A 15-kg bicycle carrying a 50-kg boy is traveling at a speed of 5 m/s. What is the KE of the bicycle and the boy?

Answers

Answer 1

Answer: Together as one moving package, their mass is (15 + 50) = 65 kg.

   Kinetic energy = (1/2) (mass) (speed)²

   = (1/2) (65 kg) (5 m/s)²

   = (32.5 kg) (25 m²/s²)

   =    812.5 kg-m²/s² = 812.5 joules .

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A toy cannon uses a spring to project a 5.24-g soft rubber ball. The spring is originally compressed by 5.01 cm and has a force constant of 8.08 N/m. When the cannon is fired, the ball moves 15.8 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.031 0 N on the ball.With what speed does the projectile leave the barrel of the cannon?

Answers

Answer:

Speed will be equal to 1.40 m/sec

Explanation:

Mass of the rubber ball m = 5.24 kg = 0.00524 kg

Spring is compressed by 5.01 cm

So x = 5.01 cm = 0.0501 m

Spring constant k = 8.08 N/m

Frictional force f = 0.031 N

Distance moved by ball d = 15.8 cm = 0.158 m

Energy gained by spring

$KE=(1)/(2)kx^2=(1)/(2)* 8.08* 0.0501^2=0.0101J$

Energy lost due to friction

$W=Fd=0.031* 0.158=0.0048J$

So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J

This energy will be kinetic energy

$(1)/(2)mv^2=0.0052$

$(1)/(2)* 0.00524* v^2=0.0052$

v = 1.40 m/sec

Determine the conversion factor between (a) km/h and mi/h, (b) m/s and ft/s, and (c) km/h and m/s.

Answers

(a) km/h and mi/h,
1 km/hr (km hr-1) = 0.278 m/s (m s-1) = 0.621mi/hr (mi hr-1) = 0.912 ft/s (ft s-1)
(b) m/s and ft/s,
Speed of sound in air at 0 oC (32 oF)= 331 m/s (m s-1) = 1090 ft/s (ft s-1) =41 mi/hr (mi hr-1)
(c) km/h and m/s. 1km/hr (km hr-1) = 0.278 m/s (m s-1)

(a) 1 mi/h = 1.60934 km/h

(b) 1 ft/s = 0.3048 m/s

(c) 1 m/s = 3.6 km/h

Further explanation

Acceleration is rate of change of velocity.

$\large {\boxed {a = (v - u)/(t) } }$

$\large {\boxed {d = (v + u)/(2)~t } }$

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem !

$1 ~ foot = 30.48 ~ cm$

$1 ~ inch = 2.54 ~ cm$

$1 ~ foot = 12 ~ inches$

$1 ~ mile = 1609.34 ~ metres$

Question (a):

$1 ~ mi/h = (1 ~ miles)/(1 ~ hour)$

$1 ~ mi/h = (1.60934 ~ km)/(1 ~ hour)$

$\large {\boxed {1 ~ mi/h = 1.60934 ~ km/h} }$

Question (b):

$1 ~ ft/s = (1 ~ foot)/(1 ~ s)$

$1 ~ ft/s = (0.3048 ~ m)/(1 ~ s)$

$\large {\boxed {1 ~ ft/s = 0.3048 ~ m/s} }$

Question (c):

$1 ~ m/s = (1 ~ m)/(1 ~ s)$

$1 ~ m/s = (1/1000 ~ km)/(1/3600 ~ h)$

$1 ~ m/s = (3600)/(1000) ~ km/h$

$\large {\boxed {1 ~ m/s = 3.6 ~ km/h} }$

Learn more

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate

What are 3 types of cells that helper T cells produce upon being an alarmed of an antigen? Thanks for your help!! :)

Answers

Whenever these are an alarming due to a foreign antigen, the Helper T cells, Memory T cells, and Cytotoxic T cells are being produced. Helper T cells to recognize the specific foreign antigen that is entering the body, Memory T cells to remember the specific foreign antigen that is attacking, and Cytotoxic T cells that are for the destruction of the foreign cells and produce cytokines to attract macrophages

A car climbs 10.0 kilometers up a hill that inclines 8.0 degrees. What is the car's vertical displacement?

Answers

$s=10km\n\alpha=8^o\n\ny=?\n\\sin\alpha=(y)/(s)\Rightarrow y=s\sin\alpha\n\ny=10\sin8^o\n\ny=10\cdot 0,1392\n\n\y=1,392km$

In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be balanced? The answer is 13.4 In the mobile what is the value for m3 to the nearest hundredth of a kilogram?

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

The first mass is $m_1 = 0.42kg$

The second mass is $m_2 = 0.47kg$

From the question we can see that at equilibrium the moment about the point where the string holding the bar (where $m_1 \ and \ m_2$ are hanged ) is attached is zero

Therefore we can say that

$m_1 * 15cm = m_2 * xcm$

Making x the subject of the formula

$x = (m_1 * 15)/(m_2)$

$= (0.42 * 15)/(0.47)$

$x = 13.4 cm$

Looking at the diagram we can see that the tension T on the string holding the bar where $m_1 \ and \ m_2$ are hanged is as a result of the masses ( $m_1 + m_2$ )

Also at equilibrium the moment about the point where the string holding the bar (where ( $m_1 +m_2$ ) and $m_3$ are hanged ) is attached is zero

So basically

$(m_1 + m_2 ) * 20 = m_3 * 30$

$(0.42 + 0.47) * 20 = 30 * m_3$

Making $m_3$ subject

$m_3 = ((0.42 + 0.47) * 20 )/(30 )$

$m_3 = 0.59 kg$

Which statement(s) correctly compare the masses of protons, neutrons, and electrons?

Answers

FULL ANSWER

Neutrons and protons make up a nucleus and they are in the middle of an atom. Electrons surround the nucleus. Opposite charges attract each other. Because of this, protons do not attract other protons and electrons do not attract other electrons. Instead, protons attract electrons and electrons attract protons. When someone looks at the elements on the periodic table, they can see how many protons are present by looking at the atomic number. Since atoms have to have an equal amount of each, if an element contains 36 protons, it also has 36 electrons.

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