A truck of mass 1600 kg is traveling at 15 m/s. In order to avoid running over a squirrel in the middle of the road, the truck driver begins to brake. What must the braking force on the truck be if the truck comes to stop in 2 s? *


a)12000n
b) 8000n
c)7000n
d)6000n

Answers

Answer 1
I think A is correct

Related Questions

Who watching all star draft? Luka better get picked first ong

Answers

Na dame Dolla fo surrrreeeeee

why type of volcano is built almost entirely from ejected lava fragments

Answers

Answer:

Shield volcanoes

Explanation:

A 20g bullet moving at 200m/s hits a bag of sand and comes to rest in 0.011s, calculate the momentum of the bullet just before hitting the bag​

Answers

Answer:

momentum = mass * velocity ... kg•m/s

acceleration = v / t = 200 m/s / 11 ms ... m/s^2

force = mass * acceleration = .02 kg * (200 m/s / 11 ms) N

Explanation:

Why might video games increase creativity while the use of cell phones, the internet, or computers do not?

Answers

Explanation:

Video games are developed around a structure that is unique from your usual media, where there is a sense purpose in mind when playing a game, as decided by the player, and it allows them to explore creative options in order to solve scenarios, depending on the genre. In the use of phones, internet, or computers, this structure is more rare and diverse from video games, which does not lean more toward a creative purpose to build from. That idea gives people the inspiration and discover skills they never knew they even had.

4. When you are holding a book, energy is stored between the book and the Earth.
This type of energy is called
potential energy.
A. Elastic potential energy
B. Chemical potential energy
C. Gravitational potential energy
D. Kinetic energy

Answers

Answer:

gravitational potential energy

Which word best completes the sentence?

Select the word from the drop-down menu

He is quite
Choose...
despite never having left his smalL TOWEN

Answers

Answer:

it’s cosmopolitan

Explanation:

k12

An object that is spinning, but not orbiting anything, has zero angular momentum.


True
False

Answers

The answer is fals:)

A physics student sits in a chair. The chair pushes up on the student's body. Identify the other force of the interaction force pair.

Answers

Answer:

The other force is the weight of the student.

Explanation:

With respect to Newton's third law of motion, for the student to sit and balance on the chair, there must be two equal and opposite forces involved. The student applies his/ her weight on the chair which acts downwards, while the chair applies an equal but opposite force to the weight of the student.

The force applied by the chair on the student's body is counter balanced by the student's weight. Note that, if the weight of the student is greater than the opposing force from the chair, the chair would collapse.

A 10 kg object moving to the left collides with and sticks to a 3 kg object moving to the right. Which of the following is true of the motion of the combined objects immediately after the collision?

Answers

Answer:

"Cannot be determined," if that's an answer choice. It depends on the velocities of both objects, since momentum=mass*velocity.

Explanation:

An elevator filled with passengers has a mass of 1663 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s2 for 3.25 s. Calculate the tension in the cable (in N) supporting the elevator. 18317 Correct: Your answer is correct. N (b) The elevator continues upward at constant velocity for 8.92 s. What is the tension in the cable (in N) during this time

Answers

Answer:

(a) T = 18309.63 N = 18.31 KN

(b) T = 16314.03 N = 16.314 KN

Explanation:

(a)

The tension in an elevator while moving upward with some acceleration is given by the following formula:

[tex]T = m(g+a)\\[/tex]

where,

T = Tension = ?

m = mass = 1663 kg

g = acceleration due to gravity = 9.81 m/s²

a = acceleration of elevator = 1.2 m/s²

Therefore,

[tex]T = (1663\ kg)(9.81\ m/s^2 + 1.2\ m/s^2)\\[/tex]

T = 18309.63 N = 18.31 KN

(b)

Constant velocity means no acceleration. So, in that case, the tension will be equal to the weight of the elevator:

[tex]T = mg\\T = (1663\ kg)(9.81\ m/s^2)\\[/tex]

T = 16314.03 N = 16.314 KN

a car moved 120km to the north. what is its displacement?

Answers

120Km. They are the same

An 5kg object is released from rest near the surface of a planet. The vertical position of the object as a function of time is shown in the graph. All frictional forces are considered to be negligible. What is the closest approximation of the weight of the object.
a) 300N
b) 30N
c) 5N
d) 150N

Answers

Answer:

The correct option is b: 30 N.

Explanation:

First, we need to find the acceleration due to gravity (a):

[tex] y_{f} - y_{0} = v_{o}t - \frac{1}{2}a(\Delta t)^{2} [/tex]   (1)

Where:

[tex]y_{f}[/tex]: is the final vertical position (obtained from the graph)

[tex]y_{0}[/tex]: is the initial vertical position (obtained from the graph)

v₀: is the initial speed = 0 (it is released from rest)

Δt: is the variation of time (from the graph)

From the graph, we can take the following values of height and time:

t₀ = 0 s → [tex]t_{f}[/tex] = 5 s

y₀ = 300 m → [tex]y_{f}[/tex] = 225 m

Now, by entering the above values into equation (1) and solving for "a" we have:

[tex] a = 2\frac{y_{0} - y_{f}}{(t_{f} - t_{0})^{2}} = 2\frac{300 m - 225 m}{(5 s - 0)^{2}} = 6 m/s^{2} [/tex]

Finally, the weight of the object is:

[tex] W = ma = 5 kg*6 m/s^{2} = 30 N [/tex]

Therefore, the correct option is b: 30 N.

I hope it helps you!                                                                                                            

An object which is dropped from a certain height has zero (0) initial velocity.

Answers

Answer:

0 m/s

Explanation:

if an object is dropped we know the initial velocity is zero when in free fall

I'm asking for a quick favor. I'm trying to understand an equation that has to do with Projectile motion, Bernoulli's principle, and Magnus Effect. Basically focused on understanding air resistance on a projectile. I would like to discuss this privately rather than have it on this public forum. I'll give you 100 of my points if you help.

Answers

Explanation:

Projectile motion, Bernoulli's principle, and Magnus Effect.

Sure I would be happy to discuss projectile motion!

I'll do it if you mark brainliest :) I need the points thanks

3. If a spring extends by 3 cm when a 4 N weight is suspended from it, find the extension
when the weight is changed to
(a) 8 N
(b) 10 N
(c) 14 N

Answers

8N - 6cm
10N - 7.5cm
14N - 10.5cm

PLEASE CLICK ON THIS IMAGE I NEED HELP

Answers

Answer:

Second option

Explanation:

"Uniform" pretty much means the same thing happens.

An object is moving with constant non-zero velocity. Which of thw following statements about it must be true

Answers

Answer:

The net force on the object is zero.

Explanation:

An object is moving with constant non-zero velocity. If velocity is constant, it means that the change in velocity is equal to 0. As a result, acceleration of the object is equal to 0. Net force is the product of mass and acceleration. Hence, the correct option is (d) "The net force on the object is zero".


Name the four layers of the atmosphere (in order starting at the bottom

Answers

Answer:

Troposphere, stratosphere, mesosphere and thermosphere. The next region is the exosphere, but that region is 500+ km from the Earth's surface.

[tex]\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}[/tex]

Actually Welcome to the Concept of the Atmosphere.

The four layers of Atmosphere starting from bottom are as follows:

1.) Troposphere - The troposphere is the lowest layer of our atmosphere. Starting at ground level, it extends upward to about 10 km (6.2 miles or about 33,000 feet) above sea level.

2.) Stratosphere - The next layer up is called the stratosphere. The stratosphere extends from the top of the troposphere to about 50 km (31 miles) above the ground.

3.) Mesosphere - Above the stratosphere is the mesosphere. It extends upward to a height of about 85 km (53 miles) above our planet. Most meteors burn up in the mesosphere.

4.) Thermosphere - The layer of very rare air above the mesosphere is called the thermosphere. High-energy X-rays and UV radiation from the Sun are absorbed in the thermosphere, raising its temperature to hundreds or at times thousands of degrees.

Select the correct answer.

A pair of pliers is an example of a simple machine with two levers. Which part of the pliers is the fulcrum?



A. A
B. B
C. C
D. D
E. E

Answers

Answer:

There isnt enough in your question to answer the question bro, like we need a picture or something bro.

Explanation:

You don't have a image attached


1. A Zambeef delivery track travels 18 km north, 10 km east, and 16 km south. What is its final displacement from the origin?

Answers

Answer:

2km

Explanation:

Given data

We are told that the direction traveled are

North>>>East>>>South

Hence the displacement is defined as the distance away from the initial position is

Initial position =18km

FInal position = 16km

The displacement = 18-16= 2km

Hence the displacement is 2km

****PLEASE HELP**** WILL MARK BRAINLIEST
Assuming that voltage remains constant, what happens to the current in a
wire if the length of the wire increases?
O A. The current decreases.
OB. The current alternates between high and low values.
O C. The current increases.
O D. The current is not affected by a change in wire length.

Answers

Answer:

The Current decreases

Explanation:

HOPE THIS HELPS!

The earth rotates through one complete revolution every 24 hours. Since the axis of rotation is perpendicular to the equator, you can think of a person standing on the equator as standing on the edge of a disc that is rotating through one complete revolution every 24 hours. Find the angular and linear velocity of a person standing on the equator. The radius of earth is approximately 4000 miles.

Answers

Answer:

ω = 7.27 x 10⁻⁵ rad/s

v = 467.99 m/s

Explanation:

First, we will find the angular velocity of the person:

[tex]Angular\ Velocity = \omega = \frac{Angular\ Distance}{Time}[/tex]

Angular distance covered = 1 rotation = 2π radians

Time = (24 h)(3600 s/ 1 h) = 86400 s

Therefore,

[tex]\omega = \frac{2\pi\ rad}{86400\ s}[/tex]

ω = 7.27 x 10⁻⁵ rad/s

Now, for the linear velocity:

[tex]v = r\omega[/tex]

where,

v = linear velocity = ?

r = radius of earth = (4000 miles)(1609.34 m/1 mile) = 6437360 m

Therefore,

[tex]v = (6437360\ m)(7.27\ x\ 10^{-5}\ rad/s)[/tex]

v = 467.99 m/s

When individuals improve their aerobic endurance, which body systems are affected?

Answers

When individuals improve their aerobic endurance, which body systems are affected?

Anwser: your heart and lungs

In a mass spectrometer, a specific velocity can be selected from a distribution by injecting charged particles between a set of plates with a constant electric field between them and a magnetic field across them (perpendicular to the direction of particle travel). If the fields are tuned exactly right, only particles of a specific velocity will pass through this region undeflected. Consider such a velocity selector in a mass spectrometer with a 0.105 T magnetic field.
Part (a) What electric field strength, in volts per meter, is needed to select a speed of 3.8 × 106 m/s?
Part (b) What is the voltage, in kilovolts, between the plates if they are separated by 0.75 cm?

Answers

Answer:

a).[tex]$3.99 \times 10^5 \ v/m$[/tex]

b). 2.9925 kV

Explanation:

Given :

For mass spectrometer

The magnetic field = B

B = 0.105 T

a).  Given speed, v = [tex]$3.8 \times 10^6 \ m/s$[/tex]

We known

[tex]$\frac{E}{B}=v$[/tex]

∴ [tex]$E= 3.8 \times 10^6 \times 0.105$[/tex]

      [tex]$=3.99 \times 10^5 \ v/m$[/tex]

b). Now spectrometer, d = 0.75 cm

                                      [tex]$d=0.75 \times 10^{-2} \ m$[/tex]

We known

[tex]$E=\frac{V}{d}$[/tex]

[tex]$V = E\times d$[/tex]

[tex]$V = 3.99 \times 10^5 \times 0.75 \times 10^{-2}$[/tex]

[tex]$V = 2.9925 \times 10^3 \ V$[/tex]

   = 2.9925 kV

 

A 125 kg mail bag hangs by a vertical rope 3.3 m long. A postal worker then displaces the bag to a position 2.2 m sideways from its original position, always keeping the rope taut.
1) What horizontal force is necessary to hold the bag in the new position?
2) As the bag is moved to this position, how much work is done by the rope?
3) As the bag is moved to this position, how much work is done by the worker?

Answers

Answer:

1) the required horizontal force F is 1095.6 N

2) W = 0 J { work done by rope will be 0 since tension perpendicular }

3) work is done by the worker is 1029.4 J

Explanation:

Given that;

mass of bag m = 125 kg

length of rope [tex]l[/tex] = 3.3 m

displacement of bag d = 2.2 m

1) What horizontal force is necessary to hold the bag in the new position?

from the figure below; ( triangle )

SOH CAH TOA

sin = opp / hyp

sin[tex]\theta[/tex] = d / [tex]l[/tex]

sin[tex]\theta[/tex] = 2.2/ 3.3

sin[tex]\theta[/tex] = 0.6666

[tex]\theta[/tex] = sin⁻¹ ( 0.6666 )

[tex]\theta[/tex]  = 41.81°

Now, tension in the string is resolved into components as illustrated in the image below;

Tsin[tex]\theta[/tex] = F  

Tcos[tex]\theta[/tex] = mg

so

Tsin[tex]\theta[/tex] / Tcos[tex]\theta[/tex] = F / mg

sin[tex]\theta[/tex] / cos[tex]\theta[/tex] = F / mg

we know that; tangent = sine/cosine

so

tan[tex]\theta[/tex] = F / mg

F = mg tan[tex]\theta[/tex]

we substitute

Horizontal force F = (125kg)( 9.8 m/s²) tan( 41.81° )

F = 1225 × 0.8944

F = 1095.6 N

Therefore, the required horizontal force F is 1095.6 N

2)  As the bag is moved to this position, how much work is done by the rope?

Tension in the rope and displacement of mass are perpendicular,

so, work done will be;

W = Tdcos90°

W = Td × 0

W = 0 J { work done by rope will be 0 since tension perpendicular }

3) As the bag is moved to this position, how much work is done by the worker

from the diagram in the image below;

SOH CAH TOA

cos = adj / hyp

cos[tex]\theta[/tex]  = ([tex]l[/tex] - h) / [tex]l[/tex]

we substitute

cos[tex]\theta[/tex]  = ([tex]l[/tex] - h) / [tex]l[/tex]  = 1 - h/[tex]l[/tex]

cos[tex]\theta[/tex] = 1 - h/[tex]l[/tex]

h/[tex]l[/tex] = 1 - cos[tex]\theta[/tex]

h = [tex]l[/tex]( 1 - cos[tex]\theta[/tex] )

now, work done by the worker against gravity will be;

W = mgh = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )

W = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )

we substitute

W = (125 kg)((9.8 m/s²)(3.3 m)( 1 - cos41.81° )

W = 4042.5 × ( 1 - 0.745359 )

W = 4042.5 × 0.254641

W = 1029.4 J

Therefore,  work is done by the worker is 1029.4 J

The chemical equation for the decomposition of potassium chlorate into potassium chloride and oxygen gas is
KCIO: |_ KCI + ___ 02
Which coefficients correctly balances the equation?
A) 4.4,3
B 3,3,2
C) 2,2,3
D
2, 1,3
E The equation is already balanced.

Answers

Answer:

Option C. 2, 2, 3

Explanation:

__KClO₃ —> __ KCl + __O₂

The above equation can be balance as illustrated below:

KClO₃ —> KCl + O₂

There are 2 atoms of O on the right side and 3 atoms on the left side. It can be balance by writing 2 before KClO₃ and 3 before O₂ as shown below:

2KClO₃ —> KCl + 3O₂

There are 2 atoms of K on the left side and 1 atom on the right side. It can be balance by writing 2 before KCl as shown below:

2KClO₃ —> 2KCl + 3O₂

Thus, the equation is balanced. The coefficients are: 2, 2, 3

A 50 kg mass is sitting on a frictionless surface. An unknown constant force called force A pushes the mass for 2 seconds until the mass reaches a velocity of 3 m/s. If the 50 kg mass is now pushed by an unknown force B and reaches the velocity of 3 m/s in 4 seconds, compare the impulse delivered to the mass when acted upon by force A with the impulse delivered to the mass when acted on by force B? *

A) The impulse delivered to the mass when acted upon by force A is greater
B) The impulse delivered to the mass when acted upon by force B is greater
C) The impulse is the same in each case
D) We need to know the value of force A and force B in order to determine this

Answers

Answer:

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

A wave with a frequency of 5Hz travels a distance of 40mm in 2 seconds.What is the speed of the wave​

Answers

Answer:

20mm per second

Explanation:

A tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 m/s2 for g. Assume r = 15 m and h = 5 m.) W = 1.59 J

Answers

Answer:

3.9 × 10^7 J

Explanation:

Given that a tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 m/s2 for g. Assume r = 15 m and h = 5 m.) W = 1.59 J

Solution

Since the tank is half full, the height = 2.5m

Pressure = density × gravity × height

Pressure = 900 × 9.8 × 2.5

Pressure = 22050 Pascal

The cross sectional area of the pump will be area of a circle.

A = πr^2

A = π × 15^2

A = 706.858 m^2

Using the formula

Density = mass/volume

Mass = density × volume

Mass = 900 × 706.86 × 2.5

Mass = 1590.435

Energy = mgh

Energy = 1590.435 × 9.8 × 2.5

Energy = 38965657.8 J

Since the work done = energy

Therefore, the work done = 3.9 × 10^7 J

We intend to measure the open-loop gain (LaTeX: A_{open}A o p e n ) of an actual operational amplifier. The magnitude of LaTeX: A_{open}A o p e n is in the range of 106 V/V. However, the signal generator in measurement setup can supply minimal voltage of 1 mV, and the oscilloscope used at amplifier output can measure maximal voltage level of 10 V. Can you design a simple measurement setup using this signal generator and oscilloscope, and accurately measure the LaTeX: A_{open}A o p e n

Answers

Answer:

voltage divider,  R₂ = 1000 R₁

measuring the output in the resistance R₁

Explanation:

Let's analyze the situation, in an op amp in open gain loop, the gain is maximum G = 10⁶ V / V

in this case the signal generator gives a minimum wave of 1 10⁻³ V, after passing through the amplified it becomes 10³ V which saturates the oscilloscope.

To solve this problem we must use a simple voltage divider, for this we use the fact that in a series circuit the voltage is the sum of the voltages of each element.

If we use two resistors whose relationship is

            R₂ / R₁ = 10³

            R₂ = 1000 R₁

When measuring the output in the resistance R₁ we have the desired divider, with a tolerance range, for the minimum output of the generator (1 10⁻³V) we have a reading of V = 1 V in the oscilloscope, for which we can use voltage up to 10V on the generator

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