Answer:
See explanation
Explanation:
Now we have;
vi = (-30, 0) m/s
vf = (0, 20) m/s
Δvav= vf - vi= (30, 20) m/s
magnitude of Δvav= √30^2 + 20^2 = 36.0 m/s
Direction = tan-1(20/30) = 33.69°
For aav
aav= Δvav/t = (30/5, 20/5) = (6,4) m/s^2
magnitude of aav = √6^2 + 4^2 = 7.2 m/s^2
direction of aav = tan-1(4/6) = 33.69°
A 2 kg toy car moves at a speed of 5 m/s. If a child applies a 5N force for 2 m in the same direction the car is already moving, what is the change in kinetic energy of the car?
Answer:
[tex]10\: \mathrm{J}[/tex]
Explanation:
The kinetic energy of an object is [tex]KE=\frac{1}{2}mv^2[/tex], where [tex]m[/tex] is the mass of the object and [tex]v[/tex] is the velocity of the object.
The toy car's initial kinetic energy is [tex]KE_{i}=\frac{1}{2}\cdot 2\cdot 5^2=25\: \mathrm{J}[/tex].
After the child applies a 5N force on it in the same direction, its velocity will increase but its mass will stay the same.
To find the final velocity of the toy car, we can use kinematic equation [tex]v_f^2=v_i^2+2a\Delta x, \\ v_f=\sqrt{v_i^2+2a\Delta x}[/tex]
We are given [tex]v_i=5\: \mathrm{m/s}[/tex] and [tex]\Delta x = 2\: \mathrm{m}[/tex].
To find acceleration:
[tex]F=ma, a=\frac{F}{m}=\frac{5}{2}=2.5\: \mathrm{m/s^2}[/tex].
Now substitute [tex]v_i=5\: \mathrm{m/s}, \: a=2\: \mathrm{m/s^2}, \: \Delta x = 2\: \mathrm{m}[/tex] into [tex]v_f=\sqrt{v_i^2+2a\Delta x}[/tex] to get [tex]v_f\approx 5.92\: \mathrm{m/s}[/tex].
Using this, we can find the final kinetic energy of the toy car is [tex]KE_f=\frac{1}{2}\cdot 2\cdot 5.92^2[/tex].
Thus, the change in kinetic energy is [tex]KE_f-KE_i=\frac{1}{2}\cdot2\cdot 5.92^2-\frac{1}{2}\cdot 2\cdot 5^2=\fbox{$10\: \mathrm{J}$}[/tex] (one significant figure).
The change in the kinetic energy of the car is 10 J.
The given parameters;
mass of the car, m = 2 kginitial velocity of the car, u = 5 m/sforce applied by the child, F = 5 Ndistance traveled, s = 2 mThe acceleration of the car is calculated as follows;
[tex]F = ma\\\\a = \frac{F}{m} \\\\a = \frac{5}{2} \\\\a = 2.5 \ m/s^2[/tex]
The final velocity of the car is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\v = \sqrt{u^2 + 2as} \\\\v = \sqrt{5^2 \ + \ 2(2.5)(2)} \\\\v = 5.92 \ m/s[/tex]
The change in the kinetic energy of the car is calculated as follows;
[tex]\Delta K.E = \frac{1}{2} m(v^2 - u^2)\\\\\Delta K.E = \frac{1}{2} \times 2 \times (5.92^2\ - \ 5^2)\\\\\Delta K.E = 10 \ J[/tex]
Learn more about kinetic energy here: https://brainly.com/question/1932411
A receiver catches a football on the 50.0 yard line and is tackled 5.42 seconds later on the 12 yard line. What
was the runner's average speed?
Answer:
7.01yard/sec
Explanation:
Given parameters:
Initial position = 50yard
Final position = 12yard
Time = 5.42s
Unknown:
Average speed of runner = ?
Solution:
To solve this problem;
Speed = [tex]\frac{distance}{time}[/tex]
Distance covered = Initial position - final position = 50 - 12 = 38yards
So;
Speed = [tex]\frac{38}{5.42}[/tex] = 7.01yard/sec
Solve each of the following problems to 3 sig figs and correct Sl units, showing all work.
1. A cart with a mass of 45.0 kg is being pulled to the right with a force of 250 N giving it an
acceleration of 1.30 m/s2. The wheels of the cart are locked and the cart must be dragged.
a) Draw a free body diagram of the cart.
b) Calculate the net force acting on the cart.
c) Create a force table and fill it in.
d) Find the coefficient of kinetic friction.
Answer:
every number to 3 sf = 1) 45.0 2) 250 3) 1.30
Explanation:
your welcome :)
Suppose it takes a constant force a time of 6.0 seconds to slow a 2500 kg truck
from 26.0 m/sec to 18.0 m/sec. What is the magnitude of the force? Give
your answer in scientific notation rounded correctly.
Answer:
[tex]3.3\cdot 10^3\:\mathrm{N}[/tex]
Explanation:
Impulse on an object is given by [tex]\mathrm{[impulse]}=F\Delta t[/tex].
However, it's also given as change in momentum (impulse-momentum theorem).
Therefore, we can set the change in momentum equal to the former formula for impulse:
[tex]\Delta p=F\Delta t[/tex].
Momentum is given by [tex]p=mv[/tex]. Because the truck's mass is maintained, only it's velocity is changing. Since the truck is being slowed from 26.0 m/s to 18.0 m/s, it's change in velocity is 8.0 m/s. Therefore, it's change in momentum is:
[tex]p=2500\cdot 8.0=20,000\:\mathrm{kg\cdot m/s}[/tex].
Now we plug in our values and solve:
[tex]\Delta p=F\Delta t,\\F=\frac{\Delta p}{\Delta t},\\F=\frac{20,000}{6}=\fbox{$3.3\cdot 10^3\:\mathrm{N}$}[/tex](two significant figures).
A bottle rocket is fired off and has an acceleration of 14.5 m/s2 for the 2.25s until it burns out. If it starts at rest, what distance does it cover?
Answer:
S = 16.3125m
Explanation:
Given the following data;
Acceleration, a = 14.5m/s²
Time, t = 2.25secs
Since the bottle rocket starts from rest, its initial velocity is 0m/s.
To find the distance S, we would use the second equation of motion.
S = ut + ½at²
Substituting into the equation, we have
S = 0(2.25) + ½*14.5*2.25
S = 0 + 7.25*2.25
S = 16.3125m
Therefore, the bottle rocket covered a distance of 16.3125 meters.
In designing buildings to be erected in an area prone to earthquakes, what relationship should the designer try to achieve between the natural frequency of the building and the typical earthquake frequencies?
A) The natural frequency of the building should be exactly the same as typical earthquake frequencies.
B) The natural frequency of the building should be almost the same as typical earthquake frequencies but slightly lower
C) The natural frequency of the building should be very different frem typical earthquake frequencies
D) The natural frequency of the building should be almost the same as typical earthquake frequencies but slightly higher.
Answer:
C) The natural frequency of the building should be very different from typical earthquake frequencies
Explanation:
We shall apply the concept of resonance in this problem .
When a body is applied an external harmonic force ( forced vibration) such that natural frequency of body is equal to frequency of external force or periodicity of external force , the body vibrates under resonance ie its amplitude of vibration becomes very high .
In the present case if natural frequency of building becomes equal to the earthquake's frequency ( external force ) , the building will start vibrating with maximum amplitude , resulting into quick collapse of the whole building . So to avoid this situation , natural frequency of building should be very different from typical earthquake frequencies .
PLS HELP ME!
A motorist is traveling 40ms-¹ and applies brakes and slow down at a rate of 2ms-² the available distance for the the motorist to stop is 400m will the motorist be able to stop?
Answer:
[tex] \underline{ \boxed{ yes}}\\[/tex]
Explanation:
[tex]given : initial \: velocity \: (u )= 40 {ms}^{ - 1} \\ given : final \: velocity \: (u )= 0 {ms}^{ - 1} \\ given : - (acceleration) \: (a_r) = 2 {ms}^{ - 2} \\ given : distance \: (s) \: = \: ? : \\ but \: {v}^{2} = {u}^{2} + 2( a)s\\ {0}^{2} = {40}^{2} + 2( - 2)s \\ - {40}^{2} = - 4s \\ s = \frac{ - {40}^{2} }{ - 4} \\ s = \frac{1600}{4} \\s = 400 \: m[/tex]
A 62 kg student, starting from rest, slide down an 10.6 m high water slide. How fast is he going at the bottom of the slide? Use g = 10 m/s2
Answer:
14.6m/s
Explanation:
Given parameters:
Mass of the student = 62kg
Initial velocity = 0m/s
Height of slide = 10.6m
g = 10m/s²
Unknown:
Speed at the bottom of the slide = ?
Solution:
The speed at the bottom of the slide is the final velocity;
v ² = u² + 2gh
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity
h is the height
v² = 0² + 2x 10 x 10.6
v² = 212
v = 14.6m/s
I need this done by tonight!! Can anyone help me please? Answer these 4 questions
Answer:
1. 14 g of chocolate mixture.
2. 24 fl oz of chocolate milk
3. 10 cups of chocolate milk.
4. 12½ cups.
Explanation:
From the question given above, the following data were obtained:
1 TBSP = 7 g
1 Cup = 8 fl oz
2 Table spoons (TBSP) for 1 cup (8 fl oz) of milk.
1. Determination of the mass of chocolate mixture in 1 cup of chocolate milk.
From the question given above,
1 Cup required 2 Table spoons (TBSP)
But
1 TBSP = 7 g
Therefore,
2 TBSP = 2 × 7 = 14 g
Thus, 1 Cup required 14 g of chocolate mixture.
2. Determination of the number fl oz of chocolate milk in 3 cups
1 Cup = 8 fl oz
Therefore,
3 Cups = 3 × 8
3 Cups = 24 fl oz
Thus, 24 fl oz of chocolate milk are in 3 cups.
3. Determination of the number of cups of chocolate milk produce from 20 TBSP.
2 TBSP is required to produce 1 cup.
Therefore,
20 TBSP will produce = 20/2 = 10 Cups.
Thus, 10 cups of chocolate milk produce from 20 TBSP.
4. Determination of the number of cups obtained from 100 fl oz chocolate milk.
8 fl oz is required to produce 1 cup.
Therefore,
100 fl oz will produce = 100 / 8 = 12½ cups.
Thus, 12½ cups is obtained from 100 fl oz chocolate milk.
You perform nine (identical) measurements of the acceleration of gravity (units of m/s2): 10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90. The true value is 9.81. Calculate the standard error of your results to ONE significant digit.
Answer:
0.01
Explanation:
Given the data:
10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, 9.90
True value = 9.81
Mean value :
Σx / n
Sample size, n = 9
(10.1 + 9.87 + 9.76 + 9.91 + 9.75 + 9.88 + 9.69 + 9.83 + 9.90) / 9
= 88.69 / 9
= 9.854
Standard deviation (σ) :
Sqrt (Σ(X - m)² / n)
[(10.1 - 9.854)^2 + (9.87 - 9.854)^2 + (9.76 - 9.854)^2 + (9.91 - 9.854)^2 + (9.75 - 9.854)^2 + (9.88 - 9.854)^2 + (9.69 - 9.854)^2 + (9.83 - 9.854)^2 + (9.90 - 9.854)^2] / 9
Sqrt(0.113824 / 9)
Sqrt(0.0126471)
σ = 0.1124593
Standard Error = σ / sqrt(n)
Standard Error = 0.1124593 / 9
Standard Error = 0.0124954
Standard Error = 0.01 ( 1 significant digit)
Which is the best explanation for why Toms technique works ?
How long ago did most Middle Eastern countries gain their independence?
A.
10-20 years ago
B.
50-100 years ago
C.
200-300 years ago
D.
400-500 years ago
Please select the best answer from the choices provided
A
B
C
D
Answer:
the correct answer is B
Explanation:
Answer:
.B
Explanation:
This is the answer on edge 2021
Have a good day!
Lisa skips 5 m North to the playground. She realizes she forgot to bring water so she turns around and goes 3 m South to the convenience store
Answer:
-2 South
Always subtract where they ended first is what my teacher said
Which image shows the difference between the speed of molecules in hot and cold water? Explain your answer choice.
HELP ME,EVERYONE!!!!!!!! :(
Answer:
the answer is B
Explanation:
I think its B because on the top it shows the molecule speed and A looks like the water is cold, C shows that the hot
water is cooler, and D shows that both are cold
A spring stretches by 15cm when a mass of 300g hangs down from it,if the spring is then stretched an additional 10cm and released, calculate;the spring constant,the angular velocity, amplitude of oscillation, maximum velocity, maximum acceleration of the mass,period, frequency
Answer:
0.1 m
Explanation:
It is given that,
Mass of the object, m = 350 g = 0.35 kg
Spring constant of the spring, k = 5.2 N/m
Amplitude of the oscillation, A = 10 cm = 0.1 m
Frequency of a spring mass system is given by :
Time period:
John runs 3 km north then walks 2 km south. What is his total distance traveled and displacement?
Answer:
the total distance is 5km and the displacement is 1km
Explanation:
The total distance would be the addition of John running both ways so 3 km, 2 km.
However since he only walked back from a distance of 3 km to 2 km, he would be displaced 1 km because displacement is more like the position from the original point.
Think about 2 km as a positive value for the first part of the question and a negative value for the second part.
what is the pressure on a swimmer 50 m below the surface of a lake
Answer:
P = 490500 [Pa]
Explanation:
The pressure at the bottom of a vessel and even of a lake or sea can be calculated by means of the following hydrostatic equation.
[tex]P=Ro*g*h[/tex]
where:
P = pressure [Pa] (units of pascal)
Ro = water density = 1000 [kg/m³]
g = gravity acceleration = 9.81 [m/s²]
h = elevation = 50 [m]
Now replacing:
[tex]P=1000*9.81*50\\P=490500[Pa][/tex]
the diameter of the wheels on your car ( including the tires) is 25 inches. you are going to drive 250 miles today. each of your wheels is goingnto turn by an angle of
What is the instantaneous velocity of a freely falling object 11 s after it is released from a position of rest
Answer:
v= -107.8 m/s
Explanation:
Since the object is in free fall, this means that is moving at an accceleration equal to the one due to gravity.Since it starts at rest, we can apply the definition of acceleration, rearranging terms as follows:[tex]v_{f} = v_{o} + a*t = a*t = -g*t = 9.8m/s2*11s = -107.8 m/s (1)[/tex]
(Assuming as positive the upward direction)What is the function
of second plate in
parallel plate capacitor?
A 71-kg swimmer dives horizontally off a 500-kg raft. If the diver's speed immediately after leaving the raft is 6m/s, what is the corresponding raft speed?
Answer:
The answer is below
Explanation:
Momentum is used to measure the quantity of motion in an object. Momentum is the product of mass and velocity.
Momentum = mass * velocity
The principle of conservation of momentum states that momentum cannot be created or destroyed but can be transferred. Therefore the momentum before and after an action is equal.
Initial momentum = Final momentum
Let m be the mass of the diver, M be the mass of the raft, u be the initial velocity of the diver, U be the initial velocity of the raft, v be the final velocity of the diver and V be the final velocity of the raft.
m = 71 kg, M = 500 kg, v = 6 m/s
Initial both the raft and diver are at rest, hence u and U is zero, hence:
mu + MU = mv + MV
71(0) + 500(0) = 71(6) + 500(V)
0 = 426 + 500(V)
500(V) = -426
V = -426/500
V = -0.852 m/s
a student lift a 25kg mass at vertical distance of 1.6m in a time of 2.0 seconds. a. Find the force needed to lift the mass (in N ). b. Find the work done by the student (in J). c. Find the power exerted by the student (in W)
Answer:
a. F = 245 Newton.
b. Workdone = 392 Joules.
c. Power = 196 Watts
Explanation:
Given the following data;
Mass = 25kg
Distance = 1.6m
Time = 2secs
a. To find the force needed to lift the mass (in N );
Force = mass * acceleration
We know that acceleration due to gravity is equal to 9.8
F = 25*9.8
F = 245N
b. To find the work done by the student (in J);
Workdone = force * distance
Workdone = 245 * 1.6
Workdone = 392 Joules.
c. To find the power exerted by the student (in W);
Power = workdone/time
Power = 392/2
Power = 196 Watts.
Can a single atom be considered a molecule?
A:only if the atom is found in water
B:no, it takes two or more atoms bonded to create a molecule
C:only if it is an oxygen atom floating in the air
D:yes, all atoms are made up of many different molecules
What relationship must exist between an applied force and the velocity of a moving object if uniform circular motion is to result?
Answer:
See explanation
Explanation:
Centripetal force is defined as the inward force required to keep an object moving with a constant speed in a circular path.
The magnitude of this force depends on the mass of the object, radius of the object and the velocity of the body.
So we can write;
F = mv^2/r
I will give brainly
Defend Democritus' work on the atom and its contribution to the modern atomic model.
help me help me help me
If an atom of oxygen has an atomic number of eight that means...…
E. there are 8 protons
F. there are 8 neutrons
G. it weighs 8 amu
H. it is in group 8
If a car is traveling at an average speed of 20 m/s, how long will it take to travel 500 meters?
A. 0.04 seconds
B. 25 seconds
C. 520 seconds
D. 10,000 seconds
Answer:
B. 25 seconds
Explanation:
500÷20=25
describe measurement in our daily life
____ is factual information not subject to bias.
Interpretation
Analysis
Data
Opinion
Answer:
Data
Explanation:
Data is factual information not subject to bias.
This ultimately implies that, data connotes fact, thus, it is an information that is credible, accurate, a statement of truth, evidential and proven.
In Computer programming, a data dictionary can be defined as a centralized collection of information on a specific data such as attributes, names, fields and definitions that are being used in a computer database system.
In a data dictionary, data elements are combined into records, which are meaningful combinations of data elements that are included in data flows or retained in data stores.
This ultimately implies that, a data dictionary found in a computer database system typically contains the records about all the data elements (objects) such as data relationships with other elements, ownership, type, size, primary keys etc. This records are stored and communicated to other data when required or needed.
