Answer:
The speed of the completed train is 0.0[tex]\bar 6[/tex] m/s
The direction of the completed train is South
Explanation:
The given parameters are;
The mass of the locomotive, m₁ = 125,000 kg
The initial speed of the locomotive, v₁ = 1.0 m/s
The mass of the gondola cars, m₂ = 1,750,000 kg
The initial sped of the gondola cars, v₂ = 0
Let v₃ represents the speed of the completed train. From the principle of conservation of linear momentum, we have;
m₁·v₁ + m₂·v₂ = (m₁ + m₂)×v₃
Substituting the known values, gives;
125,000 × 1.0 + 1,750,000 × 0 = (125,000 + 1,750,000) × v₃
∴ v₃ = (125,000 × 1.0 + 1,750,000 × 0)/(125,000 + 1,750,000) = 1/15 = 0.0[tex]\bar 6[/tex]
v₃ = 0.0[tex]\bar 6[/tex] m/s
The speed of the completed train = 0.0[tex]\bar 6[/tex] m/s
The direction of the completed train = The direction of the locomotive = South.
Charges are of two kinds—negative charges and ________________ charges. __________ amounts of both kinds of charges are found in every piece of matter.
Answer: the first is positive and second is equal
Explanation:
Answer: 1. Positive
Explanation:
Because there is negative charges and positive charges
Donny, who has a mass of 75.0 kg, is riding at 25.0 m/s in his truck when he suddenly slams on the brakes to avoid hitting a squirrel crossing the road. His seatbelt brings his body to a stop in 0.500 seconds. What force does the seatbelt exert on him?
Answer:
F = 3750 N
Explanation:
Given that,
Mass of Donny, m = 75 kg
Initial speed, u = 25 m/s
He suddenly slams on the brakes to avoid hitting a squirrel crossing the road, final speed = 0
Time, t = 0.5 s
We need to find the force the seatbelt exert on him. The force is given by :
F = ma
a is acceleration
[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{75\times (0-25)}{0.5}\\\\=-3750\ N[/tex]
So, the force is 3750 N.
The force the seatbelt exert on him is 3750 N
Using Newton's third law
f = ma
where
m = mass
a = acceleration
a = v - u / t
where
v = final velocity
u = initial velocity
t = time
Therefore,
f = m(v - u / t)
m = 75 kg
u = 25 m/s
t = 0.500 secs
v = 0 m/s (The car will stop because he applied a brake)
f = 75 × (0 - 25 / 0.5)
f = 75 × -25 / 0.5
f = 75 × -50
f = - 3750
f = 3750 N
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An object starts at 16 m/s with an acceleration of 4.5 m/s? How far does it go in 9.0 seconds?
A 52 kg person is running 2.5 m/s. What is their momentum?
Answer:
130kg/ms
Explanation:
given data mass,52kgvelocity,2.5m/smomentum?from momentumM=mvMass=52kg
Velocity=2.5m/s
Momentum=?
We know that,
Momentum=Mass × velocity
Or,momentum=52kg× 2.5m/s
Or,momentum=130 kgm/s
So,the momentum is 130 kgm/s.
What is the speed of an eagle that travels 200 meters in 4 seconds? A. 800 m/s2 B. 800 m/s C. 50 m/s2 D. 50 m/s
Answer:
The answer is D.
Explanation:
You have to apply distance formula :
[tex]distance = speed \times time[/tex]
[tex]let \: d = 200,t = 4[/tex]
[tex]200 = s \times 4[/tex]
[tex]4s = 200[/tex]
[tex]s = 200 \div 4[/tex]
[tex]s = 50 \: m {s}^{ - 1} [/tex]
An eagle travels 200 meters in 4 seconds
To find:-Speed of the eagle
Solution:-HereDistance=d=200m
Time=t=4s
As we know[tex]{\boxed{\sf Speed ={\dfrac{Distance{}_{(d)}}{Time{}_{(t)}}}}}[/tex]
Substitute the values[tex]\qquad\quad {:}\longmapsto\sf Speed=\dfrac {200}{4}[/tex]
[tex]\qquad\quad {:}\longmapsto\sf Speed=50m/s [/tex]
[tex]\therefore{\underline{\boxed{\bf Speed\:of\:eagle=50m/s}}}[/tex]
Hence Correct option is DA warehouse worker pushed a cart weighing 4.50 kg to the top of an inclined plane. Initially, the cart was 0.670 m above the floor. If the top of the inclined plane is 2.70 m above the floor, calculate the work done by gravity as the worker pushed the cart to the top of the plane.
Answer:
- 89.523J
Explanation:
We solve the above question using work done(Potential Energy) formula
The formula =
-(mgh)
Where m = Mass = 4.50J
g = Acceleration due to gravity = 9.8m/s²
We are told in the question:
Initially, the cart was 0.670 m above the floor. If the top of the inclined plane is 2.70 m above the floor
Hence, the height =
2.70m - 0.670m = 2.03m
Work done = -(2.03m × 4.5kg× 9.8m/s²)
= - 89.523J
a cannon launches a 3.0 kg pumpkin with 110J of kinetic energy. what is the pumpkin’s speed?
Answer:
v = 8.56 [m/s]
Explanation:
The kinetic energy can be calculated by means of the following equation.
[tex]E_{k}=\frac{1}{2}*m*v^{2}[/tex]
where:
m = mass = 3 [kg]
v = velocity [m/s]
Ek = kinetic energy [J]
[tex]110 = \frac{1}{2} *3*(v^{2} )\\v^{2} = 220/3\\v=\sqrt{73.333}\\v=8.56[m/s][/tex]
The normal force of a parked car is 15,000 Newtons. The coefficient of static friction between the rubber of the tires and the asphalt of the road is 0.75. What is the maximum static friction force?
Answer:
11250 N
Explanation:
From the question given above, the following data were obtained:
Normal force (R) = 15000 N
Coefficient of static friction (μ) = 0.75
Frictional force (F) =?
Friction and normal force are related by the following equation:
F = μR
Where:
F is the frictional force.
μ is the coefficient of static friction.
R is the normal force.
With the above formula, we can calculate the frictional force acting on the car as follow:
Normal force (R) = 15000 N
Coefficient of static friction (μ) = 0.75
Frictional force (F) =?
F = μR
F = 0.75 × 15000
F = 11250 N
Therefore, the frictional force acting on the car is 11250 N
PLEASE HELP ME IM TIMED
tarzans mass is 75kg. calculate his weight
Answer:
75kg=165.346697lbs
Explanation: F A T
An object that is experiencing two vertical forces (upwards and a downwards) is moving downward with a constant speed. What can be concluded about the strength of the two forces?
Answer:
Both upward and downward forces are equal
Explanation:
When an object is acted upon by two vertical forces ; upward and downward, and the object moved at a constant speed, then the net force acting upon the object is zero(upward force + downward force = 0). This is because an object will only move at a constant velocity if the opposing forces cancels each other out. Since the falling object is already in motion, and maintaining a constant velocity, there is no for required to keep the body in motion and as such the Velocity at which the body is moving remains constant.
A book that weighs 5 N sits on a table. What force does the table apply to the book?
Answer:
E =F.d =[1/2]mv^2
mad = [1/2]mv^2
d= v^2/2a ,v=u+at , v^2 = [at]^2 since u=0
So d = at^2/2
F = ma= 20a= 50 , a=5/2 and t=2
so d = [5/2][2^2]/2=5
Explanation:
Every action has an equal and opposite reaction. It is an action-reaction principle. Therefore the table exerts a force of 5 N on the book in order to be in stable condition.
What is Newton's third law of motion?Newton's third law of motion state that every action has an equal and opposite reaction. It is an action-reaction principle. It stated that the force always exists in a pair.
Therefore the table exerts a force of 5 N on the book in order to be in stable condition.
The given data in the problem is ;
W is the weight of the book sits on table = 5N
N is the normal force on the book
From the equilibrium equation ;
Weight -Normal force on the book =0
Weight =Normal force on the book
The normal force on the book =5N
Hence the table exerts a force of 5 N on the book in order to be in stable condition.
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When you mix a solution, the mass of the substances before and after you mix them should be the same.true or false
Answer:
True
Explanation:
The mixing substances can be taken to be the reactants and the resulting mixture as the product. When the substances change state during mixing, the mass of the substances does not change meaning that the mass of the product will equal the mass of the reactants .This is true due to the law of conservation of mass.
Answer:
TRUE
Explanation:
because blablablabla blablolbleep scientific talk
what is the mass and volume of 1000kg/m3 of water?
Answer: The mass would be 1000m3 and the volume would be 1000kg
Explanation:
*NEED THIS ANSWERED ASAP*
run this moving man simulation to get a visual understanding of all that you learned in this lesson about position velocity and acceleration during this activity you will explain these Concepts first click the introduction tab, and answer the question. set a small constant V on the velocity slider and check the velocity Vector box. click play, watch, and describe the motion in terms of what you see happening with the man's displacement, velocity and acceleration.
Answer:
When the velocity is set to a constant value, the displacement increases with time but the acceleration remains zero
Explanation:
plato
When the velocity is set to a constant value, the displacement increases with time but the acceleration remains zero.
What is Velocity?Velocity is the directional speed of a object in motion as an indication of its rate of change in position as observed from a particular frame of reference and as measured by a particular standard of time.What is Displacement?Displacement is defined as the change in position of an object. It is a vector quantity and has a direction and magnitude.What is Acceleration?Acceleration, rate at which velocity changes with time, in terms of both speed and direction. A point or an object moving in a straight line is accelerated if it speeds up or slows down.When the velocity is set to a constant value, the displacement increases with time but the acceleration remains zero.
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Two students are studying for a Science Test. On the table, they had a glass of water and identical cell phones. Both of their phones had run out of battery, therefore they were both charging in the same wall outlet. One student’s charging cord was 3ft and the other student’s charging cord was 9ft. which of the phones charged the fastest using Ohm’s Law. Cite evidence to support your theory.
Answer:
The student with the 3ft charger
Explanation:
Because the electricity from the 3ft charger has a shorter distance to mooves than the 9ft one
Select three effective methods for helping a friend who is showing self harming behavior
Answer:
Avoid judging or criticizing your friend.
Let your friend know that you care and want to help
Share the ways you cope with stress and other problems in your life
I had this question
At the local destruction derby a 400 kg Toyota moving at 10 m/s collides with an 800 kg Chevy. Both are at rest after the collision. What was the velocity of the Chevy before the collision?
Answer:
[tex]u_2 = -5m/s[/tex]
Explanation:
Given
Before Collision
Toyota
[tex]mass = m_1 = 400kg[/tex]
[tex]iniital\ velocity = u_1 =10m/s[/tex]
Chevy
[tex]mass = m_2=800kg[/tex]
[tex]initial\ velocity = u_2 = ??[/tex]
After Collision
Both Toyota and Chevy
[tex]final\ velocity = v = 0m/s[/tex]
Required
Determine the initial velocity of Chevy
This question will be answered using the following law of conservation of momentum which states that:
[tex]m_1u_1 + m_2u_2 = (m_1 + m_2)v[/tex]
Substitute values for m1, m2, u1 and v
[tex]400 * 10 + 800 * u_2 = (400 + 800) * 0[/tex]
[tex]4000 + 800u_2 = (1200) * 0[/tex]
[tex]4000 + 800u_2 = 0[/tex]
Collect Like Terms
[tex]800u_2 = 0 - 4000[/tex]
[tex]800u_2 = -4000[/tex]
Divide through by 800
[tex]\frac{800u_2 = -4000}{800}[/tex]
[tex]u_2 = \frac{-4000}{800}[/tex]
[tex]u_2 = -5m/s[/tex]
The velocity of Chevy before collision was 5m/s in the opposite direction of Toyota
what is the KE of a 1.00 kg hammer swinging at 20.0 m/s? 200 Joules
I know the answer I just need help understanding it.
Answer:
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 1250 kg
Velocity (v) = 20.0 m/s.
Kinetic energy (K.E) =?
Kinetic energy is simply defined as energy possed by a body in motion. Mathematically, it is expressed as:
K.E = ½mv²
Where:
K.E is the kinetic energy
m is the mass of the object
v is the velocity of the object.
Thus, we can obtain the kinetic energy of the automobile by using the above formula as illustrated below:
Mass (m) = 1250 kg
Velocity (v) = 20.0 m/s.
Kinetic energy (K.E) =?
K.E = ½mv²
K.E = ½ × 1250 × 20²
K.E = 625 × 400
K.E = 250000 J
Therefore, the kinetic energy of the automobile is 250000 J
Please help! Will mark brainliest!
Answer:
im pretty sure its d
Explanation:
A car’s brakes decelerate it at a rate of -2.20 m/s2. If the car is originally travelling at 17 m/s and comes to a stop, then how far, in meters, will the car travel during that time?
Answer:
x = 65.68 [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 = 0 (comes to stop)
Vo = initial velocity = 17 [m/s]
a = desaceleration = -2.2 [m/s²]
x = displacement [m]
Now replacing, we have:
[tex](0)^{2}=(17)^{2}-2*(2.2)*x\\4.4*x = 289\\x = 65.68 [m][/tex]
What is the period of a rotating object if it spins 24 times in 13 seconds?
Answer:
The period is 0.54 seconds
Explanation:
Period (T)
Is the time required for a rotating object to make one complete revolution around a circular path.
If it takes a time t to complete n revolutions, the period is:
[tex]\displaystyle T=\frac{t}{n}[/tex]
The rotating object spins n=24 times in t=13 seconds, thus its period is:
[tex]\displaystyle T=\frac{13}{24}[/tex]
T = 0.54 sec
The period is 0.54 seconds
a person covers equal half distance at speed V and remaining half at speed V1 and V2 in equal interval of time .find average speed.
Answer:
Average speed = ( 2V + V1 + V2)/4
Explanation:
Given that a person covers equal half distance at speed V and remaining half at speed V1 and V2 in equal interval of time .find average speed.
Since the distance is covered at equal intervals of time, and
Speed = distance/time
For the first half distance,
V = distance/t
Cross multiply
Distance = Vt
For the second half distance
(V1 + V2)/2 = distance/t
Distance = t(V1 + V2)/2
The average speed = total distance/ total time.
Average speed = [Vt + t( V1 + V2)/2] ÷ 2t
Average speed = (2Vt + V1t + V2t)/4t
Average speed = t( 2V + V1 + V2)/4t
Time t will cancel out
Average speed = ( 2V + V1 + V2)/4
a motorbike can travel 1000 meters in 10 minutes calculate how car it can travel in 1 sec.
Answer:
1.67meter
Explanation:
if it can travel 1000 meters in 10 minutes, 10 minutes are 600 secs (10×60), 1000÷600 is 1.67
Answer:
HI
Explanation:
PLEASE HELP ME! IM TIMED
Answer:
Toothpaste
Because it is viscous
according to newtons third law of motion what happened to the back of a skateboard when the person pushes down on the front of the skateboard
Answer:I think it would start moving
Explanation:
Find the kinetic energy of a 3.5 kg cat that has a momentum of 0.22 kg • m/s.
A) 3.98 x 10^-5 J
B) 4.32 x 10^-4 J
C) 6.91 x 10^-3 J
D) 7.22 x 10^-2 J
Answer:
6.19*10^-3 J
Explanation:
Just took the test. Kiss me
How much heat in kcal is required to change 0.5 kg of ice, originally at - 10 0 * C into steam at 110 C?Constants needed in the problemLatent heat of fusion=79.7 kcal/kg Specific heat of ice=0.5 kcal/kg/K ; Latent heat of vaporization ation = 539 kcal/kg ; Specific heat of water 1.0 kcal/kg/K Specific heat of ieam=0.480 kcal/kg
Answer:
Q = 364.25 kcal
Explanation:
In this question, we will have to calculate the heat absorptions for different steps of temperature rise and phase change. And then we will ad them to calculate total heat absorbed.
1. RISE IN TEMPERATURE OF ICE:
First, the temperature of ice will be increased from - 10°C to 0 °C. Heat absorbed during this process will be given as:
Q₁ = mC₁ΔT₁
where,
Q₁ = Heat absorbed while increasing temperature of ice = ?
m = mass of ice = 0.5 kg
C₁ = specific heat of ice = 0.5 kcal/kg k
ΔT₁ = change in temperature of ice = 0 - (-10) = 10 k
Therefore,
Q₁ = (0.5 kg)(0.5 kcal/kg.k)(10)
Q₁ = 2.5 kcal
2. MELTING OF ICE:
Now, the melting of ice will occur at 0°C and the heat absorbed during this process will be:
Q₂ = m(Latent Heat of Fusion of Ice)
where,
Q₂ = heat Absorbed during melting of ice = ?
Therefore,
Q₂ = (0.5 kg)(79.7 kcal/kg)
Q₂ = 39.85 kcal
3. RISE IN TEMPERATURE OF WATER:
Now, the temperature of water will be increased from 0°C to 100 °C. Heat absorbed during this process will be given as:
Q₃ = mC₃ΔT₃
where,
Q₃ = Heat absorbed while increasing temperature of water = ?
m = mass of water = 0.5 kg
C₃ = specific heat of water = 1 kcal/kg k
ΔT₃ = change in temperature of ice = 100 - 0 = 100 k
Therefore,
Q₃ = (0.5 kg)(1 kcal/kg.k)(100 k)
Q₃ = 50 kcal
4. VAPORIZATION OF WATER:
Now, the vaporization of water will occur at 100°C and the heat absorbed during this process will be:
Q₄ = m(Latent Heat of Vaporization of Water)
where,
Q₄ = heat Absorbed during vaporization of water = ?
Therefore,
Q₄ = (0.5 kg)(539 kcal/kg)
Q₄ = 269.5 kcal
5. RISE IN TEMPERATURE OF STEAM:
Now, the temperature of steam will be increased from 100°C to 110 °C. Heat absorbed during this process will be given as:
Q₅ = mC₅ΔT₅
where,
Q₅ = Heat absorbed while increasing temperature of steam = ?
m = mass of steam = 0.5 kg
C₅ = specific heat of steam = 0.48 kcal/kg k
ΔT₅ = change in temperature of ice = 110 - 100 = 10 k
Therefore,
Q₅ = (0.5 kg)(0.48 kcal/kg.k)(10 k)
Q₅ = 2.4 kcal
Hence, the total heat absorbed to change 0.5 kg of ice at - 10°C into steam at 110°C will be:
Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅
Q = 2.5 kcal + 39.85 kcal + 50 kcal + 269.5 kcal + 2.4 kcal
Q = 364.25 kcal
Which of the following are possible non-SI units to describe the rate of
acceleration? (Choose all that apply)
1) meters per second per minute
2) miles per kilometer per
second
3) miles per hour per second
4) kilometers per hour squared
5) meters per hour per kilometer
6) kilometers per minute
The possible non-SI units to describe the rate of acceleration is the kilometers per hour squared. That is option 4.
What is acceleration?Acceleration is the relationship that exists between velocity and time. It is defined as the change of velocity in both speed and direction with respect to time.
The formula for acceleration is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt.
Velocity is defined as the rate of objects position with time.
The SI units of velocity = meter/seconds
Substitute for velocity in first equation:
a = meter/sec/sec
a = meter/sec²
But 1000meters = 1 Km
3600 secs = 1 hour
Therefore, the possible non-SI units to describe the rate of acceleration is the kilometers per hour squared.
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6th grade science I mark as brainliest.
Answer:
Solution:-Distance =400m
Time=20s
We need to find speedAs we know that
[tex]{\boxed{\sf speed \dfrac {Distance {}_{(d)}}{Time {}_{(t)}}}}[/tex]
Substitute the values[tex]\LARGE\leadsto\sf speed=\dfrac {400}{20}[/tex]
[tex]\LARGE\leadsto\sf 20m/s[/tex]