Calculate the distance covered by a bus moving at a rate of 11.5km/h with a time of 2060seconds​

Answers

Answer 1

Answer:

6.56km

Explanation:

Given parameters:

Speed  = 11.5km/hr

Time  = 2060s

Unknown:

Distance covered  = ?

Solution:

Speed is distance divided by the time taken.

   Speed  = [tex]\frac{distance}{time}[/tex]

 

   Distance  = Speed x time

Let us convert the seconds to hours;

      3600s  = 1hr

      2060s = [tex]\frac{2060}{3600}[/tex]   = 0.57hr

Now

 Distance  = 11.5km/hr x 0.57hr  = 6.56km


Related Questions

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

I need emergency help we only have 3 minutes left

Answers

Answer:

Option A

Explanation:

Ast the force is equal and the diayance is equal the beam is also balanced

The chart shows Daniela’s run through her race.

A graph with horizontal axis time (seconds) and vertical axis position in (meters). The line runs straight from 0 seconds 5 meters to 10 seconds 30 meters.
What is her velocity?

1.5 m/s
2 m/s
2.5 m/s
5 m/s

Answers

Answer:

Explanation:

velocity is the rate of change of displacement with respect to time.

Velocity = displacement/time

If a  line runs straight from 0 seconds 5 meters to 10 seconds 30 meters.

The velocity will be expressed as;

v = d2-d2/t2-t1

v = 30-5/10-0

v = 25/10

v = 2.5m/s

Hence her velocity is 2.5m/s

Daniela's velocity is 2.5 m/s.

Velocity is the ratio of displacement to time taken. It is given by:

Velocity = displacement / time

The velocity of Daniela can be gotten by calculating the slope of the graph. Considering the points (0, 5) and (4, 15).

[tex]Velocity(slope)=\frac{15-5}{4-0}=2.5\ m/s[/tex]

Therefore Daniela's velocity is 2.5 m/s.

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A 4.00-kg particle moves along the x axis. Its position varies with time according to x= 5t +1 2.0t^3, where x is in meters and t is in seconds. Find:

a. the kinetic energy of the particle at any time t.
b. the acceleration of the particle and the force acting on it at time t.
c. the power being delivered to the particle at time t.
d. the work done on the particle in the interval t = 0 to t =5

Answers

Answer:

a) The kinetic energy of the particle at any time t is [tex]K = 50+120\cdot t^{2}+72\cdot t^{4}[/tex].

b) The acceleration of the particle at time t is [tex]a= 12\cdot t[/tex]. The force acting on the particle at time t is [tex]F = 48\cdot t[/tex].

c) The power being delivered to the particle at time t is [tex]\dot W = 240\cdot t +288\cdot t^{3}[/tex].

d) The work done on the particle in the interval t = 0 to t = 5 is 48000 joules.

Explanation:

a) The kinetic energy of the particle is entirely translational, whose formula is:

[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (1)

Where:

[tex]K[/tex] - Translational kinetic energy, measured in joules.

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

[tex]v[/tex] - Velocity of the particle, measured in meters per second.

The velocity of the particle is the rate of change of the position of the particle in time, that is:

[tex]v = 5+6\cdot t^{2}[/tex] (2)

Where [tex]t[/tex] is the time, measured in seconds.

By substituting on (1), we have the following expression: ([tex]m = 4\,kg[/tex])

[tex]K = 2\cdot (5+6\cdot t^{2})^{2}[/tex]

[tex]K = 2\cdot (25+60\cdot t^{2} +36\cdot t^{4})[/tex]

[tex]K = 50+120\cdot t^{2}+72\cdot t^{4}[/tex]

The kinetic energy of the particle at any time t is [tex]K = 50+120\cdot t^{2}+72\cdot t^{4}[/tex].

b) The acceleration of the particle is the rate of change of the velocity of the particle in time, that is:

[tex]a= 12\cdot t[/tex] (3)

Where [tex]a[/tex] is the acceleration of the particle, measured in meters per square second.

The acceleration of the particle at time t is [tex]a= 12\cdot t[/tex].

The force is obtained by multiplying (3) by the mass of the particle. That is to say: ([tex]m = 4\,kg[/tex])

[tex]F = m\cdot a[/tex] (4)

[tex]F = 48\cdot t[/tex]

The force acting on the particle at time t is [tex]F = 48\cdot t[/tex].

c) According to the Work-Energy Theorem, the change in kinetic energy of the particle equals the change in the net work done on the particle. In this case, the power is equal to the rate of change in kinetic energy.

[tex]\dot W = \dot K[/tex] (5)

[tex]\dot W = \frac{d}{dt}(50+120\cdot t^{2}+72\cdot t^{4})[/tex]

[tex]\dot W = 240\cdot t +288\cdot t^{3}[/tex]

The power being delivered to the particle at time t is [tex]\dot W = 240\cdot t +288\cdot t^{3}[/tex].

d) The work done on the particle ([tex]W[/tex]), measured in joules, is equal to the change of the kinetic energy of the particle:

[tex]W = K(5)-K(0)[/tex] (6)

[tex]W = [50+120\cdot (5)^{2}+72\cdot (5)^{4}]-[50+120\cdot (0)^{2}+72\cdot (0)^{4}][/tex]

[tex]W = 48000\,J[/tex]

The work done on the particle in the interval t = 0 to t = 5 is 48000 joules.

During a baseball game, a batter hits a popup to a fielder 83 m away.
The acceleration of gravity is 9.8 m/s
2
.
If the ball remains in the air for 5.7 s, how
high does it rise?
Answer in units of m

Answers

Answer:

The height risen by the ball is 39.8 m.

Explanation:

Given;

total time spent in air by the ball, t = 5.7 s

time to rise to maximum height, t = 5.7 /2 = 2.85 s

The initial velocity of the ball is given by;

v = u - gt

where;

v is the final velocity at maximum height = 0

u is the initial velocity of the ball

0 = u - gt

u = gt

u = (9.8 x 2.85)

u = 27.93 m/s

The vertical height traveled by the ball is given by;

v² = u² - 2gh

where;

v is the final velocity of the ball at maximum height = 0

g is acceleration due to gravity = 9.8 m/s²

0 = u² - 2gh

2gh = u²

h = u² / 2g

h = (27.93)² / ( 2 x 9.8)

h = 39.8 m

Therefore, the height risen by the ball is 39.8 m.

A potter's wheel is a uniform disk of mass of 10.0 kg and radius 20.0 cm. A 2.0-kg lump of clay, roughly cylindrical with radius 3.0 cm, is placed at the center of the wheel. The wheel initially rotates at 30.0 rev/min. The clay then flattens into a disk of radius 8.0 cm. What is the final angular speed of the wheel?

a. 29.6 rev/min
b. 29.2 rev/min
c. 30.8 rev/min
d. 30.4 rev/min
e. 30.0 rev/min

Answers

Answer:

b. 29.2 rev/min

Explanation:

Assuming no external torques acting during the process, total angular momentum must be conserved, as follows:

       [tex]L_{0} = L_{f} (1)[/tex]

The initial angular momentum L₀, can be expressed as follows:

        [tex]L_{0} = I_{0} * \omega_{0} (2)[/tex]

        where I₀ = initial moment of inertia = moment of inertia of the disk +

        moment of inertia of the cylinder and ω₀ = initial angular velocity  =

       30.0 rev/min.

Replacing by the values, we get:[tex]I_{0} = \frac{1}{2} * m_{d} *r_{d} ^{2} + \frac{1}{2}* m_{c} *r_{c} ^{2} = 0.2 kg*m2 +9e-4 kg*m2 = 0.2009 kg*m2 (3)[/tex]⇒ L₀ = I₀* ω₀ = 0.2009 kg*m² * 30.0 rev/min = 6.027 kg*m²*rev/minThe final angular momentum can be written as follows:

       [tex]L_{f} = I_{f} * \omega_{f} (4)[/tex]

       where If = final moment of inertia = moment of the inertia of the solid

      disk + moment of  inertia of the clay flattened on a disk, and ωf = final

      angular velocity.

Replacing by the values, we get:

   [tex]I_{f} = \frac{1}{2} * m_{d} *r_{d} ^{2} + \frac{1}{2}* m_{fd} *r_{fd} ^{2} = 0.2 kg*m2 +6.4e-3 kg*m2 = 0.2064 kg*m2 (5)[/tex]

       ⇒ Lo =Lf = If*ωf

Replacing (2) in (1), and solving for ωf, we get:

        [tex]\omega_{f} = \frac{L_{o}}{I_{f} } = \frac{6.027kg*m2*rev/min}{0.2064kg*m2} = 29.2 rev/min (6)[/tex]

The rate of change of angular displacement is defined as angular speed. The final angular speed of the wheel will be 29.2 rev/min.

What is angular speed?

The rate of change of angular displacement is defined as angular speed.  is stated as follows:

ω = θ t

Where,

θ is the angle of rotation,

t is the time

ω is the angular speed

The given data in the problem is

m is the mass of wheel = 10.0 kg

r₁ is the radius of disk = 20.0 cm=0.2

M is the mass of clay= 2.0 kg

R is the radius of cylinder =  3.0cm

[tex]\rm \omega_i[/tex] is the initial rotational speed =30.0 rev/min

r₂ is the final radius of disk=  8.0 cm.

[tex]\rm \omega_f[/tex] is the initial rotational speed=?

When the external torques act on the body is zero the total angular momentum must be conserved, as follows:

Initial momentum = Final momentum

[tex]\rm L_0=L_f[/tex]

The value of the initial angular momentum L₀ is found by

I₀ = initial moment of inertia = moment of inertia of the disk +moment of inertia of the cylinder

[tex]\rm I_0= \frac{1}{2}m_dr_d^2+ \frac{1}{2}m_cr_c^2\\\\ \rm I_0= \frac{1}{2}\times 10\times (0.2)^2+ \frac{1}{2}\times m_2(0.03)^2[/tex]

[tex]\rm L_0 = I_0\times \omega_0\\\\ L_0 = 2009\times 30\\\\ \rm L_0 =6.027 \;kgm^2rev/min[/tex]

The value of the final angular momentum [tex]I_f[/tex] is found by

[tex]\rm I_f= \frac{1}{2}m_dr_d^2+ \frac{1}{2}m_fr_fd^2\\\\ \rm I_0= 0.2064 m_2[/tex]

[tex]\rm I_f[/tex] is the final moment of inertia = moment of the inertia of the solid disk + moment of inertia of the clay flattened on a disk.

[tex]L_0 =L_f = I_f \times \omega_f[/tex]

[tex]\rm \omega_f=\frac{L_0}{I_f} \\\\ \rm \omega_f=\frac{6.027 m_2}{0.2064 m_2} \\\\ \rm \omega_f= 29.2\; rev/min[/tex]

Hence the final angular speed of the wheel will be 29.2 rev/min.

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which refers to information gathered by the systematic study of nature?

prediction
communication of facts
scientific knowledge
hypothesis

Answers

Answer:

it is scientific knowledge

Explanation:

got i right on edge

The scientific knowledge refers to information gathered by the systematic study of nature.

What is scientific knowledge?

A generalized corpus of laws and ideas developed employing the scientific method to describe an interesting occurrence or behavior is referred to as scientific knowledge.

What is systematic study?

Science would be the systematic study of the composition and dynamics of the physical as well as the natural world via experimentation and observation.

the systematic study of natural circumstances and phenomena. empirical proof. the term for the facts and figures that support an explanation in science.

Therefore, the correct answer will be option (C)

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9. What distance is a book from the floor if the book contains 196 Joules of potential energy and has a mass of 5 kg?​

Answers

Answer:

the book is 4 meters from the floor

Explanation:

Use the formula for potential energy:

U = m g h

196 J = 5kg 9.8m/s^2 h

h = 196 / (5*9.8 ) m

h = 4 m

how long does the guided discovery stage of the drawing in phase last? in training program

Answers

Answer
2 to 3 weeks
This aspect of the drawing-in phase can last 2 to 3 weeks, or as little as a few workouts, and is designed to a) reverse the effects of disuse, b) get the clients oriented to all available technologies, c) allow the clients to learn what their options are and which exercises they likes best.

Answer:

2 to 3 weeks

Explanation:

For a research project, a student needs a solenoid that produces an interior magnetic field of 0.0300 T. She decides to use a current of 1.00 A and a wire 0.500 mm in diameter. She winds the solenoid in layers on an insulating form 1.00 cm in diameter and 10.0 cm long. Determine the number of layers of wire needed. (Round your answer up to the nearest integer.)

Answers

Answer:

121m

Explanation:

Magnetic field = 0.03

Current = 1.00A

Diameter = 0.5

Length = 10vm

B = UoIN/Lcylinder * n

n = number of layer

n = 0.5x10^-3 x 0.03/4pi x 10^-7 x 1

= 0.0005x0.03/12.56x10^-7

= 11.94

~12

Average radius

= 0.008

Length of wire = 2x3.13x0.008x12x200

= 120.6m

= 121m

A radio signal has a frequency of 1.023 x 108 HZ. If the speed of the signal in air is 2.997 x 108m/s, what is the wavelength of the signals? а 7.15 m b 5.23 m C 2.93 m d 0.93 m​

Answers

Answer:

2.93 m  (which agrees with answer "C" on the list)

Explanation:

Recall that the speed of the wave equals the product of the wave's length times its frequency. Therefore, the wavelength is going to be the quotient of the speed of the signal divided its frequency:

Wavelength = 2.997  10^8 / 1.023  10^8 =  2.93 m

How are graphs used to represent and analyze acceleration? please help!

Answers

Graphs are used to show how fast it goes and how long it takes

Mr Jones launches an arrow horizontally at a rate of 40m/s off of a 78.4 m cliff towards the south, what direction and value is his acceleration air resistance is negligible.

A. 9.8 m/s/s west
b. 9.8m/s/s east
C. 9.8m/s's down
d 9.8m/s/s south​

Answers

Answer:

9.8m/s^2 down  (option C)

Explanation:

The only acceleration acting on this motion case in the acceleration due to gravity: 9.8 m/s^2 in the downwards direction.

A 1150 kg car is on a 8.70° hill.
Using X-Y axes tilted down the
plane, what is the y-component
of the weight?

Answers

Answer:

Explanation:

y-component = - mgcos(8.7)

= - (1150)(9.81)cos(8.7)

= - 11151.69378

= - 11151.69 N

The weight of the y-component is 11140.33N.

How to find the weight of the y-component?

To find the weight of the y-component:

Given,

Car weight = 1150 kg

Anfle = 8.70 degree

weight = mg = 1150 * 9.8

= 11270 N

Y-component =  mg cos∅

= 11270 * cos(8.70)

= 11140.33N

The aspect that pushes proper or left is referred to as the x-factor, and the element that pushes up or down is known as the y-component.

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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]

1. It plays a vital role for self- expression and has been part of rituals
and religious gatherings.
B. Dance
A. Aerobic activity
C. Fitness D. Zumba​

Answers

Answer:

b.dance

Explanation:

don't know the explanation

I WILL GIVE BRAILYEST!!! What is the mass of an object moving at a velocity of 5 m/s if the momentum of the object is 50 kg•m/s?
a. 250 kg
c. 10 Kg
b. .002 Kg
d. 45 Kg

Answers

Answer:

a. 250kg I think it's the right answer. hope it helps:)

Answer:

C.10

Explanation:

because when you divide 50 divided by 5 = 10

An object accelerates 12.0 m/s2 when a force of 6.0 newtons is applied to it. What is the mass of the object?

Answers

Answer:

0.5 kg

Explanation:

The mass of the object can be found by using the formula

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

f is the force

a is the acceleration

From the question we have

[tex]m = \frac{6}{12} = \frac{1}{2} \\ [/tex]

We have the final answer as

0.5 kg

Hope this helps you

What is Newton's universal law of gravitation

Answers

Answer:

an object that is in motion wont go out of motion until there is another force pushing on it

An object moves at constant speed in a circle. Which of the following is true:

A. A net force in the direction of motion acts on the object.
B. A net force pointing away from the center of the circle acts on the object.
C. A net force pointing towards the center of the circle acts on the object.
D. The net force acting on the object is zero.

Answers

The answer to your question is B

Experiment: Gravity

Potential Energy = Force x Distance

To show you how this works, study the following example.

If the washers had a mass of 2 grams, and since the force due to gravity in the metric system is 9.8 m/sec2, the weight would be:

W = mass (kg) x gravity

W = 0.002 kg x 9.8 = 0.0196 Newton's. The Newton (N) is the measure of force (or weight here) in the metric system.

The next phase of the computation is to determine the difference in potential energy between where the washers originally started from and the energy of the new position. If the washers were originally 1 meter from the floor or desktop, and the magnet raised them 1 centimeter, the difference in distance raised will be 1 cm, or 0.01 m. Therefore, the potential energy change will be:

Force x Distance = Potential Energy

(0.0196 N) x (0.01 m) = 0.000196 Nm (or Joules)

The magnet is changing the potential energy of the washers by 0.000196 J, or in scientific notation:
1.96 x 10-4 J.

Answers

Answer: the washer would slow down depending on how strong the magnet is even though the gravitational pull is the same. the change would be .6

Explanation: i don't really get it

A 5kg rock is lifted 2m. Find the amount of work done. ​

Answers

Answer:

98J

Explanation:

Given parameters:

Mass of rock  = 5kg

Height = 2m

Unknown:

Work done  = ?

Solution:

The amount of work done is given as:

 Work done  = Force x distance

 Work done = Weight x height

  Work done  = mgH

Now insert the parameters and solve;

 Work done  = 5 x 9.8 x 2 = 98J

The amount of work done on the rock is equal to 98 Nm.

Given the following data:

Mass of rock = 5 kgDistance = 2 meters

To determine the amount of work done:

First of all, we would calculate the force acting on the rock:

[tex]Force = mg\\\\Force = 5 \times 9.8[/tex]

Force = 49 Newton

Now, we can determine the amount of work done:

[tex]Work \;done = force \times distance\\\\Work \;done = 49 \times 2[/tex]

Work done = 98 Nm

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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

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

why is it more painful to walk on gravel with your shoes off then on (3 marks please)

Answers

Answer:

Because shoes protect our feet from some of the most harmful platforms

Gravel has some small pebbles on it sometimes (or other sharp objects)

Gravel is pretty hard.

What happens when the elevator is in free-fall, that is, what is the value for FN and what is the sensation experienced by the person in the elevator? Also, discuss the meaning of FN<0.

Answers

Answer:

Weightlessness

Explanation:

When the elevator is in free fall, this can only occur when the cord of the elevator breaks.

The acceleration of the elevator will be equal to the acceleration due to gravity. That is:

a = g

Then, the body will experience what we called WEIGHTLESSNESS

Where the normal reaction N of the person will tend to zero. That is

N = 0

The value of FN < 0 because the person inside the lift will experience weightlessness.

What is gravitonal force

Answers

Answer:

its something that hold the air for forceing liy by the exgen

Explanation:

Mr Jones launches an arrow horizontally at a rate of 40m/s off of a 78.4 m cliff towards the south, how far south does the arrow travel. (Step 2, you need the previous questions answer to answer correctly

a. 118.4 m south
b 1936 m south
C 2 m south
d 640 m south​

Answers

Answer:c

Explanation:its the answer because its the answer

A force of 20 N to the south is applied to each object below. Which object will undergo the greatest change in momentum?

Answers

Answer:

Hello there! The answer your would be looking for is:

A 33 kg object that is moving north at 10 m/s

Momentum can be defined as the product of mass and velocity. Thus momentum is directly proportional to both velocity and momentum. Thus, the object with greater mass as well as velocity has greater momentum. Therefore, option C is correct.

What is force?

Force can be described as an external agent acting on a body, to change its state of rest or motion. There are several kinds of forces such as magnetic force, frictional force, nuclear force, etc.

In physics, force can be described as the product of the mass and acceleration of the body. Greater mass results in greater force required to be exerted on the object to make it move or stop.

Therefore, when mass or velocity or both increases, the momentum of the object increases as well. Therefore, the larger object moving faster gain greater momentum.

Therefore, when the same force is exerted on the object then, the 41 kg object that is moving north at 12 m/s will undergo the greatest change in momentum.

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Your question was incomplete but most probably the complete question was,

A force of 20 N to the south is applied to each object below. Which object will undergo the greatest change in momentum?

A. A 33 kg object that is moving north at 12 m/s

B. A 41 kg object that is moving north at 10 m/s

C. A 41 kg object that is moving north at 12 m/s

D. A 33 kg object that is moving north at 10 m/s

A T-shirt cannon mounted at the top of an arena needs to fire a t-shirt into the first row, a horizontal distance of 39 meters away. If the cannon launches t-shirts at 12 m/s, how high is the cannon mounted?
A.) 8.9 m
B.) 3.3 m
C.) 53.4 m
D.) 16.2 m

Answers

B.)3.3 would be the answer
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