It takes an airplane nearly 3/4 of a mile to stop. Which law is being used?
A) Newton's first law
B) Newton's second law
C) Newton's third law
D) Law of conservation of momentum
E) Law of universal gravitation

Answers

Answer 1

Answer:

I think it is D

Explanation:

might be wrong


Related Questions

A 3.05 kg block is sliding across level ground and feels a 5.27 N friction force. What is the coefficient of friction?

Answers

Answer:

0.176

Explanation:

(5.27)/(3.05*9.8)

It is indeed 0.176 for the answer.

Find μsμsmu_s, the coefficient of static friction between the rod and the rails. Express the coefficient of static friction in terms of variables given in the introduction.

Answers

Answer:

The question is incomplete. However the complete question is :

A Rail Gun uses electromagnetic forces to accelerate a projectile to very high velocities. The basic mechanism of acceleration is relatively simple and can be illustrated in the following example. A metal rod of mass m and electrical resistance R rests on parallel horizontal rails (that have negligible electric resistance), which are a distance L apart. The rails are also connected to a voltage source V, so a current loop is formed.

The vertical magnetic field, initially zero, is slowly increased. When the field strength reaches the value [tex]$B_0$[/tex], the rod, which was initially at rest, begins to move. Assume that the rod has a slightly flattened bottom so that it slides instead of rolling. Use g for the magnitude of the acceleration due to gravity.

Find μs, the coefficient of static friction between the rod and the rails.

Express the coefficient of static friction in terms of variables given in the introduction.

So the answer is : [tex]$\mu_s=\frac{\Delta V LB_0}{Rmg}$[/tex]

Explanation:

It is given : B = [tex]$B_0$[/tex]

Therefore, using force on the current carrying wire is:

[tex]$\mu_s mg = B_0IL$[/tex]

[tex]$\mu_s= \frac{ILB_0}{mg}$[/tex]

Therefore,

[tex]$\mu_s=\frac{\Delta V LB_0}{Rmg}$[/tex]

The coefficient of static friction in terms of variables will be [tex]\rm \mu_s =\frac{ \triangle VLB_0}{Rmg} \\\\[/tex].

What is the cofficient of static friction?

The ratio of the greatest static friction force (F) between the surfaces in contact before movement begins to the normal (N) force is the coefficient of static friction.

The given data in the problem is;

[tex]\rm B=B_0[/tex]

The force on a current-carrying wire is balanced by the friction force.

The force on a current-carrying wire is given by;

[tex]\rm \mu_s mg= B_0IL \\\\ \rm \mu_s=\frac{ILB_0}{mg}\\\\ \rm \mu_s= \frac{\triangle VLB_0}{Rmg}[/tex]

Hence the coefficient of static friction in terms of variables will be [tex]\rm \mu_s =\frac{ \triangle VLB_0}{Rmg} \\\\[/tex].

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A half-eaten apple turns brown which of these is evidence a chemical change took place A.th3e apple is softer b.the apple changed color C.half the apple was eaten D.the apple is warmer \

Answers

Answer:

b the apple changed color

A block with mass 0.50 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.20 m . When released, the block moves on a horizontal tabletop for 1.00 m before coming to rest. The spring constant k is 100 N/m. What is the coefficient of kinetic friction between the block and the tabletop

Answers

Answer:

μ = 0.41

Explanation:

[tex]\frac{1}{2} kx^2 = \mu mgd[/tex]

[tex]\mu = (\frac{1}{2} kx^2)/mgd[/tex]

[tex]\mu = (\frac{1}{2} (100)(0.2)^2)/(0.5)(9.81)(1)[/tex]

[tex]\mu = 0.41[/tex]

The coefficient of kinetic friction between the block and the tabletop is 0.41.

The given parameters;

mass of the block, m = 0.5 kgextension of the spring, x = 0.2 mspring constant, K = 100 N/m

The coefficient of kinetic friction between the block and the tabletop by applying the principle of conservation of energy as shown below.

Fd = Uₓ

μmgd = ¹/₂kx²

where;

μ is the coefficient of kinetic friction

The coefficient of kinetic friction between the block and the tabletop;

[tex]\mu_k = \frac{kx^2}{mgd} \\\\\mu_k = \frac{100\times 0.2^2}{2\times 0.5\times9.8 \times 1 } \\\\\mu_k = 0.41[/tex]

Thus, the coefficient of kinetic friction between the block and the tabletop is 0.41.

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a mass of .4 kg is raised by a vertical distance of .450 m in the earth's gravitational field. what is the change in its gravitational potential energy

Answers

Answer:

E = 1.76 J

Explanation:

Given that,

Mass of an object, m = 0.4 kg

It moves by a vertical distance of 0.45 m in the Earth's gravitational field.

We need to find the change in its gravitational potential energy. It can be given by the formula as follow :

[tex]E=mgh\\\\E=0.4\times 9.8\times 0.45\\\\E=1.76\ J[/tex]

So, the change in its gravitational potential energy is 1.76 J.

what vehicle are runaway ramps designed for​

Answers

Answer:truck

Explanation:

truck

A truck is traveling at an initial velocity
of +39m/s and it starts slowing down smoothly for 3.3s. It covers a distance of 45m while slowing down. What is the acceleration?​

Answers

Answer:

Approximately [tex]-15.37m/s^2[/tex]

Explanation:

The acceleration can be found using the formula:

[tex]x_{f} = x_{i} + v_{i} (t)+\frac{1}{2} (a)(t^{2} )[/tex]

The work is as shown:

[tex]45m = 0 + 39 m/s (3.3s)+\frac{1}{2} a(3.3s)^2[/tex]

[tex]45m=128.7m + 5.445s^2 a[/tex]

[tex]-83.7m=5.445s^2a[/tex]

[tex]a = -15.37190083m/s^2[/tex]

a is about -15.37[tex]m/s^2[/tex]

Two identical objects of the same size and shape are dropped. One is dropped from a table, while the other is dropped from the top of a tall building. Which will hit the ground with more impact?

Answers

the one from the tall building will have more impact because it is going at a faster pace towards the ground

Answer:

These two objects hit the ground at the same time. But their impact velocity will be different

because they are dropped from different heights.

The one that is dropped from the building will have more impact when it hits the ground because gravity causes this onject to fall toward the ground at a faster and faster velocity the longer the object falls. In fact, its velocity increases by 9.8 m/s2.

Explanation:

Initial velocity of the object u=0

Final velocity of the object when it reaches the ground v=A m/s

Using v^2−u^2=2gS where g=10m/s^2

A^2 −0^2=2(10)H

A^2=20*H H- heigh from the table

Object falling from the building will have the same initial velocity but different heigh.

T>H

so

B^2=20*T

20*T>20*H

so B^2>A^2

it equal to B>A so object falling from the building will have more impact than falling from the table.

A scout hikes 2.35KM east of a campsite he takes a break for lunch and then hikes another 1.25 KM north of the location where he ate lunch. what distance is the scout from the campsite

Answers

Answer:

mk7000000.0000000000

A segment of wire of total length 3 m carries a 10 A current and is formed into a semicircle. Determine the magnitude of the magnetic field (in micro-tesla) at the center of the circle along which the wire is placed.

Answers

Answer:

[tex]B=3.29\ \mu T[/tex]

Explanation:

Given that,

Length of a wire is 3 m

Current in the wore, I = 10 A

The wire formed a semicircle.

We need to find the magnitude of the magnetic field (in micro-tesla) at the center of the circle along which the wire is placed.

The magnetic field at the center of semicircle is given by :

[tex]B=\dfrac{\mu_o I}{4R}[/tex]

R is radius of semicircle,

[tex]R=\dfrac{3}{\pi}=0.954\ m[/tex]

So,

[tex]B=\dfrac{4\pi \times 10^{-7}\times 10I}{4(0.954)}\\\\B=3.29\times 10^{-6}\ T\\\\B=3.29\ \mu T[/tex]

So, the magnitude of the magnetic field is [tex]3.29\ \mu T[/tex].

You are given two vectors A⃗ =−3.00i^+7.00j^ and B⃗ =5.00i^+2.00j^. Let counterclockwise angles be positive.
1- What angle θA, where 0∘≤θA<360∘, does A⃗ make with the +x-axis?
2-What angle θB, where 0∘≤θB<360∘, does B⃗ make with the +x-axis?
Express your answer in degrees.
3-Vector C⃗ is the sum of A⃗ and B⃗ , so C⃗ =A⃗ +B⃗ . What angle θC, where 0∘≤θC<360∘, does C⃗ make with the +x-axis?
Express your answer in degrees

Answers

Answer:

Ax = -3       By= 7         tan phi =  -7 / 3    phi = -66.8

Since theta is in the second quadrant   theta = 180 - 66.8 = 113.2 deg

Bx = 5        By = 2        tan theta = .4        theta = 21.8 deg

Cx = Ax + Bx = 2       Cy = Ay + By = 9

tan theta = 9 / 2 = 4.5      theta = 77.5 deg

1) The solution is [tex]\theta_{1} \approx 113.141^{\circ}[/tex].

2) The solution is [tex]\theta_{1} \approx 21.874^{\circ}[/tex].

3) The solution is [tex]\theta_{1} \approx 77.467^{\circ}[/tex].

1) In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:

[tex]\cos \theta = \frac{\vec A\,\bullet \, \vec u}{\|\vec A\|\cdot \|\vec u\|}[/tex] (1)

Where [tex]\|\vec A\|[/tex] and [tex]\|\vec u\|[/tex] are the norms of [tex]\vec A[/tex] and [tex]\vec u[/tex].

If we know that [tex]\vec A = (-3, 7)[/tex] and [tex]\vec u = (1, 0)[/tex], then the angle that [tex]\vec A[/tex] make with the x-axis is:

[tex]\|\vec A\| = \sqrt{(-3)^{2}+7^{2}}[/tex]

[tex]\|\vec A\| = \sqrt{58}[/tex]

[tex]\|\vec u\| = 1[/tex]

[tex]\cos \theta = \frac{(-3)\cdot (1)}{\sqrt{58} \cdot 1}[/tex]

[tex]\cos \theta \approx -0.393[/tex]

The solution is [tex]\theta_{1} \approx 113.141^{\circ}[/tex].

2) In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:

[tex]\cos \theta = \frac{\vec B\,\bullet \, \vec u}{\|\vec B\|\cdot \|\vec u\|}[/tex] (1)

Where [tex]\|\vec B\|[/tex] and [tex]\|\vec u\|[/tex] are the norms of [tex]\vec B[/tex] and [tex]\vec u[/tex].

If we know that [tex]\vec B = (5, 2)[/tex] and [tex]\vec u = (1, 0)[/tex], then the angle that [tex]\vec A[/tex] make with the x-axis is:

[tex]\|\vec B\| = \sqrt{5^{2}+2^{2}}[/tex]

[tex]\|\vec B\| = \sqrt{29}[/tex]

[tex]\|\vec u\| = 1[/tex]

[tex]\cos \theta = \frac{(5)\cdot (1)}{\sqrt{29} \cdot 1}[/tex]

[tex]\cos \theta \approx 0.928[/tex]

The solution is [tex]\theta_{1} \approx 21.874^{\circ}[/tex].

3) In this case, we need to find the value of [tex]\vec C[/tex] by vectorial sum before calculating the angles with respect to x-axis:

[tex]\vec C = \vec A + \vec B = (-3, 7) + (5, 2)[/tex]

[tex]\vec C = (2, 9)[/tex]

In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:

[tex]\cos \theta = \frac{\vec C\,\bullet \, \vec u}{\|\vec C\|\cdot \|\vec u\|}[/tex] (1)

Where [tex]\|\vec C\|[/tex] and [tex]\|\vec u\|[/tex] are the norms of [tex]\vec C[/tex] and [tex]\vec u[/tex].

If we know that [tex]\vec B = (2, 9)[/tex] and [tex]\vec u = (1, 0)[/tex], then the angle that [tex]\vec A[/tex] make with the x-axis is:

[tex]\|\vec C\| = \sqrt{2^{2}+9^{2}}[/tex]

[tex]\|\vec C\| = \sqrt{85}[/tex]

[tex]\|\vec u\| = 1[/tex]

[tex]\cos \theta = \frac{(2)\cdot (1)}{\sqrt{85} \cdot 1}[/tex]

[tex]\cos \theta \approx 0.217[/tex]

The solution is [tex]\theta_{1} \approx 77.467^{\circ}[/tex].

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Calculate the peak voltage (in Volts) of a generator that rotates its 200-turn, 0.100 m diameter coil at 3.5 x 103 rpm in a 0.7 T field. You should round your answer to the nearest integer, do not include unit.

Answers

Angular velocity of the coil :

[tex]\omega=\dfrac{2\pi n}{60}[/tex]

[tex]\omega=\dfrac{2\times \pi\times 3500}{60}\\\\\omega =366.52\ rad/s[/tex]

Now,

Peak voltage is given by :

[tex]V=NAB\omega\\\\V= 200\times \dfrac{\pi (0.1)^2}{4}\times 0.7\times 366.52\\\\V=403.01\ V[/tex]

Hence, this is the required solution.

Two points are on a disk that rotates about an axis perpendicular to the plane of the disk at its center. Point B is 3 times as far from the axis as point A. If the linear speed of point B is V, then the linear speed of point A is:_______

a. 9V.
b. 3V.
c. V.
d. V/3.
e. V/9.

Answers

Answer:

D. V/3.

Explanation:

V = rw

At point a

Linear velocity

= Va = rw

At point b

We have,

V = 3rw

So linear velocity of a

= V/Va = 3rw/re

I cross multiples and after which I got Va = v/3

Please view attachment

Answer:

The linear speed of point A is [tex]\frac{V}{3}[/tex] (Option D)

Explanation:

The relation between linear and angular velocity,

[tex]V = rw[/tex]

Now, for point A the linear velocity is

[tex]V_A = rw[/tex]

And for point B

[tex]V = 3rw[/tex]

Therefore, the linear velocity of the point A

[tex]\frac{V}{V_A} = \frac{3rw}{rw}\\\\V_A = \frac{V}{3}[/tex]

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A mass that has a force of the 10N to the right, 7N up, 3N down, and 4N to the left. What is the net external force on that mass? If the mass is 2kg find the acceleration.

Answers

Net force = square root (6^2 + 4^2) = sq root (36+16) = 7.2 N
acceleration = 7.2 / 2 = 3.6 m/s^2

I HAVE A QUESTION ABOVE CAN ANYONE PLEASE HELP ME I WOULD REALLY APPRECIATE THE HELP I HAVE BEEN ON THIS QUESTION FOREVER i will give you the brainlist !!!!!

Answers

Answer:

answer is C

Explanation:

Answer: Monetary policy used during times of recession

Explanation:

What is the minimum compression of the spring necessary for the block to make it to the top of the hill on the other side?

Answers

Answer:

sciencephysicsphysics questions and answersA 12-kg Block Is Pressed Against A Spring (spring Constant 870 N/m ), Compressing It Some Distance. ...

Question: A 12-kg Block Is Pressed Against A Spring (spring Constant 870 N/m ), Compressing It Some Distance. The Block Is Released From Rest And Slides Across A Track As Shown In (Figure 1). While Most Of The Track Is Frictionless, There Is A 55-cm Section Of Track That Has A Coefficient Of Friction With The Block Of 0.4. A Bit Further On, The Track Ascends ...

A 12-kg block is pressed against a spring (spring constant 870 N/m ), compressing it some distance. The block is released from rest and slides across a track as shown in (Figure 1). While most of the track is frictionless, there is a 55-cm section of track that has a coefficient of friction with the block of 0.4. A bit further on, the track ascends into a hill that is 40-cm tall.

What is the minimum compression of the spring necessary for the block to slide past the section with friction?

What is the minimum compression of the spring necessary for the block to make it to the top of the hill on the other side? 40cm 12kg Ww.w.wwwww 55cm

Explanation:

What would be the luminosity of the Sun if its surface temperature were 4000 K and its radius were 2.0 AU ?

Answers

Answer:

0.04558

Explanation:

Luminosity is defined as the total amount of energy a star can produce in a second. Luminosity is given by the formula

σ * A * T⁴

where;

σ = Stefan-Boltzmann constant

= 5.670367 * 10⁻⁸

A= Surface Area

=4πr²

T= Temperature in Kelvin

In the question,

Temperature= 4000K

Radius= 2.0 A.U

Substituting the variables into the equation,

0.000000056704 * 4(3.14)(2.0²) * 4000⁴

= 0.000000056704 * 50.24 * 16000

= 0.04558

All living creatures have some form of sleep that they engage in at regular periods, making sleep a(n) __________ function. A. environmental B. biological C. species-specific D. psychological


The answer is B. biological

Answers

Answer:

ans: a biological function

Answer:

B

Explanation:

EDGE2020

a force of 193 pounds makes an angle of 86.27 with a second force. The resultant of the two forces makes an angle of 31.4 to the first force. Find the magnitudes of the second force and of the resultant.

Answers

Answer:

Explanation:

Let force P = 193 pounds

second force = Q

Angle that resultant makes with force P is θ

Tanθ = Q sin86.27 / (P + Q cos86.27 )

Tan 31.4 = Q sin86.27 / (193 + Q cos86.27 )

.61 = .99 Q / (193 + .065Q )

117.73 + .04 Q = .99Q

117.73 = .95 Q

Q = 123.93  pounds .

What is the average velocity of the cart? Is the velocity increasing decreasing or constant as the cart moves along the air track?

Answers

Answer:

It is increasing .60 m/s

Explanation:

In an old-fashioned television tube, an electron () starting from rest experiences a force of 4.0 × 10–15 N over a distance of 50 cm. Ignoring the relativistic effects, the final speed of the electron is:

Answers

Answer:

The final speed of the electron is 6.626 x 10⁷ m/s

Explanation:

force applied to the electron, f = 4 x 10⁻¹⁵ N

distance traveled by the electron, d = 50 cm = 0.5 m

The work done on the electron;

W = F x d

W = (4 x 10⁻¹⁵ N) x (0.5 m)

W = 2 x 10⁻¹⁵ J

The kinetic energy of the electron is given by;

[tex]K.E = \frac{1}{2}m_ev^2[/tex]

Apply work - energy theorem;

K.E = W

[tex]\frac{1}{2}m_ev^2 = 2 *10^{-15}\\\\ v^2 = \frac{2( 2 *10^{-15})}{m_e}\\\\v= \sqrt{\frac{2( 2 *10^{-15})}{m_e}}\\\\ v = \sqrt{\frac{2( 2 *10^{-15})}{9.11*10^{-31}}}\\\\v = 6.626*10^7 \ m/s[/tex]

Therefore, the final speed of the electron is 6.626 x 10⁷ m/s

Based on these photographs, how do Mars's surface and atmosphere differ from Earth's?

Answers

Answer:

Explanation:

Mars does not appear to have any water, while Earth is mostly water. Mars also has yellow-brown sky (pink-brown at sunrise and sunset), which suggests that it has a different atmospheric makeup.

Answer:

Mars does not appear to have any water, while Earth is mostly water. Mars also has yellow-brown sky (pink-brown at sunrise and sunset), which suggests that it has a different atmospheric makeup.

Explanation:

What is the mass of an object if a force of 17N causes it to accelerate at 1.5 m/sec/sec
A. 11.38
B. 11.3 kg
C. 12.3 kg
D. 14.3 kg

Answers

Answer:

B.11.3 kg

Explanation :

F=ma here force is given as 17N and acceleration as 1.5 m/sec/sec using the formula mass can be calculated

∴ F=ma

17N =m × 1.5 m[tex]s^{-2}[/tex]

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

[tex]m= \frac{17 kg m/sec/sec}{1.5 m/sec/sec}[/tex]

∴ m=11.3 kg

The speed of light in vacuum is exactly 299,792,458 m/s. A beam of light has a wavelength of 621 nm in vacuum. This light propagates in a liquid whose index of refraction at this wavelength is 1.18.What is the speed of this light in the liquid

Answers

Answer:

2.54*10^8 m/s

Explanation:

Given that

Speed of light in space, c = 299792458 m/s

Wavelength of the beam of light, λ = 621 nm or 621*10^-9 m

Index of refraction of the liquid, n = 1.18

Speed of light in the liquid, v = ?

The index of refraction of an optical material, n is given by the following formula.

n = c/v, and if we substitute directly, we have

1.18 = 299792458 / v, if we make v subject of formula, we have

v = 299792458 / 1.18

v = 2.54*10^8 m/s

A 55kg skateboarder wants to just make it to the upper edge of a quarter pipe, a pipe that is one-quarter of a circle with a radius of 3.10m. What speed does he need at the bottom?

Answers

Answer:

Explanation:

A 52.0 kg skateboarder wants to just make it to the upper edge of a "quarter pipe," a track that is one-quarter of a circle with a radius of 2.60 m .

What speed does he need at the bottom

Let the lowest point of the circle = (0,0)

Center of circle = (0, 2.6)

The height of the circle = 2 * radius = 2 * 2.60 = 5.20 m

Highest point = (0, 5.2)

As the skateboarder moves around ¼ of a vertical circle, the skateboarder moves from the lowest position to a position that is ½ the way up to the highest position. This is the point that is 2.6 meters directly to the right of left of the center = (2.6, 2.6)

As the skateboarder has moved 2.6 m upward, the potential energy increase = m * g * ∆h = 52.0 * 9.8 * 2.6

During this same time, the kinetic energy decreased from the maximum to 0. The decrease of KE = the increase of PE

½ * m * ∆v^2 = m * g * ∆h

½ * 52 * ∆v^2 = 52.0 * 9.8 * 2.6

½ * ∆v^2 = 9.8 * 2.6

∆v = (2 * 9.8 * 2.6)^0.5 = 7.14 m/s

The velocity at highest point, (2.6,2.6) is 0 m/s

So the velocity at the lowest point must be 7.14 m/s

Let’s see if the skateboarder has enough KE to move upward 2.6 m.

KE at bottom = ½ * 52 * 7.14^2 = 1325.5

PE = 52 * 9.8 * h

52 * 9.8 * h = 1325.5

h = 2.6 m

The answer is OK

The speed needed by the skateboarder at the bottom is 7.8 m/s.

The given parameters;

mass of the skateboarder, m = 55 kgradius of the quarter pipe, r = 3.1 m

Apply the principle of conservation of mechanical energy to determine the sped of the skateboarder at the bottom;

[tex]P.E_{top} = K.E_{bottom}\\\\mgh = \frac{1}{2}mv^2\\\\gh = \frac{1}{2} v^2\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8 \times 3.1} \\\\v = 7.8 \ m/s[/tex]

Thus, the speed needed by the skateboarder at the bottom is 7.8 m/s.

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The gravity on the moon is 1/6 as strong as that of earth.An object on earth weighs 1 Newton another object on the moon also weighs 1 Newton.Which of the two objects has the greater mass

Answers

Answer:

The object on the moon has greater mass

Explanation:

Given;

weight of the object on Earth, W₁ = 1 N

wight of the object on Moon, W₂ = 1 N

gravity on Earth, g₁ = 9.8 m/s²

gravity on the Moon, g₂ = g₁/6 = 1.633 m/s²

Apply Newton's second law of motion;

Weight of the object on Earth is given by;

W₁ = m₁g₁

m₁ = W₁/g₁

m₁ = 1 / 9.8

m₁ = 0.102 kg

Weight of the object on Moon is given by;

W₂ = m₂g₂

m₂ = W₂ / g₂

m₂ = 1 / 1.633

m₂ = 0.612 kg

Therefore, the object on the moon has greater mass.

Additional Practice on Speed

1. A group of campers and one group leader left a campsite in a canoe.

They traveled at an average rate of 10 km/h. Two hours later, the other

group leader left the campsite in a motorboat. He traveled at an average

rate of 22 km/h. How long had each boat traveled when the motorboat

caught up with the canoe?

a.

canoe: 2.2 h; motorboat: 1 h

b. canoe: 2.2 h; motorboat: 10 h

c. O canoe: 3 2 h; motorboat: 12 h

3

3

d. O canoe: 4.4 h; motorboat: 2 h

Answers

Answer:

canoe = 3 h 40 min.  motorboat = 1 h 40 min

Explanation:

As both boats travel at a constant speed, we can apply the definition of average velocity to both, so we can get:

        [tex]v_{b} = \frac{\Delta x_{b} }{\Delta t_{b} } (1)[/tex]

       [tex]v_{mb} = \frac{\Delta x_{mb} }{\Delta t_{mb} } (2)[/tex]

Now, when the motorboat caught up with the canoe, distance travelled by both boats were the same, so, solving (1) and (2) for Δx, we get:

        [tex]v_{b} *\Delta t_{b} = v_{mb} *\Delta t_{mb} (3)[/tex]

if we take the initial time for the canoe to be t₀ =0, the initial time for the motorboat will be tmb₀ = 2 Hs.So, replacing in (3) we have:

        [tex]v_{b} *t_{b} = v_{mb} *(t_{b} - 2 hs.) = 10 km/h* t_{b} = 22 km/h* (t_{b} - 2 hs.)[/tex]

Solving for tb:

        [tex]t_{b} = \frac{-44 }{-12} hs = 3h 40 min.[/tex]

        tmb = tb - 2 hs. = 1h 40 min.

A dart is loaded into a compressed spring. Starting from rest the spring is released and shoots the dart up into the air. From the instant the dart is released (before it starts to move) until it reaches its maximum height, what is the Net work done on the dart

Answers

Answer:

Wnet = 0

Explanation:

The work-energy theorem says that the total net work done on an object, is equal to the change in the kinetic energy of that object.In this case, the total change in kinetic energy of the dart is zero, due to the initial status is at rest (loaded into a compressed spring) and the final is also at rest (when it reaches to its maximum height).The total work done on the dart can be written as follows:

        [tex]W_{net} = \frac{1}{2} k* \Delta x^{2} - m*g*h_{max} (1)[/tex]

Now, in absence of friction, all the elastic potential energy becomes gravitational potential energy, so the following is true:

        [tex]\frac{1}{2} k* \Delta x^{2} = m*g*h_{max} (2)[/tex]

From (1) and (2) we conclude that Wnet = 0.

A golf ball rolls up a hill toward a miniature-golf hole. Assume the direction toward the hole is positive. If the golf ball starts with a speed of 2.0 m/s and slows at a constant rate of 0.50 m/s2. What is the golf ball’s velocity if the constant acceleration continues for 6.0 s? *

Answers

Answer:

0.2533333334

Explanation:

It is simple, just do the math

Which is the right one?
(Show work)

Answers

The correct answer is the last one.
Air resistance will cause less movement both in the vertical and the horizontal direction.
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