Higher temperatures or lower activation energies result in more sufficiently energetic collisions, which increases the fraction of such collisions, increases the value of k, and increases reaction rates.
What are the four components of the collision theory?The consequences are explained by the collision theory. These variables include temperature, catalysts, surface area, concentration, and reactant type. Each of these influences raises the number and energy of collisions, which increases reaction rate.
What among the three crash theories holds true here?The hypothesis states that responding particles usually collide without reacting. For interactions to take place, reacting particles need to collide with enough energy, in the appropriate direction, and with enough mass.
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The wireless system is operating at 2GHz. A base station and a mobile unit are separated by 16km. The maximum gain of the transmitting antenna at the base station is 20dB. The input power to the transmitter is 100W, and the power received by the antenna is 5nW. The antennas are aligned, and there are no re ections or loses. What will be the received power if the distance between the mobile unit and the base station increases to 20km?
If everything else remains constant, then the received signal power is simply an inverse-square function of its distance from the transmitter.
That is, down 6dB when distance is doubled etc.
If distance increases from 16 to 20, then received power decreases by the factor of (16/20)^2.
That's (0.8)^2 = 0.64
New receive power is 5x0.64 = 3.2 nW
=. ===== ==========
Here's how I do it at my job:
Initial RSL = 5nW ~ - 53 dBm
loss = 20log(20/16)= 20log(1.25)~1.94dB
New RSL = - 54.94 dBm or ~ 3.2 nW .
The unit J/Pa is equivalent to
(a) m³
(b) cm³
(c) dm³
(d) None of these
Answer:
(a) m³
Explanation:
Joule is the unit of work and Pascal is unit of pressure.
J/Pa = work/pressure = Nm/Nm⁻² = m³
Thus, The unit J/Pa is equivalent to m³
-TheUnknownScientist
HELP 30 POINTS
The picture above shows a cross section of the Earth’s interior including both oceanic crust and continental crust. Along the seafloor, there are mid–ocean ridges where magma rises to the surface and creates new crust. There are also areas where crust is recycled as oceanic crust goes below continental crust. What type of boundary would occur at a mid-ocean ridge?
Mid-ocean ridges happen along divergent plate boundaries, where new ocean floor is created as the Earth’s tectonic plates spread apart. As the plates separate, molten rock rises to the seafloor, producing large volcanic eruptions of basalt.
PLEASE HELP! I'LL GIVE BRAINLEST
PLEASE HELP !!!!!!!!!!
Answer:
1
Explanation:
scholastic science world
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A fisherman sitting on the end of a pier notices that 6 wave crests pass him in 3 seconds. What is the frequency of the waves?
9.0 Hz
18.0 Hz.
20 Hz
4.5 Hz
The frequency of the waves will be 18 Hz. Then the correct option is B.
What is the frequency?The frequency of a repeated event is its number of instances per unit of time. It differs from angular frequency and is sometimes referred to as temporal frequency for clarification. One event occurs per second when measuring frequency in hertz.
Frequency is the number that represents the number of oscillations or intervals each second. The hertz is the SI unit for frequency (Hz). One cycle per second equals one hertz.
The number of finished waves produced each second is considered the wave frequency.
Six wave crests pass a fisherman sitting at the end of a pier in the span of three seconds.
The frequency of the waves is given as,
f = 6 x 3
f = 18 Hz
The waves will have a frequency of 18 Hz. Then, choice B is the best one.
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Which diagram best represents the gravitational forces, F, be-
tween a satellite, S, and Earth?
Answer:
Diagram (3).
Explanation:
N3L states that if object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A ([tex]F_{A} = -F_{B}[/tex]).
The diagram which best represents the gravitational forces, F, between a satellite, S, and Earth is; Choice (3).
The Newton's law of gravitation states that the Force of attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of their distance apart.The law clearly states a Force of attraction; the two objects come towards each other.
Consequently, Choice (3) best represents the gravitational forces, F, between a satellite, S, and Earth.
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Energy associated with moving objects or that could move later is?
a bag of sugar has a mass of 2.26 kg. what is its weight
Answer:
2260
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PHYSICS HELP
PLEASE HELP ITS ABOUT ATWOOD MACHINES
Answer:
7.23407 [tex]\frac{m}{s^2}[/tex]
Explanation:
(I will not include units in calculations)
I'm assuming FBD's are already drawn, so I will work from there.
Let the 2.2kg block equal [tex]m_2[/tex], and the 20kg block equal [tex]m_1[/tex].
Summation equation for [tex]m_2[/tex]: [tex]\sum F_x=F_t_2-(F_f+F_g_x)=m_2a[/tex], [tex]\sum F_y=F_n-F_g_y=0[/tex]
Summation equation for [tex]m_1[/tex]: [tex]\sum F_y=F_g-F_t_1=m_1a[/tex]
Torque Summation Equation: [tex]\sum\tau=F_t_1*r-F_t_2*r=I\alpha[/tex]
Do some plugging in with the values given: [tex]\sum\tau=F_t_1*r-F_t_2*r=.5Mr^2\alpha[/tex]
Replace [tex]\alpha[/tex] with [tex]\frac{a}{r}[/tex], and cancel out the r's.
[tex]\sum\tau=F_t_1-F_t_2=.5Ma[/tex]
This step is important: Rearrange the force summation equation to solve for each tension force.
[tex]F_t_2=m_2a+F_f+F_g_x\\F_t_1=m_1g=m_1a[/tex]
Perform Substitution: [tex]\sum\tau=m_1g-m_1a-(m_2a+F_f+F_g_x)=.5Ma[/tex]
Now, we need to find the friction force and the horizontal component of the force of gravity.
Note that [tex]F_f=[/tex]μ[tex]F_n[/tex]
And based on our earlier summation equation: [tex]F_n=F_g_y[/tex]
First, break [tex]F_g[/tex] into x and y components. [tex]F_g_y=F_g\cos(\theta)[/tex], [tex]F_g_x=F_g\sin(\theta)[/tex]
Perform substitution with this and the fact that [tex]F_g=mg[/tex].
[tex]\sum\tau=m_1g-m_1a-(m_2a+\mu*m_2g\cos(\theta)+m_2g\sin(\theta))=.5Ma[/tex]
Solving for a, plugging in numbers yields an answer of 7.23407 [tex]\frac{m}{s^2}[/tex]
Answer:
7.23407
Explanation:
easy
Niobium metal becomes a superconductor when cooled below 9K. Itssuperconductivity is destroyed when the surface magnetic fieldexceeds 0.100 T. Determine the maximum current a 2.00-mm-diameterniobium wire can carry and remain superconducting, in the absenceof any external magnetic field.
Answer:
the maximum current is 500 A
Explanation:
Given the data in the question;
the B field magnitude on the surface of the wire is;
B = μ₀i / 2πr
we are to determine the maximum current so we rearrange to find i
B2πr = μ₀i
i = B2πr / μ₀
given that;
diameter d = 2 mm = 0.002 m
radius = 0.002 / 2 = 0.001 m
B = 0.100 T
we know that permeability; μ₀ = 4π × 10⁻⁷ Tm/A
so we substitute
i = (0.100)(2π×0.001 ) / 4π × 10⁻⁷
i = 500 A
Therefore, the maximum current is 500 A
Instead of changing the focal length of the lens, the eyes of amphibians work in a different manner: a set of muscles changes the shape of the eye which increases the distance between the front of the eye and the retina. The world's largest frog, the Goliath frog of west Africa, has an eye with a maximum size similar to a human's: 2.5cm. However, unlike a human, where the focal length is also 2.5cm, the focal length of the Goliath frog's eye is 2.146 cm. What is the maximum distance this frog can see
Answer:
The answer is "15.56 cm".
Explanation:
[tex]v= 2.5 \ cm\\\\f= 2.154 \ cm[/tex]
Calculating object of length is x so:
[tex]u= -x[/tex]
Using formula:
[tex]\to \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\\\to \frac{1}{2.5}-\frac{1}{-x}=\frac{1}{2.154}\\\\\to \frac{1}{x}=\frac{1}{2.154}-\frac{1}{2.5}\\\\\to \frac{1}{2.5}-\frac{1}{-x}=\frac{1}{2.154}\\\\\to x= 15.56 \ cm[/tex]
You need to load a crate of mass m onto the bed of a truckon earth. One possibility is just to lift the crate straight up over a height h, equal to height of the truck's bed. Theforce exert in this case is F1 and the work done in this case is W1. The other possibility is to slide the crate up the frictionless ramp of length L.. In this case you exert force F2 and perform work W2. Which statement is true?
a. F1 > F2 and W1 > W2
b. F1 = F2 and W1 > W2
c. F1 = F2 and W1 = W2
d. F1 > F2 and W1< W2
e. F1 > F2 and W1 = W2
Answer:
The correct answer is - option E. e. F1 > F2 and W1 = W2
Explanation:
Case 1 - F1 = mg, m= mass and g = gravitational force
Work done W1 = F1h= mgh (h - height of bed of truck)
Case 2 - F2 = mgsinθ
work done W2 = mgsinθ×L (L=length of ramp ans sin θ angle of ramp)
sinθ = h/l, then W2 = mg(h/L)L
W2 = mgh
By comparing both,
F2 = mgsinθ and F1 = mg therefore,
F2>F1
now, W2 = mgh, and W1 = mgh
therefore, W2 = W1.
does altitude has an effect on weight? HELP
Answer: lose weight at high altitudes.
Explanation:
Answer:
Just a week at high altitudes can cause sustained weight loss, suggesting that a mountain retreat could be a viable strategy for slimming down. Overweight, sedentary people who spent a week at an elevation of 8,700 feet lost weight while eating as much as they wanted and doing no exercise
what is an axis in social
Answer:
An axis is an invisible line around which an object rotates, or spins. The points where an axis intersects with an object's surface are the object's North and South Poles.
Brian Lara is a cricketer playing in the field on the second day of a cricket test-match. He exerts a forward force on the 0.145kg cricket ball, as he catches it, to bring it to rest from a speed of 38.2m/s. During the process, his hand recoils a distance of 0.135m. Determine the acceleration of the ball and the force which is applied to it by Brian Lara.
Answer:
a = -3984.6 m/s²
F = 577.76 N
Explanation:
The acceleration of the ball can be calculated by using the third equation of motion:
[tex]2as = v_f^2 - v_i^2\\[/tex]
where,
a = acceleration of ball = ?
s = distance covered = recoil distance = 0.135 m
vf = final speed = 0 m/s
vi = initial speed = 38.2 m/s
Therefore,
[tex]2(0.135\ m)a = (0\ m/s)^2-(38.2\ m/s)^2\\[/tex]
a = -3984.6 m/s²
here negative sign shows deceleration.
Now, for the force applied by Brian Lara will be equal in magnitude but opposite in direction of the force required to stop the ball:
[tex]F = -ma\\F = -(0.145\ kg)(-3984.6\ m/s^2)\\[/tex]
F = 577.76 N
Two vectors are given as A⃗ = 2i^ + 3j^ − 3k^ and B⃗ = -1i^ + 5j^ + 3k^. Find A⃗ ⋅ B⃗
Answer:
A·B = 4
Explanation:
Given that,
Vector A = 2i+3j-3k
Vector B = -i+5j+3k
We need to find the value of A·B.
We know that,
i·i=j·j=k·k = 1 and i·j=j·k=k·i=0
So,
[tex]A\cdot B=(2i+3j-3k)\cdot (-i+5j+3k)\\\\=-2+5(3)+(-3)(3)\\\\=-3+15-9\\\\=4[/tex]
So, the value of A·B is equal to 4.
A toy car of mass 2.0kg is made to move in a circular track of radius 10.0m. If the centripetal force is 800.0N, calculate the angular velocity of the car
Answer:
w = 6.325 rad/s
Explanation:
Given the following data;
Mass = 2kg
Radius, r = 10m
Centripetal force = 800N
To find the angular velocity of the car;
First of all, we would use the following formula to solve for the speed of the car;
Centripetal force = mv²/r
800 = 2*v²/10
Cross-multiplying, we have;
8000 = 2v²
4000 = v²
Taking the square root of both sides, we have;
v = √4000
v = 63.25 m/s
Next, we find the angular velocity, w using the formula below;
w = v/r
w = 63.25/10
w = 6.325 rad/s
Chemical energy may be released during a chemical reaction in the form of heat ..... True or false?
Answer:
heat and gas sometimes so true
A wheel has an angular speed of 32 rev/s when it experiences a constant angular acceleration of 65 rev/s2 which causes it to spin FASTER. During this time the wheel completes 92 rev. Determine how long the wheel was experiencing this angular acceleration and how fast the wheel was spinning at the end of this period. Assume that the wheel doesn't change the direction of its spin.
Answer:
ωf = 113.95 rev/s
t = 1.26 s
Explanation:
We can use the third equation of motion to find out the final spinning speed of the wheel:
[tex]2\alpha \theta = \omega_f^2 -\omega_i^2\\[/tex]
where,
α = angular acceleration = 65 rev/s²
θ = No. of revolutions completed = 92 rev
ωf = final angular speed = ?
ωi = initial angular speed = 32 rev/s
Therefore,
[tex](2)(65\ rev/s^2)(92\ rev) = \omega_f^2 - (32\ rev/s)^2\\\omega_f^2 = 11960\ rev^2/s^2 + 1024\ rev^2/s^2\\\omega_f = \sqrt{12984\ rev^2/s^2}[/tex]
ωf = 113.95 rev/s
Now, for the time we can use the first equation of motion:
[tex]\omega_f = \omega_i +\alpha t\\113.95\ rev/s - 32\ rev/s = (65\ rev/s^2)t\\t = \frac{81.95\ rev/s}{65\ rev/s^2}\\\\[/tex]
t = 1.26 s
A 3 kg block collides with a massless spring of spring constant 98 N/m attached to a wall. The speed of the block was observed to be 1.5 m/s at the moment of collision. The acceleration of gravity is 9.8 m/s 2 . How far does the spring compress if the surface on which the mass moves is frictionless
Answer: 0.83 m
Explanation:
Given
mass of the block is m=3 kg
spring constant k=98 N/m
The Speed at the time of collision is v=1.5 m/s
Here, the kinetic energy of the block is converted into Elastic potential energy
[tex]\Rightarrow \dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2 \\\\\Rightarrow 3\times 1.5^2=98\times x^2\\\\\Rightarrow x^2=0.6887\\\\\Rightarrow x=0.829\approx 0.83\ m[/tex]
Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. Suppose the mass of a certain spherical neutron star is twice the mass of the Sun and its radius is 13.0 km. Determine the greatest possible angular speed the neutron star can have so that the matter at its surface on the equator is just held in orbit by the gravitational force. (The mass of the Sun is 1.99 1030 kg.)
Answer:
The required angular speed the neutron star is 10992.32 rad/s
Explanation:
Given the data in the question;
mass of the sun M[tex]_S[/tex] = 1.99 × 10³⁰ kg
Mass of the neutron star
M[tex]_N[/tex] = 2( M[tex]_S[/tex] )
M[tex]_N[/tex] = 2( 1.99 × 10³⁰ kg )
M[tex]_N[/tex] = ( 3.98 × 10³⁰ kg )
Radius of neutron star R[tex]_N[/tex] = 13.0 km = 13 × 10³ m
Now, let mass of a small object on the neutron star be m
angular speed be ω[tex]_N[/tex].
During rotational motion, the gravitational force on the object supplies the necessary centripetal force.
GmM[tex]_N[/tex] = / R[tex]_N[/tex]² = mR[tex]_N[/tex]ω[tex]_N[/tex]²
ω[tex]_N[/tex]² = GM[tex]_N[/tex] = / R[tex]_N[/tex]³
ω[tex]_N[/tex] = √(GM[tex]_N[/tex] = / R[tex]_N[/tex]³)
we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²
we substitute
ω[tex]_N[/tex] = √( ( 6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)
ω[tex]_N[/tex] = √( 2.65466 × 10²⁰ / 2.197 × 10¹²
ω[tex]_N[/tex] = √ 120831133.3636777
ω[tex]_N[/tex] = 10992.32 rad/s
Therefore, The required angular speed the neutron star is 10992.32 rad/s
Which waves can travel through space?
a. Electromagnetic waves only
b. Mechanical waves only
c. Electromagnetic and mechanical waves
d. Longitudinal and electromagnetic waves
Answer:
electromagnetic waves only
Explanation:
I just took the test, Hope it helps!
Answer:
A: Electromagnetic waves only
Explanation:
I need help please only way to put my grade up !!!!!! Would appreciate it !!! Someone who’s good at this
Answer:
Total energy = 1000J
KE = 500J
PE = 500J
Explanation:
As you may know, the equation for gravitational potential energy is mgh (weight x height)
If the skateboard is halfway down, that means it is at half the height. As the skateboard speeds up (as it goes downward), the potential energy becomes kinetic energy. Since it has 500J of kinetic energy at half way down, it means it had double that amount of Potential energy at the top (1000J). Since half of that became kinetic energy, there is only 500J of PE left.
Total energy = KE + PE = 1000J
KE = 500J
PE = 500J
can someone please help me !!!!
Answer:
it's A subduction, deep water trench
A sphere with the same mass and radius as the original cylinder, but a smaller rotational inertia, is released from rest from the top of the ramp. KS and KC are the sphere's and cylinder's total kinetic energy at the bottom of the ramp, respectively. How do KS and KC compare, and why
Answer:
The Kinetic energy of Sphere is higher than the cylinder.
( KS > KC )
Explanation:
Given - A sphere with the same mass and radius as the original cylinder, but a smaller rotational inertia, is released from rest from the top of the ramp. KS and KC are the sphere's and cylinder's total kinetic energy at the bottom of the ramp, respectively.
To find - How do KS and KC compare, and why ?
Proof -
We know that,
The total energy of an object = Potential energy + linear kinetic energy + rotational kinetic energy.
⇒E = mgh + [tex]\frac{1}{2} mv^{2}[/tex] + [tex]\frac{1}{2} l\omega^{2}[/tex]
Now,
Mass of sphere = m
Radius of sphere = r
So,
The moment of inertia of a uniform solid sphere = [tex]\frac{2}{5} mr^{2}[/tex]
Also,
Mass of cylinder = m
Radius of cylinder = r
So,
The moment of inertia of a uniform solid cylinder = [tex]\frac{1}{2} mr^{2}[/tex]
Now,
Total energy for the sphere , Es = mgh + [tex]\frac{7}{10} mv^{2}[/tex]
Total energy for the cylinder, Ec = mgh + [tex]\frac{3}{4} mv^{2}[/tex]
As they always have the same total energy,
So, for height h of the sphere's velocity has to be higher.
Therefore,
The Kinetic energy of Sphere is higher than the cylinder.
Answer:
KS < KC
Explanation:
What my fav food for 20 points if you know it!?
Answer:
pizza
Explanation:
Answer:
sea food???
Explanation:
ian pushed a piano across the room with the correct amount of force. Which of newton’s laws is this?
Newton’s 1st, 2nd, 3rd, or 1st and 2nd law?
Answer:
1st and 2nd
Explanation:
3) 4 electrons are placed - one electron per corner - at the corners of a square of side 1 meter. One fixed proton is placed in the middle of the square.The 4 electrons are held in place by some mechanism. The 4 electrons are released by the mechanism at the same time. They move and reach the corners of a square of side 0.8 meters, and keep on moving . Find the velocity of each electron at the corners of the square of side 0.8 meters.
Explanation:
3
i believe that they are all going at 3.2 meters each, I did 4 times 0.8
The velocity of each electron at the corners of the square is 15.92 m/s.
The given parameters;
charge of electron, q = 1.6 x 10⁻¹⁹ Clength of the square, L = 0.8 mThe diagonal length of the square is calculated as;
[tex]d^2 = 0.8^2 + 0.8^2\\\\d = \sqrt{0.8^2 + 0.8^2} \\\\d = 1.13 \ m[/tex]
The distance of each corner charge and the middle charge is calculated as;
[tex]r = \frac{1.13}{2} \\\\r = 0.565 \ m[/tex]
The force between each corner charge and the middle charge is calculated as;
[tex]F= \frac{kq^2}{r^2}[/tex]
The centripetal force on each charge moving around the square is calculated as;
[tex]F = \frac{mv^2}{r}[/tex]
solve the forces together;
[tex]\frac{mv^2}{r} = \frac{kq^2}{r} \\\\v^2 = \frac{kq^2}{m} \\\\v = \sqrt{ \frac{kq^2}{m} } \\\\v = \sqrt{ \frac{(9\times 10^9) \times (1.602\times 10^{-19})^2}{9.11 \times 10^{-31}} } \\\\v = 15.92 \ m/s[/tex]
Thus, the velocity of each electron at the corners of the square is 15.92 m/s.
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