A student makes the following claim, "Acceleration is when an object changes speed, so it can be discussed as a scalar quantity." The error here is that acceleration is said to be done when either speed of that object changes or direction of that object changes. Hence , acceleration is not a scaler quantity.
Scalar quantities are quantities that are described only by a magnitude. They do not have a direction of action.
A vector quantity is defined as the physical quantity that has both directions as well as magnitude.
Acceleration is said to be occurred in two cases :
when the object changes its speed
or when the object changes its direction
since , acceleration depends upon both direction and magnitude ,hence it is a vector quantity not a scaler quantity.
for example : a stone attached to a string moving in a circular motion at a constant speed will be considered in accelerated motion because it is constantly changing its direction.
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Modern roller coasters have vertical loops like the one shown in the figure. The radius of curvature is smaller at the top than on the sides so the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats.
1. What is the speed of the roller coaster, in meters per second, at the top of the loop if the radius of curvature there is 14 m and the downward acceleration of the car is 1.1g? Note that g here is the acceleration due to gravity.
2. The beginning of this roller coaster is at the top of a high hill. If it started from rest at the top of this hill, how high, in meters, above the top of the loop is this initial starting point? You may assume there is no friction anywhere on the track.
3. If it actually starts 7.5 m higher than your answer to the previous part (yet still reaches the top of the loop with the same velocity), how much energy, in joules, did it lose to friction? Its mass is 1800 kg.
1 ) The speed of the roller coaster = 12.28 m / s
2 ) Height of the hill above the top of the loop = 7 m
3 ) Energy lost due to friction = 132 KJ
1 ) The speed of the roller coaster,
[tex]a_{c}[/tex] = v² / r
[tex]a_{c}[/tex] = Centripetal acceleration
v = Linear velocity
r = Radius
r = 14 m
[tex]a_{c}[/tex] = 1.1 g = 1.1 * 9.8
[tex]a_{c}[/tex] = 10.78 m / s²
v² = [tex]a_{c}[/tex] * r
v² = 10.78 * 14
v² = 150.9
v = 12.28 m / s
2 ) Initial starting point,
Considering hill as 1 and the loop as 2,
v1 = 0
h2 = 2 r = 2 * 14
h2 = 28 m
∑ [tex]F_{y}[/tex] = m [tex]a_{c}[/tex]
[tex]F_{N}[/tex] + [tex]F_{g}[/tex] = m [tex]a_{c}[/tex]
0 + m g = m v2² / r
v2² = g r
According to law of conservation of energy,
E1 = E2
m g h1 + 1 / 2 m v1² = m g h2 + 1 / 2 m v2²
m g h1 + 0 = 28 m g + 1 / 2 m g r
h1 = 28 + 1 / 2 ( 14 )
h1 = 35 m
Height of the hill above the top of the loop = h1 - h2
Height of the hill above the top of the loop = 35 - 28
Height of the hill above the top of the loop = 7 m
3 ) Energy lost due to friction,
h1 = 35 + 7.5
h1 = 42.5 m
m = 1800 kg
v2² = g r
v2² = 9.8 * 14
v2² = 137.2 m / s
Since energy is lost due to friction,
KE1 + U1 = KE2 + U2 + W
0 + m g h1 = 1 / 2 m v2² + m g h2 + W
( 1800 * 9.8 * 42.5 ) = ( 0.5 * 1800 * 137.5 ) + ( 1800 * 9.8 * 28 ) + W
749700 = 123750 + 493920 + W
W = 749700 - 617670
W = 132030 J
W = 132 KJ
Therefore,
1 ) The speed of the roller coaster = 12.28 m / s
2 ) Height of the hill above the top of the loop = 7 m
3 ) Energy lost due to friction = 132 KJ
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A 4-kg ball traveling westward at 25 m/s hits a 15-kg ball at rest. The 4-kg ball bounces east at 8.0 m/s. What is the speed and direction of the 15-kg ball? What is the impulse of the second ball?
Given:
The mass of the first ball is,
[tex]m_1=4\text{ kg}[/tex]The initial velocity of the first ball towards West is,
[tex]u_1=25\text{ m/s}[/tex]The mass of thr second ball is,
[tex]m_2=15\text{ kg}[/tex]the second object is initially at rest.
The final velocity of the first ball is,
[tex]v_1=-8.0\text{ m/s}[/tex]we are taking West as positive.
Applying momentum conservation principle we can write,
[tex]m_1u_1+m_2\times0=m_1v_1+m_2v_2[/tex]Substituting the values we get,
[tex]\begin{gathered} 4\times25+0=4\times(-8.0)+15\times v_2 \\ v_2=\frac{100+32}{15} \\ v_2=8.8\text{ m/s} \end{gathered}[/tex]THe final velocity of the second ball is towards East and the magnitude is 8.8 m/s.
The impulse of the Second ball is,
[tex]\begin{gathered} I=m_2v_2-m_2\times0 \\ =15\times8.8 \\ =132\text{ kg.m/s} \end{gathered}[/tex]A plane is traveling with a velocity of 70 miles/hr with a direction angle of 24 degrees. The wind is blowing at 25 miles/hr with a direction angle of 190 degrees. What is the vertical component of the wind velocity? Round your answer to the nearest whole number.
Wind velocity:
25 m/h with a direction angle of 190°.
Vertical component:
25 sin 190 = -4.34 m/s = - 4 m/s
which are neutrally charged, are found in thenucleus of the atom.
Given:
Nucleus of the atom
Required:
Neutra
Timmy walks 5 m North, 3m West, and finally 1 m South. What is his displacement from his starting point?
Timmy walks 5 m North, 3m West, and finally 1 m South then his displacement from the starting point would be 5 meters in the northwest direction.
What is displacement?Displacement describes this shift in location and it is calculated with the help of the initial and the final position of the object.
As given in the problem If Timmy walks 5 m North, 3m West, and finally 1 m South ,
The resultant displacement of the Timmy = √(4² + 3²)
= 5 meters
Thus, the resultant displacement of the Timmy would be 5 meters
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A 77 kg student traveling in a car with a constant velocity has a kinetic energy of 1.7 104 J. What is the speedometer reading of the car in km/h? answer with:_____km/h
ANSWER
75.65 km/h
EXPLANATION
Given:
• The student's mass, m = 77 kg
,• The kinetic energy of the student in the car, KE = 1.7 x 10⁴ J
Find:
• The speed read in the speedometer of the car, which is the speed of the student, v (in km/h)
The kinetic energy of an object with mass m, traveling at a speed v, is,
[tex]KE=\frac{1}{2}mv^2[/tex]Solving for v,
[tex]v=\sqrt{\frac{2KE}{m}}[/tex]Replace the known values and solve,
[tex]v=\sqrt{\frac{2\cdot1.7\cdot10^4J}{77kg}}\approx21.013m/s[/tex]Note that because the kinetic energy is given in Joules - which is equivalent to kg*m²/s², the speed we found is in m/s. Now, knowing that there are 3600 seconds in 1 hour and that 1 km is equivalent to 1000 m, we can convert this to km/s,
[tex]v=21.013\frac{m}{s}\cdot\frac{3600s}{1h}\cdot\frac{1km}{1000m}\approx75.65km/h[/tex]Hence, the speedometer reading of the car is 75.65 km/h, rounded to the nearest hundredth.
12. How could extreme heat (resulting from Climate Change) affect human andanimal life?
ANSWER:
The answer is given in the step by step of the question
STEP-BY-STEP EXPLANATION:
Extreme heat can affect human and animal life in the following ways:
• Animals are sensitive to changes in temperature, as much or more than humans. Think how bad you feel when you have a fever. They also suffer this type of discomfort as a consequence of climate change.
,• Climate change is pushing many species to the limit. They lack water to drink or suffer from temperatures in which they are not comfortable.
,• Provocation of forest fires damaging the habitat of animals and humans
carts, bricks, and bands
2. Which of the following conclusions are specifically supported by the data in Table 1?
a. A constant mass causes the acceleration value to increase.
b. An increase in the number of bricks causes the acceleration to decrease.
c. An increase in the length of the rubber band causes the acceleration to increase.
d. An increase in the number of rubber bands causes an increase in the acceleration.
The conclusions that are specifically supported by the data in Table 1 is that An increase in the number of rubber bands causes an increase in the acceleration. That is option D.
What is acceleration?Acceleration is defined as the rate at which the velocity of a moving object changes with respect to time which is measured in meter per second per second (m/s²).
From the table given,
Trial 1 ----> 1 band = 0.24m/s²
Trial 2 ----> 2 bands = 0.51 m/s²
Trial 3 ----> 3 bands = 0.73 m/s²
Trial 4 -----> 4 bands = 1.00 m/s²
This clearly shows that increase in the number of bands increases the acceleration of one brick that was placed on the cart.
This is because increasing the number of rubber bands has the effect of doubling the force leading to an effective increase in velocity of the moving cart.
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What is the difference in the path of a ball tossed straight up in the air by a passenger on a a bus from the view point of the passenger, and the person on the street?
The difference in the path of a ball tossed straight up in the air by a passenger on a a bus from the view point of the passenger, and the person on the street is that the ball would appear to the passenger to be making an up and down movement when the ball is thrown up, while the stationary observer will actually see the ball moving along a parabolic path.
What is a parabolic path?A parabolic path is described as a Kepler orbit with the eccentricity equal to 1 and is an unbound orbit that is exactly on the border between elliptical and hyperbolic.
The Parabolic path is also defined as the angle of trajectory of a projectile.
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Which is negatively charged?A. protonB. nucleusC. electronD. neutron
Protons, electrons an neutrons are the particles that make up atoms.
Protons have a positive electric charge, electrons have a negative electric charge an neutrons are electrically neutral.
The nucleus of an atom is made of protons and neutrons, so, its electric charge is positive an proportional to th
In the setup of a cart pulled by a hanging mass, without friction, we have the following information: the mass of the cart is 0.53 Kg., and the hanging mass is 0.077 Kg.
Determine the acceleration of the cart (in m/s2).
The acceleration of the cart
a=1.2431m/s^2
What is acceleration?Generally, The equation for Newton's second law of motion is
2nd law of motion,
Fnet=m a
on hanging mass,
m_1 g-T=m_1 a
m_1 g-m_2 a=m_1 a
Acceleration, [tex]$a=\frac{m_1 g}{m_1+m_2}$[/tex]
[tex]&a=\frac{0.077 \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^2}{0.077 \mathrm{~kg}+0.53 \mathrm{~kg}} \\[/tex]
a=1.2431m/s^2
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21. An object m is tied to one end of a string, moves in a circle with a constant speed v
on a horizontal frictionless table. The second end of the string is connected to a big
mass M and goes through a small hole in the table. What is the value of M if it stays
in equilibrium?
I
(B) v²/rmg
(A) mv²/rg
(C) rg/mv²
(D) mv²r/g
The value of M that goes through a small hole in the table if it stays in equilibrium is mv²/rg ( A )
The force acting on object m is the centripetal force.
[tex]F_{c}[/tex] = m v² / r
Mass = m
Velocity = v
Radius = r
The force acting on object M is the gravitational force,
[tex]F_{g}[/tex] = M g
g = Acceleration due to gravity
Since the system is at equilibrium,
[tex]F_{c}[/tex] - [tex]F_{g}[/tex] = 0
m v² / r = M g
M = m v² / r g
Therefore, the value of M if it stays in equilibrium is mv²/rg ( A )
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A car travels 400 km in the first 4.5 hours of a trip. It stops for an hour and then travels final 300 km in 2.5 hours. Find the average speed of the car.
Given data:
Distance traveled by car in t_1=4.5 hr is s_1=400 km.
Distance traveled by car in t_2=1 hr is s_2=0 km (as the car was stopped).
Distance traveled by car in t_3=2.5 hr is s_3=300 km.
The average speed is given as,
[tex]\begin{gathered} v_{avg}=\frac{\text{ total distance traveled}}{\text{total time taken}} \\ =\frac{s_1+s_2+s_3}{t_1+t_2+t_3} \end{gathered}[/tex]Substitute all known values,
[tex]\begin{gathered} v_{avg}=\frac{(400\text{ km})+(0\text{ km})+(300\text{ km})}{(4.5\text{ hr})+(1\text{ hr})+(2.5\text{ hr})} \\ =87.5\text{ km/h} \end{gathered}[/tex]Therefore, the average speed of the car is 87.5 km/h.
orce and Motion Unit TestUse the following scenario to answer the question.Taj and Micah chose to go bowling. Taj rolled the ball toward the pins first, knocking them all down.Which of the following is affecting these objects?point)O Gravity is affecting these objects.O An unbalanced force is affecting the objects.O Inertia is affecting these objects.O A balanced force is affecting the objects.
So lets go through all four answer choices.
The easiest to choose is whether gravity is affecting these objects. Assuming that there is some sort of gravity that would pull the pins down, gravity does affect these objects
Second is inertia.
We know that if an object has inertia, it will try to resist moving/coming to rest. In this case, we know that the pins have inertia because the pins fell over, so we know that the pins do have inertia
The last part is whether these objects have an unbalanced or balanced force. If a balanced force did exist, there would need to a force that would equally counteract the force of the bowling ball, which there isn't. Which means there is an unbalanced force affecting these objects.
Given that the user must choose one, the correct answer would be that an unbla
A.Calculate the combined force of vector F ?B.Calculate the direction of the combined force vector F ?
Answer:
A. 282.93 N
B. 1.94 degrees
Explanation:
The combined force is found by first adding the three forces given.
We add the three forces by adding their x and y components separately and then combining the results to produce the total force,
The x component of a force is
[tex]\begin{gathered} \cos \theta=\frac{f_x}{F} \\ \Rightarrow f_x=F\cos \theta \end{gathered}[/tex]Therefore, x components of the forces is
[tex]F_x=120\cos 65+100\cos 25+200\cos (-45)[/tex]The y-component of the forces is
[tex]F_y=120\sin 120+100\sin 25+200\sin (-45)[/tex]Now evaluating the above two components gives
[tex]F_x=282.77N[/tex][tex]F_y=9.597N[/tex]Let us draw on big vector whose components are the above vectors.
The angle of the combined vector with respect to the x-axis is
[tex]\tan \theta=\frac{9.59}{282.77}[/tex][tex]\theta=\tan ^{-1}(\frac{9.59}{282.77})[/tex][tex]\boxed{\theta=1.94^o}[/tex]which is our answer!
The magnitude of the combined vector is
[tex]F=\sqrt[]{F^2_x+F^2_y_{}}[/tex][tex]F=\sqrt[]{(9.59)^2_{}+(282.77)^2_{}}[/tex][tex]\boxed{F=282.93N}[/tex]which is our answer!
Hence, to summerise:
A. 282.93 N
B. 1.94 degrees
i am not sure the best way to solve this problem
ANSWER
14.11 s
EXPLANATION
We know that in total, the runner will run a distance of 100m. He runs at constant acceleration for a while and then his velocity gets constant until the end of the track - this means that in the last part, his acceleration is zero.
So we have two parts:
For the first part, we have the acceleration and time. If we set that the initial position is zero, as shown in the diagram above, and that the runner starts from rest - therefore, his initial velocity is zero - we can find the distance of the first part of the path, which we'll call x1:
[tex]x_1=x_0+v_0t+\frac{1}{2}at^2[/tex]Since x0 and v0 are both zero, then those terms get cancelled:
[tex]x_1=\frac{1}{2}\cdot a\cdot t^2=\frac{1}{2}\cdot1.5\cdot6^2=27m[/tex]So the first part of the track, where the runner is speeding up, has a distance of 27m. Therefore, the rest of the track where the runner runs at constant acceleration is:
[tex]100-27=73[/tex]73m.
We want to find the time it took the runner to run the whole 100m. We know that he did the first part in 6 seconds. To find the time of the second part, we can use the distance we just found. Let's call it xf:
[tex]x_f-x_1=\frac{1}{2}at^2+v_0t[/tex]We know that the acceleration in this part of the track is zero and the initial velocity for this part is the velocity the runner had when he reached 6 seconds - i.e. 27m:
[tex]73m=v_1\cdot t[/tex]We don't know the time and we don't know the velocity, but we can find the second one using the formula for velocity for the first part of the track with t = 6s:
[tex]\begin{gathered} v_1=a\cdot t+v_0 \\ v_1=1.5\cdot6 \\ v_1=9m/s \end{gathered}[/tex]Now we can find the time for the second part of the track:
[tex]\begin{gathered} 73m=9m/s\cdot t \\ t=\frac{73m}{9m/s} \\ t\approx8.11s \end{gathered}[/tex]Therefore, the total time it took the runner to run 100m was:
[tex]\begin{gathered} t=6s+8.11s \\ t=14.11s \end{gathered}[/tex]14.11 s
which of the following are independent of the mass of an object falling freely near earth's surface: (may have more than 1 answer) 1) acceleration of the object 2) gravitational force acting on the object 3) gravitational force acting on the object 4) magnitude of the gravitational field
As the object is falling freely, the acceleration of the object will be equal to the acceleration due to gravity.
It is given as,
[tex]g=\frac{GM}{R^2}[/tex]Here, G is the univarshal gravitational constant and M is the mass of the Earth.
means acceleration of the object is constant and independent of the mass of the object.
so option 1 is correct.
now the gravitational force on that object is,
[tex]F=\frac{GMm}{R^2}[/tex]here this is dependent on the mass of the object(m).
NOw the gravitational field means the force per unit mass and is given by,
[tex]E=\frac{GM}{R^2}[/tex]Here we can se that this gravitational field is also independent of the mass of the object.
So, option 1 and 4 are correct.
Two sounds of 610Hz and 620Hz are played together. The frequency you hear will be ___. The beat frequency is ___
Explanation
The beat frequency is equal to the complete value of the alteration in the frequency of the two waves,it can be calculated by using the expression
[tex]f_{beat}=\lvert{{f_1}-f_2}\rvert[/tex]so
Step 1
a)let
[tex]\begin{gathered} f_1=610\text{ Hz} \\ f_2=620\text{ Hz} \end{gathered}[/tex]b) replace and calculate
[tex]\begin{gathered} f_{beat}=\lvert{{f_1}-f_2}\rvert \\ f_{beat}=\lvert610\text{ Hz-620Hz}\rvert \\ f_{beat}=10\text{ Hz} \end{gathered}[/tex]so
the beat frequency is 10 HZ
Step 2
the frequency yuo will detect is the difference of the frequency, so , teh frequency beat, so
the frequency you hear will be 10 Hz
I hope this helps you
Car A is traveling with a constant velocity of 18 [m/s]. Car B speeds up from 0 [m/s] to 10 [m/s] in 4 seconds. Which car has a greater acceleration?Car ACar BCar A and Car B have the same accelerationNeither car is accelerating
Given:
Car A is traveling with a constant velocity of,
[tex]18\text{ m/s}[/tex]The initial speed of car B is,
[tex]v_i=0\text{ m/s}[/tex]After t=4 s, carB's speed is,
[tex]v_f=10\text{ m/s}[/tex]To find:
Which car has a greater acceleration
Explanation:
The acceleration of Car A is Zero as there is no change in velocity with time.
The acceleration of car B is,
[tex]\begin{gathered} a_B=\frac{v_f-v_i}{t} \\ =\frac{10-0}{4} \\ =2.5\text{ m/s}^2 \end{gathered}[/tex]So, Car B has greater acceleration.
Hence, Car B has greater acceleration.
Popeye the Sailor man, who has a mass of 85 kg, ran ( at a constant rate ) up a flight of stairs that are 3.55 m high in 6 seconds. How many watts of power did he generate during his run ?
Given data
*The given mass of the Sailorman is m = 85 kg
*The given height is h = 3.55 m
*The given time is t = 6 s
*The value of the acceleration due to gravity is
[tex]g=9.8m/s^2[/tex]The formula for the power generated by the Sailorman during his run is given as
[tex]\begin{gathered} P=\frac{W}{t} \\ =\frac{\text{mgh}}{t} \end{gathered}[/tex]*Here W is the work done
[tex]\begin{gathered} P=\frac{85\times9.8\times3.55}{6} \\ =492.85\text{ W} \end{gathered}[/tex]If an 800 kg roller coaster is at the top of its 50 m high track, it will have a potential energy 392,000 and a kinetic energy of 0J. This means the total mechanical energy is 392,000J. If the cart drops down to a new height of 10m, how much energy does the cart have now?
ANSWER:
313600 J
STEP-BY-STEP EXPLANATION:
We have that the gravitational potential energy is given by the following equation:
[tex]E_p=m\cdot g\cdot h[/tex]We substitute and calculate the potential energy, knowing that g is the acceleration of gravity and is equal to 9.8 m/s^2:
[tex]\begin{gathered} E_p=800\cdot9.8\cdot10 \\ E_p=78400\text{ J} \end{gathered}[/tex]We know that the total energy is 392,000 joules, so the energy it now carries would be the total minus the calculated potential energy:
[tex]\begin{gathered} E_k=392000-78400 \\ E_k=313600\text{ J} \end{gathered}[/tex]The energy carried by the cart is 313600 J
Please help me with this!
The balloon goes 25 m east in first 10 s.
Then the wind blows the balloon 35 m west in 8 s.
a
The balloon travels 25 m in 10s.
b
the balloon travels a distance 35 m in next 8 s.
c
thhe total distance travelled by the balloon is,
[tex]\begin{gathered} d=25+35 \\ =60\text{ m} \end{gathered}[/tex]d
Average speed in 10 s is,
[tex]\begin{gathered} s=\frac{25}{10} \\ =2.5\text{ m/s} \end{gathered}[/tex]e
The average speed in next 8 s is,
[tex]\begin{gathered} s_8=\frac{35}{8} \\ =4.37\text{ m/s} \end{gathered}[/tex]f
the average speed for the entire trip is,
[tex]\begin{gathered} s_{av}=\frac{25+35}{10+8} \\ =3.33\text{ m/s} \end{gathered}[/tex]g
displacement during the first 10 s is.
[tex]d_1=(25m)\hat{i}[/tex]h.
displacement during next 8 s is,
[tex]d_2=(-35m)\hat{i}[/tex]i
The total displacement is,
[tex]\begin{gathered} d_1+d_2=(25-35)\hat{i}_{} \\ =-(10m)\hat{i} \end{gathered}[/tex]j
the average velocity in 10 s is,
[tex]\begin{gathered} v_1=\frac{25m}{10}\hat{i} \\ =(2.5m)\hat{i} \end{gathered}[/tex]k
The average velocity in 8 s is,
[tex]\begin{gathered} v_2=\frac{-35\text{ m}}{8\text{ s}}\hat{i} \\ =-(4.37m)\hat{i} \end{gathered}[/tex]l
The average velocity entire the whole trip is,
[tex]\begin{gathered} v_{av}=\frac{d_1+d_2}{18} \\ =\frac{-10m\hat{i}}{18\text{ s}} \\ =-(0.55\hat{m/s)i} \end{gathered}[/tex]A student on skateboard pushes off from the top of small hill with a apees of 2.0m/s, and then geos down the hill with a constant acceleration of 0.5 m/s2
After traveling a distancie 12.0m, how fast is the student going?
The final velocity of the student after travelling 12 m is 4 m/s.
What is the final velocity of the student?
The final velocity of the student is determined by applying the following Kinematic equation.
v² = u² + 2as
where;
u is the initial velocity of the studentv is the final velocity of the studenta is the acceleration of the students is the distance travelled by the studentv² = (2)² + 2(0.5)(12)
v² = 16
v = √16
v = 4 m/s
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Name the instrument which is made on the basis of expansion of heat.
The instrument is the Thermometer
Answer:
It is a thermometer and it helps to see the temperature
Two children are riding on a merry-go-round. Child A is at a greater distance from the axis of rotation than child B. Which child hasthe larger linear displacement?Select one:a. There is not enough information given to answer the question.b. Child Ac. They have the same non-zero linear displacementd. Child Be. They have the same zero linear displacement.
The linear displacement in a rotation is given by:
[tex]s=r\theta[/tex]where r is the distance from the axis of rotation and theta is the angular displacement.
Since the linear displacement is proportional to the radius we conclude that child A has a larger linear displacement.
A car is being tested for safety by colliding it with a brick wall.The 1300 kg car is initially driving towards the wall at a speedof 15 m/s, and after colliding with the wall the car movesaway from the wall at 2 m/s. If the car is in contact with thewall for 0.5 s, calculate the average force exerted on the carby the wall.
Answer:
44200 N
Explanation:
To calculate the average force exerted on the car, we will use the following equation
[tex]\begin{gathered} Ft=\Delta p \\ Ft=m(v_f-v_i) \end{gathered}[/tex]Where F is the average force, t is the time, m is the mass, vf is the final velocity and vi is the initial velocity of the car.
Replacing t = 0.5s, m = 1300 kg, vf = -2 m/s, and vi = 15 m/s and solving for F, we get
[tex]\begin{gathered} F(0.5s)=(1300\text{ kg\rparen\lparen-2 m/s - 15 m/s\rparen} \\ F(0.5s)=(1300\text{ kg\rparen\lparen-17 m/s\rparen} \\ F(0.5s)=-22100\text{ kg m/s} \\ F=\frac{-22100\text{ kg m/s}}{0.5\text{ s}} \\ F=-44200\text{ N} \end{gathered}[/tex]Therefore, the average force exerted on the car by the wall was 44200 N
Three resistors with resistances of 9 Ω, 18 Ω, and 30 Ω are in a series circuit with a 12 volt battery. What is the total resistance of the resistors in the circuit?
When resistors are in series, the total resistance is equal to the sum of the individual resistances of each resistor. Thus, the total resistance in this circuit is 9+18+30 = 57Ω
The source of the Sun’s heat and light energy is:A. combustion of helium gas.B. fusion of hydrogen nuclei.C. gravitational pressure.D. burning of fossil fuels.
To find
The source of the Sun’s heat and light energy is:
Explanation
The sun's core is very hot. So under pressure nuclear fusion takes place. Here hydrogen is changed to helium.
Conclusion
The correct option is
B. fusion of hydrogen nuclei.
A sample of unknown material weight 900N In air and and 400N when submerged in an alcohol solution with a density of 0.7 g/cm³.What is the density of the material ?
1.26 g/cm³
ExplanationStep 1
given
[tex]\begin{gathered} F_{g(air)}=900\text{ N} \\ F_{g(alchodol)}=400\text{ N} \\ \rho_{alcohol}=0.7\text{ }\frac{g}{cm^3} \end{gathered}[/tex]unknown; the density of the material, so
[tex]\begin{gathered} F_B=F_{g(air)}-F_{g(alcohol)} \\ F_B=900\text{ N-400 N=500 N} \end{gathered}[/tex]so, the proportion is
the ratio of the force equals the ratio of the density ,so
[tex]\begin{gathered} \frac{F_{g(air)}}{F_B}=\frac{\rho_{material}}{\rho_{alcholol}} \\ replace \\ \frac{900\text{ N}}{500\text{ N}}=\frac{\rho_{material}}{0.7\text{ }\frac{g}{cm^3}} \\ mutliply\text{ both sides by 0.7}\frac{g}{cm^3} \\ \frac{900\text{N}}{500\text{N}}*0.7\text{ }\frac{g}{cm^3}=\frac{\rho_{mater\imaginaryI al}}{0.7\text{\frac{g}{cm^{3}}}}*0.7\frac{g}{cm^3} \\ 1.26\frac{g}{cm^3}=\text{ density of the material } \\ \end{gathered}[/tex]so, the density of the material is
1.26 g/cm³
I hope this helps you
In a lightning discharge, 45 C of charge move through a potential difference of 1.0 x108 V in 0.030 s.A. What is the current of the lightning strike?B. How much energy is released by the lightning bolt?
Given:
Charge, Q = 45 C
Potential difference, V = 1.0 x 10⁸ V
Time, t = 0.030 s
Let's solve for the following:
• (A). What is the current of the lightning strike?
To find the current, apply the formula:
[tex]I=\frac{Q}{t}[/tex]Where:
I si the current
Q is the charge = 45 C
t is the time = 0.030 s
Thus, we have:
[tex]\begin{gathered} I=\frac{45}{0.030} \\ \\ I=1500\text{ A} \end{gathered}[/tex]Therefore, the current of the lightning strike is 1500 Amperes.
• (B). How much energy is released by the lightning bolt?
To find the amount of energy released, apply the formula:
[tex]E=V\times Q[/tex]where:
E is the Energy released
V is the potential difference, V = 1.0 x 10⁸ V
Q is the charge = 45 C
Thus, we have:
[tex]\begin{gathered} E=1.0\times10^8\ast45 \\ \\ E=4.5\times10^9\text{ J} \end{gathered}[/tex]Therefore, the energy released is 4.5 x 10⁹ Joules.
ANSWER:
(a). 1500 A
(b). 4.5 x 10⁹ J
