Answer:
All true statements are shown: a) The average speed of the car is [tex]\frac{v}{2}[/tex], c)The magnitude of the acceleration of the car is [tex]\frac{v}{t}[/tex].
Explanation:
Let prove the validity of each statement:
a) The average speed of the car is [tex]\frac{v}{2}[/tex].
The average speed ([tex]\bar v[/tex]) is defined by the following formula:
[tex]\bar v = \frac{v_{o}+v_{f}}{2}[/tex] (1)
Where:
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds of the racing car.
If we know that [tex]v_{o} = 0[/tex] and [tex]v_{f} = v[/tex], then the average speed of the racing car:
[tex]\bar v = \frac{0+v}{2}[/tex]
[tex]\bar v = \frac{v}{2}[/tex]
The statement is true.
b) The car travels a distance [tex]v\cdot t[/tex].
Since the racing car is accelerating uniformly, the distance travelled by the car is represented by the following kinematic formula:
[tex]x - x_{o}=v_{o}\cdot t + \frac{1}{2}\cdot a\cdot t^{2}[/tex] (2)
Where [tex]a[/tex] is the acceleration of the racing car, measured in meters per square second.
The statement is false.
c) The magnitude of the acceleration of the car is [tex]\frac{v}{t}[/tex].
Since the racing car is accelerating uniformly, the velocity of the racing car is represented by the following kinematic formula:
[tex]v_{f} = v_{o}+a\cdot t[/tex] (3)
Then, we clear the acceleration of the expression:
[tex]a = \frac{v_{f}-v_{o}}{t}[/tex]
If we know that [tex]v_{o} = 0[/tex] and [tex]v_{f} = v[/tex], then the acceleration of the car is:
[tex]a = \frac{v-0}{t}[/tex]
[tex]a = \frac{v}{t}[/tex]
The statement is true.
d) The velocity of the car remains constant.
Since the car accelerates uniformly, the vehicle does not travel at constant velocity.
The statement is false.
Dee Liverance shoots an arrow out of her bow. The string is stretched 57 cm, and when
released the arrow leaves the bow in 0.17 s. What is the acceleration of the arrow?
Answer:
Mr. Dececco will not be happy
Explanation:
I'll give you a hint, the answer starts with the number 3. Use the kinematics equation of [tex]delta\ x =v_{0}t+ \frac{1}{2}at^{2}[/tex]
What is the average reaction time of people
Answer:
the average reaction time is 0.25 seconds.
. Which of the following is the best description about cancer cells?
Cancer cells are cells that divide relentlessly, forming solid tumors or flooding the blood with abnormal cells. Cell division is a normal process used by the body for growth and repair.
To travel at a constant speed,a car engine provides 24 KW of useful powers. The driving force on the car is 600N.AT what speed does it travel? A 2.5 B 4.0 C.25 D 40
Answer: D. 40
Explanation:
Power = Force * Velocity
Force = 600N
Power needs to be converted to Watts:
= 24 * 1,000
= 24,000 W
24,000 = 600 * Velocity
Velocity = 24,000/600
= 40 m/s
1. The center of gravity of both cats and humans is roughly in the thoracic region. However, they each bear their weight very differently. a. Which are the major, weight-bearing structures of the two skeletons
Answer:
Major, weight-bearing structures are the bones of the body that are strong and dense to be able to bear the weight of the body. The major, weight-bearing structures of cat and human skeletons are :
Human skeleton: The body weight of an individual is on his pelvic girdles that are attached to the bones of lower limbs. Thigh bones, leg bones, and bones of feet comprise lower limbs The lower limbs consist of the thigh, the leg, and the foot.
Cat skeleton: cats are quadrupedal so it bears all the body weight on shoulders and legs that includes the Scapula and pelvis.
earth has a radius of 6370 km what is earth's mass?
Answer:
approx 6 × 10^24 kg is earth's mass
An ideal step-down transformer is needed to reduce a primary voltage of 120 V to 6.0 V. What must be the ratio of the number of turns in the secondary to the number of turns in the primary
Answer:
[tex]N_s : N_p = 20 : 1[/tex]
Explanation:
From the question we are told that
The primary voltage is [tex]V_p = 120 \ V[/tex]
The secondary voltage is [tex]V_s = 6 \ V[/tex]
Generally from the transformer equation we have that
[tex]\frac{V_p}{V_s} = \frac{N_p}{N_s}[/tex]
So
[tex]\frac{120}{6} = \frac{N_p}{N_s}[/tex]
=> [tex]\frac{N_p}{N_s} = 20[/tex]
Therefore the ratio of the number of turns in the secondary to the number of turns in the primary is
[tex]N_s : N_p = 20 : 1[/tex]
The cars in a soapbox derby have no engines; they simply coast downhill. Which of the following design criteria is best from a competitive point of view? a. The car's wheels should b. have large moments of inertia be massive c. be hoop-like wheels rather than solid disks d. have small moments of inertia
Complete question is;
The cars in a soapbox derby have no engines; they simply coast downhill. Which of the following design criteria is best from a competitive point of view?
The car's wheels should;
A. have large moments of inertia
B. be massive
C. be hoop-like wheels rather than solid disks
D. have small moments of inertia
Answer:
D. The car's wheels should have small moments of inertia
Explanation:
They are simply cast downhill due to no engine and thus the solid wheels will have very small moments of inertia and will therefore accelerate at a faster rate.
Looking at the options, the correct one that corresponds to the description I just made is Option D.
The car's wheels should have small moments of inertia.
What is Inertia?This is defined as the resistance of any physical object to a change in its velocity.
We told they were cast downhill due to no engine and the solid wheels having very small moments of inertia causing increase in acceleration depicts option D which was why it was chosen.
Read more about Inertia here https://brainly.com/question/1140505
alpha and beta rays origin is same
Answer:
Alpha particles carry a positive charge, beta particles carry a negative charge, and gamma rays are neutral. An alpha particle is made up of two protons and two neutrons bound together. Beta particles are high energy electrons. Gamma rays are waves of electromagnetic energy, or photons.
A ball has a mass of 250g. The ball is kicked and this applies a force 15 N to the ball, as shown in the diagram. How much does the ball accelerate by in the direction of the kick?
Answer:
0.06 m/s²
Explanation:
force= mass × acceleration
acceleration= force / mass
acceleration= 15/250= 0.06 m/s²
hope it helps! please mark me brainliest..
Have a good day ahead
The acceleration of the ball in the direction of the kick will be 600 m/[tex]s^{2}[/tex]
We have a Ball.
We have to determine the acceleration of the ball in the direction of the kick.
What is Force?The product of mass and acceleration of a body is called force.
Force = Mass x Acceleration.
According to the question, we have -
Mass = 250 g = 0.025 Kg
Force = 15 N
Then, the acceleration of the ball in the direction of the kick will be -
F = m x a
a = F / m
a = 15 / 0.025 = 600 m/[tex]s^{2}[/tex]
Hence, the acceleration of the ball in the direction of the kick will be 600 m/[tex]s^{2}[/tex]
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An ideal gas is confined to a container with constant volume. The number of moles is constant. By what factor will the pressure change if the absolute temperature triples
Answer:
1/3
Explanation:
Gay Lusaac's law states that "the pressure of a given mass of gas is directly proportional with the absolute temperature of the gas, provided that the volume is kept constant."
In formula, we say that
P/T = k
Where
P = pressure at different points
T = temperature at different points
k = constant of proportionality
From the stated formula, if we multiply the temperature by 3, we have
P/3T = k
P * 1/3T = k
And from this, we see the pressure will change by a value of 1/3
What is the formula to finding your target heart rate.
Solubility Curve Practice Problems Worksheet 1
Directions: Find the mass of solute will dissolve in 100mL of water at the following temperatures?
150
1. KCl at 70°C =
140
2. Ce,(SO.), at 100°C=
130
120
3. NH.Cl at 90°C=
4. Which of the above three substances is most
110
100
NaNO3
soluble in water at 15°C. =
90
KNO3
Grams of solute
per 100 g H,0
80
70
60
NH3
NHACI
50
KCI
40
Naci
30
20
KCIO
10
Ce2(SO4)3
0
10 20 30 40 50 60 70 80 90 100
Temperature (°C)
According to the solubility curve, KCl is soluble in water at a rate of about 40 grammes per 100 g. Given that we have 100 mL, or 100 grammes, of water, the mass of KCl that will dissolve is similarly 40 grammes.
[tex]Ce_2(SO_4)_3[/tex] is soluble in water at a rate of around 30 grammes per 100 grammes at 100 °C. [tex]Ce_2(SO_4)_3[/tex] will therefore dissolve in 100 mL of water at a mass of 30 grammes.
[tex]NH_4Cl[/tex] dissolves in water at a rate of around 90 grammes per 100 grammes at 90 °C. The mass of [tex]NH_4Cl[/tex] that will dissolve in 100 mL of water is similarly 90 grammes.
We must ascertain the solubilities of the abovementioned chemicals at 15°C in order to ascertain which substance is most soluble at that temperature.
[tex]NaNO_3[/tex] has the highest solubility at 15°C, as seen by the solubility curve. At 15°C, [tex]NaNO_3[/tex] dissolves in water at a rate of about 80 grammes per 100 g.
Therefore, at 15°C, NaNO3 is the most soluble substance among the three given substances.
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*WILL MARK BRAINLIEST*
A glass window (n = 1.52) has a
uniform layer of ice on it (n = 1.31).
What is the critical angle for a ray
trying to pass from glass to ice?
(Water n = 1.33, Air n = 1.00)
(Unit = deg)
Answer:
1.52
Explanation:
Answer: 59.5 deg
Explanation:
I got it right on acellus ✅
I’m in a test and I’m sort of being timed here’s a photo of question
Answer: D.
Explanation:
Answer:
i think the answer is b
Explanation:
playing sports often require both
convert 144 km/h to m/s
Answer:
40m/s
Explanation:
144km/h
1km=1000m
1hr=3600secs
144×1000/3600=
40m/s
In the graph, which two regions show the particle undergoing zero acceleration and negative acceleration respectively?
A. BC shows zero acceleration, and AB shows negative acceleration.
B. AB shows zero acceleration, and CD shows negative acceleration.
C. BC shows zero acceleration, and CD shows negative acceleration.
D. AB shows zero acceleration, and BC shows negative acceleration.
Answer:
Option C. BC shows zero acceleration, and CD shows negative acceleration.
Explanation:
To successfully answer the question given above, we must know the meaning of zero acceleration and negative acceleration respectively.
Zero acceleration simply means the object is not accelerating. This implies that velocity of the object is constant (i.e unchanged) with time.
Negative acceleration simply means the object is decelerating i.e coming to rest.
Now, let us answer the question given above.
From the question given above, the following data were obtained:
1. AB indicates that the particle is accelerating because the velocity increased from A to B with time.
2. BC indicates zero acceleration because the velocity of the Particle is constant (i.e unchanged) with time.
3. CD indicates that the object is coming to rest i.e decelerating (negative acceleration) because the velocity of the object decreased from C to D with time.
From the above illustrations, we can see that only option C gives the correct answer to the question.
Answer:
Option C. BC shows zero acceleration, and CD shows negative acceleration.
Explanation:
Is the amount of stretch of the springs proportional to the hanging mass? Explain briefly.
Derive a theoretical expression to find k-equivalent for springs in parallel and springs in series.
Answer:
When connected in parallel, the extension of both springs is the same, and the total elastic force will be equal to the sum of the forces in each spring: x=x1=x2,F=F1+F2. F=F1+F2,⇒kx=k1x1+k2x2=(k1+k2)x,⇒k=k1+k2.
Yes. The amount of stretch of a spring is proportional to the hanging mass.
According to Hooke's Law, the restoring force of the spring F is:
F = -kx
k is the spring constant and x is the stretch of spring.
The restoring force F is in opposite direction of the weight of the hanging mass which is mg
⇒ F = -mg
mg = -kx
x = mg/k
Hence x, the stretch of the spring is directly proportional to the hanging mass.
(i) If two springs are connected in parallel, their k- equialent is
[tex]k= k_{1}+k_{2}[/tex]
(ii) If two springs are connected in series, their k- equialent is
[tex]\frac{1}{k}=\frac{1}{k_{1} } +\frac{1}{k_{2} }[/tex]
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A horizontal force of 150 N is used to push a 38.0 kg packing crate a distance of 6.85 m on a rough horizontal surface. If the crate moves at constant speed, find each of the following.
a) find the work done by the 150N force:
b) find the coefficient of kinetic energy between the crate and rough surface
Answer:
0.40
Explanation:
a) Work done is expressed as
W = Force × distance
Given.
Force = 150N
Distance = 6.85m
Workdone = 150×6.85
Work done = 1027.5Joules
b) According to Newton's second law
\sumFx = ma
Fm - Ff = ma
Since speed is constant, acceleration us zero
Fm - Ff = 0
Fm = Ff = nR
Fm is the moving force
Ff is the frictional force
n is the coefficient of kinetic energy between the crate and rough surface
R is the reaction
From the formula;
Fm = nR
n = Fm/R
n = Fm/mg
n = 150/38(9.8)
n = 150/372.4
n = 0.40
Hence the coefficient of kinetic energy between the crate and rough surface is 0.40
7. A beam of light converges at a point P. Now a lens is placed in the path of the convergent
beam 12cm from P. At what point does the beam converges if the lens is a) a convex lens of
focal length 20cm and b) a concave lens of focal length 16cm.
Answer:
Lens at a distance = 7.5 cm
Lens at a distance = 6.86 cm (Approx)
Explanation:
Given:
Object distance u = 12 cm
a) Focal length = 20 cm
b) Focal length = 16 cm
Computation:
a. 1/v = 1/u + 1/f
1/v = 1/20 + 1/12
v = 7.5 cm
Lens at a distance = 7.5 cm
b. 1/v = 1/u + 1/f
1/v = 1/16 + 1/12
v = 6.86 cm (Approx)
Lens at a distance = 6.86 cm (Approx)
My dad gifted me a calculator. I have observed that very small cells are used in a calculator. What are these cells called and what are their main advantage?
What is the difference between static frication and kinetic friction
static friction is acting on stationery object (rest) but kinetic friction acting on a moving body.
The other one is static friction oppose the object to start a motion so it force is great than kinetic friction
Answer all these questions correctly and will mark brainliest!
1. In Football a field goal is kicked for how many points?
2. Football a touch down is worth how many points?
3. What part of the body can a goalie in soccer use that no one else can?
4. How many quarters are in a football game?
5. How many players are on the field in football?
6. Who invented basketball?
7. Who invented the game Nukem?
8. In volleyball can a ball (Including serve) be played off of the net?
9. When did basketball become an Olympic sport?
10.How many quarters are in a football game?
Inappropriate answers will be reported
Answer:
1. 3 points
2. 6 points
3. goalkeeper
4. four quarters
5. eleven players
6. James Naismith
7. Duke nukem ( I'm not sure)
8. no
9. 1936
10. four
HOPE IT'S HELP :)
A 15.0kg block is dragged over a rough, horizontal surface by a 70.0-N force acting at 20.0 degrees above the horizontal. The block is displaced 5.0 m, and the coefficient of kinetic friction is 0.3.
a. Draw a free body diagram. Draw your coordinate system and label the axes.
b. Calculate the work done on the block by the 70 N force.
c. Calculate the work done on the block by the normal force.
d. Calculate the work done on the block by the gravitational force.
e. Calculate the work done on the block by the force of friction.
Answer:
[tex]W=70 * 5cos20 = 328.89 J[/tex]
[tex]W_n = 0[/tex]
[tex]W_g=0[/tex]
[tex]W_f= -184.59J[/tex]
Work done is 0
Explanation:
From the question we are told that
Weight of block =15.0kg
Force acting on the block = 70.0N
At an angle of 20 degree
Displacement of block is 5m
Coefficient of kinetic friction 0.3
b) Generally work done by force is give by [tex]W=fdcos \theta[/tex]
therefore
[tex]W=70 * 5cos20 = 328.89 J[/tex]
c) there is no work done by the normal force in this scenario because
normal force in this case is perpendicular to the displacement of the motion
[tex]W_n = 0[/tex]
d) The displacement in the vertical direction is 0
Therefore the gravitational work done is 0 [tex]W_g=0[/tex]
e)Generally in finding work done by friction we first find frictional force
Mathematically the equation for frictional force is given [tex]f = \alpha N[/tex]
Given that
[tex]N=mg-Fsin20[/tex]
[tex]N= 15.0*9.8 - 70 sin20[/tex]
[tex]N=123 N[/tex]
[tex]f=0.3* 123.06 = 36.92N[/tex]
Mathematically solving to get work done by frictional force [tex]W_f[/tex]
[tex]W_f= -fd\\W_f = -36.92 * 5[/tex]
[tex]W_f= -184.59J[/tex]
the frictional force work done is [tex]W_f= -184.59J[/tex]
Determine whether each of the following statements is true or false.
a. An object's weight is always equal to its mass.
b. The force of tension always pushes.
c. The magnitude of the sum of the forces on an object is never greater than its weight. Explain.
Answer:
a) For an object with mass M, in a region with a gravitational acceleration g, is:
W = M*g
Then the weight is g times the mass of the object, this means that the weight is not always equal to the mass of the object, this first statement is false.
For example, in Earth the gravitational acceleration is 9.8m/s^2
Then there is no object in Earth with a weight equal to its mass.
b) The force of tension can be the force in a piece of string that is holding an object with mass M.
The force of tension will be always pointing in the direction to the "center" of the string, then it does not push, the tension force "pulls"
The statement is false.
c) we know that the weight is:
W = M*g
This is a force, that for an object that is in the air, will pull the object back to the ground.
Suppose that we also have that object attached to a string, and the object is in the air, now the object starts to fall due to its weight and we also pull down with the string, then the total force pulling down will be the weight plus the tension of the string, then we will have a force larger (in magnitude) than the weight, which means that the statement is false.
Two particles, an electron and a proton, move in a circular path in a uniform magnetic field of intensity B=1.23 T. Find the ratio between the time period of the proton Tp to the electron Te (i.e., find 'p)
Answer:
The ratio of the time period of the proton to the electron is 1835.16.
Explanation:
Given that,
Two particles, an electron and a proton, move in a circular path in a uniform magnetic field of intensity B=1.23 T
We need to find the ratio between the time period of the proton Tp to the electron Te.
The time period in magnetic field is given by :
[tex]T=\dfrac{2\pi m}{qB}[/tex]
For proton, time period is :
[tex]T_P=\dfrac{2\pi m_P}{q_pB}\ ....(1)[/tex]
For an electron, the time period is :
[tex]T_e=\dfrac{2\pi m_e}{q_eB}\ ....(2)[/tex]
From equation (1) and (2) :
[tex]\dfrac{T_p}{T_e}=\dfrac{\dfrac{2\pi m_p}{q_pB}}{\dfrac{2\pi m_e}{q_eB}}\\\\As\ q_e=q_p\\\\\dfrac{T_p}{T_e}=\dfrac{m_p}{m_e}\\\\=\dfrac{1.67\times 10^{-27}}{9.1\times 10^{-31}}\\\\=1835.16[/tex]
So, the ratio of the time period of the proton to the electron is 1835.16.
Remaining Time: 1 hour, 49 minutes, 34 seconds.
Question Completion Status:
QUESTION 8
If the magnification is calculated to be +0.5 in a particular situation, what does that tell you about the image?
O A. It is bigger by 0.5
OB
It is half the size of the object.
O C. It is twice the size of the object.
O D. It is smaller by 0.5
O E. It tells you nothing about the image.
QUESTION 9
Explanation:
i think C . it is twice the size of the object
Assume the mass of a shuttlecraft is 2,200 kilograms [kg]. If the shuttlecraft requires 3 minutes [min] to rise from the surface of the Moon to an altitude of 19 kilometers [km], and the shuttle's engine is rated at 500 kilowatts [kW], what is the efficiency of the engines
Answer:
The efficiency of the engine is 75.94 %
Explanation:
Given;
mass of the shuttlecraft, m = 2,200 Kg
time required by the shuttlecraft, t = 3 minutes = 180 s
height risen by the shuttle, h = 19 km = 19,000 m
Input power of the shuttle, Pi = 500 kW = 500,000 W
The potential energy due to height risen by the Shuttle on Moon surface is given as;
E = mgh
where;
g is the acceleration due to gravity on moon, = ¹/₆ x 9.81 m/s² = 1.635 m/s²
E = 2200 x 1.635 x 19,000
E = 68343000 J
Output power of the shuttle, is given as Energy / time
Output power = (68343000) / (180)
Output power = 379,683.33 W
The efficiency of the shuttle is given as;
Efficiency = (Output power) / (Input power)
Efficiency = (379,683.33) / (500,000)
Efficiency = 0.7594
Efficiency(%) = 75.94 %
Therefore, the efficiency of the engine is 75.94 %
An airplane is flying through a thundercloud at a height of 2000 m (This is a very dangerous thing to do because of updrafts, turbulence, and the possibility of electric discharge.) If there is a charge concentration of 40 C at height 3000 m with the cloud and -40 C at height 1000 m, what is the electric field at the aircraft
Answer:
[tex]400000\ \text{N/C}[/tex]
Explanation:
[tex]q_1[/tex] = Charge at 3000 m = 40 C
[tex]q_2[/tex] = Charge at 1000 m = -40 C
[tex]r_1[/tex] = 3000 m
[tex]r_2[/tex] = 1000 m
k = Coulomb constant = [tex]9\times10^9\ \text{Nm}^2/\text{C}^2[/tex]
Electric field due to the charge at 3000 m
[tex]E_1=\dfrac{k|q_1|}{r_1^2}\\\Rightarrow E_1=\dfrac{9\times 10^9\times 40}{3000^2}\\\Rightarrow E_1=40000\ \text{N/C}[/tex]
Electric field due to the charge at 1000 m
[tex]E_2=\dfrac{k|q_2|}{r_2^2}\\\Rightarrow E_2=\dfrac{9\times 10^9\times 40}{1000^2}\\\Rightarrow E_2=360000\ \text{N/C}[/tex]
Electric field at the aircraft is [tex]E_1+E_2=40000+360000=400000\ \text{N/C}[/tex].
A 2000 kg car moving at 100 km/h crosses the top of a hill with a radius of curvature of 100 m. What is the normal force exerted by the seat on the driver if the mass of the driver is 60 kg?
Answer:
The normal force the seat exerted on the driver is 125 N.
Explanation:
Given;
mass of the car, m = 2000 kg
speed of the car, u = 100 km/h = 27.78 m/s
radius of curvature of the hill, r = 100 m
mass of the driver, = 60 kg
The centripetal force of the driver at top of the hill is given as;
[tex]F_c = F_g - F_N[/tex]
where;
Fc is the centripetal force
[tex]F_g[/tex] is downward force due to weight of the driver
[tex]F_N[/tex] is upward or normal force on the drive
[tex]F_N = F_g-F_c\\\\F_N = mg - \frac{mv^2}{r} \\\\F_N = (60 \times 9.8) -\frac{60 \ \times \ 27.78^2 \ }{100} \\\\F_N = 588 \ N - 463 \ N\\\\F_N = 125 \ N[/tex]
Therefore, the normal force the seat exerted on the driver is 125 N.
The normal force exerted by the seat on the driver if the mass of the driver is 60 kg is of 125 N.
Given data:
The mass of car is, m' = 2000 kg.
The speed of car is, v = 100 km/h = 100 × 5/18 = 27.77 m/s.
The radius of curvature of path is, r = 100 m.
The mass of driver is, m = 60 kg.
In this case, the normal force on the driver is equal to the difference between weight of the driver and the centripetal force on the driver. Then the expression is given as,
[tex]N'= W - F\\\\N '= mg-\dfrac{mv^{2}}{r}[/tex]
Solving as,
[tex]N' = (60 \times 9.8)-\dfrac{60 \times 27.77^{2}}{100}\\\\N' = 125\;\rm N[/tex]
Thus, we can conclude that the normal force exerted by the seat on the driver if the mass of the driver is 60 kg is of 125 N.
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