Answer:
v = 26.54 m/s
Explanation:
Given that,
The mass of a hockey puck, m = 0.21 kg
The kinetic energy of the hockey puck, E= 74 J
We need to find the speed of the hockey puck. Let the speed is v. We know that, the kinetic energy is given by :
[tex]E=\dfrac{1}{2}mv^2[/tex]
Where
v is the speed
So,
[tex]v=\sqrt{\dfrac{2E}{m}} \\\\v=\sqrt{\dfrac{2\times 74}{0.21}} \\\\v=26.54\ m/s[/tex]
So, the speed of the hockey puck is 26.54 m/s.
The mass of a brick is 2kg. Find the mass of water displaced by it when it is completely immersed in water. (Density of the bricks is 2.5 g/cm^3)
Answer:
2000g
Explanation:
volume=mass/density
=2000/2.5
=800cm³
mass=density×volume
=800×2.5
=2000g
A basketball is shot by a player at a height of 2.0m. The initial angle was 53° above the horizontal. At the highest point, the ball was travelling 6 m/s. If he scored (the ball went through the rim that is 3.00m above the ground), what was the player's horizontal distance from the basket?
At the ball's highest point, it has no vertical velocity, so the 6 m/s is purely horizontal. A projectile's horizontal velocity does not change, which means the ball was initially thrown with speed v such that
v cos(53°) = 6 m/s ==> v = (6 m/s) sec(53°) ≈ 9.97 m/s
The player shoots the ball from a height of 2.0 m, so that the ball's horizontal and vertical positions, respectively x and y, at time t are
x = (9.97 m/s) cos(53°) t = (6 m/s) t
y = 2.0 m + (9.97 m/s) sin(53°) t - 1/2 gt ²
Find the times t for which the ball reaches a height of 3.00 m:
3.00 m = 2.0 m + (9.97 m/s) sin(53°) t - 1/2 gt ²
==> t ≈ 0.137 s or t ≈ 1.49 s
The second time is the one we care about, because it's the one for which the ball would be falling into the basket.
Now find the distance x traveled by the ball after this time:
x = (6 m/s) (1.49 s) ≈ 8.93 m
The direction equivalent to {40° W of S} is:
A. 40 ° E of S
B. 40° W of N
C. 40° E of N
D. 50° S of W
E. 50° E of N
Answer:
c
Explanation:
please answer quick for brainlist ; )
Answer:
The diagram assigned B
explanation:
Check the direction of the two vectors, their resultant must be in the same direction.
Example to measure the interval of time of a small stone dropped from 1m height.
Answer:
The time required is 0.45 s.
Explanation:
Height, h = 1 m
initial velocity, u = m/s
Let the time is t.
Use second equation of motion
[tex]h = u t + 0.5 at^2\\\\1 = 0 +0.5 \times 9.8 \times t^2\\\\t = 0.45 s[/tex]
2 What are(i) free fall , (ii) acceleration due to gravity, (iii) escape velocity , (iv) centripetal force?
Answer:
Explanation:
1. Free fall implies an object falling under the gravitational influence only. During the flight, no other force acts on it except the gravitational pull.
2. Acceleration due to gravity is the earth's natural force of pull on all objects on its surface, close to its surface, or in the region where the force can be felt. This force pulls object to the surface of the earth.
3. Escape velocity is the required minimum velocity for an object to leave the gravitational influence of the earth. It has a constant value which can be determined by;
Escape velocity = [tex]\sqrt{2gR}[/tex]
where g is the gravitation force of the earth and R is the radius of the earth.
4. Centripetal force is the force of pull that is required to keep a rotation object in its curved path.
Question 4(Multiple Choice Worth 3 points)
(03.02 MC)
Which statement best reflects a change in weather?
Today is cloudy, but tomorrow will be clear and sunny.
The average rainfall has decreased over the past five years.
Ocean temperatures are projected to increase over time.
Glaciers are melting more rapidly now than in the past 100 years.
plzzz helpppp asap
Answer:
a is your answer.............
Which statement should he place in the region marked X?
Explanation:
The answer is Electrons move between objects.
Compared with dim light, what do light waves that look bright tend to have the subject just says science but the picker doesn't have that
Answer:
The brightness of a light depends on the amplitude of the light wave, which is the extent the waves moves from their equilibrium position. The brightness is also related to the amount of light that is emitted or reflected by an object
Therefore, compared to dim light, light that look bright have a higher amplitude and emit or reflect more light energy (photons)
Explanation:
When an object is in a gravitational field, it has energy in its __________ __________ energy store.
Answer:
Gravitational potential
Explanation:
Any object that is not on the surface of the Earth, but at a height instead has potential energy. Eventually, this can become kinetic energy once the object falls.
When an object is in a gravitational field, it has energy in its gravitational potential energy store.
An egg is dropped onto a wood floor and breaks. When a different egg (same mass) is dropped the same distance onto a foam pad on the wood floor, it does not break. Why (in terms of physics) does this happen?
A. The pad extends the time so the impulse changes.
B. An egg requires a harder surface so it won’t break.
C. The pad pulls the egg to the floor.
D. Wood is harder than the pad.
Answer:
A.
Explanation:
The pad extends the time so the impulse changes.
Answer:
A. The pad extends the time so the impulse changes
Explanation:
in term of physics it has same mass:)
An unbalanced 6.0-newton force acts eastward on an object for 3.0 seconds. The impulse
produced by the force is
18 N:s east
2 Nós east
18 N:s west
2 Nis west
PLEASE HEEEEEEELP
Assume that the velocity of the soda bottle falling from a height of 0.8 m will be 4 m/s. Record this velocity for each mass in Table A, and use it in calculating the predicted kinetic energy of the soda bottle for the masses of 0.125 kg, 0.250 kg, 0.375 kg, and 0.500 kg using the equation: KE=1/2 mv^2 When solving for kinetic energy (KE), m is mass, and v is the speed (or velocity).
KE = (0.5) m v²
given that : v = speed of the bottle in each case = 4 m/s when m = 0.125 kg
KE = (0.5) m v² = (0.5) (0.125) (4)² = 1 J
when m = 0.250 kg KE = (0.5) m v² = (0.5) (0.250) (4)² = 2 J
when m = 0.375 kg KE = (0.5) m v² = (0.5) (0.375) (4)² = 3 J
when m = 0.0.500 kg KE = (0.5) m v² = (0.5) (0.500) (4)² = 4 J
The moon is 250,000 miles away. How many feet is it from Earth? (dimensional analysis)
Answer:
1.261 billion
Explanation:The distance between the earth and the moon in miles is 250,000,so in feet,The distance is 1.261 billion
The gravitational force acting on various masses is measured on different planets. Measured values for the forces acting on the corresponding masses are shown in the data table. Analyze the data and develop a method for comparing the gravitational field strengths on the different planets. Use your method to compare the gravitational field strengths, and report your conclusions.
The summer camps had a field trip from the campus to Fragrance Hill. They traveled at an average speed of 65 km/h in the first 2 hours. After that, traveled at another average speed of 78 km/h. If the distance between the campus and Fragrance Hill is 364 km, what was the total time for the field trip?
Answer:
Explanation:
They traveled this distance in 2 parts, essentially. Part 1 had an average speed for a certain number of hours, part 2 had an average speed for a certain number of hours, and those 2 parts taken together took them a distance of 364 km. In equation form, that looks like this:
km/hr part 1 + km/hr part 2 = 364 km
Now we need to find each part on the left side of that equation. Part 1 first:
We traveled 65 km/hr for 2 hours, so that took us
[tex]65\frac{km}{hr}*2hr[/tex] and canceling out the hour label, we have that in part 1 we got
65(2) = 130 km. Good. Now onto the second part, where our unknown is.
We traveled 78 km/hr the second part for x hours, so that took us
[tex]78\frac{km}{hr}*xhr[/tex] and canceling out the hour label, we have that in part 2 we got
78x km. Now we can fill in the main equation (the one in bold print)
130 km + 78x km = 364 km and subtracting 130 km from both sides:
78x km = 234 km and dividing by 78 km:
x = 3 hours. Part 2 took 3 hours. Part 1 took 2 hours, so the whole trip took 5 hours.
Raquel establece que la rapidez del sonido en el aire en un día es de 346 m/s . Días después hace la misma medición obteniendo una rapidez de 340 m/ s . ¿Cuál será la temperatura del aire en cada día?
Answer:
La ecuación para la velodiad del sonido en aire esta dada por:
[tex]v = \sqrt{ \frac{\gamma*R*T}{M} }[/tex]
Resolviendo esto para T, obtenemos:
[tex]T = v^2*(\frac{M}{\gamma*R} )[/tex]
donde:
T = temperatura del aire en grados Kelvin
γ = constante adiabatica = 1.4
R = constante del gas ideal =
M = masa molar del aire = 29*10^(-3) kg/mol
Si sabemos que el primer día la velocidad es 346 m/s, tenemos:
v = 346 m/s
Ahora podemos reemplazar todos esos valores en la ecuación para T, y asi obtener:
[tex]T = (346m/s)^2*(\frac{29*10^{-3}kg/mol}{1.4* 8.31 J/mol*K} ) = 298.42 K[/tex]
Para el segundo día la velocidad es 340 m/s, entonces ese día la temperatura va a ser:
[tex]T = (340m/s)^2*(\frac{29*10^{-3}kg/mol}{1.4* 8.31 J/mol*K} ) = 288.16 K[/tex]
List out the fundamental and derived units
Drag each label to the correct location on the image. Identify the particles and characteristics on this model of an atom.
Positively charged
Electron
Proton
Neutron
Negatively charged
Answer:
cant see picture
Explanation:
Answer:
please add picture so i can help you
Explanation:
a uniform meter ruler is balanced at its midpoint
Answer:
a) i) x = 0.25 m, ii) x = 0.10 m, iii) x = 0.050 m
b) i) x = 0.40 m
Explanation:
a) For this exercise we use the rotational equilibrium equation, where we assume that the anticlockwise rotations are positive.
1) L = 2W
we set our reference system in the center of the bar where the fulcrum is
∑τ = 0
W 0.50 - L x = 0
x = 0.50 W / L
we substitute the value
x = 0.50 W / 2W
x = 0.25 m
ii) L = 5W
we calculate
x = 0.50 W / 5W
x = 0.10 m
iii) L = 10 W
x = 0.50 W / 10W
x = 0.050 m
b) a new weight is placed at x₂ = 30 cm on the left side
W 0.50 + W 0.30 - L x = 0
x = (0.50 + 0.30) W / L
x = 0.80 W / L
we calculate
i) L = 2W
x = 0.80 w / 2w
x = 0.40 m
49. A block is pushed across a horizontal surface with a
coefficient of kinetic friction of 0.15 by applying a
150 N horizontal force.
(a) The block accelerates at the rate of 2.53 m/s2
Find the mass of the block.
(b) The block slides across a new surface while
experiencing the same applied force as before.
The block now moves with a constant speed.
What is the coefficient of kinetic friction between
the block and the new surface?
Answer:
(a) 37.5 kg
(b) 4
Explanation:
Force, F = 150 N
kinetic friction coefficient = 0.15
(a) acceleration, a = 2.53 m/s^2
According to the newton's second law
Net force = mass x acceleration
F - friction force = m a
150 - 0.15 x m g = m a
150 = m (2.53 + 0.15 x 9.8)
m = 37.5 kg
(b) As the block moves with the constant speed so the applied force becomes the friction force.
[tex]F = \mu m g \\\\150 = \mu\times 37.5\\\\\mu = 4[/tex]
Does anyone know the answer ? I forgot my calculator.
12 x sin50
12 × sin50 = 9.192533317........
The boiling point of water is 1000 C at sea level. The boiling point of butane is -1.50C… If we leave liquid butane in a bowl on a table in a room where the temperature is 240C, butane will
A. evaporate.
B. condense.
C. freeze.
D. melt.
Answer: If we leave liquid butane in a bowl on a table in a room where the temperature is [tex]24^{o}C[/tex], butane will evaporate.
Explanation:
A temperature at which the the liquid and gaseous phase of a substance of a substance are present in equilibrium with each other is called boiling point.
For example, the boiling point of butane is -1.5 degree Celsius.
This means that at a temperature above -1.5 degree Celsius, butane will exist is gaseous state. That is, at a temperature of 24 degree Celsius butane will evaporate.
Thus, we can conclude that if we leave liquid butane in a bowl on a table in a room where the temperature is [tex]24^{o}C[/tex], butane will evaporate.
HELPPPPPPPPPPP PLEASEEEEEEEEEEE
Complete this sentence. The solubility of a sample will ____________ when the size of the sample increases.
stay the same
decrease
increase
be unable to be determined
the answer is not decrease
The solubility of the sample will decrease
Can the magnitude of a vector ever (a) be equal to one of its components, or (b) be less than one of its components? 9. Can a particle with constant speed be accelerating? What if it has constant velocity?
Answer:
a) the other components are zero, in the direction of one of the coordinate axes
b) the magnitude is less than the value of one of its components, it must occur when the vector is in some arbitrary direction
9) constant velocity the acceleration must necessarily be zero,
constant speed can be accelerated since it may be changing the direction of the velocity vector
Explanation:
Vectors are quantities that have modulo (scalar) direction and sense.
a) If in a vector its magnitude is equal to one d its components implies that the other components are zero, therefore the vector must be in the direction of one of the coordinate axes
b) if the magnitude is less than the value of one of its components, it must occur when the vector is in some arbitrary direction, other than the direction of the axes, that is
R² = x² + y²
where R is the magnitude of the vector e x, and are the components
9) When a particle has a constant velocity, the acceleration must necessarily be zero,
v = vo + a t
The bold letters indicate vectors If a = 0 implies that v = vo
If a particle has constant speed it can be accelerated since it may be changing the direction of the velocity vector, this type of acceleration has the name of centripetal acceleration
What does the m stand for in the enthalpy equation?
Answer:
Use the formula ∆H = m x s x ∆T to solve.
Explanation:
Once you have m, the mass of your reactants, s, the specific heat of your product, and ∆T, the temperature change from your reaction, you are prepared to find the enthalpy of reaction. Simply plug your values into the formula ∆H = m x s x ∆T and multiply to solve.
If we go on the top of the Mountain we will our weight increase or decrease?
Answer:
increase
Explanation:
guesshejwoowsnnsnakalalapqjhdjd
Answer:
our weight would decrease.
Explanation:
because of you burning all those calories and fat from hiking up the mountain.
What is the volume of a cone with a height of 27 cm
and a radius of 13 cm? Round your answer to the
nearest tenth.
Use the button on your calculator to complete this
problem.
V =
cm?
Explanation:
→ Volume of cone = πr² × h/3
Here,
Radius (r) = 13 cmHeight (h) = 27 cm→ Volume of cone = π(13)² × 27/3 cm³
→ Volume of cone = 169π × 9 cm³
→ Volume of cone = 1521π cm³
→ Volume of cone = 1521 × 22/7 cm³
→ Volume of cone = 33462/7 cm³
→ Volume of cone = 4780.28 cm³
Answer:
4,778.4 is correct
Explanation:
If Cl− is the only anion in the solution, what is the Cl− concentration in milliequivalents per liter?
Answer:
155mEq/L Cl-
Explanation:
A Ringer’s solution contains the following concentrations of cations: 146 mEq/L of Na+, 5 mEq/L of K+, and 4 mEq/L of Ca2+.
As Cl- is the only counterion of those cations:
For Na, the molecule is NaCl and the mEq/L of Cl- = mEq of Na+. The Cl- of the first ion is 146mEq/L Cl-
For K+, The molecule is KCl, mEq Cl- = 5mEq/L Cl-
And, for Ca2+, The molecule is CaCl2 but the equivalents of Ca2+ = Equivalents of Cl- = 4mEq/L Cl-
The total concentration of Cl- are:
146 + 5 + 4 =
155mEq/L Cl-Which labels are correct for the regions marked? a. X: Slower in gases than liquids Y: Faster in solids than gases Z: Velocity depends on medium b. X: Faster in gases than liquids Y: Slowest in solids Z: Faster in liquids than gases c. X: Slower in solids than liquids Y: Velocity depends on medium Z: Faster in liquids than gases d. X: Velocity depends on medium Y: Fastest in gases Z: Slower in liquids than solids
Answer:
a. X: Slower in gases than liquids Y: Faster in solids than gases Z: Velocity depends on medium.
Explanation:
Speed of sound is fastest in solids. Sound waves travel more quickly in solid, than of liquid and gases. Sound waves travel most slowest in gases. Speed of sound varies significantly and it depends upon medium it is travelling through. In more rigid medium sounds velocity will be faster.
