How do unbalanced forces acting on an object affect its motion when the object is at rest? What if it is moving?

Answers

Answer 1

Answer:

It pushes it because an unbalanced force is pushing more newtons than something that isn't even moving. Even if it is moving, it depends which side is pushing/pulling the most force.

Explanation:

Answer 2

Answer:

It pushes it because an unbalanced force is pushing more newtons than something that isn't even moving. Even if it is moving, it depends which side is pushing/pulling the most force.


Related Questions

2

10 points

Find the total displacement of each of the motions.

a) You walk 45 m W, then 34 mW

b) You drive 5 km N, then 7 km S

c) You cycle 350 m E, then 800 m W, then 200 m E

d) You fly 850 km N then 850 km S

Answers

Answer:

a) s = 79 m W

b) s = 2 km S

c) s = 250 m W

d) s = 0 km  

Explanation:

We take the following sign convention for the directions:

North (N)  ---> positive

South (S)  ---> negative

East (E) ---> negative

West (W)  ---> positive

a)

45 m W, 34 m W

s = 45 m + 34 m

s = 79 m W

b)

5 km N, 7 km S

s = 5 km - 7 km

s = - 2 km

s = 2 km S

c)

350 m E , 800 m W, 200 m E

s = -350 m + 800 m - 200 m

s = 250 m

s = 250 m W

d)

850 km N, 850 km S

s = 850 km - 850 km

s = 0 km

An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s2. The takeoff speed for this plane will be 65 m/s. All airplanes will start from rest(0m/s). Assuming this minimum acceleration, what is the minimum allowed length for the runway for take off?

Answers

Answer:

x = 704 [m]

Explanation:

To solve this problem we must use the following equation of kinematics.

[tex]v_{f} ^{2} =v_{o} ^{2} +2*a*x[/tex]

where:

Vf = final velocity = 65 [m/s]

Vo = initial velocity = 0 (starts from rest)

a = acceleration = 3 [m/s²]

x = distance [m]

Now replacing we have:

65² = 0 + 2*3*x

4225 = 6*x

x = 704 [m]

A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of 50 m. If he completes the 200 m dash in 29.6 s and runs at constant speed throughout the race, what is the magnitude of his centripetal acceleration (in m/s2) as he runs the curved portion of the track?

Answers

Answer:

The centripetal acceleration of the runner as he runs the curved portion of the track is 0.91 m/s²

Explanation:

Given;

distance traveled in the given time = 200 m

time to cover the distance, t = 29.6 s

speed of the runner, v = d / t

v = 200 / 29.6

v = 6.757 m/s

The centripetal acceleration of the runner is given by;

[tex]a_c = \frac{V^2}{r}[/tex]

where;

r is the radius of the circular arc, given as 50 m

Substitute the givens;

[tex]a_c = \frac{V^2}{r}\\\\a_c = \frac{(6.757)^2}{50}\\\\a_c = 0.91 \ m/s^2[/tex]

Therefore, the centripetal acceleration of the runner as he runs the curved portion of the track is 0.91 m/s².

if the motor m rotates in the direction shown by the arrow what is going on

Answers

The question is incomplete,I will complete the question and provide the answer.

Due to the nature of the question,I will sketch an answer/solution to the question and submit it as an attachment.

So you will be having two attachments,

1) The question

2) The solution

From the options given in the first attachment with is the question,the correct answer is option C.

1 AND 2 ARE GOING UP.

Proper explanation using sketch and arrows is given in the second attachment which shows the solution to the question.

a car accelerates at a constant rate from 15 m/s to 25 m/s while it travels a distance of 125 m. How long does it take to achieve this speed?

Answers

T=Vf-Vi/s
25m/s -15m/s/ 125m
10m/s /125m
=0.08s
I hope it’s correct !

The time taken by the car to achieve the final speed is 6.25 seconds.

What is the equation of motion?

The equations of motion can be defined as the equation that represents the relationship between the time, velocity, acceleration, and displacement of a moving object.

The mathematical expressions for the equations of motions can be written as:

[tex]v= u+at\\S=ut+(1/2)at^2\\v^2-u^2=2aS[/tex]

Given, the initial speed of the car, u = 15 m/s

The final speed of the given car, v = 25m/s

The distance covered by car, S = 125 m

From the third equation of motion: v² = u²+ 2aS

(25)² = (15)² + 2×a× 125

a = 1.6 m/s²

From the first equation of motion we can find the time to achieve the final speed:

v = u+ at

25 = 15 + (1.6) × t

t = 6.25 sec

Therefore, 6.25 seconds will be taken by the car to catch the final speed.

Learn more about the equation of motion, here:

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A satellite was in two separate crashes. In both crashes, the satellite had the same mass. Engineers want to know about the speed and direction of the satellite after the crashes. Why would the crash affect the motion of the satellite, and which crash caused a greater change in motion for the satellite?

WILL GIVE BRAINLIEST

Answers

Answer:

The fastest satellite must change orbit

The most massive body (m₁) transfers more momentum to the satellite,

Explanation:

For this problem we consider a system formed by the satellite and each of the bodies with which it collides, in this system the forces during the collision are internal, the amount of movement must be conserved. Let's write the momentum is two instants

Most massive body (m1)

initial. Before the crash

      p₀₁ = M v + m₁ v₁

after the crash

      [tex]p_{f1}[/tex] = M v´ + m₁ v₁´

how momentum is conserved

     p₀ = p_{f}

Lighter body (m2)

      p₀₂ = M v + m₂ v₂

       p_{f2} = M v´ + m₂ v₂´

           

Let's clarify that the speed of the satellite and the object do not have the same direction, in general these shocks are elastic.

We can see that  p₀₁> p₀₂

Let us analyze the two cases when the body collides, The most massive body (m₁) transfers more momentum to the satellite, therefore there must be a greater change in its momentum and velocity.

The fastest satellite must change orbit, thus rotating at a different distance from Earth

20 POINTS
In order to maximize the acceleration of an object, one should
maximize the mass
maximize the force
minimize the velocity
maximize the inertia

Answers

I think it’s the first one- maximise the mass

A child whose weight is 287 N slides down a 7.20 m playground slide that makes an angle of 31.0° with the horizontal. The coefficient of kinetic friction between slide and child is 0.120. (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of 0.559 m/s, what is her speed at the bottom?

Answers

Answer:

a

[tex]H =212.6 \ J[/tex]

b

[tex]v = 7.647 \ m/s[/tex]

Explanation:

From the question we are told that

   The child's weight is  [tex]W_c = 287 \ N[/tex]

    The length of the sliding surface of the playground is  [tex]L = 7.20 \ m[/tex]

    The coefficient of friction is  [tex]\mu = 0.120[/tex]

      The angle is [tex]\theta = 31.0 ^o[/tex]

      The initial  speed is  [tex]u = 0.559 \ m/s[/tex]

Generally the normal force acting on the child is mathematically represented as

=>    [tex]N = mg * cos \theta[/tex]

Note  [tex]m * g = W_c[/tex]

Generally the frictional force between the slide and the child is    

         [tex]F_f = \mu * mg * cos \theta[/tex]

Generally the resultant force acting on the child due to her weight and the frictional  force is mathematically represented as

      [tex]F =m* g sin(\theta) - F_f[/tex]

Here  F is the resultant force and it is represented as  [tex]F = ma[/tex]

=>   [tex]ma = m* g sin(31.0) - \mu * mg * cos (31.0)[/tex]

=>   [tex]a = g sin(31.0)- \mu * g * cos (31.0)[/tex]

=>  [tex]a = 9.8 * sin(31.0) - 0.120 * 9.8 * cos (31.0)[/tex]

=>[tex]a = 4.039 \ m/s^2[/tex]

So

   [tex]F_f = 0.120 * 287 * cos (31.0)[/tex]

=> [tex]F_f = 29.52 \ N[/tex]

Generally the heat energy generated by the frictional  force which equivalent tot the workdone by the frictional force  is mathematically represented as

     [tex]H = F_f * L[/tex]

=>  [tex]H = 29.52 * 7.2[/tex]

=>  [tex]H =212.6 \ J[/tex]

Generally from kinematic equation we have that

    [tex]v^2 = u^2 + 2as[/tex]

=>  [tex]v^2 = 0.559^2 + 2 * 4.039 * 7.2[/tex]

=>  [tex]v = \sqrt{0.559^2 + 2 * 4.039 * 7.2}[/tex]

=>  [tex]v = 7.647 \ m/s[/tex]

   

As a rough model of the impact of walking/running, consider that half the mass of the body falls from a height of 4.77-cm onto a single foot. (During a typical stride, an adult's center-of-mass moves approximately this distance vertically). Use the kinematic equations to calculate the speed of an object falling from this height at the moment of impact with the ground under the influence of gravity.A. As a rough model of the impact of walking, consider that half of the mass of the entire body strikes the ground with a downward velocity of 1.0 m/s and comes to a full vertical stop over an impact duration of 20 ms. Calculate the force associated with this single step for a person with a mass of 74.2 kg. B. Calculate the stress (solid pressure) of a force of 1880 N applied across the 0.4 cm^2 cross-sectional area of the typical Achilles tendon. For reference, the maximum rupture stress of tendons has been reported in the range of 100-150 MPa.

Answers

Answer:

0.967 m/s

1855 N

[tex]46.375\ \text{MPa}[/tex]

Explanation:

v = Final velocity

u = Initial velocity = 0

s = Displacement = 4.77 cm

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

From the kinematic equations

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.0477+0}\\\Rightarrow v=0.967\ \text{m/s}[/tex]

The velocity of the object at the moment of impact is 0.967 m/s

Now

[tex]\Delta v[/tex] = Change in velocity = 1 m/s

t = Time taken = 20 ms

m = Half mass of the person = [tex]\dfrac{74.2}{2}=37.1\ \text{kg}[/tex]

[tex]F=\dfrac{m}{t}\\\Rightarrow F=\dfrac{37.1\times 1}{20\times 10^{-3}}\\\Rightarrow F=1855\ \text{N}[/tex]

The force associated with a single step of the person is 1855 N

A = Area = [tex]0.4\ \text{cm}^2[/tex]

Stress is given by

[tex]\sigma=\dfrac{F}{A}\\\Rightarrow \sigma=\dfrac{1855}{0.4\times 10^{-4}}\\\Rightarrow \sigma=46375000\ \text{Pa}=46.375\ \text{MPa}[/tex]

The stress on the tendon is [tex]46.375\ \text{MPa}[/tex]

The speed of object during falling is 0.967 m/s.

(A)  The magnitude of force  associated with this single step for a person is 1855 N.

(B) The required value of stress at tendons is [tex]4.70 \times 10^{7} \;\rm Pa[/tex].

Given data:

The height of fall is, h = 4.77 cm = 0.0477 m.

The magnitude of downward velocity is, v' = 1.0 m/s.

The duration of impact is, [tex]t = 20 \;\rm ms =20 \times 10^{-3} \;\rm s[/tex].

The mass of person is, m = 74.2 kg.

The magnitude of force is, F' = 1880 N.

The cross-sectional area is, [tex]A =0.4 \;\rm cm^{2} = 0.4 \times 10^{-4} \;\rm m^{2][/tex].

The problem has several parts using different concepts. First obtain the final speed of object to fall by using the second kinematic equations of motion as,

[tex]v^{2}=u^{2}+2gh[/tex]

Solving as,

[tex]v^{2}=0^{2}+(2 \times 9.8 \times 0.0477)\\\\v = \sqrt{(2 \times 9.8 \times 0.0477)} \\v = 0.967 \;\rm m/s[/tex]

Thus, the speed of object during falling is 0.967 m/s.

(A)

Now coming to next part, the half of mass means, m' = m/2 = 74.2/2 = 37.1 kg.

Apply the expression of average force as,

[tex]F =\dfrac{m'v'}{t}[/tex]

Solving as,

[tex]F =\dfrac{37.1 \times 1}{20 \times 10^{-3}}\\\\F = 1855 \;\rm N[/tex]

Thus, the magnitude of force  associated with this single step for a person is 1855 N.

(B)

The expression for the stress is given as,

[tex]\sigma = \dfrac{F'}{A}[/tex]

Solving as,

[tex]\sigma = \dfrac{1880}{0.4 \times 10^{-4}}\\\\\sigma =4.70 \times 10^{7} \;\rm Pa[/tex]

Thus, the required value of stress at tendons is [tex]4.70 \times 10^{7} \;\rm Pa[/tex].

Learn more about the Stress force here:

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Having established that a sound wave corresponds to pressure fluctuations in the medium, what can you conclude about the direction in which such pressure fluctuations travel?A) The direction of motion of pressure fluctuations is independent of the direction of motion of the sound wave.B) Pressure fluctuations travel perpendicularly to the direction of propagation of the sound wave.C) Pressure fluctuations travel along the direction of propagation of the sound wave.D) Propagation of energy that passes through empty spaces between the particles that comprise the mediumDoes air play a role in the propagation of the human voice from one end of a lecture hall to the other?a) yesb) no

Answers

Answer:

None of them: the direction of the pressure fluctuations is parallel to the direction of motion of the wave

Explanation:

If the weight of a person is 500 newton what is his mass on the earth ?

Answers

Answer:

The person is on the Moon having a weight of 500 N. The acceleration of gravity on the Moon is approximately 1.6 m/s2. What is your his, which includes his space suit?

f= Force (of gravity)=500N

g=acceleration of gravity=1.6m/s^2

m=mass=312kg

m=f/a= 500N/1.6 m/s^2 = 500 (kg-m/1.6m/s^2) = 500/1.6kg = 312kg

his mass is 312kg

Which statement explains how it is possible to carry books to school without changing the kinetic or potential energy of the books or doing any work?
a. by moving the book without acceleration and keeping the height of the book constant
b. by moving the book with acceleration and keeping the height of the book constant
c. by moving the book without acceleration and changing the height of the book
d. by moving the book with acceleration and changing the height of the book

Answers

Answer:

a. by moving the book without acceleration and keeping the height of the book constant

Explanation:

FOR CONSTANT KINETIC ENERGY:

The kinetic energy of a body depends upon its speed according to its formula:

ΔK.E = (1/2)mΔv²

So, for Δv = 0 m/s

ΔK.E = 0 J

So, for keeping kinetic energy constant, the books must be moved at constant speed without acceleration.

FOR CONSTANT POTENTIAL ENERGY:

The potential energy of a body depends upon its height according to its formula:

ΔP.E = mgΔh

So, for Δh = 0 m/s

ΔP.E = 0 J

So, for keeping potential energy constant, the books must be moved at constant height.

So, the correct option is:

a. by moving the book without acceleration and keeping the height of the book constant

In the Bohr model of the hydrogen atom, an electron in the 1st excited state moves at a speed of 2.19 106 m/s in a circular path having a radius of 5.29 10-11 m. What is the effective current associated with this orbiting electron?

Answers

Answer:

I = 1.05x10⁻³ A

Explanation:

By definition, an electric current is the rate of charge flow at a given time:

[tex] I = \frac{q}{t} [/tex]

Where:

q: is the electrons charge = 1.602x10⁻¹⁹ C

t: is the time

In a circular motion, the time is given by:

[tex] t = T = \frac{2\pi}{\omega} = \frac{2\pi}{v/r} = \frac{2\pi r}{v} [/tex]

Where:

ω: is the angular speed = v/r

v: is the speed = 2.19x10⁶ m/s

r: is the radius = 5.29x10⁻¹¹ m

[tex] t = \frac{2\pi r}{v} = \frac{2\pi 5.29 \cdot 10^{-11} m}{2.19 \cdot 10^{6} m/s} = 1.52 \cdot 10^{-16} s [/tex]

Now, the effective current is:

[tex] I = \frac{q}{t} = \frac{1.602 \cdot 10^{-19} C}{ 1.52 \cdot 10^{-16} s} = 1.05 \cdot 10^{-3} A [/tex]  

Therefore, the effective current associated with this orbiting electron is 1.05x10⁻³ A.

I hope it helps you!                                

pls help quick. the number line shows the starting and ending velocities for ball 1 what's the change in velocity of ball 1 calculate the value mathematically and check using the number line​

Answers

Answer:

The starting velocity for ball 1 is 1.00 meter/second. Its ending velocity is 0.25 meter/second.

The change in velocity for ball 1 is 0.25 – 1.00 = -0.75 meter/seconds

The particle accelerator at CERN can accelerate an electron through a potential

difference of 80 kilovolts. Calculate

(a) The kinetic energy (in keV) of the electron​

Answers

Answer:

K.E = 1.28 × 10^-17 KeV

Explanation:

Given that a particle accelerator at CERN can accelerate an electron through a potentialdifference of 80 kilovolts.

To Calculate the kinetic energy (in keV) of the electron​, let us first find the electron charge which is 1.60 × 10^-19C

The kinetic energy = work done

K.E = e × kV

Substitute e and the voltage into the formula

K.E = 1.60 × 10^-19 × 80

K.E = 1.28 × 10^-17 KeV

Therefore, the kinetic energy is approximately equal to 1.28 × 10^-17 KeV

A medicine ball has a mass of 5kg and is thrown with a speed of 3 m/sec what is it's kinetic energy

Answers

KE = 1/2mv^2

KE = 1/2 (5kg)(3m/s)

KE = 22.5 J

Which term refers to the ability to do work?
o energy
power
force
motion
PLEAEE HURRY IM DOING MY EXAM PLEASE HELP WILL MARK BRAINLIEST​

Answers

Answer:

Energy

Explanation:

work is actually a transfer of energy. When work is done to an object , energy is transferred to that object.

The ability to do work is called energy.

What is work ?

"Work is the energy transferred to or from an object via the application of force along a displacement. In its simplest form, it is often represented as the product of force and displacement."

What is energy ?

"Energy is defined as the “ability to do work, which is the ability to exert a force causing displacement of an object.”  energy is just the force that causes things to move. Energy is divided into two types: potential and kinetic."

Know more about energy here

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match the variables to its definition​

Answers

Answer:

See connections below

Explanation:

1   [tex]\Rightarrow[/tex] b

2   [tex]\Rightarrow[/tex] a

3   [tex]\Rightarrow[/tex] d

4   [tex]\Rightarrow[/tex] i

5   [tex]\Rightarrow[/tex] g

6   [tex]\Rightarrow[/tex]  h

7   [tex]\Rightarrow[/tex]  c

8   [tex]\Rightarrow[/tex]  e

9   [tex]\Rightarrow[/tex] f

please help i will mark brainliest
what does a speedometer measure?
a. accerlation
b. velocity
c. speed
d. average speed.

Answers

im pretty sure the answer is c; velocity is usually measure by a vector :)

Answer:

c. speed

Explanation:

Speedometer is a device used to measure the speed of a vehicle. I am pretty sure this is the correct option.

If a ball rolls down an incline with a starting velocity of 0m/s and a final velocity of 6m/s
and it takes a total of 1.4 seconds, calculate its acceleration.

Answers

Answer:

If a ball rolls down an incline with a starting velocity of 0m/s and a final velocity of 6m/s

and it takes a total of 1.4 seconds, calculate its acceleration.

Answer:

Acceleration is 4.28 m/s²

Explanation:

Acceleration is change of speed in time. To solve this, we will assume that the acceleration is constant, meaning that every second the velocity increases for the same constant value.

a = ∆v/t

∆v is the difference between two measured velocities:

a = (v2 - v1) / t

v1 = 0m/s

v2 = 6m/s

t = 1.4 s

Now, we only plug in the given values:

a = (6 - 0) / 1.4

a = 6 m/s / 1.4 s

a = 4.28 m/s²

that delivers oxygen to your body and In the video your blood is compared to a picks up CO2 to be released out when you breath. PLEASE I NEED A ANSWER ​

Answers

I’m a bit confused. What’s the question?

an object falls from a hovering helicopter and hits the ground at a speed of 30m per seconds. how long does it take the object to reach the ground and how far does it fall? sketch a velocity-time graph for the object ( ignore air resistance

Answers

Answer:

45.9m

Explanation:

Given parameters:

Final velocity  = 30m/s

Initial velocity  = 0m/s

Unknown:

Time it takes for the object of fall  = ?

Height of fall = ?

Solution:

For the first problem, we use the equation below to solve for t;

      V = U + gt

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

t is the time taken

          30  = 0 + 9.8 x t

          30  = 9.8t

            t  = [tex]\frac{30}{9.8}[/tex]  = 3.1s

Now, height of fall;

  V² = U²  + 2gH

   30² = 0² + 2 x 9.8 x H

   900  = 19.6H

      H  = 45.9m

A 8.45μC particle with a mass of 6.15 x 10^-5 kg moves perpendicular to a 0.493-T magnetic field in a circular path of radius 34.1 m. How much time will it take for the particle to complete one orbit?

a. 92.7 s
b. 0.0927 s
c. 9.27 s
d. 927 s

Answers

This question is incomplete, the complete question is;

A 8.45μC particle with a mass of 6.15 x 10⁻⁵ kg moves perpendicular to a 0.493-T magnetic field in a circular path of radius 34.1 m.

How much time will it take for the particle to complete one orbit?

a. 92.7 s

b. 0.0927 s

c. 9.27 s

d. 927 s

Answer:

it will take 92.7 seconds for the particle to complete one orbit.

Option a) 92.7 s is the correct option

Explanation:

Given that;

mass m =  6.15 x 10⁻⁵ kg

q = 8.45μC = 8.45 × 10⁻⁶ C

B = 0.493

we know that

Time period T = 2πr / V

where r = mv/qB

so T = 2πm/qB

we substitute

T = (2 × 3.14 × 6.15 x 10⁻⁵) / ( 8.45 × 10⁻⁶ × 0.493)

T = 0.0003862 / 0.000004165

T = 92.7 sec

Therefore it will take 92.7 seconds for the particle to complete one orbit.

Option a) 92.7 s is the correct option

when is thermal equilibrium achived between two identical objects



need help ASAP

Answers

Answer: When two objects in contact with each other are at different temperatures, they are said to be in thermal equilibrium.

Explanation: . When two objects not in contact with each other are at the same pressure, they are said to be in thermal equilibrium.

How do we use energy transformation in our daily lives?

Answers

Answer:hat are some examples of energy transformation?

The Sun transforms nuclear energy into heat and light energy.

Our bodies convert chemical energy in our food into mechanical energy for us to move.

An electric fan transforms electrical energy into kinetic energy.

Explanation:

We pick up cups and place them in different places

A hydraulic car jack needs to be designed so it can lift a 2903.57 lb car assuming that a person can exert a force of 24.41 lbs. If the piston the person is pushing on had a radius of 3.26 cm, what should the diameter of the piston be that is used to raise the car?

Answers

Answer:

Diameter of the piston would be 0.71 m (71.1 cm)

Explanation:

From the principle of pressure;

[tex]\frac{F_{1} }{A_{1} }[/tex] = [tex]\frac{F_{2} }{A_{2} }[/tex]

Let [tex]F_{1}[/tex] = 2903.57 lb, [tex]F_{2}[/tex] = 24.41 lbs, [tex]r_{2}[/tex] = 3.26 cm = 0.0326 m.

[tex]A_{2}[/tex] = [tex]\pi r^{2}[/tex]

    = [tex]\frac{22}{7}[/tex] x [tex](0.0326)^{2}[/tex]

    = 0.00334 [tex]m^{2}[/tex]

So that:

[tex]\frac{2903.57}{A_{1} }[/tex] = [tex]\frac{24.41}{0.00334}[/tex]

[tex]A_{1}[/tex] = [tex]\frac{2903.57*0.00334}{24.41}[/tex]

    = 0.3973

[tex]A_{1}[/tex] = 0.4 [tex]m^{2}[/tex]

The radius of the piston can be determined by:

[tex]A_{1}[/tex] = [tex]\pi r^{2}[/tex]

0.3973 = [tex]\frac{22}{7}[/tex] x [tex]r^{2}[/tex]

[tex]r^{2}[/tex] = [tex]\frac{0.3973*7}{22}[/tex]

   = 0.1264

r = [tex]\sqrt{0.1264}[/tex]

 = 0.3555

r = 0.36 m

Diameter of the piston = 2 x r

                                     = 2 x 0.3555

                                     = 0.711

Diameter of the piston would be 0.71 m (71.1 cm).

is 2/2 1 or 0? please help lol

Answers

Answer:

1.

Explanation:

Hello!

In this case, for such mathematical operations, we can wee that the slash represents a fraction or a division, say 8 ÷ 4 = 2, 6 ÷ 3 = 2, 20 ÷ 4 = 5, etc. In such a way, since the operation 2/2, represents 2 ÷ 2, it is clear that two is once in 2, therefore, the result is:

2 ÷ 2 = 1.

Best regards!

Robin would like to shoot an orange in a tree with his bow and arrow. The orange is hanging yf=5.00 myf=5.00 m above the ground. On his first try, Robin looses the arrow at v0=35.0 m/sv0=35.0 m/s at an angle of θ=30.0°θ=30.0° above the horizontal. The arrow has an initial height of y0=1.50 m,y0=1.50 m, and its tip is x=60.0 mx=60.0 m away from the target orange. Treating the arrow as a point projectile and neglecting air resistance, what is the height of the arrow once it has reached the horizontal position xx of the orange? Use g=9.81 m/s2g=9.81 m/s2 for the acceleration due to gravity.

Answers

Answer:

h' = 55.3 m

Explanation:

First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:

s = vt

where,

s = horizontal distance between arrow and orange = 60 m

v = initial horizontal speed of the arrow = v₀ Cos θ

θ = launch angle = 30°

v₀ = launch speed = 35 m/s

Therefore,

60 m = (35 m/s)Cos 30° t

t = 60 m/30.31 m/s

t = 1.98 s

Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:

h = Vi t + (1/2)gt²

where,

Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°

Vi = 17.5 m/s

Therefore,

h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²

h = 34.6 m + 19.2 m

h = 53.8 m

since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:

h' = h + y

h' = 53.8 m + 1.5 m

h' = 55.3 m

Which interaction contributes to the greenhouse effect?
A (Gases in the atmosphere absorb heat.
B (Dust particles scatter and reflect light from the Sun.
C (Green visible light is trapped in Earth’s atmosphere.
D (Light travels through a thick part of the atmosphere at sunrise and sunset.

Answers

Answer:

A

Explanation:

Answer:

A

Explanation:

Gases in the atmosphere absorb heat.

You know when you have a blanket around you and some heat gets trapped in but some still gets out. Thats basically what it is.

Plus I got it right in multiple questions including the test!

I hope that reassured you!

Have a good night!

What specific changes in two climate variables are expected to lead to major decreases in soil moisture southern Africa and the Mediterranean region?

Answers

Answer:

Less precipitation, droughts9: How might agriculture in southern Europe change by the end of the century if conditions follow the RCP8.

Explanation:

Precipitation and droughts are the specific changes in two climate variables that are expected to lead to major decreases in soil moisture.

What is drought?

Drought is defined as a period of protracted water scarcity, whether it is due to atmospheric surface water, or groundwater constraints.

Droughts can last months or years, although they can be proclaimed in as little as 15 days.

It has the potential to have a significant influence on the afflicted region's ecology and agriculture as well as harm the local economy.

Precipitation and droughts are the specific changes in two climate variables that are expected to lead to major decreases in soil moisture in southern Africa and the Mediterranean region.

Hence Precipitation and droughts are the specific changes in two climate variables.

To learn more about the drought refer to the link;

https://brainly.com/question/26693108

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