Answer:
every 4 years
Explanation:
Answer:
Every four years
Explanation:
Objects accelerate because
why do we consider market demand as indicator of harvesting raised animal/fish?
Answer:
Following are the solution to this question:
Explanation:
In the given question, the substantial growth throughout stocks and also in agricultural productivity, combined with a growing public understanding of both the important importance of seafood as a food item in a healthy, diversified diet, has led to the upward rise in fish consumption in the last fifty years.
Thorium^+2
Chemical symbol:
Atomic Number:
Mass: 232
# of protons
# of neutrons
Group #
Period #
Answer:
chemical symbol: Th
atomic number:90
protrons :90
neutrons:142
group#:4
period#: 9
Explanation:
you take the atomic weight (232.038)and subtract the atomic number to get (90) which is your neutrons
A monatomic ideal gas with an initial pressure of 500 kPa and an initial volume of 1.80 L expands isothermally to a final volume of 5.20 L. How much work is done on the gas in this process?
A) 1700J
B) 875J
C) 1570J
D) 900J
E) 955J
Answer:
955 J
Explanation:
PV = nRT
500 x 10³ x 1.8 x 10⁻³ = nRT
= 900 J
work done by gas in isothermal expansion
= nRT lnV₂ / V₁
= 900 ln 5.2 / 1.8
= 900 x ln 2.89
= 900 x 1.06
= 955 J
Which of the following is true regarding the speed of earthquake waves?
OA.
S waves travel faster than P waves and surface waves.
ОВ.
Surface waves travel faster than P waves and S waves.
OC.
P waves, S waves, and surface waves all have the same speed.
OD.
P waves travel faster than S waves and surface waves.
Answer:
p waves travel faster than s waves and surface waves
Answer:
p waves travel faster than s waves and surface waves
Explanation:
I took a quiz and got this right.
A 1430 kg is moving at 25.6 m/s when a force is applied, in the direction of the cars motion. The car speeds up to 31.3 m/s. If the force is applied for 5.4 s what is the magnitude of the force
The car accelerates with magnitude a such that
31.3 m/s = 25.6 m/s + a (5.4 s)
→ a = (31.3 m/s - 25.6 m/s) / (5.4 s) ≈ 1.056 m/s²
Then the applied force has a magnitude F of
F = (1430 kg) a ≈ 1500 N
Which change in an object would increase the force needed to move the object?
А.
decreasing the velocity of an object
B
increasing the volume of an object
с
decreasing the mass of an object
D
increasing the mass of an object
Answer:
D i think
Explanation:
Heavier objects (objects with more mass) are more difficult to move and stop.
Answer:
Increasing the mass of the object (option D in the list of answers)
Explanation:
Recall that F = m x a
therefore, if the mass increases, the force increases
Two cylinders each with a 60 cm diameter, thatare closed at one end, open at the other, are joined to form asingle cylinder, then the air inside is removed.
How much force does the atmosphere exert onthe flat end of each cylinder?
Suppose one cylinder is bolted to a sturdy ceiling. How many 90 kg football players would need to hang from the lower cylinder to pull the two cylinders apart
Answer:
a
The force is [tex]F = 2864561.4 \ N[/tex]
b
The number is [tex]N = 3248 \ players[/tex]
Explanation:
From the question we are told that
The of each cylinder is [tex]d = 60 \ cm = 6 \ m[/tex]
The mass of the players is [tex]m = 90 \ kg[/tex]
Generally the cross-sectional area of the cylinder is mathematically represented as
[tex]A = \pi * \frac{d^2}{4}[/tex]
=> [tex]A = 28.3 \ m^2[/tex]
Generally force exerted on the flat end of each cylinder is mathematically represented as
[tex]F = A * P[/tex]
Here P is the atmospheric pressure with value [tex]P = 101300 \ Pa[/tex]
So
[tex]F = 28.3 * 101300[/tex]
=> [tex]F = 2864561.4 \ N[/tex]
Generally the weight of a single football player is
[tex]W = m * g[/tex]
=> [tex]W = 90 * 9.8[/tex]
=> [tex]W = 882\ N[/tex]
Generally the number of player required to pull the two cylinders apart is mathematically represented as
[tex]N = \frac{ F }{W}[/tex]
=> [tex]N = \frac{ 2864561.4 }{882}[/tex]
=> [tex]N = 3248 \ players[/tex]
n
Question 4
1 pts
A bus travels on an interstate highway at an average speed of 90 km/hrs. How far does it take to travel
in 30 mins? The distance equals speed times time, or d = st.
O 45 km
O 98 Km
O 56 km
O 432 Km
[tex]d = s \times t \\ d = 90 \times \frac{30}{60} \\ d = 90 \times \frac{1}{2 } \\ d = 45km[/tex]
A 5.0 kg block is pushed 2.0 m at a con-
stant velocity up a vertical wall by a constant
force applied at an angle of 30.0° with the
horizontal, as shown in the figure.
The acceleration of gravity is 9.81 m/s2.
F
30°
2 m
5 kg
Drawing not to scale.
If the coefficient of kinetic friction between
the block and the wall is 0.40, find
a) the work done by the force on the block.
Answer in units of J.
Answer:
[tex]W_F=127.64283 J[/tex]
Explanation:
Information Given:
[tex]m = 5kg[/tex] [tex]v=constant[/tex]
Key: μ = Kinetic Friction (Kf) θ = Theta α = 180° N = Normal Force
[tex]W_F=F_ydcos[/tex]θ
[tex]W_F=Fdsin[/tex]θ
[tex]_{net}F_y = sin[/tex]θ-μ[tex]N-mg=0[/tex]
[tex]_{net}F_x = 0[/tex]
[tex]N=Fcos[/tex]θ
[tex]Fsin[/tex]θ-μ[tex]N=mg[/tex]
[tex]Fsin[/tex]θ-μ[tex]Fcos[/tex]θ[tex]=mg[/tex]
[tex]F=\frac{mgdsin(theta)}{sin(theta)-(Kf)cos(theta)}[/tex] →[tex]W_F=\frac{mgdsin(theta)}{sin(theta)-(Kf)cos(theta)}[/tex]
[tex]W_F=\frac{(2)(9.81)(2)sin(30)}{sin(30)-(0.40)cos(30)}[/tex]
[tex]W_F=127.64283 J[/tex]
What is the mass number for the following Bohr Model?
e-
e-
e'
P = 11
N = 12
e'
e
Answer:
[tex]\boxed {\boxed {\sf B. \ 23}}[/tex]
Explanation:
The mass number is found by adding up the nucleons in an atom.
The nucleons are the subatomic particles found in the nucleus, so just protons and neutrons.
There are 11 protons and 12 neutrons.
Add them together.
[tex]mass \ number = protons + neutrons[/tex]
[tex]mass \ number= 11+12[/tex]
[tex]mass \ number= 23[/tex]
The mass number for this atom is 23.
A 500 kg car is moving at 30 m/s. The driver sees a barrier ahead. If the car takes 100 m to come to rest, what is the magnitude of the force necessary to stop the car?
How do you solve this question?
Answer:
F = 2250 [N]
Explanation:
In order to solve this problem, we must first use the following equation of kinematics.
[tex]v_{f}^{2} =v_{o}^{2}-2*a*x[/tex]
where:
Vf = final velocity = 0 (come to rest)
Vo = initial velocity = 30 [m/s]
a = acceleration or desaceleration [m/s²]
x = distance = 100 [m]
[tex](0)=30^{2} -2*a*100\\900 = 200*a\\a = 4.5 [m/s^{2}][/tex]
Now we must use the following equation of kinetics, which is based on Newton's second law that explains that the sum of forces on a body is equal to the product of mass by acceleration.
∑F = m*a
where:
F = force [N]
m = mass = 500 [kg]
a = acceleration = 4.5 [m/s²]
[tex]F = 500*4.5\\F = 2250 [N][/tex]
There is a bell at the top of a tower that is 45 m high. The bell weighs 190 N. The bell has ____________ energy.
Answer:
The bell has 8,550 Joule energy.
Explanation:
Gravitational Potential Energy
Gravitational potential energy is the energy stored in an object because of its height in a gravitational field.
It can be calculated with the equation:
U=m.g.h
Where:
m = mass of the object
h = height
g = acceleration of gravity, or [tex]9.8 m/s^2[/tex]
Since the weight of an object of mass m can be calculated as:
W = m.g
The gravitational potential energy is:
U = W.h
The bell of weight W=190 N at the top of a tower is h=45 m high. Thus its energy is:
U = 190 N . 45 m
U = 8,550 Joule
The bell has 8,550 Joule energy.
A 30 N force toward the west is applied to an object. The object moves 50 m east during the time the force is applied. What is the change in kinetic energy of the object?
a) 1.0 J
b) 750 J
c) 1.7 J
d) -1500 J
Answer:
D.-1500Joules
Explanation:
The change in kinetic energy of the object s equivalent to the workdone by the body in the west direction (negative x direction)
Workdone = Force * Distance
Given
Force = 30N
Distance moved by the object = 30m
Required
Kinetic energy
Kinetic energy = 30 * 50
Kinetic energy = 1500Joules
Since the body moves in the negative direction, hence the kinetic energy will be -1500Joules
As the building collapses, the volume of air inside the building decreases, while the mass of the air stays the same. This means that the _____ of the air inside the building will increase.
Answer:
Density
Explanation:
What would its weight be on Jupiter?
24.9N
Answer:
1.898 × 10^27 kg
Explanation:
thats how much it ways
A 100kg couch is being pushed with 196N of force. As it slides along the ground it experiences a coefficient of friction of 0.1. What is the net force in this situation?
A 300N
B 202N
C 398N
D 98N
Answer:98
Explanation:hope this helps!
The scientific term describing the ball changing position as it goes from
Steph Curry's hand, into the air, and through the hoop.
A= velocity
B= speed
C= vector
D= motion
Answer:
Vector
Explanation:
Vector is a quantity that shows the direction or path through which a body travels with as it changes position.
As body travels, the direction sometimes changes and this is described by the vector of the body.
Velocity is a vector quantity that describes the displacement per unit of time of a body.
Speed is a scalar that deals with the distance covered per time
So, a vector specifies the magnitude of a physical quantity and also the direction through which it travels.
Professional baseball pitchers deliver pitches that can reach the blazing speed of 100 mph (miles per hour). A local team has drafted an up-and-coming, left-handed pitcher who can consistently pitch at 42.24 m/s (94.50 mph).
A. Assuming a pitched ball has a mass of 0.1420 kg and has this speed just before a batter makes contact with it, how much kinetic energy does the ball have?
B. How high would the ball need to be dropped from to attain the same energy (neglect air resistance)?
Answer:
A. ) K =126. 7 J
B. ) h= 91.1 m.
Explanation:
A)
Assuming no air resistance, once released by the pitcher, the speed must keep constant through all the trajectory, so the kinetic energy of the ball can be expressed as follows:[tex]K = \frac{1}{2}*m*v^{2} = \frac{1}{2}*0.142 kg*(42.24m/s)^{2} = 126.7 J (1)[/tex]
B)
Neglecting air resistance, total mechanical energy must be the same at any point, so, if we choose the ground level as the zero reference level for the gravitational potential energy, and assuming that the ball attains this kinetic energy just before striking ground, this value must be equal to the gravitational potential energy just before be dropped, so we can write the following equality:[tex]U_{o} = K_{f} = 126. 7 J (2)[/tex]
⇒ m*g*h = 126. 7 J
Solving for h, we get:[tex]h = \frac{K_{f}}{m*g} = \frac{126.7J}{0.1420kg*9.8m/s2} = 91.1 m (3)[/tex]
When liquid water gets into cracks of rock and freezes, it __ and ___.
What is the minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any? The distance from the handle to the bottom of the bucket is 35 cm.
Complete Question
What is the minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any? The distance from the handle to the bottom of the bucket is 35 cm. The student has 70-cm-long arms
Answer:
The value is [tex]w__{rpm} } = 29.17 \ rpm[/tex]
Explanation:
From the question we are told
The distance from the handle to the bottom of the bucket is [tex]d = 35 \ cm = 0.35 \ m[/tex]
The length of the students arm is L = 70 cm = 0.70 m
Generally the acceleration due to gravity experienced by the bucket of water is mathematically represented as
[tex]g = w^2 * r[/tex]
Here is is the radius of the circle which swinging of the bucket makes and this is mathematically represented as
[tex]r = L + d[/tex]
So
[tex]g = w^2 * ( L + d )[/tex]
= > [tex]w = \sqrt{\frac{g }{ L + d } }[/tex]
= > [tex]w = \sqrt{\frac{ 9.8}{ 0.7 + 0.35} }[/tex]
= > [tex]w = 3.055 \ rad/s[/tex]
Generally the angular speed in revolution per minute is mathematically represented as
[tex]w__{rpm} } = \frac{w * 60 }{2 \pi }[/tex]
=> [tex]w__{rpm} } = \frac{3.055 * 60 }{2 * 3.142 }[/tex]
=> [tex]w__{rpm} } = 29.17 \ rpm[/tex]
3) A 10kg object rests on a frictionless surface when it is struck by a 300N force. At what rate will it accelerate?
3m/s/s
30m/s/s
0.3m/s/s
300m/s/s
Answer: 0.3m/s/s
(i'm really sorry if i'm wrong)
:(
As a bicycle is ridden west in a straight line with decreasing speed,the acceleration of the bicycle must be
Answer:
Decreasing
Hope this helps! :)
Help ASAP plz and thx u
Answer:
a). a = F/m
Explanation:
Formula is F=ma
What is the distance between a 900 kg compact car and a 1600 kg pickup truck if the gravitational force between them is about 0.0001 N?
Answer:
The distance is 0.96m
Explanation:
Given
m1= 900kg
m2= 1600kg
Force F= 0.0001nN
G=6.67430*10^-11 Nm^2/kg^2
Required
The distance r
Step two:
the formula for the force is given as
F = Gm1m2/r2
make r subject of the formula
[tex]r= \sqrt{\frac{Gm1m2}{F} }[/tex]
[tex]r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m[/tex]
Answer:
The distance is 0.96m
Explanation:
Given
m1= 900kg
m2= 1600kg
Force F= 0.0001nN
G=6.67430*10^-11 Nm^2/kg^2
Required:
The distance r
Step two:
the formula for the force is given as
F = Gm1m2/r2
make r subject of the formula
[tex]r= \sqrt{\frac{Gm1m2}{F} }[/tex]
[tex]r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m[/tex]
Answer:
The distance between the compact car and pickup truck is 0.96048 m
Explanation:
The gravitational force is directly proportional to the product of the masses of the interacting object, it is also inversely proportional to the square of the distance between them. This is shown in equation 1;
[tex]F =G \frac{m_{1} X m_{2} }{d^{2} }[/tex]............ 1
Where F is the gravitational force = 0.0001 N
G is the gravitational constant = 6.673 x [tex]10^{-11} Nm^{2} kg^{-2}[/tex]
[tex]m_{1}[/tex] is the mass of the compact car = 900kg
[tex]m_{2}[/tex] is the mass of the pickup truck = 1600kg
d is the distance and its unknown ?
Let us make d the subject formula in equation 1
[tex]d = \sqrt{G\frac{m_{1} m_{2} }{F } }[/tex] .... 2
Substituting into equation 2 we have
[tex]d = \sqrt{\frac{6.673x10^{-11} x 900 x 1600}{0.0001N} }[/tex]
d = 0.96048m
Therefore the distance between the compact car and pickup truck is 0.96048 m
Question 1 of 20
Which statement best describes the effect of the magnet on the block of
material next to it?
is N
O
A. The magnet has magnetized the center of the block.
U
B. The magnet has magnetized the right side of the block.
C. The magnet has magnetized the whole block.
ОО
D. The magnet has magnetized the left side of the block.
Answer:
B. The magnet has magnetized the right side of the block.
Explanation:
a pe x
How can you prove that the potential energy of a stretched spring turns into kinetic energy when you release the spring?
Potential energy+Kinetic energy=Total energy
When you release a spring the velocity increases, therefore the kinetic energy increases ke=1/2*mv^2 and the displacement decreases therefore the potential energy decreases pe=1/2*kx^2.
A diffraction grating with 68 slits per cm is used to measure the wavelengths emitted by hydrogen gas.
A. At what angles in the fourth-order spectrum would you expect to find the two violet lines of wavelength 434 nm and of wavelength 410 nm?
B. What are the angles if the grating has 12,800 slits per cm?
Answer:
a
[tex]\theta _1 =0.687 ^o[/tex]
[tex]\theta _2 =0.630 ^o[/tex]
b
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.22 will not be valid
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.1 will not be valid
Explanation:
From the question we are told that
The slit grating is [tex]N = 68 \ slits / cm = 6800 \ slits / m[/tex]
The order of spectrum is [tex]n = 4[/tex]
Generally the width of the slit is mathematically represented as
[tex]a = \frac{1}{ 6800}[/tex]
=> [tex]a = 0.000147 \ m[/tex]
Generally the condition for constructive interference is
[tex]asin\theta = n * \lambda[/tex]
Now for the first wavelength the angle is evaluated as
[tex]\theta _1 = sin ^{-1} [ \frac{n \lambda_1 }{a} ][/tex]
=> [tex]\theta _1 = sin ^{-1} [ \frac{4* 434 *10^{-9} }{ 0.000147 } ][/tex]
=> [tex]\theta _1 =0.687 ^o[/tex]
Now for the second wavelength the angle is evaluated as
[tex]\theta _2 = sin ^{-1} [ \frac{n \lambda_2 }{a} ][/tex]
=> [tex]\theta _2 = sin ^{-1} [ \frac{4* 410 *10^{-9} }{ 0.000147 } ][/tex]
=> [tex]\theta _2 =0.630 ^o[/tex]
Gnerally if grating is [tex]N = 12800 \ slits per cm = 1280000 \ slits / m[/tex]
Generally the width of the slit is mathematically represented as
[tex]a = \frac{1}{ 1280000}[/tex]
=> [tex]a = 7.813 *10^{-7} \ m[/tex]
Generally the condition for constructive interference is
[tex]asin\theta = n * \lambda[/tex]
Now for the first wavelength the angle is evaluated as
[tex]\theta _1 = sin ^{-1} [ \frac{n \lambda_1 }{a} ][/tex]
[tex]\theta _1 = sin ^{-1} [ \frac{4* 434 *10^{-9} }{ 7.813*10^{-7} } ][/tex]
=> [tex]\theta _1 = sin ^{-1} [ 2.22][/tex]
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.22 will not be valid
=> [tex]\theta _1 =0.687 ^o[/tex]
Now for the second wavelength the angle is evaluated as
[tex]\theta _2 = sin ^{-1} [ \frac{n \lambda_2 }{a} ][/tex]
=> [tex]\theta _2 = sin ^{-1} [ \frac{4* 410 *10^{-9} }{ 7.813*10^{-7} } ][/tex]
=> [tex]\theta _2 = sin ^{-1} [2.1 ][/tex]
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.22 will not be valid
An athlete stretches a spring an extra 40.0 cm beyond its initial length. How much energy has he transferred to the spring, if the spring constant is 52.9 N/cm?
a) 4230 kJ
b) 4230 J
c) 423 kJ
d) 423 J
Answer:
b) 4230 JExplanation:
Step one:
given data
extension= 40cm
Spring constant K= 52.9N/cm
Step two:
Required
the Kinetic Energy KE
the expression to find the kinetic energy is
KE= 1/2ke^2
substituting our data we have
KE= 1/2*52.9*40^2
KE=0.5*52.9*1600
KE= 42320Joules
The answer is b) 4230 J
Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular acceleration of 0.311 rad/s2 for 4.13 s. What is the drill's angular displacement during that time interval?
Answer:
The drill's angular displacement during that time interval is 24.17 rad.
Explanation:
Given;
initial angular velocity of the electric drill, [tex]\omega _i[/tex] = 5.21 rad/s
angular acceleration of the electric drill, α = 0.311 rad/s²
time of motion of the electric drill, t = 4.13 s
The angular displacement of the electric drill at the given time interval is calculated as;
[tex]\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad[/tex]
Therefore, the drill's angular displacement during that time interval is 24.17 rad.