Answer:
Th average force impact is [tex]F = 168.298 \ N[/tex]
Explanation:
From the question we are told that
The mass of the golf ball is [tex]m_g = 26.7 \ g = 0.0267 \ kg[/tex]
The angle made is [tex]\theta = 33.6 ^o[/tex]
The range of the golf ball is [tex]R = 190 \ m[/tex]
The duration of contact is [tex]\Delta t = 7.13 \ ms = 7.13 *10^{-3} \ s[/tex]
Generally the range of the golf ball is mathematically represented as
[tex]R = \frac{v^2 sin2(\theta)}{g}[/tex]
Here v is the velocity with which the golf club propelled it with, making v the subject
[tex]v = \sqrt{\frac{R * g}{sin 2 (\theta)} }[/tex]
=> [tex]v = \sqrt{\frac{190 * 9.8}{sin 2 (33.6)} }[/tex]
=> [tex]v = 44.94 \ m/s[/tex]
Generally the change in momentum of the golf ball is mathematically represented as
[tex]\Delta p = m * (v - u )[/tex]
here u is the initial velocity of the ball before being stroked and the value is 0 m/s
[tex]\Delta p = 0.0267 * ( 44.94 - 0 )[/tex]
=> [tex]\Delta p = 1.19996 \ kg \cdot m/s[/tex]
Generally the average force of impact is mathematically represented as
[tex]F = \frac{\Delta p }{\Delta t}[/tex]
=> [tex]F = \frac{1.19996 }{7.13 *10^{-3}}[/tex]
=> [tex]F = 168.298 \ N[/tex]
If the velocity of Homer the astronaut (mass =200 kg) is 5 m/s and he runs into and grabs his stationary pal Larry (mass = 150 kg), what is the new velocity of the astronauts after the collision?
We are given:
Homer the Astronaut:
Mass of Homer the astronaut(m1) = 200 kg
initial velocity of Homer the astronaut(u1) = 5 m/s
Larry the Pal:
Mass of Larry the Pal (m2) = 150 kg
initial velocity of Larry the Pal (u2) = 0 m/s
Since they will move together after the collision, they will have the same velocity:
v1 = v2 = V
Solving for the Final velocity:
from the law of conservation of momentum:
m1u1 + m2u2 = m1v1 + m2v2
since v1 = v2 = V:
m1u1 + m2u2 = V(m1 + m2)
replacing the variables with the given values
200 * 5 + 150 * 0 = V(200 + 150)
1000 = 350V
V = 1000 / 350
V = 2.86 m/s
In football we see unbalanced forces. When 1 player exerts an unbalanced force on another player and causes a player to
Answer:
Fall
Explanation:
An oil refinery uses a Venturi tube to measure the flow rate of gasoline. The density of the gasoline is
ρ = 7.40 ✕ 102 kg/m3,
the inlet and outlet tubes, respectively, have a radius of 3.74 cm and 1.87 cm, and the difference in input and output pressure is
P1 − P2 = 1.20 kPa.
a) find the speed of the gasoline as it leaves the hose
b) find the fluid flow rate in cubic meters per second
Answer:
(a) V₂ = 1.86 m/s
(b) Q = 5.1 x 10⁻⁴ m³/s
Explanation:
(a)
The formula derived for Venturi tube is as follows:
P₁ - P₂ = (ρ/2)(V₂² - V₁²)
where,
P₁ - P₂ = Difference in Pressure of Inlet and Outlet = 1.2 KPa = 1200 Pa
ρ = Density of Gasoline = 7.4 x 10² kg/m³
V₂ = Exit Velocity = ?
V₁ = Inlet Velocity
Therefore,
1200 Pa = [(7.4 x 10²kg/m³)/2](V₂² - V₁²)
V₂² - V₁² = (1200 Pa)/(3.7 x 10² kg/m³)
V₂² - V₁² = 3.24 m²/s² ------------------- equation (1)
Now, we will use continuity equation:
A₁V₁ = A₂V₂
where,
A₁ = Inlet Area = πd₁²/4 = π(0.0374 m)²/4 = 1.098 x 10⁻³ m²
A₂ = Exit Area = πd₂²/4 = π(0.0187 m)²/4 = 2.746 x 10⁻⁴ m²
Therefore,
(1.098 x 10⁻³ m²)V₁ = (2.746 x 10⁻⁴ m²)V₂
V₁ = (2.746 x 10⁻⁴ m²)V₂/(1.098 x 10⁻³ m²)
V₁ = 0.25 V₂
using this value in equation (1):
V₂² - (0.25 V₂)² = 3.24 m²/s²
0.9375 V₂² = 3.24 m²/s²
V₂² = (3.24 m²/s²)/0.9375
V₂ = √(3.456 m²/s²)
V₂ = 1.86 m/s
(b)
For fluid flow rate we use the following equation:
Flow Rate = Q = A₂V₂ = (2.746 x 10⁻⁴ m²)(1.86 m/s)
Q = 5.1 x 10⁻⁴ m³/s
The formula for finding variables in a Venturi tube is shown below:
The speed of the gasolineP₁ - P₂ = (ρ/2)(V₂² - V₁²)
where, P₁ - P₂ is difference in pressure of Inlet and outlet, ρ = density, V₂ = exit velocity and V₁ is inlet velocity
P₁ - P₂ = 1.2 KPa = 1200 Pa
ρ = 7.4 x 10² kg/m³
V₂ = Exit Velocity = ?
V₁ = Inlet Velocity
We then substitute the variables into this equation.
P₁ - P₂ = (ρ/2)(V₂² - V₁²)
1200 Pa = [(7.4 x 10²kg/m³)/2](V₂² - V₁²)
V₂² - V₁² = (1200 Pa)/(3.7 x 10² kg/m³)
V₂² - V₁² = 3.24 m²/s² ------ equation (1)
The continuity equation A₁V₁ = A₂V₂ is then used
where,A₁ = Inlet area = πd₁²/4 = π(0.0374 m)²/4 = 1.098 x 10⁻³ m²
A₂ = Exit Area = πd₂²/4 = π(0.0187 m)²/4 = 2.746 x 10⁻⁴ m²
(1.098 x 10⁻³ m²)V₁ = (2.746 x 10⁻⁴ m²)V₂
V₁ = (2.746 x 10⁻⁴ m²)V₂/(1.098 x 10⁻³ m²)
V₁ = 0.25 V₂
We then substitute the value into equation 1
V₂² - (0.25 V₂)² = 3.24 m²/s²
0.9375 V₂² = 3.24 m²/s²
V₂² = (3.24 m²/s²)/0.9375
V₂ = √(3.456 m²/s²)
V₂ = 1.86 m/s
The fluid flow rate we use the following equation:This can be calculated using the formula
Flow Rate = Q = A₂V₂
= (2.746 x 10⁻⁴ m²)(1.86 m/s)
= 5.1 x 10⁻⁴ m³/s
What must happen to an atom of magnesium in order to become a magnesium ion Mg+2?
It must lose two neutrons and become a different isotope.
It must gain two neutrons and become a different isotope.
It must lose two electrons and become an ion.
It must gain two electrons and become an ion.
Answer:
Answer is: c. It must lose two electrons and become an ion.
Magnesium (Mg) is metal from 2. group of Periodic table of elements and has low ionisation energy and electronegativity, which means it easily lose valence electons (two valence electrons).
Magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.
Electron configuration of magnesium ion: ₁₂Mg²⁺ 1s² 2s² 2p⁶.
Explanation:
Answer:
the third option
Explanation:
How does the abundance of hydrogen and helium support the Big Bang Theory?
It is the proportion predicted to be present in the early universe.
The hydrogen and helium abundance helps us to model the expansion rate of the early universe.
In the abundance of hydrogen and helium, we can say that they account for nearly all the nuclear matter in today's universe.
In big Bang model, the universe is mostly light or protons.
This abundance of hydrogen and helium is consistent with this big bang model. The process of forming this hydrogen and helium is often called big bang nucleosynthesis.The Schramm's model for relative abundances indicate that helium is about 25% by mass and hydrogen about 73% with all other elements constituting less than 2%.
Several proponents of big Bang theory has proposed similar relative abundance for hydrogen and helium. In all it is clear that hydrogen and helium constitute of more than 98% of the ordinary matter in the universe.
Finally, the hydrogen and helium abundance helps us to model the expansion rate of the early universe.
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HELP ASAP !!!!!!!!!!!!!!
Answer:
big bang theory
Explanation:
Edwin Hubble is credited for the initial development of the Big Bang theory, an idea which helps to explain the formation of the universe over 15 billion years ago.
PLS HURRY!!!! 15 PTS!!! The pictures represent three different states of matter.
Which order of pictures places molecules with the most
amount of motion to the least amount of motion?
O X Y Z
O ZYX
O Y Z
O Y - X - Z
A ball has a diameter of 3.77 cm and average density of 0.0839 g/cm3. What force is required to hold it completely submerged under water?
magnitude _________ N
The force required to hold it completely submerged under water is 0.252 N
As a result of the low density (ρ1 = 0.0839 g/cm3 = 83.9 kg/m3)of the ball compared to that of water (ρ2 =1000 kg/m3), the buoyant force that is acting on the ball is greater than its weight.
Therefore, the minimum force required to hold the ball submerged under water can be calculated using the relation
F = Buoyant force - weight of sphere
Radius = 3.77/2 cm = 0.0377/2 m = 0.01885 m
Volume of sphere = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ = 2.805 e-5 m³
Mass of sphere = 4/3 π r³ ρ1 = 4/3 * 3.142 * 0.01885³ * 83.9 = 0.0023 kg
Weight of sphere = 4/3 π r³ ρ1 g = 4/3 * 3.142 * 0.01885³ * 83.9 * 9.8 = 0.023 N
Volume of water displaced = 4/3 π r³ = 2.805 e-5
Buoyant force = weight of water displaced = 4/3 π r³ ρ2 g = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ * 1000 * 9.8 = 0.275 N
F = 0.275 - 0.023 = 0.252 N
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The force required to hold it completely submerged under water is 0.25 N
The density of the ball ([tex]\rho_b[/tex]) = 0.0839 g/cm³ = 83.9 kg/m³
The density of water [tex]\rho_w[/tex] = 1000 kg/m³
Diameter = 3.77 cm = 0.0377 m
radius of ball = 0.0377/2 = 0.01885 m
The volume (V) = [tex]\frac{4}{3} \pi r^3=\frac{4}{3}*\pi*0.01885^3=2.8*10^{-5}\ m^3[/tex]
Let us assume the acceleration due to gravity (g) = 9.8 m/s², Hence:
The force is required to hold it completely submerged under water (F) is:
[tex]F=\rho_w Vg-\rho_b Vg=1000*(2.8*10^{-5})*9.81-83.9*(2.8*10^{-5})*9.81\\\\[/tex]
F = 0.25 N
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what is environment
Answer:
Explanation:
all biotic and abiotic factors around a living organism is its environment
Answer: An environment is the surroundings or conditions in which a person, animal, or plant lives or operates. the natural world, as a whole or in a particular geographical area, especially as affected by human activity. The best synonym for environment would be nature.
Explanation: Environment is a place where different things are such as a swampy or hot environment. They constantly interact with it and adapt themselves to conditions in their environment. In the environment there are different interactions between animals, plants, soil, water, and other living and non-living things. Environment plays an important role in healthy living and the existence of life on planet earth. Earth is a home for different living species and we all are dependent on the environment for food, air, water, and other needs. Therefore, it is important for every individual to save and protect our environment. To divide environments' sorts we can mention 3 kinds of environments Natural, industrial, and social environment. Natural environment: Include water, light, land, air and all organisms that live in nature.
HELP ASAP !!! !!!!!!!
Answer:
they are cooler than the rest if the sun
Catching a wave, a 77 kg surfer starts with a speed of 1.3 m/s, drops through a height of 1.65 m, and ends with a speed of 8.2 m/s. How much non-conservative work was done on the surfer?
Answer:
Explanation:
The total work done by the wave is expressed as;
Workdone = Potential energy + Kinetic energy
Workdone = mgh + 1/2mv²
m is the mass = 77kg
g is the acceleration due to gravity = 9.8m/s²
v is the velocity = 8.2m/s
h is the height = 1.65m
Substitute into the formula;
Workdone = 77(9.8)(1.65) + 1/2(77)8.2²
Workdone = 1245.09 + 2588.74
Workdone = 3833.83Joules
Hence the amount of non conservative work done on the sofa is 3833.83Joules
Given:
Velocity, v = 8.2 m/sHeight, h = 1.65 mMass, m = 77 kgWe know,
→ [tex]Work \ done = Potential \ energy +Kinetic \ energy[/tex]
or,
[tex]= mgh +\frac{1}{2} mv^2[/tex]
By putting the values,
[tex]= 77\times 9.8\times 1.65+\frac{1}{2}\times 77\times (8.2)^2[/tex]
[tex]= 1245.09+2588.74[/tex]
[tex]= 3833.83 \ Joules[/tex]
Thus the above approach is right.
Learn more about work done here:
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Please help me with this question guys.
Answer:
The average speed is 22.2 km/h
Explanation:
Average Speed
Given an object travels a total distance d and took a total time t, then the average speed is:
[tex]\displaystyle \bar v=\frac{d}{t}[/tex]
The mailman first drives d1=7 km at v1=15 km/h. The time taken to drive is:
[tex]\displaystyle t1=\frac{d1}{v1}=\frac{7}{15}=0.467\ h[/tex]
Then he drives d2=7 km at v2=43 km/h taking a time of:
[tex]\displaystyle t2=\frac{d2}{v2}=\frac{7}{43}=0.163\ h[/tex]
The total time is
t=0.467 h + 0.163 h = 0.63 h
The total distance is
d = 7 km + 7 km = 14 km
The average speed is:
[tex]\displaystyle \bar v=\frac{14}{0.63}=22.2\ km/h[/tex]
The average speed is 22.2 km/h
Fast and safe heart rate for workouts is called muscular strength? True or false
Answer:
False
Explanation:
Answer:
False
Explanation:
Hope this helped, Have a Wonderful Day/Night!!
Romeo is a 68 kg astronaut. Juliet is a beautiful cosmonaut who is standing on the balcony of a 4.58 x 10^5kg space station that is at rest and out of gas. Romeo is floating 25 meters away from the space station’s center of mass, how strong is the force between Romeo and Juliet?
Answer:
F = 3.32 x 10⁻⁶ N
Explanation:
The force of attraction between two masses is given by Newton's Law of Gravitation, as follows:
F = Gm₁m₂/r²
where,
F = Force between Romeo and Juliet = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
m₁ = mass of Romeo = 68 kg
m₂ = mass of space station = 4.58 x 10⁵ kg
r = distance = 25 m
Therefore,
F = (6.67 x 10⁻¹¹ N.m²/kg²)(68 kg)(4.58 x 10⁵ kg)/(25 m)²
F = 3.32 x 10⁻⁶ N
Two identical conducting spheres each have a charge of 2 C. They have a radius of 0.1 m and are separated by 0.5 meters. If you were to increase the radius of these spheres to 0.2 meters, the electrostatic force between them would
Answer:
The electrostatic force will remain the same
Explanation:
From the question we are told that
The charge on the each conducting sphere is [tex]q = 2 \ C[/tex]
The radius each sphere is [tex]r_1 = r_2 = r = 0. 1\ m[/tex]
Generally electrostatic force between the sphere is mathematically represented as
[tex]F = \frac{k * q_1 q_2 }{d^2}[/tex]
Here k is the coulomb constant ,
d is the distance between the two sphere which is measured from one center of the sphere to the other center of the sphere
Now from the question we are told that the radius of the spheres is doubled (i.e from 0.10 m to 0.2 m ) , this will not affect the distance between the sphere because their center are still in the same position
and given there is no change in the distance between the spheres , the electrostatic force will remain the same
A shell traveling with speed, v0 , exactly horizontally and due north explodes into two equal mass fragments. It is observed that just after the explosion one fragment is traveling vertically up with speed v0 . What is the velocity of the other fragment? Hint: Velocity has both magnitude and direction.
Answer:
yeah
Explanation:
yeah yeah yeah yeah
A group of students collected the data shown below while attempting to measure the coefficient of static friction (of course, it looks like this group varied the amount of mass sitting on the block with each trial - this is not recommended). Nonetheless, what is their average coefficient of static friction?
Trial Mass of block (g) Hanging mass (kg)
1 105 0.053
2 165 0.081
3 220 0.118
4 280 0.149
5 315 0.180
6 385 0.198
Answer:
0.130
Explanation:
From the given data, the coefficient of static friction for each trial are:
1. 0.053
2. 0.081
3. 0.118
4. 0.149
5. 0.180
6. 0.198
The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198
= 0.779
So that;
the average coefficient of static friction = [tex]\frac{sum of coefficient of static friction}{number of trials}[/tex]
= [tex]\frac{0.779}{6}[/tex]
= 0.12983
The average coefficient of static friction is 0.130
The average coefficient of static friction is 0.13.
The coefficient of static friction is obtained using the formula; μ = F/R
Where;
F = force acting on the body
R = reaction
μ = coefficient of static friction
The average of measurements is given as; ∑summation of measurements/number of measurements
We can see from the question that there were 6 measurements of the coefficient of static friction. Hence, the average coefficient of static friction is obtained from;
0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198/6
= 0.13
The average coefficient of static friction is 0.13
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Our school needs to offer healthier options in the lunchroom. Elever High School has recently updated its cafeteria menu to include whole wheat pasta and breads, a fresh salad bar, and other healthy menu items. Students there claim that they have more energy and focus throughout their school day. Let's encourage healthier menus in our lunchroom!
What type of evidence does the writer of this passage use to support her claim?
statistics
statistics
an expert's opinion
an expert's opinion
examples
examples
the writer's opinion
Answer:
b
Explanation:
A 2.0 cm thick brass plate (k_r = 105 W/K-m) is sealed to a glass sheet (kg = 0.80 W/K m), and both have the same area. The exposed face of the brass plate is at 80°C, while the exposed face of the glass is at 20 °C. How thick is the glass if the glass brass interface is at 65 C? Ans. 0.46 mm
A car travels at a speed of 55 km/hr and slows down to 10 km/hr in 20 seconds. What is the acceleration?
Answer:
a = 0.62 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics. But first, we must convert speeds from kilometers per hour to meters per second.
[tex]v_{f} =v_{o} -a*t[/tex]
[tex]55[\frac{km}{hr}]*\frac{1hr}{3600s}*\frac{1000m}{1km} =15.27[\frac{m}{s} ]\\10[\frac{km}{hr} ]*\frac{1hr}{3600s}*\frac{1000m}{1km} = 2.77[\frac{m}{s} ][/tex]
where:
Vf = final velocity = 2,77 [m/s]
Vo = initial velocity = 15.27 [m/s]
t = time = 20 [s]
a = acceleration [m/s²]
Now replacing:
[tex]2.77=15.27-a*(20)\\20*a=12.49\\a = 0.62[m/s^{2}][/tex]
Answer:
It would be 10.00
Explanation:
Hope this helps its different for everyone what was it for u it was D for me
If a 25 kg lawnmower produces 347 w and does 9514 J of work, for
how much time did the lawnmower run?
Steps 1 and 2)
The variables are W = work, P = power, and t = time. In this case, W = 9514 joules and P = 347 watts.
The goal is to solve for the unknown time t.
-----------------------
Step 3)
Since we want to solve for the time, and we have known W and P values, we use the equation t = W/P
-----------------------
Step 4)
t = W/P
t = 9514/347
t = 27.4178674351586
t = 27.4 seconds
-----------------------
Step 5)
The lawn mower ran for about 27.4 seconds. I rounded to three sig figs because this was the lower amount of sig figs when comparing 9514 and 347.
-----------------------
Note: we don't use the mass at all
An arrow in a bow has 357 J of elastic potential energy How much Winette enere
Will the arrow have after it has been shot assuming there is no sir restoran
Answer:
357 J
Explanation:
The elastic potential energy of arrow in the stretched bow is 357 J.
The kinetic energy of the arrow after it has been shot is given by half of the product of the arrow's mass and velocity of the arrow.
Here there are no other forms of energy at play here. Only potential and kinetic energy.
As we know that in any system the energy is conserved accordingly the elastic potential energy of the arrow will be equal to the kinetic energy of the bow after it is released i.e., 357 J.