Answer:
Period = 5 seconds
Frequency = 0.2 Hz
Explanation:
The period is the time per cycle. So, if a wave hits the beach every 5 seconds, the period will be 5 seconds.
Additionally, the frequency is the inverse of the period, so the frequency of the waves can be calculated as:
[tex]\text{frequency = }\frac{1}{Period}=\frac{1}{5}=0.2\text{ Hz}[/tex]So, the answers are:
Period = 5 seconds
Frequency = 0.2 Hz
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
When you apply a force F on an object of mass of m, it would produce acceleration a. If you apply the same force on another object of mass 2m, how would be the acceleration of the second object?
ANSWER
[tex]\frac{a}{2}[/tex]EXPLANATION
When a force is applied on an object of mass m, it produces an acceleration of a.
We can represent this relationship using Newton's second law of motion:
[tex]\begin{gathered} F=ma \\ \Rightarrow a=\frac{F}{m} \end{gathered}[/tex]Now, the same force is applied on an object with a mass of 2m.
Let the acceleration experienced by the object be a1. This implies that:
[tex]\begin{gathered} F=\left(2m\right)\left(a_1\right) \\ \Rightarrow a_1=\frac{F}{2m} \end{gathered}[/tex]We can write this new acceleration in terms of a as follows:
[tex]\begin{gathered} a_1=\frac{1}{2}\left(a\right) \\ a_1=\frac{a}{2} \end{gathered}[/tex]That would be the acceleration of the second object.
The answer is the third option.
DQuestion 111 ptsAn object has a mass of 12 kg. Assume the acceleration due to gravity is 10 m/s². If it is lifted to aheight of 20 m, what is its gravitational potential energy?2400 JoulesO 2400 NewtonsO 200 JoulesO 120 Newtons
Given:
• Mass, m = 12 kg
,• Acceleration due to gravity, g = 10 m/s²
,• Height, h = 20 m
Let's find the gravitational potential energy.
To find the gravitational potential energy, apply the formula:
[tex]GPE=m*g*h[/tex]Where GPE is the gravitational potential energy.
Thus, we have:
[tex]\begin{gathered} GPE=12*10*20 \\ \\ GPE=2400\text{ J} \end{gathered}[/tex]Therefore, the gravitational potential energy is 2400 Joules.
ANSWERl
A 50 kW tractor moves at a speed of 2.5 m/s. What is the traction force of the tractor?
In order to calculate the force, we can use the formula below:
[tex]P=F\cdot v[/tex]Where P is the power (in W), F is the force (in N) and v is the velocity (in m/s).
So we have:
[tex]\begin{gathered} 50000=F\cdot2.5\\ \\ F=\frac{50000}{2.5}\\ \\ F=20000\text{ N} \end{gathered}[/tex]Therefore the traction force is 20 kN.
Question 10 of 10If one of two interacting charges is doubled, the force between the chargeswillO A. decrease by 4 timesO B. increaseC. decreaseD. stay the sameSUBMIT
The force between the two charges is
[tex]Force\text{ }\propto\text{ charge}[/tex]Thus, if the charge is doubled, then its force will also be doubled.
Hence the correct option is increase.
cassy can get more force on the bricks she breaks with a blow of her bare hand when _______.
Answer:Her hand is made to bounce from the bricks
Explanation:
Cassy can get more force on the bricks that she breaks with a blow of her bare hand when her hand is made to bounce from the bricks,
What is Force?A force in physics is an effect that has the power to alter an object's motion. A mass-containing object's speed can vary, or accelerate, as a result of a force. Intuitively, a pull or a push can also be used to describe force. Being a vector quantity, a force also has magnitude and direction. The SI unit of newton is used to measure it (N). The letter F stands for force.
According to Newton's second law's original formulation, an object's net force is equal to the speed at which momentum is changing over time.
Objects' velocities can be altered by the concepts of push, drag, and torque. Thrust causes an object's velocities to increase, while torque causes an object's velocities to decrease. Each part of an extended body typically exerts pressures on its neighboring sections, and the internal mechanical force is determined by how these stresses are distributed.
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Problem Try to answer the following questions:(a) What is the maximum height above ground reached by the ball?(b) What are the magnitude and the direction of the velocity of the ball just before it hits the ground? Show Your Problem Solving Steps: Show these below:1) Draw a Sketch2) Choose origin, coordinate direction3) Inventory List – What is known?4) Write the kinematics equation(s) and solution of Part (a):5) Write the kinematics equation(s) and solution of Part (b):Problem 3 A small ball is launched at an angle of 30.0 degrees above the horizontal. It reaches a maximum height of 2.5 m with respect to the launch position. Find (a) the initial velocity of the ball when it’s launched and (b) its range, defined as the horizontal distance traveled until it returns to his original height. As always you can ignore air resistance.(a) Initial velocity [Hints: How is v0 related to vx0 and vy0. How can you use the information given to calculate either or both of the components of the initial velocity?](b) Range [Hints: This problem is very similar to today’s Lab Challenge except that for the challenge the ball will land at a different height.]
3.
[tex]\begin{gathered} \theta=30^{\circ} \\ y_{\max }=2.5m \end{gathered}[/tex]a)
[tex]\begin{gathered} y_{\max }=\frac{v^2\sin ^2(\theta)}{g} \\ \end{gathered}[/tex]Solve for v:
[tex]\begin{gathered} v=\sqrt[]{\frac{y_{\max }\cdot g}{\sin ^2(\theta)}} \\ v=98\cdot\frac{m}{s} \end{gathered}[/tex]b)
[tex]\begin{gathered} r=\frac{v^2}{9}\sin (2\theta) \\ r=\frac{98^2}{9.8}\cdot\sin (2\cdot30) \\ r=\frac{98^2}{9.8}\sin (60) \\ r=848.7m \end{gathered}[/tex]Will give brainliest!
A pair of fuzzy dice is hanging by a string from your rearview mirror. While you are accelerating from a stop-light to 28 m/s in 6.0 s, what angle theta does the string make with the vertical?
For the pair of fuzzy dice hanging by a string from the rearview mirror, while accelerating from a stop-light to 28 m/s in 6.0 s, the string makes an angle of 25.5 degrees with the vertical.
What is an angle?An angle is the figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle.
Parameters given include:
velocity: 28,/s time= 6.0 seconds
a = v/t = 28m/s / 6.0s = 4.667 m/s²
Taking into account the forces acting on the dice...there is a force of gravity acting straight down with a magnitude of mg.
so we have that
Angle = tan-1(a/g)
Angle = tan-1(4.667 / 9.8) = 25.5 degrees.
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a student drops a pebble from the edge of a vertical cliff. the pebble hits the ground 4 s after it was dropped. what is the height of the cliff? a. 20 m b. 40 m c. 60 m d. 80 m
The object's speed shortly before it lands on the earth is 40 m/s.
What is an example of velocity?The speed at which something moves in a specific direction is known as its velocity. as the speed of a car driving north on a highway or the pace at which a rocket takes off. Because the velocity vector is scalar, its absolute value magnitude will always equal the motion's speed.
The parameters are as follows: the pebble's time, t = 4 s; the object's velocity right before impact;
The kinematic equation is as follows;
v = in which
v = 0+10 (4)
The object's speed right before impact with the earth is v = 40 m/s2, where g is the acceleration caused by gravity and an is a constant of 10 m/s2. As a result,
the object's final velocity before impact is 40 m/s.
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Hi can you help me understand how to do this?
From the information given,
initial velocity of parachute = 198 ft/s
We want to convert 198 ft/s to m/s
Recall,
1 m = 3.3 ft
x m = 198 ft
By crossmultiplying, we have
3.3x = 198
x = 198/3.3
x = 60
Thus,
198 ft/s = 60 m/s
Rate = 60 m/s
To determine the distance that the parachute will fall in 10 seconds, we would apply one of Newton's equations of motion which is expressed as
s = ut + 1/2gt^2
where
s = distance covered
u = initial velocity
t = time
g = acceleration due to gravity and its value is - 9.8m/s^2
From the information given,
t = 10
u = 60
By substituting these values into the formula, we have
s = 60 x 10 - 1/2 x 9.8 x 10^2
s = 600 - 490
s = 110
The distance covered by the parachute in 10s is 110 m
27. Scientists have observed an increase in global temperatures over the past 100 years. Which phenomena do scientists believe contributes to the increase in temperatures? A. an increase in undersea volcanic activity B. a decrease in the distance between Earth and the Sun C. an increase in certain gases released during the use of fossil fuels D. a decrease in the amount of water on Earth due to overconsumption
The answer is letter C) An increase in certain gases released during the use of fossil fuels. Although the others do cause an increase in temperature, their scale cannot be compared to the one caused by fossil fuels.
The highest frequencies humans can hear isabout 20000 Hz.What is the wavelength of sound in airat this frequency?The speed of sound is310 m/s.
Given
The highest frequency human can hear is f=20000 Hz
Speed of the sound,v=310 m/s
To find
The wavelength
Explanation
We know,
The wavelength is given by
[tex]\begin{gathered} \lambda=\frac{v}{f} \\ \Rightarrow\lambda=\frac{310}{20000} \\ \Rightarrow\lambda=0.0155\text{ m} \end{gathered}[/tex]Conclusion
The wavelength is 0.0155 m
When the reflecting wave flips upside-down on a stretchedstring, which of the following is correct?a The stretched string is with a fixed boundaryb The stretched string is with a free boundaryсThe stretched string may have a fixed boundary ora free.d The given information is not enough.e None of the above is correct.
We are given a reflecting wave on a string. This can be exemplified in the following diagram:
What is the displacement and distance of the car from t=0s to t=10s
We are given a graph of distance vs time. To determine the displacement we must add up the total distance covered by the car.
We notice from the graph that the distance from t = 0 and t =2 is:
[tex]d_{0-2}=2m[/tex]From t = 2 and t = 3 there is no change
A 10.0 cm tall object is placed 6.00 cm in front of a curved mirror and produces an image 2.00 cm behind the mirror. What is the focal length of the mirror?0.667 cm1.50 cm-3.00 cm-0.333 cm
Given data:
The height of object is h₀=10.0 cm.
The object distance is u=6 cm.
The image distance is v=-2.00 cm.(negative because the image is behind the mirror)
The focal length can be calculated by the mirror's formula as,
[tex]\begin{gathered} \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \\ \frac{1}{f}=\frac{1}{6}+\frac{1}{-2} \\ f=-3.00\text{ cm} \end{gathered}[/tex]Thus, the focal length of the mirror is -3.00 cm.
If an airplane is running low on fuel, the pilot may decide to dump unneededweight. As the airplane gets lighter, the engines need less fuel to generate thesame amount of acceleration for flight. The pilot has taken advantage ofNewton's law of motion.A. fourthB. thirdO C. firstD. second
According to Newton's second law we have that the force is equivalent to the product of the mass and the acceleration:
[tex]F=ma[/tex]This means that if we decrease the mass "m" of the object the required force is decreased also. Since the force is proportional to the required energy and the energy comes from the fuel consumption this means
A student pushes a baseball of m = 0.13 kg down onto the top of a vertical spring that has its lower end fixed to a table, compressing the spring a distance of d = 0.18 meters from its original equilibrium point. The spring constant of the spring is k = 970 N/m. Let the gravitational potential energy be zero at the position of the baseball in the compressed spring.
A. What is the maximum height, h, in meters, that the ball reaches above the equilibrium point?
B. What is the ball’s velocity, in meters per second, at half of the maximum height relative to the equilibrium point?
The maximum height reached by the baseball above the equilibrium point is 12.14 m.
The ball’s velocity, in meters per second, at half of the maximum height relative to the equilibrium point is 458.8 m/s.
What is the maximum height attained by the ball?The maximum height reached by the ball is determined as follows:
Data given:
compression of the spring is d = 0.18 m.mass of the baseball is m = 0.13 kgthe spring constant, K = 970 N/mThe gravitational potential energy at the compressed position is zero.
Based on the law of conservation of energy, the total energy of the system is conserved.
Let the final height from the bottom be h
m * g* h = ¹/₂ * K * d²
h = (k * d²) / (2 * m * g)
h = 970 * 0.18² / (2 * 0.13 * 9.81)
h = 12.32 m
Height above the equilibrium point = 12.32 - 0.18
Height above the equilibrium point = 12.14 m
Velocity is calculated1 as follows:
Half of the maximum height relative to the equilibrium point = 12.14/2 + (0.18) = 6.25 m
¹/₂ * m * v² = ¹/₂ * K * d²
m * v² = K * d²
v = √(K * d² / m)
v = √(970 * 6.25² / 0.13)
v = 458.8 m/s
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Look at Figure 13-17 to answer the question.Will the acceleration of the piano be greater in A or in B? Use Newton's second law ofmotion to explain your answer.
Answer:
Explanation:
Newton's second law states that the applied force is directly proportional to the rate of change of momentum
momentum = mass x velocity
Rate of change in momentum = (final momentum - initial momentum)/time
Rate of change in momentum = m(v - u)/t
where
v is final velocity
u is initial velocity
t is times
Recall,
Acceleration = (v - u)/t
Thus, the equation becomes
Force = ma
This means that as the force increases, the acceleration increases.
Considering the given scenarios,
First piano is pushed by force from the man
For second piano, there is additional force from the woman and the mass of the piano remains the same. Since there is more force, there is more acceleration. Thus,
Acceleration is greater in B
In a system of 2 large round objects, R1 and R2 (R1 is larger), what properties will affect the force of gravity between them? (select all that apply)
According to Newton's Law of Universal Gravitation, if two particles with masses M and m are located a distance r from each other, an attractive force between them appears, and its magnitude is given by:
[tex]F=G\frac{Mm}{r^2}[/tex]Where G is the gravitational constant:
[tex]G=6.67\times10^{-11}N\frac{m^2}{\operatorname{kg}^2}[/tex]From the equation, we can see that the properties that affect the force between them are their masses and the distance between them.
Then, the correct choices are:
- Mass of R1
- Mass of R2
- Distance from the center of R1 to the center of R2.
Which label goes through the horizontal axis?
A. Distance
B. Acceleration
C. Force
D. Mass
Answer:
A: Distance
Explanation:
You can not change gravity with acceleration or force in a weird way. Only mass and distance really affect gravity in normal physics.
Gravity is stronger when two objects are closer. Therefore gravity would decrease as distance increased.
Mass increases gravity.
The image below shows that gravity (depth) increases as the distance from the black hole decreases.
A convex spherical mirror has a radius of curvatureof 9 40 cm. A) Calculate the location of the image formed by an 7.75mm tall object whose distance from the mirror is 17.5 cmCalculate the size of the imageC) Calculate the location of the image formed by an 7.75mm tall object whose distance from the mirror is 10.0cmE) Calculate the location of the image formed by an 7.75mm tall object whose distance from the mirror is 2.65cmG) Calculate the location of the image formed by an 7.75mm tall object whose distance from the mirror is 9.60m
0.We are asked to determine the location of an image formed by an 7.75mm tall object that is located a distance of 17.5 cm from a convex mirror.
First, we will calculate the focal length using the following formula:
[tex]f=-\frac{R}{2}[/tex]Where:
[tex]\begin{gathered} f=\text{ focal length} \\ R=\text{ radius} \end{gathered}[/tex]Substituting the values we get:
[tex]f=-\frac{9.40cm}{2}[/tex]Solving the operations:
[tex]f=-4.7cm[/tex]Now, we use the following formula:
[tex]\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}[/tex]Where:
[tex]\begin{gathered} d_0=\text{ distance of the object} \\ d_i=\text{ distance of the image} \end{gathered}[/tex]Now, we substitute the known values:
[tex]\frac{1}{17.5cm}+\frac{1}{d_i}=-\frac{1}{4.7cm}[/tex]Now, we solve for the distance of the image. First, we subtract 1/17.5 from both sides:
[tex]\frac{1}{d_i}=-\frac{1}{4.7cm}-\frac{1}{17.5cm}[/tex]Solving the operation:
[tex]\frac{1}{d_i}=-0.27\frac{1}{cm}[/tex]Now, we invert both sides:
[tex]d_i=\frac{1}{-0.27}cm=-3.7cm[/tex]Therefore. the location of the image is -3.7 centimeters.
The other parts are solved using the same procedure.
Part B. To calculate the size of the image we will use the following relationship:
[tex]\frac{h_i}{h_o}=-\frac{d_i}{d_0}[/tex]Where:
[tex]h_i,h_0=\text{ height of the image and height of the object}[/tex]Substituting we get:
[tex]\frac{h_i}{7.75mm}=-\frac{-3.7cm}{17.5cm}[/tex]Solving the operations on the right side:
[tex]\frac{h_i}{7.75mm}=0.21[/tex]Now, we multiply both sides by 7.75:
[tex]h_i=(7.75mm)(0.21)[/tex]Solving the operations:
[tex]h_i=1.64mm[/tex]Therefore, the height of the iamge is 1.64 mm.
Two plastic blocks of the same plastic material have the following characteristics:
Block A: Mass 10 g, Volume 25 mL, and
Block B: Mass 20 g, Volume 50 mL.
Which of the statements below is true? (Density of water is 1g/mL at room temperature.)
Group of answer choices
Both blocks will sink equally under water.
Block A will sink deeper under water.
Both blocks will float equally over water.
Block B will float more than Block A over water.
The true statement, given the data from the question is: Both blocks will float equally over water.
How to determine the true statementTo know which statement is true, we shall obtain the density of each blocks. Details below
For block A:
Mass of block A = 10 gramsVolume of block A = 25 mL Density of block A = ?Density = mass / volume
Density of block A = 10 / 25
Density of block A = 0.4 g/mL
For block B:
Mass of block B = 20 gramsVolume of block B = 50 mL Density of block B = ?Density = mass / volume
Density of block B = 20 / 50
Density of block B = 0.4 g/mL
From the above calculation, we can see that the two blocks have the same density.
Thus, we can conclude that the true statement is both blocks will float equally over water.
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An unwary football player collides head-on with a padded goalpost while running at 7.5 m/s and comes to a full stop after compressing the padding and his body by 0.27 m. Take the direction of the player’s initial velocity as positive.1.assuming constant acceleration calculate the his acceleration during the collision in meters per second squared.2 how long does the collision last in seconds.
Answers:
1. a = -104.16 m/s²
2. t = 0.072
Explanation:
To find the acceleration, we will use the following equation:
[tex]v^2_f=v^2_i+2ax[/tex]where vf is the final velocity, vi is the initial velocity, a is the acceleration and x is the distance. So, replacing vf by 0 m/s, vi by 7.5 m/s, and x by 0.27m, we get:
[tex]\begin{gathered} 0^2=7.5^2+2a(0.27) \\ 0=56.25+0.54a \end{gathered}[/tex]Then, solving for a, we get:
[tex]\begin{gathered} 0-56.25=56.25+0.54a-56.25 \\ -56.25=0.54a \\ \frac{-56.25}{0.54}=\frac{0.54a}{0.54} \\ -104.16m/s^2=a \end{gathered}[/tex]Therefore, the acceleration during the collision is -104.16 m/s²
Then, to calculate how long the collision last, we will use the following equation:
[tex]v_f=v_i+at[/tex]So, replacing the values and solving for t, we get:
[tex]\begin{gathered} 0=7.5-104.16t \\ 104.15t=7.5 \\ t=\frac{7.5}{104.15}=0.072s \end{gathered}[/tex]Therefore, the collision last 0.072 seconds
upon leaving her club, the golf ball moved upward to a height above the surrounding trees. is the ke and pe increasing, decreasing, or staying the same?
ANSWER
PE increases and KE decreases
EXPLANATION
As described, the golf ball is moving and changing its height, like in the following diagram,
By the law of conservation of energy, the total energy when the ball starts moving and during the whole motion until it stops, must be the same. This total energy is the sum of the potential energy and the kinetic energy.
When the club hits the ball, it gives it a certain amount of kinetic energy but no potential energy. As the ball starts going uphill, the potential energy starts to increase, since it depends on the height of the object. Therefore, to maintain the total energy constant, the kinetic energy must decrease.
(20%) Problem 5: Two identical springs, A and B, each with spring constant k = 54.5 N/m, support an object with a weight W = 11.6 N. Each spring makes an angle of 0 = 20.6 degrees to the vertical, as shown in the diagram. Create an expression for the tension in spring A
The tension in spring A is T = W/(2cosθ)
What is tension?Tension is the stretching force in a spring.
How to find the expression for the tension in the spring?Let
T = the tension in the each spring, W = weight andθ = angle each spring makes with the verticalResolving the tension in each spring vertically, so we can have that
for spring A, the tension is Tcosθ and for spring B, the tension is TcosθNow the vertical component of the tension in each equals the weight. So, we have that
Tcosθ + Tcosθ = W
Adding them together, we have that
2Tcosθ = W
Dividing both sides by 2cosθ, so, we can have that
T = W/2cosθ
Thus, the tension in each spring is T = W/(2cosθ)
So, the tension in spring A is T = W/(2cosθ)
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A car is traveling at a speed of 62mph. The radius of its tires are 17 inches. What is the angular speed of the tires in rad/min? Round to the nearest whole number.
To find the angular velocity of the car tires, we can use the angular velocity formula:
[tex]v=rw[/tex]First, we will need to derive from this equation the equation to find angular velocity as this gives us the speed.
[tex]\begin{gathered} v=rw \\ w=\frac{v}{r} \end{gathered}[/tex]Now, we need to convert our values from mph to m/s.
1 mph = 0.44704 m/s
62 * 0.44704 = 27.71648 m/s = v
Then, we need to find the radius of the tire in meters.
1 inch = 0.0254 meters
17 * 0.0254 = 0.4318 m = r
Now, we will replace these values with the ones in the equation.
[tex]w=\frac{27.71648}{0.4318}=64.18823529[/tex]Finally, we can round this to be easier to interpret:
The angular speed of the tire is 64.19 radians/minute, or more compactly: 64.19 rad/min.
If a space ship traveling at a 1000 miles per hour enters and area free of Gravitational forces,it’s engine must run at some minimum level in order to maintain the ships velocity. is this statement true or false
The given statement is false.
If a spaceship traveling at 1,000 miles per hour enters an area free of gravitational forces, its engine must run at some maximum level in order to maintain the ship’s velocity
A Tour de France cyclist at a rate of 6.5 m/s and is 333 m
ahead of a crew van with powerdrink refills. The van is
traveling at 15 m/s and is accelerating at a constant rate of
0.4 m/s2
How much time will it take for the crew van to catch up
with the cyclist?
Given data:
* The speed of the cyclist is 6.5 m/s.
* The initial distance of the cyclist from the van is 333 m.
* The initial velocity of the van is 15 m/s.
* The acceleration of the van is,
[tex]a=0.4ms^{-2}[/tex]Solution:
Let x be the distance from the van initial position at which the cyclist and van meet.
Let the cyclist meet the van at time t.
By the kinematics equation, the position of the cyclist at time t is,
[tex]x-333=u_ct+\frac{1}{2}a_ct^2[/tex]where u_c is the speed of the cyclist, a_c is the acceleration of the cyclist, t is the time taken and 333-x is the distance traveled by the cyclist at time t,
The acceleration of the cyclist is zero.
Substituting the known values,
[tex]\begin{gathered} x-333=6.5t+0 \\ x-333=6.5t \end{gathered}[/tex]By the kinematics equation, the position of the van after time t is,
[tex]x=u_vt+\frac{1}{2}a_vt^2[/tex]where u_v is the velocity of the van, a_v is the acceleration of the van, and t is the time taken,
Substituting the known values,
[tex]\begin{gathered} x=15t+\frac{1}{2}\times0.4\times t^2 \\ x=15t+0.2t^2 \end{gathered}[/tex]Substituting this value of x in the kinematics equation of the cyclist,
[tex]\begin{gathered} (15t+0.2t^2)-333=6.5t \\ 15t+0.2t^2-6.5t-333=0 \\ 0.2t^2+8.5t-333=0 \end{gathered}[/tex]By solving the quadratic equation,
[tex]\begin{gathered} t=\frac{-8.5\pm\sqrt[]{8.5^2-(4\times0.2\times(-333))}}{2\times0.2} \\ t=\frac{-8.5\pm\sqrt[]{^{}8.5^2+(4\times0.2\times333)}}{2\times0.2} \\ t=\frac{-8.5\pm18.4}{0.4} \\ t=24.8\text{ s or-67.25 s} \end{gathered}[/tex]As the value of time cannot be negative.
Thus, the time at which the cyclist and van meet is 24.8 seconds.
Calculate the total capacitance of three capacitors 30µF, 20µF & 12µF connected in parallel across a d.c supply The answer is :Consider that the equivalent capacitance of three capacitors C1, C2 and C3 in parallel is given by:C=C1+C2+C3In this case:C1 = 30µFC2 = 20µFC3 = 12µFReplace the previous values into the formula for C and simplify:C=30μF+20μF+12μF=62μFHence, the total capacitance is 62µFQuestion 5 : Calculate the total charge on the capacitors connected in parallel if the supply voltage is 500V. Sketch a circuit diagram and label this to show how the charges are located
The circuit diagram is shown below:
From the diagram we notice that the same voltage will flow in every capacitor, this will be helpful later.
We know that this three capacitors are equivalent to a single equivalent capacitor with 62µF capacitance. The charge in this equivalent capacitor is:
[tex]Q_{eq}=(62\times10^{-6})(500)=0.031[/tex]Now, as we mentioned, the voltage is the same in each capacitor then the charge in each of them is:
[tex]\begin{gathered} Q_1=(30\times10^{-6})(500)=0.015 \\ Q_2=(20\times10^{-6})(500)=0.01 \\ Q_3=(12\times10^{-6})(500)=0.006 \end{gathered}[/tex]To check if this is correct we need to remember that the charge in the equivalent capacitor is equal to the sum of the charge in each capactior; for this case this conditon is fulfil; therefore we conclude that:
• The charge in the first capacitor is 0.015 C
,• The charge in the second capacitor is 0.01 C
,• The charge in the third capacitor is 0.006 C
The diagram with the labels is shown below:
What is the gravitational force between two trucks, each with a mass of 2.0 x 10^4 kg, that are 2.0m apart? (G=6.673 x 10^-11 N•m^2/kg^2)
Firs we will use the next formula
[tex]F=G\text{ }\frac{m_1\cdot m^{}_2}{r^2}[/tex]Where G is the gravitational force, m nd m2 are the masses and r is the distance between the masses
In our case
m1=m2= 2.0 x 10^4 kg
r= 2m
G=6.673 x 10^-11 N•m^2/kg^2
We substitute
[tex]F=6.673\times10^{-11}\cdot\frac{2.0x10^4\cdot2.0x10^4}{2^2}=0.0066743N=6.7\times10^{-3}[/tex]ANSWER
The gravitational force is 0.00667=6.7x10^-3N