Answer:
37.5 m/s
Explanation:
Using,
Formula
v = ωr....................... Equation 1
Where ω = instantaneous angular velocity, v = instantaneous linear velocity, r = radius or distance from the sweet spot of the bat to the axis of rotation.
From the question,
Given: ω = 30 rad/s, r = 1.25 m
Substitute these values into equation 1
v = 30(1.25)
v = 37.5 m/s.
Hence the instantaneous linear velocity of the sweet spot at the instant of ball contact is 37.5 m/s
why food cook faster with salt water than cook with pure water
Answer:
oil heats faster
Explanation:
A force of 10 N is applied at right angles to the handle of a spanner, 0.5 m from the centre of a nut. The
moment on the nut is:
20 Nm
50 Nm
5 Nm
Explanation:
the movement of the nut is 20Nm
A force of 12 N changes the momentum of a toy car from 3kgm/s t0 10kgm/s. Calculate the time the force took to produce this change in momentum.
Answer:
Time = 0.58 seconds
Explanation:
Given the following data;
Initial momentum = 3 kgm/s
Final momentum = 10 kgm/s
Force = 12 N
To find the time required for the change in momentum;
First of all, we would determine the change in momentum.
[tex] Change \; in \; momentum = final \; momentum - initial \; momentum [/tex]
[tex] Change \; in \; momentum = 10 - 3 [/tex]
Change in momentum = 7 kgm/s
Now, we can find the time required;
Note: the impulse of an object is equal to the change in momentum experienced by the object.
Mathematically, impulse (change in momentum) is given by the formula;
[tex] Impulse = force * time [/tex]
Making "time" the subject of formula, we have;
[tex] Time = \frac {impulse}{force} [/tex]
Substituting into the formula, we have;
[tex] Time = \frac {7}{12} [/tex]
Time = 0.58 seconds
5. How much heat is needed to warm .052 kg of gold from 30°C to 120°C? Note: Gold has a specific heat of 136
J/kg °C
Answer:
Q = 636.48 J
Explanation:
Given that,
The mass of gold, m = 0.052 kg
The temperature increase from 30°C to 120°C.
The specific heat of gold is 136 J/kg °C.
We need to find the heat needed to warm the gold. The formula for heat needed is given by :
[tex]Q=mc\Delta T\\\\Q=0.052\times 136\times (120-30)\\\\Q=636.48\ J[/tex]
So, 636.48 J of heat is needed to warm gold.
2. A test reveals that 150 J of work is required to lift an object 3 m at a
constant speed. What is the weight of the object?
0 50 N
25N
55N
O 75N
Answer:
50N
Explanation:
W=Fd
150=F(3)
50N=F
write the definition
Heredity, Dominant allele, recessive allele, probability, genotype, phenotype
Explanation:
Heredity: The passing on of physical or mental characteristics genetically from one generation to another.
Dominant allele: In genetics, dominance is the phenomenon of one variant of a gene on a chromosome masking or overriding the effect of a different variant of the same gene on the other copy of the chromosome. The first variant is termed dominant and the second recessive.
Answer:
Heredity: the passing on of physical or mental characteristics genetically from one generation to another.
Dominant Allele: a variation of a gene that will produce a certain phenotype, even in the presence of other alleles.
Recessive Allele: a variety of genetic code that does not create a phenotype if a dominant allele is present.
Probablity: used to measure the chance or likelihood of an event to occur.
Genotype: an organism's complete set of genetic material
Phenotype: the set of observable characteristics of an individual resulting from the interaction of its genotype with the environment.
A physics professor wants to demonstrate the large size of the henry unit. On the outside of a 16-cm-diameter plastic hollow tube, she wants to wind an air-filled solenoid with self-inductance of 1.0 H using copper wire with a 0.79-mm diameter. The solenoid is to be tightly wound with each turn touching its neighbor (the wire has a thin insulating layer on its surface so the neighboring turns are not in electrical contact).
Required:
a. How long will the plastic tube need to be?
b. How many kilometers of copper wire will be required?
c. What will be the resistance of this solenoid?
Answer:
a) the plastic tube need to be 24.7 m long
b) the kilometer of copper wire required is 15.7
c) the resistance of this solenoid is 5538.8 2 ohms
Explanation:
Given the data in the question;
we determine the length of the plastic tube. assuming the solenoid is long.
the self inductance of a long solenoid is;
L = μ₀n²πr²l
μ₀ = 4π × 10⁻⁷ T-m/A
where
n = number of turns per unit length
r = radius of the solenoid = 8cm (as the diameter of the plastic hollow tube is 16 cm)
l = length of the solenoid or the length of the plastic tube
we find n = number of turns per unit length
given that, the copper wire to be wound around the solenoid is 0.79 mm in diameter
number of turns per meter = n = 1 / ( 0.79 × 10⁻³ m ) = 1265.8 turns/meter
So from our previous formula, we find l
L = μ₀n²πr²l
we substitute
1.0 H = (4π × 10⁻⁷ T-m/A)( 1265.8 )²(3.14)(0.08)² ( l)
1 = 0.04048 × l
l = 1 / 0.04048
l = 24.7 m
Therefore, the plastic tube need to be 24.7 m long
b)
n = number of turns per unit length = 1265.8 turns/metre
so, the length of the plastic tube over which the copper wire is to be wound,
number of turns of copper wire required = n × l
= 1265.8 turns/meter × 24.7 m
= 31,265.26 turns
Now each turn of the copper wire is to be wound across the 18cm diameter of the plastic tube.
so for each turn length of copper wire required = 2π × r
= 2π × 0.08 m
= 0.5026548 m
So copper wire required for 31,265.26 turns will be;
⇒ 31,265.26 × 0.5026548 = 15715.63m = 15.7 km
Therefore, the kilometer of copper wire required is 15.7
c)
p = resistivity of copper = 1.68 × 10⁻⁸ ohm-m
Resistance = pl/a
where l is length of copper wire, a is cross sectional area;
diameter of copper wire is 0.79-mm
radius of copper wire is 0.79/2 = 0.395 mm = 0.000395 m
area of cross section of copper wire a = πr² = π( 0.00395)² = 4.9 × 10⁻⁷ m²
Resistance = pl/a
we substitute
Resistance = [(1.68 × 10⁻⁸ ohm-m)( 15715.63m )] / [ 4.9 × 10⁻⁷ m² ]
Resistance = 5538.8 2 ohms
Therefore, the resistance of this solenoid is 5538.8 2 ohms
Two 20kg spheres are placed with their
Centres 50cm apart. What is the magnitude of
gravitational force each exerts on the other?
Answer:
F = 1.07 x 10⁻⁷ N
Explanation:
The gravitational force of attraction between two objects can be found by the use of Newton's Gravitational Law:
[tex]F = \frac{Gm_{1}m_{2}}{r^2}\\\\[/tex]
where,
F = Gravitational Force of attraction = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
m₁ = m₂ = mass of spheres = 20 kg
r = distance between the objects = 50 cm = 0.5 m
Therefore,
[tex]F = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(20\ kg)(20\ kg)}{(0.5\ m)^2}\\\\[/tex]
F = 1.07 x 10⁻⁷ N
A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 38.0 ∘ above the horizontal. The glider has mass 9.00×10−2 kg. The spring has 590 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.70 m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring.
Required:
a. What distance was the spring originally compressed?
b. When the glider has traveled along the air track 0.80 m from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?
Answer:
x = 0.056 m
ΔKE = 0.489 J
Explanation:
Given that
Angle, θ = 38°
Length, L = 1.7 m
Mass, m = 0.09 kg
Spring constant, K = 590 N/m
If we use the Work-Energy theorem, then we know that Potential Energy, PE = Kinetic Energy, KE
This is mathematically written as
1/2kx² = mgH
The height, H we can get by using the relation
H = L.Sinθ
H = 1.7 * Sin 38
H = 1.7 * 0.6157
H = 1.047 m
Next, we use the Work-Energy theorem
1/2kx² = mgH
1/2 * 590 * x² = 0.09 * 9.8 * 1.047
295 * x² = 0.9234
x² = 0.9235 / 295
x² = 0.00313
x = √0.00313
x = 0.056 m
If the spring loses contact at x = 0.056, definitely, it will also lose contact at x = 0.8
Then we use the formula
ΔKE = mg(H - H1)
ΔKE = mg(xsinθ - x2.sinθ)
Where, x = 1.7 , x2 = 0.8
ΔKE = 0.09 * 9.8 (1.7 * sin 38 - 0.8 * sin 38)
ΔKE = 0.882(1.047 - 0.493)
ΔKE = 0.882 * 0.554
ΔKE = 0.489 J
Calculate the first and second order angles for light of wavelength 400. nm and 700. nm of the grating contains 1.00 x 104 lines/cm.
Answer:
[tex]23.58^{\circ}[/tex] and [tex]53.13^{\circ}[/tex]
[tex]44.43^{\circ}[/tex], second order does not exist
Explanation:
n = Number of lines grating = [tex]1\times10^4\ \text{Lines/cm}[/tex]
[tex]\lambda[/tex] = Wavelength
m = Order
Distance between slits is given by
[tex]d=\dfrac{1}{n}\\\Rightarrow d=\dfrac{1}{1\times 10^4}\\\Rightarrow d=10^{-6}\ \text{m}[/tex]
[tex]\lambda=400\ \text{nm}[/tex]
m = 1
We have the relation
[tex]d\sin\theta=m\lambda\\\Rightarrow \theta=\sin^{-1}\dfrac{m\lambda}{d}\\\Rightarrow \theta=\sin^{-1}\dfrac{1\times 400\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=23.58^{\circ}[/tex]
m = 2
[tex]\theta=\sin^{-1}\dfrac{2\times 400\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=53.13^{\circ}[/tex]
The first and second order angles for light of wavelength 400 nm are [tex]23.58^{\circ}[/tex] and [tex]53.13^{\circ}[/tex].
[tex]\lambda=700\ \text{nm}[/tex]
m = 1
[tex]\theta=\sin^{-1}\dfrac{1\times 700\times 10^{-9}}{10^{-6}}\\\Rightarrow \theta=44.43^{\circ}[/tex]
m = 2
[tex]\theta=\sin^{-1}\dfrac{2\times 700\times 10^{-9}}{10^{-6}}[/tex]
Here [tex]\dfrac{2\times 700\times 10^{-9}}{10^{-6}}=1.4>1[/tex] so there is no second order angle for this case.
The first order angle for light of wavelength 700 nm are [tex]44.43^{\circ}[/tex].
Second order angle does not exist.
A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with a second particle, which has an initial linear momentum of4.00 kg-m/s, directed along the positive y-axis. The final momentum of the first particle is 3.00 kg-m/s, directed 45.0 above the positive x-axis.
a. the magnitude and direction (angle expressed counter-clockwise with respect to the positive x-axis) of the final momentum for the second particle
b. assuming that these particles have the same mass, % loss of their total kinetic energy after they collided
Answer:
a) p₂ = 1.88 kg*m/s
θ = 273.4 º
b) Kf = 37% of Ko
Explanation:
a)
Assuming no external forces acting during the collision, total momentum must be conserved.Since momentum is a vector, their components (projected along two axes perpendicular each other, x- and y- in this case) must be conserved too.The initial momenta of both particles are directed one along the x-axis, and the other one along the y-axis.So for the particle moving along the positive x-axis, we can write the following equations for its initial momentum:[tex]p_{o1x} = 2.00 kg*m/s (1)[/tex]
[tex]p_{o1y} = 0 (2)[/tex]
We can do the same for the particle moving along the positive y-axis:[tex]p_{o2x} = 0 (3)[/tex]
[tex]p_{o2y} = 4.00 kg*m/s (4)[/tex]
Now, we know the value of magnitude of the final momentum p1, and the angle that makes with the positive x-axis.Applying the definition of cosine and sine of an angle, we can find the x- and y- components of the final momentum of the first particle, as follows:[tex]p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)[/tex]
[tex]p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s (6)[/tex]
Now, the total initial momentum, along these directions, must be equal to the total final momentum.We can write the equation for the x- axis as follows:[tex]p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x} (7)[/tex]
We know from (3) that p₀₂ₓ = 0, and we have the values of p₀1ₓ from (1) and pf₁ₓ from (5) so we can solve (7) for pf₂ₓ, as follows:[tex]p_{f2x} = p_{o1x} - p_{f1x} = 2.00kg*m*/s - 2.12 kg*m/s = -0.12 kg*m/s (8)[/tex]
Now, we can repeat exactly the same process for the y- axis, as follows:[tex]p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y} (9)[/tex]
We know from (2) that p₀1y = 0, and we have the values of p₀₂y from (4) and pf₁y from (6) so we can solve (9) for pf₂y, as follows:[tex]p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)[/tex]
Since we have the x- and y- components of the final momentum of the second particle, we can find its magnitude applying the Pythagorean Theorem, as follows:[tex]p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} } = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)[/tex]
We can find the angle that this vector makes with the positive x- axis, applying the definition of tangent of an angle, as follows:[tex]tg \theta = \frac{p_{2fy} }{p_{2fx} } = \frac{1.88m/s}{(-0.12m/s} = -15.7 (12)[/tex]
The angle that we are looking for is just the arc tg of (12) which measured in a counter-clockwise direction from the positive x- axis, is just 273.4º.b)
Assuming that both masses are equal each other, we find that the momenta are proportional to the speeds, so we find that the relationship from the final kinetic energy and the initial one can be expressed as follows:[tex]\frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)[/tex]
So, the final kinetic energy has lost a 37% of the initial one.An astronaut has a mass of 75 kg and is floating in space 500 m from his 125,000 kg spacecraft. What will be the force of gravitational attraction between the two? Since there is no force opposing him, he will accelerate toward the ship. Find his acceleration.
Answer:
1. 2.5×10¯⁹ N
2. 3.33×10¯¹¹ m/s²
Explanation:
1. Determination of the force of attraction.
Mass of astronaut (M₁) = 75 Kg
Mass of spacecraft (M₂) = 125000 Kg
Distance apart (r) = 500 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Force of attraction (F) =?
The force of attraction between the astronaut and his spacecraft can be obtained as follow:
F = GM₁M₂ /r²
F = 6.67×10¯¹¹ × 75 × 125000 / 500²
F = 2.5×10¯⁹ N
Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N
2. Determination of the acceleration of the astronaut.
Mass of astronaut (m) = 75 Kg
Force (F) = 2.5×10¯⁹ N
Acceleration (a) of astronaut =?
The acceleration of the astronaut can be obtained as follow:
F = ma
2.5×10¯⁹ N = 75 × a
Divide both side by 75
a = 2.5×10¯⁹ / 75
a = 3.33×10¯¹¹ m/s²
Thus, the acceleration the astronaut is 3.33×10¯¹¹ m/s²
What is the potential energy, if a
body of mass 250 kg is at a height of 30
metre?
12. Identify the Leader
Cells use nutrients and oxygen to supply the body with the energy it needs. What three-body systems are working together in this situation?
A
nervous, digestive, and circulatory systems
B
digestive, circulatory, and excretory systems
C
circulatory, immune, and respiratory systems
D
digestive, respiratory, and circulatory systems
Answer:
The respiratory system provides oxygen for cells, while the circulatory system transports oxygen to cells.
Explanation:
so the answer is D
Which of these is the BEST answer for why science is important?
Science can take us to other planets, even if it’s only through a telescope.
Science is part of human nature; it helps answer questions about how the world works.
Science helps us learn to think more critically and weigh evidence better.
Science gives us better tablet computers and games.
QQQUUUUCCCKKKK!!!!!!
When6-2 He He-6 undergoes beta decay, the daughter is?
Answer: The daughter is named Susie.
Explanation: LIL SUSIE!!!
HUH? DIDN'T UNDERSTAND THE QUESTION!
HAVE A GREAT DAY!!!!!
Answer:6/3 Li
Explanation:
I’m not sure what the person under me is talking about but yeah
What is the acceleration of a 4,000 kg car pushed with a
force of 12,000 N?
Answer:
3 m/s
Explanation:
A= F/m
12,000/ 4000 = 3
Answer:
3 m/s^2
Explanation:
The equation you have to use is F=ma because the problem is a Newton's 2nd law problem.
Our known values are:
F ( Force ) = 12,000 N
m ( mass ) = 4,000 kg
a ( acceleration ) = ?
Now we plug in the known values into the equation and solve
F=ma
12,000=4,000a
We have to divide 4,000 by both sides to isolate the a value
12,000/4,000=4,000/4,000a
The 4,000s on the right of the equation cancel.
And 12,000 divided by 4,000 equals 3
The acceleration (a) is 3 meters per second squared (m/s^2)
Next, check to make sure 3 does work by plugging it back into the equation.
12,000=4,000*3
12,000=12,000 ✔
As you can see, the acceleration will be 3 m/s^2
g 1. To see why an MRI utilizes iron to increase the magnetic field created by a coil, calculate the current needed in a 400-loop-per-meter circular coil 0.660 m in radius to create a 1.20-T field (typical of an MRI instrument) at its center with no iron present. The magnetic field of a proton is approximately like that of a circular current loop in radius carrying . What is the field at the
Answer:
I = 2387.32 A
Explanation:
Given that,
Number of turns in the loop, N = 400
The radius of the circular coil, r = 0.66 m
The magnetic field inside the MRI, B = 1.2 T
We need to find the current in the loop. The magnetic field inside the solenoid is given by :
[tex]B=\mu_o NI\\\\I=\dfrac{B}{\mu_o N}\\\\I=\dfrac{1.2}{4\pi \times 10^{-7}\times 400}\\\\I=2387.32\ A[/tex]
So, the required current is equal to 2387.32 A.
Time of the day when the Sun does not shine (___time)
N____N
A ceiling fan with 90-cm-diameter blades is turning at 64 rpm . Suppose the fan coasts to a stop 28 s after being turned off. What is the speed of the tip of a blade 10 s after the fan is turned off
Answer:
the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.
Explanation:
Given;
diameter of the ceiling fan, d = 90 cm = 0.9 m
angular speed of the fan, ω = 64 rpm
time taken for the fan to stop, t = 28 s
The distance traveled by the ceiling fan when it comes to a stop is calculated as;
[tex]d = vt = \omega r\times t= ( \frac{64 \ rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1 \min}{60 \ s} \times 0.9 \ m) \times 28 \ s\\\\d = 168.89 \ m[/tex]
The speed of the tip of a blade 10 s after the fan is turned off is calculated as;
[tex]v = \frac{d}{t} \\\\v = \frac{168.89}{10} \\\\v = 16.889 \ m/s[/tex]
Therefore, the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.
Discuss how the pressure cooker is designed to achieve temperatures above 100°C.
With rising heat, the steam pressure inside the pot builds up beyond atmospheric pressure, allowing the temperatures to rise well above boiling point. This design enables to save time, energy, and resources. The temperature inside a pressure cooker can well go beyond 110° C, which reduces the time needed to cook food.
The amount of light that enters the pupil is controlled by the:
retina.
lens.
inis.
Answer: The amount of light that enters the pupil is controlled by the Iris
Explanation:
PLEASE HELP
Section 1 - Question 6
Wave Movement Through Media
What could be happening to the wave as it travels from left to right?
A
It's moving through a medium whose density stays the same
B
It's moving from a low density medium to a high density medium.
С
It's moving from a high density medium to a low density medium.
D
It's moving from a low density medium, to a high density medium, and then back to a low density medium
Answer: B
Explanation:
If matter cannot be created or destroyed, then how do you end up with
rust? Below is the equation for rust.
4Fe + 302 → 2Fe203
oxygen from the air
water in the atmosphere
oxygen from in the metal
there shouldn't be any oxygen
Your question is a "non sequitur", which means "it doesn't follow".
Your "then" doesn't contradict your "If", so no mystery is implied.
Maybe you're trying to say that matter is somehow not conserved in the equation . . . 4Fe + 302 → 2Fe203 . But it is. There are 4 Irons and 6 Oxygens on each side, so conservation is not violated here.
I looked up "rust" on Floogle, and got slapped with pages and pages of chemistry that I don't completely understand. But what it's saying is that rusting is a very complex chemical process, AND it doesn't happen unless there's some water involved.
So the bottom line is that there's a lot more going on than simply
4Fe + 302 → 2Fe203 ,
there's water going in and out of the process at every stage, and when it's all over, you have rusty iron, and mass has been conserved.
Soap bubbles can display impressive colors, which are the result of the enhanced reflection of light of particular wavelengths from the bubbles' walls. For a soap solution with an index of refraction of 1.21, find the minimum wall thickness that will enhance the reflection of light of wavelength 711 nm in air.
Answer:
the minimum wall thickness that will enhance the reflection of light is 146.9 nm
Explanation:
Given the data in the question;
At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.
At the second interface, no shift occurs,
condition for constructive interference;
t = ( m + 1/2) × λ/2n
where m = 0, 1, 2, 3 . . . . . .
now, the condition for the constructive interference;
t = mλ/2n
where t is the thickness of the soap bubble, λ is the wavelength of light and n is the refractive index of soap bubble.
so the minimum thickness of the film which will enhance reflection of light will be;
t[tex]_{min[/tex] = ( m + 1/2) × λ/2n
we substitute
t[tex]_{min[/tex] = ( 0 + 1/2) × 711 /2(1.21)
t[tex]_{min[/tex] = 0.5 × 711/2.42
t[tex]_{min[/tex] = 0.5 × 293.80165
t[tex]_{min[/tex] = 146.9 nm
Therefore, the minimum wall thickness that will enhance the reflection of light is 146.9 nm
Investigator Daniels is working on the scene of a suspected arson on a rainy evening. The site of the fire was a mechanic's garage, so there are plenty of accelerants
on the scene. According to the owner of the garage, there had been an argument with a customer the day before about his motorcycle. All the employees left by
6:00 pm. The office was unharmed but there is much damage in the mechanic's bay area and toolboxes. The flames look to have been most concentrated around a
bucket that the owner says was filled with oily rags. She is trying to sift through the evidence and find the relevant facts of the case so far. In order to effectively
Identify the relevant facts, Investigator Daniels creates a word bank. Which word would not fit into her word bank?
Argument
Rainy
Olly rags
Toolboxes
Answer:
Rainy
Explanation:
It is the only one that has nothing to do with the case
Name one similarity and one difference between a set and a bump in volleyball??
One similarity is the use of physical body whereas one difference is that one is exercise and the other is a sport.
One similarity and one differenceOne similarity between a set and a bump in volleyball is the movement and use of legs and hands.
Whereas one difference between a set and a bump in volleyball is that completing several reps of a particular exercise in a row is called a set while on the other hand, the basic pass in volleyball is known as bump.
Learn more about set here: https://brainly.com/question/1090891
What type of weather would a continental Polar air mass bring
Answer:
Continental polar ( cp):
Explanation:
Cold and dry, originating from high latitudes, typically as air flowing out of the polar highs. This air mass often brings the rattleing cold, dry and clear weather on a perfect winter day and also dry and warm weather on a pleasant day in summer.
A 1500-kg car travelling 90 km/h[N] collides with a 1200-kg minivan travelling 40 km/h[S]. After the collision, the two vehicles stick together.
a. Calculate the initial momentum of the car and the minivan.
b. Using the law of conservation of momentum, determine the total momentum of the two vehicles after the collision.
c. Calculate the final velocity of the two vehicles after the collision in metres per second.
Answer:
A) car - 37500 kg*m/s, minivan - 13332 kg*m/s
B) 50832 kg*m/s
C) 18.83 m/s
Explanation:
Realize that sticky collisions are modeled by: m1v1+m2v2=(m1+m2) vf
conevert to m/s....car going 25 m/s, minivan going 11.11 m/s
A) p=mv
p(car)=(1500)(25)
p(car)=37500 kg*m/s
p(minivan)=(1200)(11.11)
p(minivan)=13332 kg*m/s
B) 37500+13332=50832 kg*m/s
C) 37500+13332=(1500+1200) vf
50832=2700(vf)
18.83 m/s = vf