Answer:
The length of rod A will be greater than the length of rod B
Explanation:
We, know that the formula for final length in linear thermal expansion of a rod is:
L' = L(1 + ∝ΔT)
where,
L' = Final Length
L = Initial Length
∝ = Co-efficient of linear expansion
ΔT = Change in temperature
Since, the rods here have same original length and the temperature difference is same as well. Therefore, the final length will only depend upon the coefficient of linear expansion.
For Rod A:
∝₁ = 12 x 10⁻⁶ °C⁻¹
For Rod B:
∝₂ = β₂/3
where,
β₂ = Coefficient of volumetric expansion for rod B = 24 x 10⁻⁶ °C⁻¹
Therefore,
∝₂ = 24 x 10⁻⁶ °C⁻¹/3
∝₂ = 8 x 10⁻⁶ °C⁻¹
Since,
∝₁ > ∝₂
Therefore,
L₁ > L₂
So, the length of rod A will be greater than the length of rod B
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
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
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)
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
On 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
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].
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]
- 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
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.
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.
how we will solve this question?
36.45cL=______=μL
Answer:
0.364
I believe... Good luck!
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:
would someone just answer this now?
brainliest please help
show the resultant force
8n downwards
since downwards force is greater than upwards just minus
what do humming birds eat? and how do they get their food?
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
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.
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.
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
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
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.
Heading into a stop light, a car is traveling with a velocity of 44mph when the driver starts to app breaks. The car takes 15 seconds (or .0041hrs) to come to a complete stop. What is the car's acceleration/deceleration?
Answer:
the car slowed at a rate of 2.93 mph
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.
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]
Four examples of second class levers
Answer:
they are :
Explanation:
wheelbarrow bottle opener an oarThe first step in a star formation is ______?
Si unit of thermal conductivity is
Answer: Watts
Explanation: In the International System of Units (SI), thermal conductivity is measured in watts per meter-kelvin (W/(m•k)).
How to properly hide a body?
NEED ANSWER NOW!!!
WORTH LIFE SENTENCE!!
WILL PAY FOR MOST USEFUL ANSWER!!
Answer:
The septic tank of an abandoned property.
Almost anywhere at a closed refinery (most refineries will never be torn down due to the massive remediation costs involved when they are. Find a good hiding spot there and place the body in the open or bury it)
Numerous quarries are around and most are filled with water. If you ensure that the body will not float to the surface, they are often great disposal locations.
Inside the wall of a home or building - If the body is allowed to desiccate (mummify) then there will be little or no odor. Simply open a wall,insert the body and replaster or drywall it and paint over it.
Underneath an area to be used as a pen or feed lot for domesticated animals. The weight of the animals, plus their manure and the straw placed down to control it will form its own "cap" over the burial site.
In plain sight - The roof of tall abandoned building (hidden from overhead viewing) , an open area in a remote part of the country (after staying there for 2-3 days to see if anyone comes along), or in a storm drain along a lightly used state or county road.
Importance of a cell
Explanation:
Cells are the basic building blocks of all living things. The human body is composed of trillions of cells. They provide structure for the body, take in nutrients from food, convert those nutrients into energy, and carry out specialized functions.
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].