Answer:
(a) The aspect of the upper string supporting the weight and the applied force
Student 1 is correct because the upper string is the source of support of the large weight and the force applied to the short string reacts at the support of the long string
(b) The aspect of Student's (2) reasoning that is correct is that the shorter piece of string will always break first, however, the statement is only true for sudden pull due to the increased force experienced by the shorter string from a more rapid change in momentum
(c) The aspect of Student 1's statement that is incorrect is the that the upper string will always break first
The aspect of Student 2's statement that is incorrect is the that the shorter piece of string will always break first
(d) A string will break when subject to a force equivalent to its breaking force. The force experienced by the string increases as the rate of pull (suddenness) increases and the suddenness increases inversely with the length of the string, as such the shorter lower string will break first from a sudden pull before the force of the pull is completely transmitted to the upper string. Whereby the lower string is slowly pulled, the force is evenly transmitted to the upper string which is then taking up the load of the weight and the applied force together and is likely to break first
Explanation:
A submarine is built to dive to a depth of 90 m where the pressure in the water is about 1 MPa. The engineers want to build a window in the submarine, but the window can only take a force of 10 kN safely. What is the largest surface area window that can be used?
Answer:
The largest surface area window that can be used is [tex]A=0.01\ m^2[/tex]
Explanation:
Pressure
The pressure is defined as force per unit area. The SI unit for pressure is the Pascal (Pa), defined as Newton per square meter.
If a force F acts on a surface area A, the pressure is calculated as:
[tex]\displaystyle P=\frac{F}{A}[/tex]
The pressure at a depth of 90 m is P=1 Mpa= 1,000,000 Pa and the submarine's window can only take a force of F=10 kN=10,000 N.
We need to calculate the largest surface area of the window. We can solve the equation for A:
[tex]\displaystyle A=\frac{F}{P}[/tex]
Substituting:
[tex]\displaystyle A=\frac{10,000}{1,000,000}[/tex]
[tex]A=0.01\ m^2[/tex]
The largest surface area window that can be used is [tex]\mathbf{A=0.01\ m^2}[/tex]
Parker (73.2 kg) is being dragged down the hall with an applied force of 123 N. If the frictional force is 27.4 N, what is the coefficient of friction in the hall?
Answer:
The coefficient of friction in the hall is 0.038
Explanation:
Given;
mass of the Parker, m = 73.2 kg
applied force on the parker, F = 123 N
frictional force, Fs = 27.4 N
the coefficient of friction in the hall = ?
frictional force is given by;
Fs = μN
Where;
μ is the coefficient of friction
N is normal reaction = mg
Fs = μmg
μ = Fs / mg
μ = (27.4) / (73.2 x 9.8)
μ = 0.038
Therefore, the coefficient of friction in the hall is 0.038
different between si unit and derived unit
Answer:
There are seven base SI units corresponding to different parameters and are considered independent of each other. Derived units are derived from these 7 base units. Derived units are dependent on the base units and are not independent of each other. For example, let us derive the SI units for force. By Newtons Second Law of Motion, the force is same as the product of mass and acceleration of a body. that is, force = mass x acceleration. Also, acceleration is defined as the rate of change of velocity. And velocity is defined as the rate of change of displacement. Thus, force = mass x acceleration = mass x dv/dt = mass x dt2 Mass has SI units of kg, distance is measured in m and t has the SI unit of second. Thus, Sl unit of force is kg.m/s^2 (also known as Newton).
A) 750gm
11. Ifa rectangular solid box of aluminum is heated, which of the following is not true?
A) Its mass remains constant
Cits density increase
B) Its volume increase
Do none of the above
Given :
A rectangular solid box of aluminum is heated.
To Find :
Which of the following is not true.
A) Its mass remains constant
C) its density increase
B) Its volume increase
D) none of the above
Solution :
We know, when an object is heated it expands.
Objects expands means that the volume increases but mass remains same by conservation of mass.
Since mass remains constant and volume increases.
So, density will decrease.
Hence, this is the required solution.
A 58-kg boy swings a baseball bat, which causes a 0.140-kg baseball to move toward 3rd base with a velocity of 38.0 m/s.
Calculate the kinetic energy of the baseball (rounding your answer to the integer).
Answer:
101 J
Explanation:
Answer:
[tex]\boxed {\boxed {\sf 101 \ Joules}}[/tex]
Explanation:
Kinetic energy can be found using the following formula:
[tex]KE=\frac{1}{2}mv^2[/tex]
The mass of the baseball is 0.140 kilograms and the velocity is 38.0 meters per second.
[tex]m= 0.140 \ kg \\v= 38.0 \ m/s[/tex]
Substitute the values into the formula.
[tex]KE=\frac{1}{2} [0.140 \ kg][(38.0 \ m/s)^2][/tex]
First, evaluate the exponent.
(38.0 m/s)²= 38.0 m/s * 38.0 m/s = 1444 m²/s²[tex]KE=\frac{1}{2}(0.140 \ kg)(1444 \ m^2/s^2)[/tex]
Multiply the two numbers in parentheses together.
[tex]KE=\frac{1}{2}(202.16 \ kg*m^2/s^2)[/tex]
Multiply the fraction by the number, or divide the number by 2.
[tex]KE=101.08 \ kg*m^2/s^2[/tex]
Round to the nearest whole number. The 0 in the tenth place tells us we can leave the number as is.
[tex]KE\approx 101 \ kg*m^2/s^2[/tex]
1 kg*m²/s² is equal to 1 Joule. Therefore, our answer of 101 kg*m²/s² is equal to 101 Joules (J).
[tex]KE\approx 101 \ J[/tex]
The kinetic energy of the baseball is about 101 Joules.
Which statement describes an action-reaction pair?
A. You push against a wall, and the wall applies a force to your
hands.
B. The Sun pulls on the Earth, and the Earth pulls on the Moon.
C. You push a book off the edge of a shelf, and the book falls to the
floor
G
D. You throw a ball straight up, and Earth's gravity pulls it downward.
You push against a wall, and the wall applies a force to your hands. (A)
1 Una pelota rueda hacia la derecha siguiendo una trayectoria en línea recta de modo que recorre una distancia de 10 m en 5 s. Calcular la velocidad y la rapidez.
Answer:
Entonces la velocidad es 2 [tex]\frac{m}{s}[/tex] hacia la derecha, mientras que la rapidez es 2 [tex]\frac{m}{s}[/tex].
Explanation:
La rapidez es una magnitud escalar que relaciona la distancia recorrida con el tiempo mientras que la velocidad es una magnitud vectorial que relaciona el cambio de posición (o desplazamiento) con el tiempo.
Es decir, la rapidez es una magnitud física que expresa el valor numérico y la unidad de una distancia en relación con el tiempo.
Matemáticamente, es expresada como el cociente entre el camino recorrido y el tiempo transcurrido:
[tex]rapidez=\frac{distancia}{tiempo}[/tex]
Por otro lado, la velocidad expresa cómo se está moviendo un objeto en cada momento, informando la dirección, el sentido y la rapidez del movimiento (la velocidad se refiere al intervalo de tiempo que le toma a un objeto desplazarse hacia una dirección determinada y, al involucrar la dirección o sentido del movimiento, es una magnitud vectorial).
Matemáticamente, la velocidad es calculada como el cociente entre el desplazamiento que realizó un cuerpo y el tiempo total que le llevó realizarlo.
[tex]velocidad=\frac{desplazamiento}{tiempo}[/tex]
En este caso, sabes que:
distancia= desplazamiento= 10 mtiempo= 5 sReemplazando obtenes que la aceleración es:
[tex]rapidez=\frac{10m}{5 s}[/tex]
rapidez= 2 [tex]\frac{m}{s}[/tex]
y la velocidad es:
[tex]velocidad=\frac{10m}{5s}[/tex]
velocidad= 2 [tex]\frac{m}{s}[/tex] hacia la derecha.
Entonces la velocidad es 2 [tex]\frac{m}{s}[/tex] hacia la derecha, mientras que la rapidez es 2 [tex]\frac{m}{s}[/tex].
Calculate the power of a crane if it lifts a load of 2400N through a height of 10m in 15 seconds.Also express the power in horse power.
Answer:
2.08 hpExplanation:
The power of the crane can be found by using the formula
[tex]p = \frac{f \times d}{t} \\ [/tex]
f is the Force
d is the distance
t is the time taken
From the question we have
[tex]p = \frac{2400 \times 10}{15} = \frac{24000}{15} \\ [/tex]
We have
power = 1600 W
But 1 W = 0.0013 hp
So 1600 = 2.08 hp
We have the final answer as
2.08 hpHope this helps you
A 1500 kg car starts from rest,
and rolls down a frictionless 5.09°
slope that is 17.5 m long. How
much time does it take to reach
the bottom?
(Unit = s)
The time taken for the heavy car to reach the bottom of the frictionless slope is 0.56 s.
Height of the slope
The height of the incline is calculated as follows;
sin θ = h/L
h = L sinθ
where;
L is length of slopeh is height of the slopeh = 17.5 x sin(5.09)
h = 1.55 m
Time of motion of the cart = √(2h/g)
where;
t is time of motiong is acceleration due to gravityt = √(2 x 1.55 /9.8)
t = 0.56 s
Learn more about time of motion here: https://brainly.com/question/2364404
#SPJ1
Answer:
T= 6.34
Explanation:
This is the answer. This answer also works on acellus
A horse gallops for 3.5 minutes around the raceing track. During this time it changes velocity from 543 min to 628 min. Calculate the acceleration of the horse during this 3.5min interval
Answer:
The acceleration of the horse during this time interval is 24.286 m/min²
Explanation:
Given;
initial velocity of the horse during the gallop, u = 543 m /min
final velocity of the horse during the gallop, v = 628 m /min
time of motion, t = 3.5 minutes
The acceleration of the horse is given by the change in velocity per change in time;
[tex]a = \frac{v-u}{t}\\\\a = \frac{628-543}{3.5}\\\\a = 24.286 \ m/min^2[/tex]
Therefore, the acceleration of the horse during this time interval is 24.286 m/min²
A satellite is orbiting the Earth in a circular orbit of radius r. Its frequency is independent of its height above the surface of the Earth. a. TRUE b. FALSE
Answer:
a. TRUE
Explanation:
When a satellite is launched to orbit around earth, it has to produce its own artificial gravity by performing rotations. The frequency of this rotation is given by the following formula:
f = √[ac/4πR²]
where,
f = frequency
ac = centripetal acceleration
R = Radius of the satellite
Therefore, it is clear from this formula that the frequency of rotation of the satellite is independent of its height above the surface of earth. So, the correct option is:
a. TRUE
- A ball is thrown at a wall 50 meters away. It takes the ball 5 seconds to get to
the wall it is thrown at What is the average speed of the ball? SPEED
Answer:
The answer is 10 m/sExplanation:
The average speed of the ball can be found by using the formula
[tex]v = \frac{d}{t} \\ [/tex]
d is the distance
t is the time taken
From the question we have
[tex]v = \frac{50}{5} \\ [/tex]
We have the final answer as
10 m/sHope this helps you
5,000 joules of thermal energy were applied to 1-kg aluminum bar. What was the temperature increase?
Answer:
ΔT = 4.9°C
Explanation:
The thermal energy of the bar can be given as follows:
Thermal Energy = mCΔT
where,
m = mass of bar = 1 kg
C = specific heat capacity of aluminum = 1020 J/kg.°C
ΔT = Change in Temperature = ?
Therefore,
5000 J = (1 kg)(1020 J/kg.°C)ΔT
ΔT = (5000 J)/(1020 J/°C)
ΔT = 4.9°C
what is electricity
Answer:
Electricity is the flow of electrical power or charge. It is a secondary energy source which means that we get it from the conversion of other sources of energy, like coal, natural gas, oil, nuclear power and other natural sources, which are called primary sources.
The SI unit of average speed m.s. True or False. But is there difference between m/s and m.s.
Answer:
false
Explanation:
there is difference in m/s and m.s
A block of mass m=2.20m=2.20 kg slides down a 30.0^{\circ}30.0
∘
incline which is 3.60 m high. At the bottom, it strikes a block of mass M=7.00M=7.00 kg which is at rest on a horizontal surface, (Assuming a smooth transition at the bottom of the incline.) If the collision is elastic, and friction can be ignored, determine the speeds of the two blocks (v_mv
m
and v_Mv
M
) after the collision.
Answer:
[tex]v_m \approx -4.38\; \rm m \cdot s^{-1}[/tex] (moving toward the incline.)
[tex]v_M \approx 4.02\; \rm m \cdot s^{-1}[/tex] (moving away from the incline.)
(Assumption: [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex].)
Explanation:
If [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex], the potential energy of the block of [tex]m = 2.20\; \rm kg[/tex] would be [tex]m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J[/tex] when it was at the top of the incline.
If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this [tex]m = 2.20\; \rm kg\![/tex] block right before the collision would also be approximately [tex]77.695\; \rm J[/tex].
Calculate the velocity of that [tex]m = 2.20\; \rm kg[/tex] based on its kinetic energy:
[tex]\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}[/tex].
A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.
Initial momentum of the two blocks:
[tex]p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}[/tex].
[tex]p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}[/tex].
Sum of the momentum of each block right before the collision: approximately [tex]18.489\; \rm kg \cdot m \cdot s^{-1}[/tex].
Sum of the momentum of each block right after the collision: [tex](m\cdot v_m + m \cdot v_M)[/tex].
For momentum to conserve in this collision, [tex]v_m[/tex] and [tex]v_M[/tex] should ensure that [tex]m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}[/tex].
Kinetic energy of the two blocks right before the collision: approximately [tex]77.695\; \rm J[/tex] and [tex]0\; \rm J[/tex]. Sum of these two values: approximately [tex]77.695\; \rm J\![/tex].
Sum of the energy of each block right after the collision:
[tex]\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right)[/tex].
Similarly, for kinetic energy to conserve in this collision, [tex]v_m[/tex] and [tex]v_M[/tex] should ensure that [tex]\displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J[/tex].
Combine to obtain two equations about [tex]v_m[/tex] and [tex]v_M[/tex] (given that [tex]m = 2.20\; \rm kg[/tex] whereas [tex]M = 7.00\; \rm kg[/tex].)
[tex]\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right.[/tex].
Solve for [tex]v_m[/tex] and [tex]v_M[/tex] (ignore the root where [tex]v_M = 0[/tex].)
[tex]\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right.[/tex].
The collision flipped the sign of the velocity of the [tex]m = 2.20\; \rm kg[/tex] block. In other words, this block is moving backwards towards the incline after the collision.
DO boys cheat more than girls?
Answer:
I'm not sure.
Explanation:
Look up the statistics
Answer: honestly girls cheat more than guys, its statically proven that females cheat more than guys its because they dont believe in value in a relationship
Explanation:
If a chicken has a weight of 300 N on Earth, what would be its weight on Mercury if the gravity is 0.38.
Answer:
114N
Explanation:
gravity on earth: 10m/s (exact: 9.82)
mass if chicken = 300/10 = 30Kg
weight of chicken on mercury = 3.8 x 30 = 114N (you wrote 0.38 but Mercury's gravity is about 3.7)
Answer:
113.15 g
Explanation:
That's because the planets weigh different amounts, and therefore the force of gravity is different from planet to planet. For example, if you weigh 100 pounds on Earth, you would weigh only 38 pounds on Mercury. That's because Mercury weighs less than Earth, and therefore its gravity would pull less on your body.
On Planet X, an astronaut has maximum walking speed of 1.9 m/s. The astronaut's legs are each 1.3 m long. What is the constant of acceleration due to gravity on Planet X?
A. 8.3 m/s2
B. 2.8 m/s2
C. 3.8 m/s2
D. 1.1 m/s2
Answer:
Acceleration due to Gravity:
Explanation:
Here g is acceleration due to gravity. So C. 3.8 m/s2
when climbing a steep hill, would you want a larger or smaller bicycle gear?
Answer:
Usually when climbing, it's best to be in the small front ring and the largest back ring. If your cadence is about 100 rpm, then whatever gear you're in is fine. It depends on the road, but as long as your pedaling is at a level you're comfortable with, you're fine.
Explanation:
Google answer by the way.
Which two fundamental forces are only attractive?
O A. Gravitational and strong nuclear
O B. Electromagnetic and weak nuclear
O C. Electromagnetic and strong nuclear
O D. Gravitational and weak nuclear
The two fundamental forces that are only attractive is A. Gravitational and strong nuclear.
What connection exists between the strong nuclear force and gravity?The "strong" nuclear force outperforms gravity by a factor of 10 to the 38th. While gravitational and electromagnetic forces have relatively large ranges of action, the strong and weak nuclear forces operate at close ranges. All objects are affected by the strong and weak nuclear forces, whereas smaller objects are affected by gravitational and electromagnetic forces.
Nuclear forces are the most enticing of the four fundamental forces. There was no explanation provided as to how the nucleus in the atom is kept together by electromagnetism, which maintains matter together.
Therefore, option A is correct.
Learn more about forces at:
https://brainly.com/question/12970081
#SPJ1
Zero is the freezing temperature of water on which temperature scale?
Kelvin
Fahrenheit
Pascal
Celsius
Answer:
Celsius. i think! I might be wrong
Four examples of second class levers
Answer:
they are :
Explanation:
wheelbarrow bottle opener an oarOn top of a 3 m tall shelf sits a lonely 4.5 kg toy snake. How much GPE does this snake have if the shelf is on Earth? (g = 9.8 m/s2
Answer:
hi
Explanation:
how r u
how we will solve this question?
36.45cL=______=μL
Answer:
0.364
I believe... Good luck!
Bathroom scales read the normal force that is exerted against the floor. What would a scale read when a 100 kg man is in an elevator accelerating upward at 1.2 m/s2? What would it read when the man is accelerated downward at 1.8 m/s2?
Answer:
a
When the lift is moving upward [tex]F = 1120 \ N[/tex]
b
When the lift is moving downward [tex]F = 820 \ N[/tex]
Explanation:
From the question we are told that
The mass of the man is [tex]m = 100 \ kg[/tex]
The upward acceleration is [tex]a_u = 1.2 \ m/s^ 2[/tex]
The downward acceleration is [tex]a_d = 1.80 \ m/s^2[/tex]
Generally the force which the scale will read when the man is moving downward is according to Newton second law represented as
[tex]F + mg = ma[/tex]
[tex]F = m (g - a_d)[/tex]
Here [tex]g = 10 m/s^2[/tex]
=> [tex]F = 100 (10 - 1.8)[/tex]
=> [tex]F = 820 \ N[/tex]
Generally the force which the scale will read when the man is moving upward is according to Newton second law represented as
[tex]F - mg = ma[/tex]
[tex]F = m (g + a_u)[/tex]
Here [tex]g = 10 m/s^2[/tex]
=> [tex]F = 100 (10 + 1.2)[/tex]
=> [tex]F = 1120 \ N[/tex]
GIVING THE REST OF MY POINTS!
Energy can be:
A. destroyed or converted to another form.
B. destroyed.
C. created, but never destroyed.
D. converted to another form.
Why does a liquid thermometer work
Light is reflected from a crystal of table salt with an index of refraction of 1.544. An analyser is placed to intercept the reflected ray, and is able to completely absorb the reflected light. What is the angle of incidence?
Answer:
hola me llamo bruno y tu?
Explanation:
pero yo soy de mexico
Two students are measuring the Gravitational Attraction between each other. One student weighs 120 kg. The other student weighs 108 kg. Calculate the gravitational pull they have on each other.
Answer:
[tex]F=8.64\times 10^{-9}\ N[/tex]
Explanation:
Given that,
Mass of student 1, m₁ = 120 kg
Mass of student 2, m₂ = 108 kg
Let they are at a distance of 10 m.
We need to find the gravitational pull they have on each other. The gravitational force acting between two objects is given by :
[tex]F=G\dfrac{m_1m_2}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{120\times 108}{(10)^2}\\\\=8.64\times 10^{-9}\ N[/tex]
So, the gravitational pull between students is [tex]8.64\times 10^{-9}\ N[/tex].