The work done is defined as,
[tex]W=F\cdot d[/tex]where F isthe force and d is the displacement (distance in magnitude).
The principal of a local high school gave all students breakfast for one week. Teachers commented that students were more alert that week in the morning.
Students commented they were more alert and could concentrate in the morning after eating breakfast. What can be concluded based on the observations?
All principals should give free breakfast every morning.
If students eat breakfast they will be able to concentrate in school in the afternoon.
Teachers should serve breakfast to their students.
Students should eat breakfast to be more alert at school during the morning.
Answer:
D Students should eat breakfast to be more alert at school during the morning.
Explanation:
Because based on the information given they said the principal gave students breakfast and teacher's said the students were more alert and the only common sense answer is D because they are more active in class in the morning not afternoon.
Jennifer hits a stationary 0.20-kg ball, and it leaves her racket at 40 m/s. Time-lapse photography shows that the ball was in contact with the racket for 40 ms.What average force did the ball exert on the racket?Express your answer using two significant figures.What is the ratio of this force to the weight of the ball?Express your answer using two significant figures.
Given:
The mass of the stationary ball is: m = 0.20 kg.
The velocity of the ball when it leaves the rocket is: Vf = 40 m/s.
The time for which the ball was in contact with the rocket is: t = 40 ms = 0.04 s.
To find:
The average force the ball exerts on the rocket.
The ratio of the force on a rocket to the weight of the ball.
Explanation
As the ball is initially stationary, its initial velocity Vi is zero. After the ball was hit, it moves with the velocity Vf = 40 m/s for time t = 0.04 s.
Thus, using Newton's second law, we get:
[tex]\begin{gathered} F=ma \\ \\ F=m\times\frac{V_f-V_i}{t} \end{gathered}[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} F=0.20\text{ kg}\times\frac{40\text{ m/s}-0\text{ m/s}}{0.04\text{ s}} \\ \\ F=200\text{ kg.}^\text{m/s}^2 \\ \\ F=200\text{ N} \end{gathered}[/tex]The weight of the ball on the earth is the product of its mass and acceleration due to gravity g. The value of the acceleration due to earth is: g = 9.8 m/s^2.
The weight W of the ball on earth is given as:
[tex]W=mg[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} W=0.20\text{ kg}\times9.8\text{ m/s}^2 \\ \\ W=1.96\text{ N} \end{gathered}[/tex]The ratio of the force F exerted by the ball on the rocket and the weight of the ball is:
[tex]\begin{gathered} \frac{F}{W}=\frac{200\text{ N}}{1.96\text{ N}} \\ \\ \frac{F}{W}=\frac{102.04}{1} \end{gathered}[/tex]Thus, F : W = 102.04 : 1.
Final answer:
The force exerted by the ball on the rocket is 200 N. The ratio of this force and the weight of the ball is 102.04 : 1.
What will be it’s final speed and kinetic energy after it has been moving for t=5sec
Given data:
* The force applied on the block is,
[tex]F_a=12.5\text{ N}[/tex]* The frictional force acting on the block is,
[tex]F_r=4.2\text{ N}[/tex]* The acceleration of the block is,
[tex]a=1.2ms^{-2}[/tex]Solution:
The net force acting on the block is,
[tex]\begin{gathered} F_{\text{net}}=F_a-F_r \\ F_{\text{net}}=12.5-4.2 \\ F_{\text{net}}=8.3\text{ N} \end{gathered}[/tex]According to Newton's second law, the net force in terms of the mass and acceleration of the block is,
[tex]\begin{gathered} F_{\text{net}}=ma \\ m=\frac{F_{\text{net}}}{a}_{} \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} m=\frac{8.3}{1.2} \\ m=6.92\text{ kg} \end{gathered}[/tex]As the block is initially at rest, thus, the initial speed of the block is u = 0 m/s.
By the kinematics equation, the final speed of the block after t = 5 seconds is,
[tex]\begin{gathered} v-u=at \\ v-0=1.2\times5 \\ v=6\text{ m/s} \end{gathered}[/tex]Thus, the final speed of the block after t = 5 seconds is 6 meters per second.
The kinetic energy of the block after t = 5 seconds is,
[tex]\begin{gathered} K=\frac{1}{2}mv^2 \\ K=\frac{1}{2}\times6.92\times6^2 \\ K=124.56\text{ J} \end{gathered}[/tex]Thus, the kinetic energy of the block after t = 5 seconds is 124.56 joule.
Two particles each with charge -Q are a fixed distance L apart, as show above. Each particle experiences a net electric force F. A particle with a charge +q is now fixed midway between the two particles. As a result the net electric force experienced by each negatively charged particle is reduced to F/2. What is the value of +q?QQ/2Q/4Q/8Q/16
The electric force between two charges can be calculated with the formula below (Coulomb's law):
[tex]F_e=\frac{K_e\cdot|q_1|\cdot|q_2|}{d^2}[/tex]Where Ke is the Coulomb's constant, q1 and q2 are the charges and d is the distance between them.
So, for Fe = F, q1 = q2 = Q and d = L, we have:
[tex]F=\frac{K_e\cdot Q^2}{L^2}[/tex]Now, after the addition of a positive charge in the middle of the charges, each negative charge will suffer another force, acting on the opposite direction of force F:
Since the new resulting force on the negative charges is F/2, the new force created by the positive charge addition is also F/2, so we have:
[tex]\begin{gathered} \frac{F}{2}=\frac{K_e\cdot Q\cdot q}{(L/2)^2} \\ F=\frac{2\cdot K_e\cdot Q\cdot q}{L^2/4} \\ F=\frac{8\cdot K_e\cdot Q\cdot q}{L^2} \\ \frac{K_e\cdot Q^2}{L^2}=\frac{8\cdot K_e\cdot Q\cdot q}{L^2} \\ Q=8q \\ q=\frac{Q}{8} \end{gathered}[/tex]Therefore the correct option is the fourth one: q = Q/8.
A football kicker is attempting a field goal from 44m out. The ball is kicked and just clears the lower bar with a time of flight of 2.9s. If the angle of the kick was 45°, what was the initial speed of the ball, assuming no air resistance?Group of answer choices
Given,
Distance travelled, d=44 m.
The time of flight t=2.9 s
The angle is
[tex]\theta=45^o[/tex]Let the initial velocity be v
The motion is projectile.
Thus the horizontal distance travelled is
[tex]\begin{gathered} d=(v\cos \theta)t \\ \Rightarrow44=v\times\cos 45^o\times2.9 \\ \Rightarrow v=21.5\text{ m/s} \end{gathered}[/tex]Thus the answer is 21.5 m/s
If Rodger wants to determine whether a baseball thrown at 100 m/s or a bowling ball thrown at 100 m/s will have more momentum, what is one measurement device he could use to figure this out, without actually throwing either ball?A. a rulerB. a stopwatchC. a balance scaleD. an accelerometer I chose C, I am not completely sure that I’m right
The momentum of an object is given by the product of its mass and its velocity:
[tex]p=mv[/tex]If both objects have the same velocity (in this case, 100 m/s), the object with the greater mass will have the greater momentum.
In order to measure the mass of an object, we can use a balance scale.
Therefore the correct option is C.
Professor Stauffer conducts basic research on the psychological, biological, and behavioral
factors that contribute to heart disease. She is most likely a
High blood pressure, high LDL cholesterol, diabetes, smoking, exposure to secondhand smoke, obesity, a poor diet, and inactivity are the main risk factors for heart disease and stroke.
Explain about the contribute to heart disease?High blood pressure, high cholesterol, and smoking are the three main risk factors for heart disease that at least half of all Americans (47%) have. Age and family history are two uncontrollable risk factors for heart disease.
The most frequent cause of coronary artery disease is atherosclerosis, which is a buildup of fatty plaques in the arteries. Poor diet, lack of exercise, obesity, and smoking are risk factors. Choosing a healthy lifestyle can help reduce the risk of atherosclerosis.
The most prevalent type of heart disease is coronary heart disease (CHD). It happens when plaque deposits cause the arteries that carry blood to the heart to constrict or stiffen. Fat, cholesterol, and other components found in blood make up plaque. The term "atherosclerosis" also refers to this plaque accumulation.
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A 3500 kg van hits a 2500 kg car with a force of 1480 N [E].a) What force does the van experience?b) Calculate the acceleration of both vehicles after the collision.
Given
m1 = 3500 kg (van)
m2 = 2500 kg (car)
F = 1480 N
Procedure
a) In a collision between object 1 and object 2, the force exerted on object 1 (F1) is equal in magnitude and opposite in direction to the force exerted on object 2 (F2). In equation form:
[tex]F1=-F2[/tex]The above statement is simply an application of Newton's third law of motion to the collision between objects 1 and 2.
b) Accelaration
[tex]\begin{gathered} F=ma \\ a=\frac{F}{m} \\ a=\frac{1480N}{3500+2500} \\ a=0.246\text{ m/s} \end{gathered}[/tex]The acceleration would be 0.246 m/s
Two coils , held in fixed positions , have a mutual inductance of 100 mH . What is the peak emf in one coil when the current in the other coil is i ( t ) = 10 sin ( 1000 t ) , where is in amperes and t is in seconds ?
The peak emf in one coil at the given current in the other coil is 1,000 A.
What is mutual inductance?Mutual Inductance between the two coils is defined as the property of the coil due to which it opposes the change of current in the other coil.
The peak emf in one coil when the current in the other coil is i ( t ) = 10 sin ( 1000 t ) depends on the mutual inductance and the peak current.
The peak emf in one coil is calculated as follows;
emf₀ = M di/dt
where;
M is the mutual inductancedi/dt is the rate of change of currentdi/dt = (10 x 1000) cos (1000 t)
di/dt = 10,000 cos(1000 t)
The peak rate of change of current from the derivative of di/dt = 10,000 A/s
emf₀ = (100 x 10⁻³) x (10,000 A/s)
emf₀ = 1,000 V
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What is the spring constant for a spring that stretches by 27 cm when a load of 200 N is suspended from it?1) 0.135 N/m2) 54 N/m3) 740 N/m4) 7.4 N/m
Given data:
Force;
[tex]F=200\text{ N}[/tex]Stretching;
[tex]\begin{gathered} x=27\text{ cm} \\ =0.27\text{ m} \end{gathered}[/tex]The force is given as,
[tex]F=kx[/tex]Here, k is the force constant.
The expression for the force constant is given as,
[tex]k=\frac{F}{x}[/tex]Substituting all known values,
[tex]\begin{gathered} k=\frac{200\text{ N}}{0.27\text{ m}} \\ =740.74\text{ N/m} \end{gathered}[/tex]Therefore, the force constant of the spring is 740.74 N/m. Hence, option (3); 740 N/m is the best choice.
Experiments and Data Collections and Interpreting Data
While studying paintball equipment, you may have learned that changes in temperature can cause paintballs to change size. At high temperatures, they swell, and at low temperatures, they shrink. This changes the way they fit in the marker barrel. You might affect their performance; if not, you can make an educated guess.
Include the following essential content:
* State a specific question about temperature and paintball performance.
Using the data provided, draw or create a simple graph that can be used to interpret the results.
Be sure to include these points:
*descriptive title
*dependent variable
*independent variable
*legend
At high temperatures, they swell, and at low temperatures, they shrink. This adjustment is the way they are in shape within the marker barrel. we may affect their overall performance.
Data Collection is the system of collecting and measuring statistics on focused variables in an established machine, which then allows one to reply to relevant questions and compare outcomes. statistics collection is a research issue in all study fields, which include physical and social sciences, humanities, and business.
Data Collection is the method of gathering and measuring facts on ation series variables of hobby, in a longtime systematic fashion that permits one to answer stated research questions, test hypotheses, and compare results.
Surveys, questionnaires and bureaucracy, completed online, in character or with the aid of smartphone, electronic mail or ordinary mail; awareness organizations and one-on-one interviews; and. direct observation of participants in a studies look at.
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A. 125 m
B. 27.5 m
C. 62.5 m
D. 55 m
(i don’t understand how to get the answer)
55m is the magnitude of the average distance of the car during this time interval.
The correct option is D.
How is distance measured in physics?The metre is the Si derived unit for distance (m). Long distances can be measured in kilometres, whereas little distances can be gauged in millimetres (cm) (km). For instance, you might measure the length in centimetres between the bottom and top of a piece of paper and the kilometres from your home and school.
What is Distance?The total movement of an object, independent of direction, is its distance. No matter where an object starts or ends, distance can be defined as the amount of space it has covered.
Briefing:The equation for the Total distance is mathematically given as
d=initial distance -final distance
d=90-35
d=55m
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The complete question is-
A car travels 90 meters due North in 10 seconds. The car turns around and travels 35 meters due south in 5 seconds. What is the magnitude of the average distance of the car during this time interval?
A. 125 m
B. 27.5 m
C. 62.5 m
D. 55 m
Answer:
27.5°
Subtract the summation of the angles from 180°
Explanation:
A crate is given a big push, and after it is released, it slides up an inclined plane which makes an angle 0.49 radians with the horizontal. The frictional coefficients between the crate and plane are (s = 0.66, k = 0.21 ). What is the magnitude of the acceleration (in meters/second2) of this crate as it slides up the incline?
The acceleration of the crate as it slides up the incline is 2.06 m/s².
What is the acceleration of the crate?
The acceleration of the crate as it slides up the inclined plane is calculated as follows;
Apply the principle of Newton's second law of motion to determine the acceleration of the crate.
Ff = ma
kmg = ma
kg = a
where;
k is the coefficient of kinetic frictiong is acceleration due to gravitya is the acceleration of the cratea = (0.21 x 9.8 m/s²)
a = 2.06 m/s²
Thus, the acceleration of the crate as it slides up the incline depends on the coefficient of kinetic friction since the crate is in motion.
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To a mass m1 = 100 g of water at θ1 = 10°C we add a mass m2 = 60 g of water at θ2 = 55°C. Calculate the final temperature of the mixture? Data: Specific heat capacity of water in the liquid state: 4.18 kJ.kg -1 .K
Given:
The mass of water is m1 = 100 g = 0.1 kg
The temperature of the water is
[tex]\begin{gathered} \theta1=10^{\circ}\text{ C} \\ =10+273 \\ =283\text{ K} \end{gathered}[/tex]The mass of water is m2 = 60g = 0.06 kg
The temperature of the water is
[tex]\begin{gathered} \theta2=\text{ 55 }^{\circ}\text{C} \\ =55+273 \\ =328\text{ K} \end{gathered}[/tex]To find the final temperature of the mixture.
Explanation:
The final temperature of the mixture can be calculated by the formula
[tex]\begin{gathered} T=\frac{(m1\theta1+m2\theta2)}{(m1+m2)} \\ \text{ } \end{gathered}[/tex]On substituting the values, the final temperature will be
[tex]\begin{gathered} T=\frac{(0.1\times283)+(0.06\times328)}{(0.1+0.06)} \\ =293.625\text{ K} \\ =20.625\text{ }^{\circ}C \end{gathered}[/tex]Thus, the final temperature of the mixture is 20.625 degrees Celsius.
evaluate when 8.1p + 7.5r when p=6 and r=7
You have the following expression:
8.1p + 7.5r
if p=6 and r=7, you obtain for the previous expression:
8.1(6) + 7.5(7) = 48.6 + 52.5 = 101.1
Hence, the result is 101.1
A 2 kg object and a 1 kg object are dropped from the same height.
Given that the inertia
masses of these objects are not the same, what hypothesis explains
why the two objects
fall with the same acceleration?
Select one:
O a. The force of gravity is the same for the 1 kg object as for the 2 kg object
O b. The force of friction is the same for the 1 kg object as for the 2 kg object.
O c. The force of gravity on the 1 kg object is twice as great as on the 2 kg object.
Od. The force of friction on the 2 kg object is twice as great as on the 1 kg object.
c. The force of gravity on the 1 kg object is twice as great as on the 2 kg object hypothesis explains why the two objects fall with the same acceleration.
Why do heavier and lighter objects fall at the same pace when dropped from the same height?If you disregard air resistance, things falling close to the surface of the Earth accelerate roughly at 9.8 m/s2 (9.8 m/s2, or g) because of gravity. As a result, the objects' acceleration is the same, and as a result, their velocity is likewise rising steadily.
No matter how heavy the object is, if there is no air resistance, the rate of descent simply relies on how far the object has dropped. This indicates that if two things are dropped simultaneously from the same height, they will both touch the ground at the same moment.
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PLS HELP MANY POINTS
If W = F x d, which of the following equations shows work being calculated using the correct units? (1 po
O 113 N= (17.4 J) x (6.51 ft)
O 113 J= (17.4 N) x (6.51 m)
O 113 N= (17.4 J) x (6.51 m)
O 113 J= (17.4 N) x (6.51 ft)
The equation that shows work being calculated using the correct units is as follows: 113J = (17.4 N) x (6.51 m) {option B}.
What is work done?Work done is a measure of energy expended in moving an object. It is most commonly calculated by multiplying force by distance. No work is done if the object does not move.
Work done is measured in Joules, which is equivalent to Newtons/metres (N/m).
Joules is the derived unit of energy, work and heat; the work required to exert a force of one newton for a distance of one metre.
Newton is the unit of measurement for force while metres is the unit of measurement for distance, hence, to calculate a work done of 113 Joules, 17.4 Newtons of force over a distance of 6.51 metres is needed.
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Answer:
It's 113 j = (17.4N) x (6.51)
Explanation:
Your welcome
One tugboat pulls on the barge with a force of magnitude 4000 units of force at 15° above the line AB (see the figure) and the other tugboat pulls on the barge with a force of magnitude 5000 units of force at 12° below the line AB. Resolve the pulling forces to their scalar components and find the components of the resultant force pulling on the barge.
The resultant components of force pulling on the barge is 8753.7 N.
What is the resultant force on the barge?
The resultant force on the barge is calculated by resolving the force into y and x-components.
The x and y-component of forces on the 4000 unit force is calculated as;
Fx = F cosθ
Fy = F sinθ
Fx = 4000 x cos(15)
Fx = 3863.7 N
Fy = 4000 x sin(15)
Fy = 1035.27 N
The x and y-component of forces on the 5000 unit force is calculated as;
Fx = 5000 x cos(12)
Fx = 4890.7 N
Fy = -5000 x sin(12)
Fy = -1039.6 N
Net horizontal force;
∑Fx = 3863.7 N + 4890.7 N
∑Fx = 8753.7 N
Net vertical force;
∑Fy = 1035.27 N - 1039.6 N
∑Fy = -4.33 N
The resultant force on the barge;
F = √[(-4.33²) + (8753.7²)]
F = 8753.7 N
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While standing on a 3-foot ladder, a grapefruit is tossed straight up with an initial velocity of 45 fu/sec. The initial position of the grapefruit is 7 feet above the ground when it is released.Its height at time t is given by y = h(t) = -16t^2 + 45t + 7.A) How high does it go before returning to the ground? Round time to 2 decimal places to compute height. ___feet.b) How long does it take the grapefruit to hit the ground? Round time to 3 decimal places. ___ seconds
ANSWER:
A) 38.64 feet.
B) 2.960 seconds
STEP-BY-STEP EXPLANATION:
We have that the function that models the situation of the statement is the following:
[tex]h(t)=-16t^2+45t+7[/tex]A)
We can calculate this height, which would be the maximum height it can reach before it begins to decay, as follows:
[tex]\begin{gathered} t_v=-\frac{b}{2a} \\ \\ \text{ in this case b = 45, a = -16, we replacing:} \\ \\ t_v=-\frac{45}{-16\cdot2}=\frac{45}{32} \\ \\ t_v=1.406\text{ sec} \\ \\ \text{ Now, we replace in the function this time like this: } \\ \\ h(t)=-16\left(1.406\right)^2+45\left(1.406\right)+7 \\ \\ h(t)=38.64\text{ ft} \end{gathered}[/tex]Therefore, the height is equal to 38.64 feet.
B)
To determine the time when the ground, we must make the height 0 and solve for t, just like this:
[tex]\begin{gathered} 0=-16t^2+45t+7 \\ \\ -16t^2+45t+7=0 \\ \\ \text{ We use the general formula for quadratic equations} \\ \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ a=-16,b=45,c=7 \\ \\ \text{ We replacing} \\ \\ t=\frac{-45\pm\sqrt{45^2-4(-16)(7)}}{2(-16)} \\ \\ t=\frac{-45\pm\sqrt{2473}}{-32} \\ \\ t_1=\frac{-45+\sqrt{2473}}{-32}=-0.148 \\ \\ t_2=\frac{-45-\sqrt{2473}}{-32}=\:2.960 \end{gathered}[/tex]Therefore, the time is equal to 2.960 seconds
When the term quantized is use to refer to a charge it meansall charges are multiples of a protons chargethat the maximum value of a charge is 1.602x10-1⁹℃all charges are multiples of an electrons chargethat the minimum value of a charge is 1.602x10-1⁹℃O
all charges are multiples of an electrons charge.
Two objects collide and stick together. What will happen to the total kinetic energy of the objects (increase, decrease, stays the same)? Explain what happens to the kinetic energy.Effect on total kinetic energy: Description of Kinetic Energy changes:
Given:
Two objects collide and stick together.
To find:
Effect on the kinetic energy due to the collision.
Explanation:
When perfectly inelastic bodies moving along the same line collide, they stick together. An inelastic collision is a collision in which there is a loss in kinetic energy.
Let m1 and m2 be the masses and velocity v1 and v2 be the velocities of objects along the same line before the collision.
Let V be the velocity of objects after they collide and stick together.
By the law of conservation of linear momentum,
[tex]m_1v_1+m_2v_2=m_1V+m_2V[/tex]Rearranging the above equation, we get:
[tex]V=\frac{m_1v_1+m_2v_2}{m_1+m_2}\text{ ......\lparen1\rparen}[/tex]The total kinetic energy of objects before the collision is:
[tex]\text{Kinetic Energy \lparen before collision\rparen}=m_1v_1^2+m_2v_2^2\text{ ......\lparen2\rparen}[/tex]The kinetic energy after collision is:
[tex]\text{Kinetic energy \lparen after collision\rparen}=\frac{1}{2}(m_1+m_2)V^2\text{ ......\lparen3\rparen}[/tex]Using equations (1), (2), and (3), the loss in kinetic energy is given as:
[tex]\begin{gathered} \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2-\frac{1}{2}(m_1+m_2)V^2=\frac{1}{2}\lbrack m_1v_1^2+m_2v_2^2-\frac{m_1v_1+m_2v_2}{m_1+m_2}\rbrack \\ \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2-\frac{1}{2}(m_1+m_2)V^2=\frac{1}{2}\lbrack\frac{m_1m_2\left(v_1^2+v_2^2-2v_1v_2\right)}{m_1+m_2}] \\ \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2-\frac{1}{2}(m_1+m_2)V^2=\frac{m_1m_2\left(v_1-v_2\right)^2}{2(m_1+m_2)} \end{gathered}[/tex]From the above result, we see that the energy loss in the kinetic energy is positive.
Thus, the kinetic energy of decreases, when two objects collide and stick together.
Final answer:
The kinetic energy of decreases, when two objects collide and stick together.
caracteristicas del sonido
In general, intensity, tone, timbre, and duration are the four subjective qualities used to describe musical sounds. Each of these characteristics depends on one or more physically measurable parameters.
What is sound and what are its characteristics?In physics, sound is any phenomenon that involves the transmission of mechanical waves, whether audible or not, through a fluid or solid medium while causing a body to vibrate.
The sensation of sound is one that is brought on by the vibrations of objects and occurs in the ear. An object vibrating generates airborne waves that are transmitted. People communicate with one another through sound, doing so by expressing their thoughts, emotions, and desires.
The sounds are divided into three categories based on their frequency: agudos (high frequency), medios (medium frequency), and graves (baja frecuencia).
There is no change in phase in the reflection when the airborne sound waves (also known as pressure waves) come into contact with a hard surface. To put it another way, when the high pressure portion of a sound wave strikes a wall, it is reflected as a high pressure rather than a low pressure, which would be a phase change.
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2. A plastic container weighs 56.7 g. When cooked pasta is placed in the container, the total weight is 170.1 g. What is the weight of the pasta?
The weight of the pasta is 113.4g.
How to find the weight of the pasta?The weight of an object is defined as the force of gravity acting on it, and it can be calculated by multiplying the mass by the acceleration of gravity, w = mg.
The formula for calculating an object's mass based on its weight is Mass is equal to Weight divided by Gravity's Acceleration Convert the weight in pounds to the equivalent weight in Newtons. Mass is measured in Newtons in the formula for calculating mass based on weight.
Given,
The plastic container weighs 56.7 g
Total weight is 170.1 g
Solution:
The weight of the pasta = Total weight - Plastic container's weight
= 170.1 - 56.7
= 113.4g
Therefore,
The weight of the pasta is 113.4g.
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HELP! Sketch the given vector and its horizontal and vertical components. Use trig ratios to find the magnitude of each component.
16 km 34 degrees N of W
Horizontal component = ?
Vertical component = ?
The x-component and the y-component of the given vector are -
[x] = - 13.36 a[x]
[y] = 8.947 a[y]
What is a vector quantity?A vector quantity is any physical quantity with both magnitude and direction. Examples - velocity, acceleration, displacement etc. Any vector quantity can be resolved into its components along the [x] and [y] axis as follows -
[x] component = AcosФ a[x]
[y] component = AsinФ a[y]
The resultant angle will be -
tanФ = |[y]|/|[x]|
Given is 16 km 34 degrees North of West
Now, in the image attached, it can be seen that a vector making an angle of 34 degrees is directed along the north of west. Now, the x-component will be -
[x] = 16 cos (34) -a[x] = - 13.36 a[x]
And, the y-component will be -
[y] = 16 sin (34) a[y] = 8.947 a[y]
Therefore, the x-component and the y-component of the given vector are -
[x] = - 13.36 a[x]
[y] = 8.947 a[y]
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To solve more questions on vectors, visit the link below-
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A 96.3 kg cannon at rest contains a 9.5 kg cannon ball. When fired, the cannon ball leaves the cannon with a speed of 84.58 m/s. What is the recoil speed of the cannon?
Using conservation of momentum:
[tex]p1=p2[/tex]Since the cannon is at rest before shots the ball:
[tex]\begin{gathered} p1=0 \\ _{\text{ }}and \\ p2=mv+MV \end{gathered}[/tex]Where:
[tex]\begin{gathered} m=9.5kg \\ v=84.58\frac{m}{s} \\ M=96.3kg \end{gathered}[/tex]So:
[tex]0=9.5\cdot84.58+96.3V[/tex]Solve for V:
[tex]\begin{gathered} 96.3V=-9.5\cdot84.58 \\ 96.3V=-803.51 \\ V=-\frac{803.51}{96.3} \\ V\approx-8.34\frac{m}{s} \end{gathered}[/tex]Answer:
The recoil speed of the cannon is approximately 8.34 m/s
Which sentence describes equilibrium in supply and demand? A golf dub manufacturing company decided to use a price-skimming strategy to enter the market. It set the price of its clubs at $450. However, a range of other premium clubs was already available in the market, and thus the product did not achieve a good demand. Eventually, when the price of the clubs fell to $300. demand for the product increased and profits began to soar. Now, market research suggests that if the company reduced its price to $275, the quantity demanded would also rise. The rise in demand was predicted to be 30 percent However, a profit study revealed that to increase demand, it would have to compromise on profits. Thus, the company decided to leave the price at $275. This would help them gain enough customers to make a profit while also manufacturing enough golf clubs to meet demand.
Given
A teenage boy was 4 feet 9 inches tall.
Since then, he has grown 9 inches.
To find:
The height of the boy.
Explanation:
It is given that,
A teenage boy was 4 feet 9 inches tall.
Since then, he has grown 9 inches.
That implies,
[tex]\begin{gathered} The\text{ }height\text{ }of\text{ }the\text{ }boy=4ft\text{ }9in+9in \\ =4ft\text{ }(9+9)in \\ =4ft\text{ }18in \\ =4ft\text{ }+1.5ft \\ =5ft\text{ }6in \end{gathered}[/tex]Hence, the height of the boy is 5feet 6inches.
Three-Step Conversion:Significant Figures are requiredHow many cm is 4.0 mi? (Remember your answer need to have the correct amount of significant figures and the correct units written).(1 mi = 5280 ft, 1 in=2.54 cm) (hint: cm -> in -> ft -> mi)
First, we need to transform 4mi to ft
we know
1mi=5280ft
[tex]4mi=4(5280)=21120ft[/tex]then we got 21120 ft to inches
1ft=12 in
[tex]21120ft=21120(12)=253440in[/tex]then we got 253440in to cm
1 in=2.54 cm
[tex]253440in=253440(2.54)=643737.6\text{ cm}[/tex]therefore the solution is
4.0mi= 643737.6 cm
Some electric power companies use water to store energy.Water is pumped from a low reservoir to a high reservoir.To store the energy produced in 1.0 hour by a 180-MWelectric power plant, how many cubic meters of water willhave to be pumped from the lower to the upper reservoir?Assume the upper reservoir is an average of 380 m abovethe lower one. Water has a mass of for every 1.00 x 10^3 kg for every 1.0m^3.
Answer:
1.74 x 10⁵ m³
Explanation:
The power is equal to:
[tex]P=\frac{E}{t}\rightarrow E=Pt[/tex]Where P is the power, E is the energy and t is the time. So, we can calculate the energy replacing
P = 180 MW = 180 x 10⁶ W
t = 1 hour = 3600 s
So, the energy is
[tex]\begin{gathered} E=(180\times10^6\text{ W\rparen\lparen3600 s\rparen} \\ E=6.48\times10^{11}\text{ J} \end{gathered}[/tex]On the other hand, the energy stored will be potential, so
[tex]E=mgh[/tex]Where m is the mass, g is the gravity and h is the height. Solving for m and replacing E = 6.48 x 10^11 J, g = 9.8 m/s², and h = 380 m, we get:
[tex]\begin{gathered} m=\frac{E}{gh} \\ \\ m=\frac{6.48\times10^{11}\text{ J}}{9.8\text{ m/s}^2\text{ \lparen380 m\rparen}} \\ \\ m=1.74\times10^8\text{ kg} \end{gathered}[/tex]Finally, we can calculate the volume of the water in cubic meters using the given ratio that Water has a mass of 1.00 x 10^3 kg for every 1.0m^3, so
[tex]1.74\times10^8\text{ kg }\times\frac{1\text{ m}^3}{1\times10^3\text{ kg}}=1.74\times10^5\text{ m}^3[/tex]Therefore, the answer is 1.74 x 10⁵ m³
how the water falls connected to gravity?
Answer:
At the top of the waterfall, the water is higher in the gravitational field of the Earth and has gravitational potential energy. When it falls, the potential energy turns into kinetic energy. So energy is not created, it was there at the beginning, stored as potential energy.
I hope this helps :)
What is the ratio of the force exerted on the floor by the ground to the persons body weight
We will have the following:
We are given:
m = 67kg
F = 2300N
Now, we have that the average are of a human foot is approximately 76 cm^2.
So, we will deterine the ratio of force excerted on the floor in terms of pressure, that is:
[tex]P=\frac{2300N}{0.0076m^2}\Rightarrow P=302631.5789Pa[/tex][tex]\Rightarrow P\approx302.632kPa[/tex]So, the ratio of the force exerted on the floor by the ground is approximately 302.632 Kilopascals.
