A bug starts at point A, crawls 8.0 cm east, then 5.0 cm south, 3.0 west, and 4.0 cm north to point B.

Question:
How far north and east is B from A?

Find the displacement of the turtle from the origin at the point.​

Answer:

The first scenario!

Explanation:

W=F*d

a) 55*8= 440J

b) 60*6= 360J

c) 50*5= 250J

d) 40*10= 400J

The scenario with the greatest amount of work done on a wagon is "A force of 55 N moves it 8 m" and the Work done is 440N.m.

Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;

W = F × d

Where F is force applied and d is distance travelled.

From the question;

A force of 55 N moves it 8 m

W = 55N×8m = 440N.m

A force of 60 N moves it 6 m.

W = 60N×6m = 360N.m

A force of 50 N moves it 5 m.

W = 50N×5m = 250N

A force of 40 N moves it 10 m.

W = 40N×10m = 400N.m

Therefore, the scenario with the greatest amount of work done on a wagon is "A force of 55 N moves it 8 m" and the Work done is 440N.m.

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Answer:

a=0.5g ms^-2

Explanation:

Let tension be T

acceleration be a

g be gravitational acceleration 9.81ms^-2

They share same T and a

by force diagram the net force on m:

T-2.1g=2.1a

net force on M:

6.3g-T=6.3a

solve:

(T-2.1g)/(6.3g-T)=2.1a/6.3a

3T+T=12.6g

T=3.15g newton

3.15g-2.1g=2.1a

a=0.5g ms^-2

Answer:

pf = 198.8 kg*m/s

θ = 46.8º N of E.

Explanation:

       [tex]p_{ox} = p_{oAx} + p_{oBx} (1)[/tex]

       [tex]p_{oy} = p_{oAy} + p_{oBy} (2)[/tex]

       [tex]p_{fx} = (m_{A} + m_{B} ) * v_{fx} (3)[/tex]

       [tex]p_{fy} = (m_{A} + m_{B} ) * v_{fy} (4)[/tex]

       [tex]v_{fx} = \frac{p_{oAx}}{(m_{A}+ m_{B)}} = \frac{m_{A}*v_{oAx} }{(m_{A}+ m_{B)}} =\frac{17.0kg*8.00m/s}{46.0kg} = 2.96 m/s (5)[/tex]

       [tex]v_{fy} = \frac{p_{oBy}}{(m_{A}+ m_{B)}} = \frac{m_{B}*v_{oBy} }{(m_{A}+ m_{B)}} =\frac{29.0kg*5.00m/s}{46.0kg} = 3.15 m/s (6)[/tex]

      [tex]v_{f} = \sqrt{v_{fx} ^{2} + v_{fy} ^{2}} = \sqrt{(2.96m/s)^{2} + (3.15m/s)^{2}} = 4.32 m/s (7)[/tex]

       [tex]p_{f} = (m_{A} + m_{B})* v_{f} = 46 kg * 4.32 m/s = 198.8 kg*m/s (8)[/tex]

       [tex]tg \theta = \frac{v_{fy}}{v_{fx}} = \frac{3.15 m/s}{2.96m/s} = 1.06 (9)[/tex]

       ⇒ θ = tg⁻¹ (1.06) = 46.8º N of E

Answer:

Please see below as the answer is self-explanatory.

Explanation:

       [tex]v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)[/tex]

       [tex]t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s (2)[/tex]

       [tex]\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (3)[/tex]

       [tex]v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)[/tex]

       [tex]\Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)[/tex]

Answer:

I think its object 1

Explanation:

Because the object that has more weight has a greater momentum and the lightest object that has a less momentum will be easier to change because its lighter.

Answer:

The acceleration is 2.448 meters per square second and is vertically upward.

Explanation:

The Free Body Diagram of the plastic ball in the liquid is presented in the image attached below. By Second Newton's Law, we know that forces acting on the plastic ball is:

[tex]\Sigma F = F - m\cdot g = m\cdot a[/tex] (1)

Where:

[tex]F[/tex] - Buoyant force, measured in newtons.

[tex]m[/tex] - Mass of the plastic ball, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]a[/tex] - Net acceleration, measured in meters per square second.

If we know that [tex]F = 5\,N[/tex], [tex]m = 0.408\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the net acceleration of the plastic ball is:

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

[tex]a= 2.448\,\frac{m}{s^{2}}[/tex]

The acceleration is 2.448 meters per square second and is vertically upward.

Answer:

The reckage after collision is motionless (E)

Explanation:

The first law of thermodynamics states that energy is neither created nor destroyed but is converted from one form to another.

The kind of collision described in the question above is known as a perfectly inelastic collision, and in this type of collision, the maximum kinetic energy is lost because the objects moving in opposite directions have a resultant momentum that is equal, but in opposite directions hence they cancel each other out.

The calculation is as follows:

m₁v₁ + m₂v₂

where:

m₁ = m₂ = 1000kg

v₁ = 20 m/s

v₂ = -20 m/s ( in the opposite vector direction)

∴ resultant momentum = (1000 × 20) + (1000 × -20)

= 20000 - 20000 = 0

∴ The reckage after collision is motionless

Answer:

The wreckage after collision is moving at the speed 18 m/s to the south.

Explanation:

Answer:

towards the north.

Explanation:

When an object is raised vertically, work is done against _gravitational force _________

The force of attraction between all masses in the universe; especially the attraction of the earth's mass for bodies near its surface is called gravitational force .

On every body there is a force acting named  gravitational force which attracts the body downward .

If a object / body is thrown up it always comes downward if only gravitational force is acting on it which act due to earth's gravity .

When an object is raised vertically, work is done against _gravitational force _________

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Answer:

Explanation:

answer C looks good

there isn't really a "centrifugal " force. :/    when we are pushed "back" in a car seat.. it's not because there is a force pushing us backwards... but a force pushing us forwards.. just like when turning too,   a force pushes us into the corner,  not a force pushing out of the corner.  :)  

Answer:

Examples of compound machines include bicycles, cars, scissors, and fishing rods with reels. Compound machines generally have lower efficiency but greater mechanical advantage than simple machines

Brilianst

Answer:

screws, inclined planes , etc

Explanation:

no explanation needed

Answer: Option b, the final velocity is half of the initial velocity.

Explanation:

Here we will use the conservation of the total momentum of a system.

This means that the total momentum at the beginning must be the same as the final momentum.

Where momentum is:

P = M*v

Initially, we have two cars, both with the same mass M, and only one of them has a velocity v.

Then the initial momentum is:

P = M*v + M*0 = M*v

After the collision, the two cars move together. Then the total mass that is moving is equal to the sum of the masses of the cars, this is 2*M

and we can suppose that the two cars move at a final velocity v'

Then the final momentum is:

P' = (2*M)*v'

Now we use the conservation of momentum, then:

P = P'

M*v = (2*M)*v'

Now we need to solve this for v'

(M*v)/(2*M) = v'

v/2 = v'

This means that the final velocity is half of the initial velocity.

Then the correct option is option b.

Answer:

rotates

Explanation:

I'm so bored

yrfgggghhgghhyuj

Answer:

Spring constant in N / m = 6,000

Explanation:

Given:

Length of spring stretches = 5 cm = 0.05 m

Force = 300 N

Find:

Spring constant in N / m

Computation:

Spring constant in N / m = Force/Distance

Spring constant in N / m = 300 / 0.05

Spring constant in N / m = 6,000

Answer:

coefficient of viscosity of 8.4 poison denotes that the tangential frictional force acting per unit area when divided by the velocity gradient as a result of streamline flow conditions gives 8.4.

Explanation:

Viscosity is defined as the extent to which a fluid can resist flow when a force is applied to it.

Now, coefficient of viscosity is the term in which viscosity is calculated. It is basically the tangential frictional force acting per unit area which is divided by the velocity gradient as a result of streamline flow conditions.

Thus, coefficient of viscosity of 8.4 poison denotes that the tangential frictional force acting per unit area when divided by the velocity gradient as a result of streamline flow conditions gives 8.4.

Answer:

B. holding a coffee mug

Explanation:

Something must move a distance for work to be done.

Answer:

The correct answer is - 1240 newtons; 278.763 lbs.

Explanation:

Answer:

[tex]31.035^{\circ}[/tex]

Explanation:

x = Displacement in x direction = 5.34 m

t = Time taken to travel the displacement

y = Displacement in y direction = 0

u = Initial velocity of ball = 7.7 m/s

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

Displacement in x direction is given by

[tex]x=u\cos\theta t\\\Rightarrow t=\dfrac{5.34}{7.7 \cos\theta}[/tex]

Displacement in y direction is given by

[tex]y=u\sin\theta t-\dfrac{1}{2}gt^2\\\Rightarrow 0=7.7\sin\theta \dfrac{5.34}{7.7\cos\theta}-\dfrac{1}{2}\times 9.81 (\dfrac{5.34}{7.7\cos\theta})^2\\\Rightarrow 0=7.7\sin\theta-4.905\times \dfrac{5.34}{7.7\cos\theta}\\\Rightarrow 0=7.7^2\sin\theta \cos\theta-4.905\times 5.34\\\Rightarrow 0=7.7^2\dfrac{\sin2\theta}{2}-4.905\times 5.34\\\Rightarrow 0=7.7^2\sin2\theta-4.905\times5.34\times 2\\\Rightarrow \sin2\theta=\dfrac{4.905\times 5.34\times 2}{7.7^2}\\\Rightarrow 2\theta=\sin^{-1}\dfrac{4.905\times 5.34\times 2}{7.7^2}[/tex]

[tex]\Rightarrow \theta=\dfrac{62.07}{2}\\\Rightarrow \theta=31.035^{\circ}[/tex]

The angle at which the ball was thrown is [tex]31.035^{\circ}[/tex].

Answer:

27 ms^-1

Explanation:

by using v= u + at

u = 0 ( because the object id starting from rest)

v= 0 + 1.5 x 18

v = 27 ms^-1

Answer:

[tex]3.031\times 10^{-5}\ \text{m}[/tex]

Explanation:

[tex]y[/tex] = Distance between central maxima and first minimum

m = Order = 1

d = Thickness of hair

[tex]\lambda[/tex] = Wavelength = 632.8 nm

L = Distance between light source and screen = 1.25 m

Width of central maximum is given by

[tex]2y=5.22\times 10^{-2}\\\Rightarrow y=\dfrac{5.22\times 10^{-2}}{2}\\\Rightarrow y=0.0261\ \text{m}[/tex]

Distance between central maxima and first minimum is given by

[tex]y=L\tan\theta_{min}\\\Rightarrow \tan\theta_{min}=\dfrac{y}{L}\\\Rightarrow \tan\theta_{min}=\dfrac{0.0261}{1.25}\\\Rightarrow \theta_{min}=\tan^{-1}0.02088\\\Rightarrow \theta_{min}=1.1962^{\circ}[/tex]

Since [tex]\theta[/tex] is small [tex]\tan\theta_{min}=\sin\theta_{min}[/tex]

[tex]\sin\theta_{min}=\dfrac{m\lambda}{d}\\\Rightarrow d=\dfrac{m\lambda}{\sin\theta}\\\Rightarrow d=\dfrac{1\times 632.8\times 10^{-9}}{\sin1.1962^{\circ}}\\\Rightarrow d=3.031\times 10^{-5}\ \text{m}[/tex]

The strand of hair is [tex]3.031\times 10^{-5}\ \text{m}[/tex] thick.

Answer:

2500 m

Explanation:

Given that,

Each lap is 500 m

Gareth takes 5 laps.

We need to find distance traveled by Gareth. The distance covered by him is given by :

d = 5×500

d = 2500 m

Hence, he will travel 2500 m.

Answer:

An object with no net forces acting on it which is initially at rest will remain at rest. If it is moving, it will continue to move in a straight line with constant velocity. Forces are "pushes" or "pulls" on the object, and forces, like velocity and acceleration are vector quantities.

Answer:

Distillation separates a liquid from a solution. For example, water can be separated from salty water by simple distillation. This method works because the water evaporates from the solution, but is then cooled and condensed into a separate container. The salt does not evaporate and so it stays behind.

Answer:

1) v= 90km/h  d = 70 m,  2)  x₁ = v t_r,  x₁ = 6.25 m, 3) x₁=6.25 no change

4) x = 22 m

Explanation:

1) for the first part, you are asked to find the minimum safety distance with the vehicle in front

The internet is searched for the stopping distance for two typical speeds on the highway

v (km/ h)      v (m/s)      d (m)

90                  25           70

100                 27.78      84

the safe distance is this distance plus the distance traveled during the person's reaction time, which can be calculated with infirm movement

              v = x / t_r

               x₁ = v t_r

the average reaction time is t_r = 0.25s for a visual stimulus and t_r 0.17 for an auditory stimulus

therefore the safe distance is

              x_total = x₁ + d

2) The distance is the sum of the distance traveled in the reaction

              x₁ = v t_r

for v = 90 km / h

              x₁ = 25 0.25

              x₁ = 6.25 m

for v = 100 km / h

              x₁ = 27.78 0.25

              x₁ = 6.95 m

the total distance is

               x_total = x₁ + d

for v = 90 km / h

             x_total = 25 0.25 + 70

             x_total = 76.25 m

this is the distance until the cars stop and do not collide

3) the stopping distance of a truck is

   v = 90 km / h       d = 100 m

in this case we see that the braking distance is much higher,

the safe distance is given by the distance traveled during the reaction, as the truck brakes slower than the car this distance does not change

4) let's analyze the empirical rule: maintain the length of a car for each increase in speed of v = 10 m / h = 4.47 m / s

for the car case at v = 90km / h = 25 m / s

according to this rule we must this to

           x = 25 / 4.47 = 5.6 cars

each modern car is about 4 m long so the distance is

           x = 22 m

we see that this distance is much greater than the reaction  distance so it does not make much sense

During the "waxing" phases, the two weeks immediately following the New Moon.

New Moon ==> waxing crescent ==> First Quarter ==> waxing gibbous ==> Full Moon

The phase of the moon decides the moon's visibility. The sequence of phases does the moon's visibility increases is; the waxing crescent, first quarter, waxing gibbous, full Moon.

The moon changes shape every day. This is due to the fact that the celestial body has no light of its own and can only reflect sunlight.

Only the side of the moon facing the sun can reflect this light and seem bright. The opposite side appears black. this is a full moon.

We can only see the black section when it lies between the sun and the earth when a new moon occurs. We witness intermediate phases like a half-moon and crescent in between these two extremes.

Following are the sequence of phases does the moon's visibility increase is;

1. Waxing crescent

2. First Quarter

3. Waxing gibbous

4. Full Moon

The phase of the moon decides the moon's visibility. Hence the visibility changes with the change in the phase of the moon.

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