5
C. 1, 4, and 5 tie, then 2 and 3 tie
D. 2 and 3 tie, then 1 and 4 tie, then 5
E. 2 and 3 tie, then 1, 4, and 5 tie

Answer:

The centripetal acceleration of the traveler is [tex]40.98\ m/s^2[/tex]

Explanation:

It is given that, A carousel has a diameter of 6.0-m and completes one rotation every 1.7 s.

We need to find the centripetal acceleration of the traveler. It is given by the formula as follows :

[tex]a=\dfrac{v^2}{r}[/tex]

r is radius of carousel

[tex]v=\dfrac{2\pi r}{T}[/tex]

So,

[tex]a=\dfrac{4\pi ^2r}{T^2}[/tex]

Plugging all the values we get :

[tex]a=\dfrac{4\pi ^2\times 3}{(1.7)^2}\\\\a=40.98\ m/s^2[/tex]

So, the centripetal acceleration of the traveler is [tex]40.98\ m/s^2[/tex].

Answer: [tex]x=\frac{2}{3}t^3+3t+1[/tex]

Explanation:

Given

velocity of object is given by

[tex]v(t)=3+2t^2[/tex]

and we know change of position w.r.t time is velocity

[tex]\Rightarrow \dfrac{dx}{dt}=v[/tex]

[tex]\Rightarrow \dfrac{dx}{dt}=3+2t^2[/tex]

[tex]\Rightarrow dx=(3+2t^2)dt[/tex]

Integrating both sides we get

[tex]\Rightarrow \int_{1}^{x}dx=\int_{0}^{t}(3+2t^2)dt[/tex]

[tex]\Rightarrow x\mid _{1}^{x}=(3t+\frac{2}{3}t^3)\mid _{0}^{t}[/tex]

[tex]\Rightarrow x-1=3(t-0)+\frac{2}{3}(t^3-0)[/tex]

[tex]\Rightarrow x=\frac{2}{3}t^3+3t+1[/tex]

Answer:

a. 900 J

b. 0.383

Explanation:

According to the question, the given data is as follows

Horizontal force = 150 N

Packing crate = 40.0 kg

Distance = 6.00 m

Based on the above information

a. The work done by the 150-N force is

[tex]W = F x = \mu N x = \mu\ m\ g\ x[/tex]

[tex]W = 150 \times 6[/tex]

= 900 J

b. Now the coefficient of kinetic friction between the crate and surface is

[tex]\mu = \frac {F}{m\timesg}[/tex]

[tex]= \frac{150}{40\times 9.8}[/tex]

= .383

We simply applied the above formulas so that each one part could calculate

We want to find the work and kinetic friction for the given situation. The solutions are:

Here we have a horizontal force of 150N pushing a 40.0 kg packing crate a distance of 6.00m at a constant speed.

a) First we want to find the work, it is given by the force applied times the distance moved, so the work is just:

W = 150N*6.00m = 900 N*m

b) Now we want to find the coefficient of kinetic friction, it must be such that the kinetic friction force is equal to the pushing force, in this way there is no net force, and then there is no acceleration.

Remember that the friction force is:

F = m*g*μ

Where:

Then we must solve:

150N = 40kg*(9.8 m/s^2)*μ = 392N*μ

150N/392N = 0.38

So the coefficient of kinetic friction between the crate and the surface is 0.38

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Answer:

The answer is option 2.

Explanation:

Both sides are pulling the rope with equal force where the rope doesn't move. So they have a balanced forces.

Answer: B, Companies passed on production and transportation costs to consumers

Explanation:

A higher oil price occurred when companies passed on production and transportation costs to consumers.

The oil producing companies spend so much money in producing crude oil from the reservoirs to the surface. They also spend money in processing and transporting the crude oil to the end users or consumers.

The final price of the oil depends on the total amount spent by these companies in producing the hydrocarbons.

Thus, a higher oil price occurred when companies passed on production and transportation costs to consumers.

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Soundproofing material is required for blocking sound during some works like recording voice in the studio. Image D represents the interaction of a sound wave with soundproofing material in a recording studio.

Soundproofing is done by absorbing the sound. A very much used material for this is a dense foam.

Foam and like materials absorbs sound and it travels directly into the soft surface resulting in soundproofing.

Thus, the correct option is C, as the D image is showing the absorption.

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#SPJ2

Answer: C.D

Explanation:...

Answer:

a) [tex]F = 210.803\,N[/tex], b) [tex]W_{F} = 779.971\,J[/tex], c) [tex]W_{f} = 235.683\,J[/tex], d) [tex]W_{N} = 0\,J[/tex]; [tex]W_{g} = 544.289\,J[/tex], e) [tex]W_{net} = 0\,J[/tex]

Explanation:

a) The net force exerted on the crate is:

[tex]\Sigma F = F - m\cdot g \cdot \sin \theta - \mu_{k}\cdot m\cdot g \cdot \cos \theta = 0[/tex]

The magnitud of the force that the work must apply to the crate is:

[tex]F = m\cdot g \cdot \sin \theta + \mu_{k}\cdot m\cdot g \cdot \cos \theta[/tex]

[tex]F = (30\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \sin 30^{\circ} + 0.25 \cdot (30\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 30^{\circ}[/tex]

[tex]F = 210.803\,N[/tex]

b) The work done on the crate due to the external force is:

[tex]W_{F} = (210.803\,N)\cdot (3.7\,m)[/tex]

[tex]W_{F} = 779.971\,J[/tex]

c) The work done on the crate due to the external force is:

[tex]W_{f} = (63.698\,N)\cdot (3.7\,m)[/tex]

[tex]W_{f} = 235.683\,J[/tex]

d) The work done on the crate due the normal force is zero, since such force is perpendicular to the motion direction.

[tex]W_{N} = 0\,J[/tex]

And, the work done by gravity is:

[tex]W_{g} = (147.105\,N)\cdot (3.7\,m)[/tex]

[tex]W_{g} = 544.289\,J[/tex]

e) Lastly, the total work done is:

[tex]W_{net} = W_{F} - W_{f} - W_{g} - W_{N}[/tex]

[tex]W_{net} = 779.971\,J - 235.683\,J - 0\,N - 544.289\,J[/tex]

[tex]W_{net} = 0\,J[/tex]

Answer:

Explanation:

moment of inertia of the rod = 1/3 mL² , m is mass and L is length of rod.

1/3 x .154 x .35²

= .00629

moment of inertia  of putty about the axis of rotation

= m d² , m is mass of putty and d is distance fro axis

= .011 x( .35 / 3 )²

= .00015

Total moment of inertia I = .00644 kgm²

angular momentum of putty about the axis of rotation

= mvRsinθ

m is mass , v is velocity , R is distance where it strikes the rod and θ is angle  with the rod at which putty strikes

= .011 x 9 x .35 / 3 x sin 29

= .0056

Applying conservation of angular momentum

angular momentum of putty = angular momentum of system after of collision

.0056 =  .00644 ω where ω is angular velocity of the rod after collision

ω = .87 rad /s .

Rotational energy

= 1/2 I ω²

I is total moment of inertia

=  .5 x .00644 x .87²

= 2.44 x 10⁻³ J .

Answer:

a. the initial velocity of the cat is 5.95 m/s at 60.2° from the horizontal

b. 0.847 m

Explanation:

a. Using v² = u² + 2as, we find the initial vertical velocity of the cat. Now at the peak height, v = final velocity = 0, u = initial velocity and a = -g = 9.8 m/s², s vertical distance travelled by the cat from its position on the cabinet = Δy = 3.130 m - 1.765 m = 1.365 m.

Substituting these variables into the equation, we have

0² = u² + 2(-9.8m/s²) × 1.365 m

-u² = -26.754 m²/s²

u = √26.754 m²/s²

u = 5.17 m/s

To find its initial horizontal velocity, u₁ we first find the time t it takes to reach the peak height from

v = u + at. where the variables mean the same as above.

substituting the values, we have

0 = 5.17 m/s +(-9.8m/s²)t

-5.17 m/s = -9.8m/s²t

t = -5.17 m/s ÷ (-9.8m/s²)

= 0.53 s

Now, the horizontal distance d = u₁t = 1.560 m

u₁ = d/t = 1.560 m/0.53 s = 2.96 m/s

So, the initial velocity of the cat is V = √(u² + u₁²)

= √((5.17 m/s)² + (2.96 m/s)²)

= √(26.729(m/s)² + 8.762(m/s)²)

= √(35.491 (m/s)²)

= 5.95 m/s

its direction θ = tan⁻¹(5.17 m/s ÷ 2.96 m/s) = 60.2°

So, the initial velocity of the cat is 5.95 m/s at 60.2° from the horizontal

(b)

First, we find the time t' it takes the cat to land on the bed from d' = u₁t'

where d' = horizontal distance of cabinet from bed = 3.582 m

u₁ = horizontal velocity = 2.96 m/s

t' = d'/u₁

= 3.582 m/2.96 m/s

= 1.21 s

The vertical between the bed and cabinet which is the vertical distance moved by the cat is gotten from Δy = ut' +1/2at'²

substituting u = initial vertical velocity = 5.17 m/s, t' = 1.21 s and a = -g = -9.8 m/s² into Δy, we have  

Δy = ut' +1/2at'² = 5.17 m/s × 1.21 s +1/2(- 9.8 m/s²) × (1.21 s)² = 6.256 - 7.174 = -0.918 m

Δy = y₂ - y₁

Since our initial position is the position of the cabinet above the ground = y₁ = 1.765 m

y₂ = position of bed above ground.

Δy = y₂ - y₁ = -0.918 m

y₂ - 1.765 m = -0.918 m

y₂ = 1.765 m - 0.918 m

= 0.847 m

Answer:

Everytime you do an experiment you need something that is regular. For example if you try and measure how much germs spread in bread. you need 1 bread thats clean and 3 different breads for different molds. So thats called a CONTROL

AAAAAAAAAAAA is the answer

Answer:

The  normal force N1 exerted by the floor is  [tex]N_1 = 951 \ N[/tex]

The  normal force N2 exerted by the wall is  [tex]N_2= 616.43 \ N[/tex]

The frictional force exerted by the wall is  [tex]f = N_2 = 616.43 \ N[/tex]  

Explanation:

From the question we are told that

    The length of the ladder is  [tex]L = 4.6 \ m[/tex]

    The weight of the ladder  is

    The distance of the ladder position on the wall from the floor is  [tex]D = 3.75 \ m[/tex]

     The mass of the person is  [tex]m = 90 kg[/tex]

Applying Pythagoras theorem

The length of the position the ladder on the ground from the base of the wall is

    [tex]A = \sqrt{L^ 2 - D^2}[/tex]

substituting values

    [tex]A = \sqrt{(4.6^2)-(3.75^2)}[/tex]

    [tex]A = 2.66 \ m[/tex]

  In order the for the ladder not to shift from the ground the sum of the moment about the position of the ladder on the ground must be equal to zero this is mathematically represented as

        [tex]\sum M = 0 = N_2 * D - [\frac{1}{2} * W_L ] * [(mg) *A ][/tex]

         [tex]\sum M = 0 = N_2 * 3.75 - [\frac{1}{2} * 69.0 ] * [(90*9.8) * \frac{4.6}{2.66} ][/tex]

        [tex]N_2 * 3.75 =2311.62[/tex]

        [tex]N_2 * 3.75 =2311.62[/tex]

        [tex]N_2= 616.43 \ N[/tex]

Now the force exerted by the floor on the ladder is mathematically represented as

           [tex]N_1 = W_L + (m * g )[/tex]

substituting values

          [tex]N_1 = 951 \ N[/tex]

Now the horizontal forces acting on the ladder are [tex]N_2 \ and \ f[/tex] and they are in opposite direction so

     [tex]f = N_2 = 616.43 \ N[/tex]  

Answer:

1. 25 Km

2. zero

3. 27.7 m/s

Explanation:

Data provided in the question:

Circumference of the track = 12.5 km

Speed of the car = 100 Km/h

Time for which car travels = 15 minutes = [tex]\frac {15}{60}[/tex] hr

Now,

1. Distance traveled = Speed × Time

= 100 × [tex]\frac{15}{60}[/tex]

= 25 Km

2. The distance traveled is 2 times the circumference of the track (i.e 2 × 12.5 = 25 Km)

Which means that the car is again at the initial position

Therefore, The displacement is zero.

3. Speed of car in Km/hr = 100 Km/h

now,

1 Km = 1000 m

1 hr = 3600 seconds

therefore,

100 Km/h = [tex]100\times\frac{1000}{3600}[/tex] m/s

= 27.7 m/s

Hence, the speed of car in m/s = 27.7

Answer:

Coefficient of kinetic friction = 0.235

Explanation:

Given:

Mass of crate = 330 kg

1st force = 430 N

2nd force = 330 N

Find:

Coefficient of kinetic friction.

Computation:

We know that, velocity is constant.

So, acceleration (a) = 0

So, net force (f) = 430 N + 330 N

Net force (f) = 760 N

F = μmg

μ = f / mg                                   [∵ g = 9.8]

μ = 760 / [330 × 9.8]

μ = 760 / [3,234]

μ = 0.235

Coefficient of kinetic friction = 0.235

Answer:

(A) object B has the largest mass because it has the least acceleration

(B) object C has the smallest mass because it has the largest acceleration

(C) mass A : mass B = 3 : 5

Explanation:

Given;

acceleration of object A = 5 m/s²

acceleration of object B = 3 m/s²

acceleration of object C = 7 m/s²

A constant force, F

According to Newton's second law of motion;

F = ma

m = F / a

Mass of object A:

m = F / 5

Mass of object B:

m = F / 3

Mass of object  C:

m = F / 7

(A). Which object has the largest mass:

object B has the largest mass because it has the least acceleration

(B). Which object has the smallest mass:

object C has the smallest mass because it has the largest acceleration

(C). What is the ratio of mass A to mass B;

mass A = F / 5

mass B = F / 3

[tex]mass \ A : \ mass \ B = \frac{F}{5} : \frac{F}{3} \\\\\frac{mass \ A}{mass \ B} = \frac{F}{5} * \frac{3}{F}= \frac{3}{5} \\\\mass \ A : \ mass \ B = 3: 5[/tex]

A. The Object B has largest mass.

B. The Object A has smallest mass.

C. The ratio of mass A to mass B is, [tex]\frac{3}{5}[/tex]

The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object.

                   [tex]F=ma\\\\m=\frac{F}{a}[/tex]

                     [tex]\frac{m_{A}}{m_{B}} =\frac{a_{B}}{a_{A}} =\frac{3}{5}[/tex]

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Answer:

d = 41.91 m

Explanation:

In order to calculate the distance traveled by the train while superman passes it, you write the equations of motion for both superman and train:

For train, you have a motion with constant speed. You write the equation of motion of the position of the front of the train:

[tex]x=x_o+v_1t[/tex]    (1)

xo: initial position of the front of the train = 500m

v1: speed of the train = 50km/h

For superman, you take into account that the motion is an accelerated motion (you assume superman is at the origin of coordinates):

[tex]x'=v_2t+\frac{1}{2}at^2[/tex]   (1)

v2: initial speed of superman = 100km/h

a: acceleration = 10m/s^2

When superman passes the train, both positions x and x' will be equal. Hence, you equal the equations (1) and (2) and you calculate the time t. But before you convert the units of the velocities v1 and v2 to m/s:

[tex]v_1=50\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=13.88\frac{m}{s}\\\\v_2=100\frac{km}{h}=\frac{1000m}{1km}*\frac{1h}{3600s}=27.77\frac{m}{s}[/tex]

Thus, you equal x=x'

[tex]x=x'\\\\x_o+v_1t=v_2t+\frac{1}{2}at^2\\\\500m+(13.88m/s)t=(27.77m/s)t+\frac{1}{2}(10m/s^2)t^2\\\\(50\frac{m}{s^2})t^2+(13.89\frac{m}{s})t-500m=0[/tex]

You solve the last equation for t by using the quadratic formula:

[tex]t_{1,2}=\frac{-13.89\pm \sqrt{(13.89)^2-4(50)(-500)}}{2(50)}\\\\t_{1,2}=\frac{-13.89\pm 316.53}{100}\\\\t_1=3.02s\\\\t_2=-3.30s[/tex]

You only use t1 = 3.02s because negative times do not have physical meaning.

Next, you replace this value of t in the equation (1) to calculate the position of the train (for when superman just passed it):

[tex]x=500m+(13.88m/s)(3.02s)=541.91m[/tex]

x is the position of the front of the train, then, the dstance traveled by the train is:

d = 541.91m - 500m = 41.91 m

Answer:

0

Explanation:

if the volume remains constans, the works is 0 because the equation

W = P . ∆V

P = pressure

∆V = change in volume

Answer:

The time when the ball strikes the ground is closest to  [tex]t_t = 9.4 \ s[/tex]

Explanation:

From the question we are told that

  The time of projection is t = 0.0 s

   The  distance of the point  from the ground  is  [tex]d = 90 \ m[/tex]

    The  initial velocity of the ball is  [tex]v _i = 36 .2 \ m/s[/tex]

generally the time required to reach maximum height is  

      [tex]t_r = \frac{g}{v}[/tex]

Where is the acceleration due to gravity  with value  [tex]g = 9.8 \ m/s^2[/tex]

Substituting values

        [tex]t_r = \frac{36.2}{9.8}[/tex]

        [tex]t_r = 3.69 s[/tex]

when returning the time and velocity at the roof level is  t =  3.69 s and  u = 36.2 m/s this due to the fact that  air resistance is negligible

   The final velocity at which it  hit the ground is

      [tex]v_f^2 = u^2 + 2ag[/tex]

So  

    [tex]v_f = \sqrt{ u^2 + 2gs}[/tex]

substituting values

    [tex]v_f = \sqrt{ 3.69^2 + 2* 9.8 * 90}[/tex]

     [tex]v_f = 55.45 \ m/s[/tex]

The time taken for the ball to move from the roof level to the ground is  

     [tex]t_g = \frac{v-u}{a}[/tex]

substituting values

    [tex]t_g = \frac{55.45 -36.2}{9.8}[/tex]

     [tex]t_g = 1.96 \ s[/tex]

The total time for this travel is  

    [tex]t_t = t_g + 2 t_r[/tex]

     [tex]t_t = 1.96 + 2(3.69)[/tex]

      [tex]t_t = 9.4 \ s[/tex]

The question is incomplete. Here is the complete question.

Calculate the maximum deceleration  of a car that is heading down a 14° slope (one that makes an anlge of 14° with the horizontal) under the following road conditions. You may assum that the weight of the car is evenlydistributed on all four tires and that the sttic coefficient of friction is involved - that is, the tires are not allowed to slip during the deceleration. (Ignore rolling) Calculate for a car: (a) On a dry concrete. (b) On a wet concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice.

Answer: (a) a = - 11.05 m/s²; (b) a = - 10.64 m/s²; (c) a = - 9.84m/s²

Explanation: The image in the attachment describe the forces acting on the car. Observing that, we know that:

[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]

The [tex]W_x[/tex] is a x-component of force due to gravity (W) and, in this case, is given by: [tex]W_x[/tex] = W.sin(14)

W is described as: W = m.g

Force due to friction ([tex]f_s[/tex]) is given by: [tex]f_s[/tex] = μs.N

N is the normal force and, in the system, is equivalent of [tex]W_y[/tex], so:

[tex]W_y[/tex] = m.g.cos(14)

Therefore, the formula will be:

[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]

m.a = - (m.g.sin14) - (μs.mg.cos14)

a = - g (sin14 + μscos 14)

a) For dry concrete, μs = 1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 1.cos14)

a = - 11.05 m/s²

b) For wet concrete, μs = 0.7:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin 14 + 0.7.cos14)

a = - 10.64 m/s²

c) For ice, μs = 0.1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 0.1cos14)

a = - 9.84 m/s²

Answer:

false

Explanation:

Answer:

Explanation:

Givens

vi = 20 m/s

vf = 28 m/s

t = 2 seconds

Formula

a = (vf - vi) / t

Solution

a = (28 - 20)/2

a = 8/2

a = 4 m/s^2

Answer:

A , D , E

Explanation:

Solution:-

- Consider the two identical objects with mass ( m ).

- The stiffness of the springs are ( k1 and k2 ).

- Both the spring store 55.0 J of potential energy.

- We will apply the principle of energy conservation on both the systems. In both cases the spring stores 55.0 Joules of energy. Once released, the objects gain kinetic energy with a consequent loss of potential energy in either spring.

- The maximum speed ( v ) is attained when all the potential energy is converted to kinetic energy.

- Apply Energy conservation for spring with stiffness ( k1 ).

                         ΔU = ΔEk

                         55.0 = 0.5*m*v^2

                         v = √ ( 110 / m )

- Apply Energy conservation for spring with stiffness ( k2 ).

                         ΔU = ΔEk

                         55.0 = 0.5*m*v^2

                         v = √ ( 110 / m )

Answer: Both objects will have the same maximum speed ( A )

- We are told that one spring is more stiff as compared to the other one. The measure of stiffness is proportionally quantified by the spring constant. To mathematically express we can write it as:

                         k1 > k2

Where,

                 k1: The stiff spring

                 k2: The flexible spring

Answer: The stiff spring has a larger spring constant than the flexible spring. ( D )

- We will assume that the spring with constant ( k1 ) undergoes a displacement ( x1 ) and the spring with constant ( k2 ) undergoes a displacement ( x2 ). The potential energy stored in both spring is the same. Hence,

                      U1 = U2

                      0.5*( k1 ) * ( x1 )^2 = 0.5*( k2 ) * ( x2 )^2

                      [ k1 / k2 ] = [ x2 / x1 ]^2

Since,

                     k1 > k2 , then [ k1 / k2 ] > 1    

Then,

                     [ x2 / x1 ]^2 > 1

                     [ x2 / x1 ] > 1

                     x2 > x1                  

Answer: The flexible spring ( x2 ) was compressed more than the stiff spring ( x1 ). ( E )

Answer:

Increased blood circulation to the body.

Explanation:

plato/edmentum

Answer:

Explanation:

Time taken to accelerate to 28 m /s

= 28 / 2 = 14 s

a ) Total length of time in motion

= 14 + 41 + 5

= 60 s .

b )

Distance covered while accelerating

s = ut + 1/2 at²

= 0 + .5 x 2 x 14²

= 196 m .

Distance covered while moving in uniform motion

= 28 x 41

= 1148 m

distance covered while decelerating

v = u - at

0 = 28 - a x 5

a = 5.6 m / s²

v² = u² - 2 a s

0 = 28² - 2 x 5.6 x s

s = 28² / 2 x 5.6

= 70 m .

Total distance covered

= 196 + 1148 + 70

= 1414 m

total time taken = 60 s

average velocity

= 1414 / 60

= 23.56 m /s .

Answer:

To determine wavelength of a wave on a string.

Explanation:

The Standing Waves lab study the parameters that affect standing waves in various strings. The effects of string tension and density on wavelength and frequency will be studied.

Answer:

107 N, option d

Explanation:

Given that

mass of the ball, m = 0.2 kg

initial velocity of the ball, u = 20 m/s

final velocity of the ball, v = -12 m/s

time taken, Δt = 60 ms

Solving this question makes us remember "Impulse Theorem"

It states that, "that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object"

Mathematically, it is represented as

FΔt = m(v - u), where

F = the average force

Δt = time taken

m = mass of the ball

v = final velocity of the ball

u = initial velocity of the ball

From the question we were given, if we substitute the values in it, we have

F = ?

Δt = 60 ms = 0.06s

m = 0.2 kg

v = -12 m/s

u = 20 m/s

F = 0.2(-12 - 20) / 0.06

F = (0.2 * -32) / 0.06

F = -6.4 / 0.06

F = -106.7 N

Thus, the magnitude is 107 N

Answer:

[tex]\Delta p=1.3475\ kg-m/s[/tex]

Explanation:

The computation of magnitude of the change in the ball's momentum in kg · m/s is shown below:-

We represent

The ball mass =  m = 275 g = 0.275 kg

Thus it goes to the floor and resurfaces upward.

The ball hits the ground at 3.30 m/s speed that is

u = -3.30 m/s which represents the Negative since the ball hits the ground)

It rebounds at a speed of 1.60 m / s i.e. v = 1.60 m/s (positive as the ball rebounds upstream)

[tex]\Delta p=p_f-p_i[/tex]

[tex]\Delta p=m(v-u)[/tex]

[tex]\Delta p=0.275\ kg(1.60\ m/s-(-3.30\ m/s))[/tex]

[tex]\Delta p=1.3475\ kg-m/s[/tex]

Answer:

D. Velocity of the car

Explanation:

The centripetal force acting on the car is given by the following formula:

[tex]F_c=ma_c=m\frac{v^2}{r}[/tex]    (1)

m: mass of the car  = 888 kg

v: tangential speed of the car = 7 m/s

r: radius of the flat circular track = 59 m

By the form of the equation (1) you can notice that the greatest change in the centripetal force is obtained when the velocity v is twice. In fact, you have:

[tex]F_c=m\frac{(2v)^2}{r}=4m\frac{v^2}{r}=4F_c[/tex]

Then, the greatest values of the centripetal force is:

[tex]F_c=4(888kg)\frac{(7m/s)^2}{59m}=2949.96N[/tex]

The greatest change in Fc is obtained by changing the value of the speed

answer

D. Velocity of the car

Answer:7.5V

Explanation:

Ohm's law, V=IR

so, V=0.3×25

V=7.5V

Answer:

7.5 V

Explanation:

Answer:

6,650 newtons

Explanation:

The computation of the force exerted on it by static friction is shown below:

Data provided in the question

Mass of car = m = 1,900 kg

speed = v = 14 m/s

radius = r = 56 m

Let us assume friction force be f

And, the Coefficient of friction = [tex]\mu[/tex]= 0.88

As we know that

[tex]f = \frac{mv^2}{r}[/tex]

[tex]= \frac{1,900 \times 14^2}{56}[/tex]

= 6,650 newtons

We simply applied the above formula so that the force exerted could come

Answer:

They had the same speed.

Explanation:

It won't be velocity, because velocity is a vector quantity. Speed is scalar.

Velocity is the rate of change of displacement. During their periods of travel, the cars definitely had the same velocity.

Velocity is the directional speed of a moving object as an indicator of its rate of change in location as perceived from a certain frame of reference and measured by a specific time standard.

Given that the first car travels 40 meters due east in 5 seconds. Therefore, we can write,

Distance = 40 meters

Time = 5 seconds

Velocity = Distance / Time = 40 meter/ 5 sec = 40 m/sec

Also, given that the second car travels 64 meters due west in 8 seconds. Therefore, we can write,

Distance = 64 meters

Time = 8 seconds

Velocity = Distance / Time = 64 meter/ 8 sec = 8 m/sec

Hence, During their periods of travel, the cars definitely had the same velocity.

Learn more about Velocity here:

https://brainly.com/question/18084516

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