Answer:
Explanation:
Blue: 10/30
Red: 5/30
Yellow: 15/30
The probability of finding the blue red and yellow chips is 1/3, 1/6 and 1/2
What is the probability?
The extent to which an event is likely to occur, measured by the ratio of the favorable cases to the whole number of cases possible.
Given that
Blue chips =10
Red chips = 5
Yellow chips = 15
Number of the total samples =10+15+5=30
Probability of choosing Blue chips = [tex]\dfrac{10}{30}= \dfrac{1}{3}[/tex]
Probability of Red chips =[tex]\dfrac{5}{30}=\dfrac{1}{6}[/tex]
Probability of Yellow chips =[tex]\dfrac{15}{30}=\dfrac{1}{2}[/tex]
Thus the probability of finding the blue red and yellow chips is 1/3, 1/6 and 1/2
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Which of the following accurately describes circuits
Answer:
In a series circuits, the current can flow through only one path from start to finish - accurately describes circuits.
Answer:
It is a series circuit, the current can flow through only one path from start to finish - accurately describes circuits.
Explanation:
I don't know the option but it is also the correct one.
Newton's law of cooling states that the rate of change of temperature of an object in a surrounding medium is proportional to the difference of the temperature of the medium and the temperature of the object. Suppose a metal bar, initially at temperature 50 degrees Celsius, is placed in a room which is held at the constant temperature of 40 degrees Celsius. One minute later the bar has cooled to 40.18316 degrees . Write the differential equation that models the temperature in the bar (in degrees Celsius) as a function of time (in minutes). Hint: You will need to find the constant of proportionality. Start by calling the constant k and solving the initial value problem to obtain the temperature as a function of k and t . Then use the observed temperature after one minute to solve for k .
Answer:
Newton's law of cooling says that the temperature of a body changes at a rate proportional to the difference between its temperature and that of the surrounding medium (the ambient temperature); dT/dt = -K(T - Tₐ) where T = the temperature of the body (°C), t = time (min), k = the proportionality constant (per minute),
Explanation:
PLEASE HELP MEE THIS IS DUE IN 45 MINS
Answer:
The distance travelled does not depend on the mass of the vehicle. Therefore, [tex]s = d[/tex]
Explanation:
This deceleration situation can be analyzed by means of Work-Energy Theorem, where change in translational kinetic energy is equal to the work done by friction:
[tex]\frac{1}{2}\cdot m\cdot v^{2}-\mu\cdot m\cdot g \cdot s = 0[/tex] (1)
Where:
[tex]m[/tex] - Mass of the car, in kilogram.
[tex]v[/tex] - Initial velocity, in meters per second.
[tex]\mu[/tex] - Coefficient of friction, no unit.
[tex]s[/tex] - Travelled distance, in meters.
Then we derive an expression for the distance travelled by the vehicle:
[tex]\frac{1}{2}\cdot v^{2} = \mu \cdot g \cdot s[/tex]
[tex]s = \frac{v^{2}}{\mu\cdot g}[/tex]
As we notice, the distance travelled does not depend on the mass of the vehicle. Therefore, [tex]s = d[/tex]
2. The given graph shows that the object is
(a) in non-uniform motion
(b) in uniform motion
(c) at rest
(d) in an oscillatory motion.
distance
time
Answer:
(c) at rest
Explanation:
Given
See attachment for the distance time graph
Required
What does the graph illustrate?
From the graph, we can see that the line of distance is a horizontal line.
This suggests that a time increases, the distance remains unchanged
When distance remains unchanged over time, then it means the object is at rest.
Hence, (c) is correct
A truck collides with a car on horizontal ground. At one moment during the collision, the magnitude of the acceleration of the truck is 12.7 m/s^2. The mass of the truck is 2490 kg and the mass of the car is 890 kg. Assume that the only horizontal forces on the vehicles during the collision are the forces they exert on one another.
Required:
What is the acceleration of the car at the same moment?
Answer:
The magnitude of the acceleration of the car is 35.53 m/s²
Explanation:
Given;
acceleration of the truck, [tex]a_t[/tex] = 12.7 m/s²
mass of the truck, [tex]m_t[/tex] = 2490 kg
mass of the car, [tex]m_c[/tex] = 890 kg
let the acceleration of the car at the moment they collided = [tex]a_c[/tex]
Apply Newton's third law of motion;
Magnitude of force exerted by the truck = Magnitude of force exerted by the car.
The force exerted by the car occurs in the opposite direction.
[tex]F_c = -F_t\\\\m_ca_c = -m_t a_t\\\\a_c =- \frac{m_ta_t}{m_c} \\\\a_c = -\frac{2490 \times 12.7}{890} \\\\a_c = - 35.53 \ m/s^2[/tex]
Therefore, the magnitude of the acceleration of the car is 35.53 m/s²
Answer:
The magnitude of the acceleration of the car is 35.53 m/s²
Explanation:
Given;
acceleration of the truck, = 12.7 m/s²
mass of the truck, = 2490 kg
mass of the car, = 890 kg
let the acceleration of the car at the moment they collided =
Apply Newton's third law of motion;
Magnitude of force exerted by the truck = Magnitude of force exerted by the car.
The force exerted by the car occurs in the opposite direction.
Therefore, the magnitude of the acceleration of the car is 35.53 m/s²
The greatest force in momentum will be produced by
Equal masses of two different liquids are put into identical beakers.
Liquid 1 is heated for 100s and liquid 2 is heated for 200s by heaters of the same power.
The temperature of both liquids increases by the same amount.
Which statement is correct?
A Both liquids receive the same amount of energy.
B. Liquid 1 receives more energy than liquid 2.
C. Both liquids have equal thermal capacity.
D. The thermal capacity of liquid 1 is less than the thermal capacity of liquid 2.
Answer:
C
Explanation:
Because they both received the same temperature
A toy car of mass 600g moves through 6m in 2 seconds. The average kinetic energy of the toy car is
Answer:
12
Explanation:
I'm a beginner so am not sureeeeee
a 1600 kg car rounds a curve of radius 71 m banked at an angle of 15, What is the magnitude of the friction force required for the car to travel at 86 km/h
Answer:
The frictional force required for the car to travel is 8,365.01 N
Explanation:
Given;
mass of the car, m = 1600 kg
radius of the curved road, r = 71 m
banking angle, θ = 15⁰
velocity of the car, v = 86 km/h = 86/3.6 = 23.89 m/s
The two forces acting on the are:
1. the parallel force to the banked plane
2. the centripetal force pushing the car up the banked plane
To keep the car traveling at 86 km/h;
frictional force + parallel force to the plane = centripetal force pushing the car up the banked plane
The parallel force to the banked plane:
F = mgsinθ
F = 1600 x 9.8 x sin(15⁰)
F = 4,057.98 N
The centripetal force pushing the car up the banked plane:
[tex]F_c= (\frac{mv^2}{r} )cos(\theta)\\\\F_c = (\frac{1600 \times 23.89^2}{71} )cos(15^0)\\\\F_c = 12,422.99 \ N[/tex]
The frictional force required for the car to travel:
[tex]F_k = F_c - F\\\\F_k = 12,422.99 \ N - 4,057.98 \ N\\\\F_k = 8,365.01 \ N[/tex]
Therefore, the frictional force required for the car to travel is 8,365.01 N
Consider a sample containing 1.70 mol of an ideal diatomic gas.
(a) Assuming the molecules rotate but do not vibrate, find the total heat capacity of the sample at constant volume. nCv = J/K
(b) Assuming the molecules rotate but do not vibrate, find the total heat capacity of the sample at constant pressure. nCp = J/K
(c) Assuming the molecules both rotate and vibrate, find the total heat capacity of the sample at constant volume. nCv = J/K
(d) Assuming the molecules both rotate and vibrate, find the total heat capacity of the sample at constant pressure. nCp = J/K
I don't know
because I don't know
A car is moving at 10 m/s on a horizontal road with friction on a dry day. The car can travel around a traffic circle with a minimum radius of 4.8 meters. It rains and the car around a traffic circle with a minimum radius of 11.8 meters. What is the percentage of the coefficient of static friction on the rainy day compared to the dry day
Answer:
[tex]\mu_w=86\%[/tex]
Explanation:
From the question we are told that:
Velocity on Dry road [tex]V_d=10m/s[/tex]
Radius Dry [tex]r_d=4.8[/tex]
Radius wet [tex]r_w=11.8[/tex]
Generally the equation for coefficient of static friction on the dry day is mathematically given by
[tex]\mu mg=\frac{mv^2}{r_d}[/tex]
[tex]\mu g=\frac{v^2}{r_d}[/tex]
[tex]\mu_d 9.8=\frac{10^2}{4.8}[/tex]
[tex]\mu_d=2.125[/tex]
Generally the equation for the relationship between Radius & coefficient of static friction is mathematically given by
[tex]\frac{\mu_d}{\mu_w}=\frac{r_d}{r_w}[/tex]
[tex]\frac{\mu_w}{2.125}=\frac{4.8}{11.8}[/tex]
[tex]\mu_w=0.86[/tex]
Therefore
[tex]\mu_w=86\%[/tex]
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. How much of the sample is left after 1.98 x 10^4 seconds?
a) 8.03 x 10^16 nuclei
b) 4.01 x 10^16 nuclei
c) 2.02 x 10^16 nuclei
d) 1.61 x 10^17 nuclei
Answer:
c) 2.02 x 10^16 nuclei
Explanation:
The isotope decay of an atom follows the equation:
ln[A] = -kt + ln[A]₀
Where [A] is the amount of the isotope after time t, k is decay constant, [A]₀ is the initial amount of the isotope
[A] = Our incognite
k is constant decay:
k = ln 2 / Half-life
k = ln 2 / 4.96 x 10^3 s
k = 1.40x10⁻⁴s⁻¹
t is time = 1.98 x 10^4 s
[A]₀ = 3.21 x 10^17 nuclei
ln[A] = -1.40x10⁻⁴s⁻¹*1.98 x 10^4 s + ln[3.21 x 10^17 nuclei]
ln[A] = 37.538
[A] = 2.01x10¹⁶ nuclei remain ≈
c) 2.02 x 10^16 nucleiKomal found that her vial of isopropyl alcohol showed a much better surface
tension bubble shape (a higher bubble) than her vial of water. Her peer group
suggested some experimental errors that may have caused this to happen.
Which three experimental errors are most likely to have occurred?
Answer:
A B C
Explanation:just took the test
Answer:
Komal could have mixed up the labels on the vials.
Someone might have jiggled the table and made the water surface tension bubble spill over.
The water could actually have been saltwater instead of pure water.
Explanation:
...........................
An object is in free fall Group of answer choices Anytime it has an acceleration equal to 9.8 m/s/s Anytime it feels the force of gravity Only if the weight is the only force acting on the object Only when it's moving vertically down
Answer:
Anytime it feels the force of gravity.
Explanation:
Gravity is considered to be a universal force of attraction which acts between all objects that has both mass, energy and can occupy space. Therefore, it acts in such a way as to bring objects together i.e causing the objects to fall down towards the Earth.
This ultimately implies that, an object is in free fall anytime it feels the force of gravity i.e regardless of how fast the object moves or the direction it moves, the only force acting on the object is the force of gravity (g).
For example, when you throw any object up, it will naturally fall down due to the gravitational force between the Earth and the object.
Additionally, the gravity of earth makes it possible for all physical objects to possess weight.
On planet Earth, the acceleration due to gravity that all physical objects experience is 9.8 meters per seconds square.
A teacher performs a demonstration to show the properties of an unknown substance. The teacher cuts off a piece of gray shiny substance and drops it in water. The substance floats and reacts violently with the water. Based on this Information, what type of element the unknown substance?
A. metalloid
B. metal
C. nonmental
D. halogen
Answer:
nonmental
Explanation:
Option B is correct. The substance floats and reacts violently with the water. Based on this Information, metal is the unknown substance.
What is metal?A metal, a glossy substance that transmits electricity and heat reasonably efficiently when newly manufactured, polished, or broken.
Metals are either malleable or ductile. Metals can be chemical elements like iron.
Metal is substance floats and reacts violently with the water. When particles collide, energy is transferred. A change in temperature can be used to identify this.
The substance floats and reacts violently with the water. Based on this Information, metal is the unknown substance.
Hence, option B is correct.
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Two objects are separated by a distance D. Which force, in theory, will act on the objects for any value of D?
tension
gravity
normal
friction
Answer:
here is ur answer...
Explanation:
gravity .......is ur answer
and the value of D is DE³S NU + s
An attractive force that exist between all object with mass attract another objects the magnitude of the force is directly proportional..
Hope this help!
A 56 kg pole vaulter falls from rest from a height of 5.1 m onto a foam rubber pad. The pole vaulter comes to rest 0.29 s after landing on the pad.
Required:
a. Calculate the athlete's velocity just before reaching the pad
b. Calculate the constant force exerted on the pole vaulter due to the collision
a. The athlete's velocity just before reaching the pad is [tex]35.21m/s[/tex]
b. The constant force exerted on the pole vaulter is 6799.52 N
a. We use Newton's equation of motion,
[tex]v=u+at\\\\S=ut+\frac{1}{2}at^{2}[/tex]
Where u is initial velocity, v is final velocity, a is acceleration , t is time and S represent distance.
Given that, s = 5.1 m , t = 0.29s, u = 0
Substitute in above equation.
[tex]5.1=\frac{1}{2}*a*(0.29)^{2} \\\\a=\frac{5.1*2}{0.084}=121.42m/s^{2}[/tex]
the athlete's velocity, [tex]v=0+121.42*(0.29)=35.21m/s[/tex]
b. The constant force exerted on the pole vaulter due to the collision is given as, [tex]Force=mass*acceleration[/tex]
[tex]Force=56*121.42=6799.52N[/tex]
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A body is dropped from a height H. In how much time will it reach the ground?
Answer:
[tex]s = ut + \frac{1}{2} g {t}^{2} \\ t(u + \frac{1}{2} gt) = H \\ u + \frac{1}{2} gt = H \\ t = 2(H - u) \div g \\ t = \frac{(H - u)}{5} \\ u \: is \: speed \: or \: velocity[/tex]
Using pascals principle, F1/A1=F2/A2, solve this like a proportion.
A force of 50 N is applied to an area of 200 sq feet, how much force will be applied of the area to be covered is 50 sq feet?
Answer:
NO CLUE
Explanation:
GOOD LUCK THOUGH
Erica (37 kg ) and Danny (45 kg ) are bouncing on a trampoline. Just as Erica reaches the high point of her bounce, Danny is moving upward past her at 4.7 m/s . At that instant he grabs hold of her. What is their speed just after he grabs her?
Answer:
V = 2.58 m/s
Explanation:
Below is the calculation:
Given the weight of Erica = 37 kg
The weight of Danny = 45 kg
Danny's speed to move upward = 4.7 m/s
Use below formula to find the answer.
m1 * u1 = (m1+m2) * V
V = m1*u1 / (m1+m2)
Here, m1 = 45
u1 = 4.7
m1 = 45
m2 = 37
Now plug the values in formula:
V = m1*u1 / (m1+m2)
V = (45*4.7)/(45+37)
V = 2.58 m/s
Is the following chemical reaction balanced?
2H202-H2O + O2
yes
no
if a 230V television takes a current of 3A. calculate it's power
Answer:
Formula
P = I x V
Information
V = 230
A = 3
P = x
Calculation
x = 230 x 3
x = 690 W
Explanation:
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what is newtons 2nd law
According to the Newton's second law :- The acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.
If a person pulls back a rubber band on a slingshot without letting to go of it, what type of energy will the rubber band have? A. Potential energy B. Rotational energy C. Kinetic energy D. Translational energy
Answer:
Potential energy. Releasing it, the potential energy would convert into motion, kinetic energy.
Potential energy is when an object has some sort of potential eg. for motion such as in this example.
Answer:
A. Potential energy
Explanation:
Potential energy can be thought of as stored energy, which has the potential of becoming kinetic energy once it has been released.
PLEASE HELP!! URGENTT!
Answer:
120 Newton
Explanation:
Given the following data;
Mass = 12 kg
Angle = 4°
We know that acceleration due to gravity is equal to 10 m/s
To find the minimum force to stop the block from sliding;
Force = mgCos(d)
Where;
m is the mass of an object.
g is the acceleration due to gravity.
d is the angle of inclination (theta).
Substituting into the formula we have;
F = 12*10*Cos(4°)
F = 120 * 0.9976
F = 119.71 ≈ 120 Newton
A slingshot obeys Hooke's law and is used to launch pebbles vertically into the air. You observe that if you pull a pebble back 20 cm against the elastic
band, the pebble goes 6 m high. Assuming that air drag is negligible, if you pull a pebble that is twice as heavy back 20 cm, how high will it go?
4 m
5 m
3 m
6 m
Answer:
24 m
Explanation:
By applying the law of conservation of energy.
K.E = P.E
[tex]\dfrac{1}{2}kx^2 = mgh[/tex]
By rearrangement;
[tex]\dfrac{k}{m} = \dfrac{2*g*h}{x^2}[/tex]
where;
k = spring constant
m = mass of pebble
h = height reached
x = stretched length
[tex]\dfrac{k}{m} = \dfrac{2*9.8*6}{(20 \times 10^{-2})^2}[/tex]
[tex]\dfrac{k}{m} =0.2940 \times 10^4[/tex]
Since the drag force is negligible;
[tex]\dfrac{1}{2}kx^2 = mgh[/tex]
[tex]h = \dfrac{k}{m}\dfrac{x^2}{g}[/tex]
[tex]h = \dfrac{1}{2}\times 0.2940 \times 10^4 \times \dfrac{(40\times 10^{-2})^2}{9.8}[/tex]
h = 24 m
Please need in hurry
Explanation:
i) center of gravity (or mass)
ii) m = W/g = (160 N)/(9.8 m/s^2)
= 16.3 kg
A ceramic tile measuring 50 cm x50cm has been designed to bear a pressure of 40 N/in . Will it with stand a force of 5 N?
Answer:
Yes the tile can withstand a force 5 N
Explanation:
We'll begin by converting the dimensions from cm to in. This can be obtained as follow:
Dimension = 50 cm x 50cm
Recall
2.54 cm = 1 in
Therefore,
50 cm = 50 cm × 1 in / 2.54 cm
50 cm = 19.685 in
Thus, the dimension becomes:
Dimension = 19.685 in × 19.685 in
Next, we shall determine the area. This can be obtained as follow:
Dimension = 19.685 in × 19.685 in
Area = 19.685 in × 19.685 in
Area = 387.5 in²
Next, we shall determine the force to which the tile can withstand. This can be obtained as follow:
Pressure (P) = 40 N/in²
Area (A) = 387.5 in²
Force (F) =?
P = F/A
40 = F/387.5
Cross multiply
F = 40 × 387.5
F = 15500 N
Thus, the tile can withstand a force up to 15500 N.
Therefore, the answer to the question is:
Yes the tile can withstand a force 5 N
Suppose a boy is pulling a 10 kg wagon at a speed of 1m/s what is the kinetic energy?
Explanation:
Answer:At any instant, the object's kinetic energy is (5) x (its speed in meters per second)2 .
how to calculate sound of an echo
by an echo meter
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