If a truck has a mass of 100 kg and
a velocity of 35 m/s, what is its
momentum?
A surfer travels for 45 min at an average speed of 0.6 kilometers per min. How far does he travel in that
time?
Answer:
you would do multiplication, d=rt. (distance=d rate=r time=t).
Explanation:
You already have all the variables figure out so just plug in the numbers.
For an easier equation you would just do .6*45=27, the 27 is the total kilometers in 45 minutes.
Select the correct answer from each drop-down menu.
Brenda is a crime scene investigator. She believes there is blood at the scene of the crime but needs to confirm her initial findings. Which of the
following techniques should she use to find probable evidence of blood and confirm that it is indeed blood?
Brenda should uses a
to confirm that it is indeed blood.
test to find probable evidence of blood. If she does so, she should use a
✓ test
Answer:
it is kastle Meyer fasure
calculate your power in watts if you do 500 N*m of work picking up a dumbbell,and takes 5 s to perform that task?
Answer:
100watts
Explanation:
Given parameters:
Workdone = 500Nm
Time taken = 5s
Unknown:
Power in watts = ?
Solution:
Power is the rate at which work is done;
Power =[tex]\frac{workdone }{time}[/tex]
Input the parameters and solve;
Power = [tex]\frac{500}{5}[/tex] = 100watts
what is a force that always resists the relative motion of objects or surfaces?
Friction is the resistance to motion of one object moving relative to another. It is not a fundamental force, like gravity or electromagnetism. Instead, scientists believe it is the result of the electromagnetic attraction between charged particles in two touching surfaces.
Consider a 400g baseball thrown from a height of 1.8m at 35m/s:
a. What is the gravitational potential energy of the baseball?
b. What is the elastic potential energy of the baseball?
c. What is the kinetic energy of the baseball?
d. What is the total mechanical energy of the baseball?
Answer:hhahe
Explanation:
Which characteristic is related to kinetic energy but not potential energy?
A. An object's shape
B. An object's mass
C. An object's speed
D. An object's height
Answer:
c
Explanation:
An object's speed is the characteristic which is related to kinetic energy but not the potential energy. Thus, the correct option is C.
What is Kinetic energy?The kinetic energy of an object is the energy which it possesses due to the motion of the object. Kinetic energy is defined as the work which is needed to accelerate an object of a given mass from rest position to its stated velocity. Having gained this amount of energy during the acceleration of the object, the object maintains this kinetic energy unless its speed changes or it experiences acceleration.
Kinetic energy is directly proportional to the mass of the object and to the square of the velocity of the object, the expression is:
K.E. = 1/2 m × v²
where, KE = Kinetic energy,
m = mass of the object,
v = velocity of the object.
Therefore, the correct option is C.
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What type of System interact with its environment
Answer:
System management
Explanation:
find weight of 45 kg at height 800 km from earth's surface?
Answer:
The weight is 348.4 N
Explanation:
Effect of the Altitude on the Acceleration of Gravity
The effect of gravity decreases with altitude because greater altitudes mean greater distances from the Earth's center.
The practical formula to calculate the gravity as a function of the altitude h is:
[tex]{\displaystyle g_{h}=g_{0}\left({\frac {R_{\mathrm {e} }}{R_{\mathrm {e} }+h}}\right)^{2}}[/tex]
Where go is the conventional gravitational acceleration = [tex]9.80665\ m/s^2[/tex]
R is the Earth's mean radius at the equator = 6,378 Km
h is the altitude.
At a height of h=800 Km, the acceleration of gravity is:
[tex]{\displaystyle g_{h}=9.80665 m/s^2\left(\frac {6,378 Km }{6,378 Km+800Km}}\right)^{2}}[/tex]
Calculating:
[tex]g_h=7.74\ m/s^2[/tex]
The weight is the product of the acceleration of gravity by the mass of the object, thus:
[tex]W=45\ Kg\cdot 7.74\ m/s^2=348.4\ N[/tex]
The weight is 348.4 N
Which expression can be used to calculate centripetal acceleration?
Answer:
B, 4π^2r/T^2
Explanation:
Im on the same quiz, and I'm pretty sure this is the right answer cause I found it on 4 other websites. Good luck
Alguien ayudeme con estos problemas de fisica porfa 1. Una persona recorre 4200 m en 25 minutos, luego 4.54 Km en un hora y finalmente 9000 m en 55 minutos. Halle, el tiempo total empleado en minutos, su velocidad y rapidez. 2. Una partícula en una circunferencia de 40 cm de diámetro da 30 vueltas completas en dos minutos. Halle la velocidad y rapidez de la partícula. 3. Una persona da 500 pasos de 40 cm al oeste, luego 400 pasos de 35 cm al norte y finalmente 300 pasos de 30 cm al este utilizando en total 2 min y 120 segundos. Halle la distancia recorrida, desplazamiento, velocidad y rapidez. 4. Una persona recorre una pista circular de 250 metros de radio, ¿Cuál es su distancia, desplazamiento, velocidad y rapidez en dar media vuelta a la pista si el tiempo empleado es de 90 segundos
Answer:
1) t_total = 140 min , v = 126.7 m / min , 2) v = 0.628 m / s ,
3) v = 0.592 m / s , θ = -39.3 º
Explanation:
Here we have some short kinematics problems
1) d₁ = 4200 m at T1 = 25 min
d₂ = 4.54 km at t2 = 1 h
d₃ = 9000 m at t3 = 55 min
of the statement the direction of each route is the same
Let's reduce the distances to the SI system
d₂ = 4.54 km (1000 m / 1 km) = 4540 m
t₂ = 1 h (60 min / 1 h) = 60 min
Total time is
t_total = t1 + t2 + t3
t_total = 25 + 60 +55
t_total = 140 min
the distance is
d_total = 4200 + 4540 +9000
d_total = 17740 m
with movement it is in one dimension and all displacement is in the same direction, distance equals displacement.
Speed is
v = 17740/140
v = 126.7 m / min
in the direction of travel
Speed is the modulus of speed
| v | = 126.7 m / min
2) the angular velocity is
w = θ / t
In this exercise
t = 2 min (60 s / 1min) = 120 s
θ = 30 rev (2π rad / 1 rev) = 60π rad
w = 60π / 120
w = 1.57 rad / s
linear velocity is
v = w r
v = 1.57 0.40
v = 0.628 m / s
velocity is the speed tangential to the trajectory
3) the data for this exercise are
distance = (number of steps) (distance of one step)
d₁ = 500 0.40 m = 200 m West
d₂ = 400 0.35 m = 140 m North
d₃ = 300 0.30 m = 90 m East
in a total time of t = 2 min and 120 s
t = 2 60 + 120 = 240 s
the distance traveled is
x axis we assume the direction towards + x (East) positive
x_total = d₃ -d₁
x_total = 90-200 = -110 m
y axis
y_total = d₂
y_total = 140 m
we use the Pythagorean theorem to find the distance0
D = √ (x_total² + y_total²)
D = √ (110² + 90²)
D = 142.13 m
The speed is
v = D / t
v = 142.13 / 240
v = 0.592 m / s
speed is
v = 0592 m / s
tan θ = Y / X
θ = tan -1 y / x
θ = tan -1 (90 / (- 110))
θ = -39.3 º
4) This angle is measured on the positive side of the x axis in a clockwise direction
θ = 1.5 rev (2π rad / 1 rev) = 3π rad
t = 90 s
the angular velocity is
w = θ / t
w = 3πi / 90
w = 0.1047 rad / s
the speed is
v = w R
v = 0.14047 250
v = 26.17 m / s
velocity is tangent to path
4. Calculate the momentum for each scenario:
a. .02 kg mass moving at 300 m/s
b. 2 kg mass moving at 40 m/s
c. 200 kg mass moving at 4 m/s
a) Answer:
Momentum = 6 kgm/s
Explanation:
Momentum = mass × velocity
Given that,
mass = 0.02 kg velocity = 300 m/s
Therefore, momentum = (0.02 × 300)kgm/s
Momentum = 6 kgm/s
b) Answer:
Momentum =80 kgm/s
Explanation:
Momentum = mass × velocity
Given that,
mass = 2 kg velocity = 40 m/s
Therefore, momentum = (40 × 2)kgm/s
Momentum = 80 kgm/s
c) Answer:
Momentum = 800 kgm/s
Explanation:
Momentum = mass × velocity
Given that,
mass = 200 kg velocity = 4 m/s
Therefore, momentum = (200 × 4)kgm/s
Momentum =800 kgm/s
many atoms of each element are in one molecule of 2-heptanone?
A chemical structure of a molecule includes the arrangement of atoms and the chemical bonds that hold the atoms together. The 2-HEPTANONE molecule contains a total of 21 bond(s) There are 7 non-H bond(s), 1 multiple bond(s), 4 rotatable bond(s), 1 double bond(s) and 1 ketone(s) (aliphatic).
:/ hope this helps
Please help my test is going on. Please answer both (i) and (ii).
Answer:
i. 7.5 m
ii. 15000 N
Explanation:
Area under vt graph shows the displacement
so area of triangle ABE= 1/2×15×2
=7.5 m
ii) F=ma
here, m=1000kg
and a=v-u/t
=15-0/1
a=15
F=1000×15=15000N
If a train goes at constant speed of 40 miles per hour for one hour, then at constant
speed of 55 miles per hour for two hours, what is its average speed at the train?
Answer: because the train goes 72 mph for 2 hours
Explanation:
does the train go 72mph for 2 hours as well? if so the answer would be 324 because you have to add 90 twice scene it goes at that speed for 2 hours thats 180. assuming the train goes 72 mph for 2 hours it would be 144. then add 180 and 144 and you get 324 merry christmas and a happy newyear to you and your family
Classify each of the following chemical reactions. Upper S plus upper O Subscript 2 right arrow upper S upper O subscript 2. Upper C a upper C l subscript 2 plus 2 upper A g upper N upper O subscript 3 right arrow upper C a (upper n upper O subscript 3) subscript 2 plus 2 upper A g upper C l. Upper Z n plus upper C u upper S upper O subscript 4 right arrow upper C u plus upper Z n upper S upper O subscript 4. 2 upper N a subscript 2 upper O right arrow 4 upper N a plus upper O subscript 2.
Answer:
synthesis and combustion
double replacement
single replacement
decomposition only
in that exact order edge 2020
The presented reactions are:
S + O₂ ⇒ SO₂ Synthesis or combination
CaCl₂ + AgNO₃ = Ca(NO₃)₂ + 2 AgCl Double displacement
Zn + CuSO₄ = Cu + ZnSO₄ Single displacement
2 Na₂O = 4 Na + O₂ Decomposition
Let's consider and classify the following reactions.
S + O₂ ⇒ SO₂
This is a synthesis or combination reaction. 2 compounds combine to form another.
CaCl₂ + AgNO₃ = Ca(NO₃)₂ + 2 AgCl
This is a double displacement reaction. Both compounds exchange their cations and anions.
Zn + CuSO₄ = Cu + ZnSO₄
This is a single displacement reaction. A metal displaces other metal from its salt.
2 Na₂O = 4 Na + O₂
This is a decomposition reaction. A compound decomposes into 2 other compounds.
The presented reactions are:
S + O₂ ⇒ SO₂ Synthesis or combination
CaCl₂ + AgNO₃ = Ca(NO₃)₂ + 2 AgCl Double displacement
Zn + CuSO₄ = Cu + ZnSO₄ Single displacement
2 Na₂O = 4 Na + O₂ Decomposition
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A land rover drops from rest a vertical distance of 1000 m to the surface of planet in 15 seconds what is the magnitude of the acceleration due to gravity on the planet
Answer:
66.6 or it could be 66.66 one of those
Explanation:
Select the correct answer.
Type O negative blood is considered the universal red blood cell donor because it
OA.
Has all 3 types of antigens.
OB.
Has all 3 types of antibodies.
Lacks all 3 types of antigens.
OD
Lacks all 3 types of antibodies.
Answer:
The answer is C.) Lacks all 3 types of antigens.
You and your friends are going on a picnic in a very small car. On your way, the car breaks down and you try to push it to get it to move but in vain. You ask your friends to help you and the car starts moving and revs up. How are you and your friends applying Newton's second law of motion here? What if the car you were traveling in was a large SUV?
Answer:
The pushing of the car by you and your friends is the applied force and when the car moves, it means that the velocity has changed thus causing the movement or acceleration.
Explanation:
Newton's work on forces regarding motion can never be neglected by scientists. Sir Isaac Newton when he was alive, among several of his works he proposed the three laws guiding the forces of motion. In this question we are only going to be treating only one out of the three Newton's Law of motion and that is the second Law Of Newton's laws of motion.
The second Law Of Newton's laws of motion states that the acceleration of an object is directly proportional to the applied force and inversely proportional to the object's mass.
(1). Now, to the question: " How are you and your friends applying Newton's second law of motion here? "
The pushing of the car by you and your friends is the applied force and when the car moves, it means that the velocity has changed thus causing the movement or acceleration.
According to the law, the more the Force, the more the acceleration.
(2). For the second part of the question, " What if the car you were traveling in was a large SUV?"
From the law stated above we see that the acceleration is inversely proportional to the mass, thus if the car is a large SUV, It means that more force is needed to change the car's velocity.
Answer:
When a friend joins you in pushing the small car, the net force on it increases. Therefore, it starts to move. The mass of a SUV is much greater than that of a small car. You would thus need the help of more friends to help push it.
Explanation:
Plato/Edmentum
If the x-component of velocity is 27m/s and the y-component of velocity is -23 m/s, what is the resultant vector?
Answer:
35.47 m/s
Explanation:
[tex]r = \sqrt{( - 23)^{2} + (27) ^{2} }[/tex]
=35.57
An elevator ascends at a constant speed of 4 m/s. How much time is required for the elevator in order to travel 200m upwards?
Answer:
Explanation:
Hope this helps, cheers
..................ok okokok
Answer:
ok
Explanation:
A truck is traveling at a constant speed of 20 m/s through a school zone. At time t = 0 seconds, he passes a hidden police car that is at rest. Five seconds after the truck passes, the police car begins accelerating at a constant rate of 2 m/s in order to catch the truck.
Determine the time at which the speed of the truck is equal to the speed of the police car. Mark this time as t↓1. How will the positions of the police car and the truck compare when they have the same speed and why?
Answer:
Let's define:
The position zero will be in the first point where the truck and police car meet. (So the initial position of both vehicles is zero)
Then, the position equation for the truck is:
T(t) = 20m/s*t
Where t is our variable, time in seconds.
Now, at t = 5s, the car starts accelerating.
Ac(t) = 2m/s^2 for( t ≥ 5s)
For the velocity of the car, we must integrate that.
V(t) = (2m/s^2)*(t - 5) for (t ≥ 5s)
Where i introduced a little change in the variable because the velocity of the car starts to increase for t larger than 5 seconds.
For the position of the car we integrate again.
C(t) = (1m/s)*(t - 5)^2 for (t ≥ 5s)
Now, let's answer the questions:
Determine the time at which the speed of the truck is equal to the speed of the police car.
Then we must have:
V(t) = (2m/s^2)*(t - 5) = 20m/s. (Remember that we only can use times requal or larger than 5 seconds).
2m/s^2*t - 10m/s = 20m/s
2m/s^2*t = 30m/s
t = (30/2) s = 15s
The velocities of both vehicles will be the same after 15 seconds.
t1 = 15s
How the positions will compare at this time?
The easier thing will be to evaluate the position equation of each vehicle in this time:
T(15s) = 20m/s*15s = 300m
C(15s) = (1m/s^2)*(15s - 5s)^2 = (1m/s^2)*(10s)^2 = 100m.
Then we can see that the truck is 200m ahead of the car.
But remember that the police car is accelerating, so the velocity will keep increasing meaning that eventually, the car will catch the truck.
why can't you run from momentum ?
lists the 4 states of matter from most amount of thermal energy to least amount of thermal energy?
Answer:
Explanation:
plasma, gas, liquid, solid
HOW LONG WILL IT TAKE SHINTA TO REACH HIS BROTHERS LOCATION? PLS HELP
Answer:
2 minutes or 120 seconds
Explanation:
The answer is 2 minutes or 120 seconds because 36 km is 36000 m and 36000 divided by 300 is 120, so the answer is 120 seconds, and converted to 2 minutes. Hope it helps!
A driver of a car traveling at -15m/s applies the brakes, causing a uniform acceleration of +2.0m/s2. If the brakes are applied for 2.5s, what is the velocity of the car at the end of the braking period? How far has the car moved during the braking period?
Given :
Initial velocity, u = -15 m/s.
Acceleration , a = 2 m/s².
Time taken to applied brake, t = 2.5 s.
To Find :
The velocity of the car at the end of the braking period.
How far has the car moved during the braking period.
Solution :
By equation :
[tex]v = u+at\\\\v=-15 + 2\times 2.5\\\\v=-10 \ m/s[/tex]
Now, distance covered by car is :
[tex]s=ut+\dfrac{at^2}{2}\\\\s=(-15)(2.5)+\dfrac{2(2.5)^2}{2}\\\\s=-31.25\ m[/tex]
Hence, this is the required solution.
6. The race car in the previous problem slows from
36 m/s to 15 m/s over 3.0 s. What is its average
acceleration?
Answer:
a_AV - 15 m/s – 36 m/s =-70 m/s2. At. 3.0 s.
Explanation:
The average acceleration of the car as it slows down is -5 m/s².
Given is a race car that slows down from 36 m/s to 15 m/s in 3 seconds.
initial velocity [u] = 36 m/s
final velocity [v] = 15 m/s
time taken [Δt] = 3 s
We can calculate the acceleration as follows -
a = Δv/Δt
a = (v - u) / Δt
a = (15 - 30)/3
a = -15/3
a = -5 m/s²
Therefore, the average acceleration of the car as it slows down is
-5 m/s².
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Two classmates, Aisha and Brandon, want to attend two school activities
over the coming weekend. They have one parking pass between them. The
probabilities that the classmates will attend each event are shown in the
table.
Alsha
Brandon
0.65
0.94
Probability of attending
the Saturday activity
Probability of attending
the Sunday activity
0.80
0.43
They decide to let the person more likely to attend both events have the
parking pass. Assuming that attendance at one activity is independent of
attendance at the other, who is more likely to attend both activities?
A. Brandon. He has a 0.40 probability of attending both activities
O B. Brandon. He has a 0.69 probability of attending both activities
C. Aisha. She has a 0.52 probability of attending both activities.
Answer:
The answer is D.
Explanation:
3. What are the two parts of a shadow?
O opaque area and dark area
the umbra and penumbra
O the umbra and
antunghaya
O the illuminated opaque object and the umbra
Answer:
the umbra and penumbra
Explanation:
The two parts of a shadow are the umbra and penumbra.
A shadow is an expression formed when light from a source is cut blocked by an opaque body.
Most shadows are made up of two parts, umbra and penumbra.
The umbra is the darkest, and the innermost part of a shadow. In the penumbra, only a portion of the light source is blocked. It is more of like a partial shadow.The umbra is directly formed by light which impinges on an opaque body and it completely cut off.