Answer: A negatively-charged ion always has more electrons than protons
Explanation:
First, we know that the elementary negative charge is the electron, while the positive one is the proton. Such that both have the same charge in magnitude, but a different sign. Such that if we have the same number of electrons and protons in an atom, the charge of this atom will be neutral.
And an ion is an atom with a different number of electrons and protons, so the charge of the atom is not neutral.
Then if we have a negatively-charged ion, the charge of this atom is negative. Then we must have a larger number of electrons (the negative ones) than protons (the positive ones)
Then the correct option is:
A negatively-charged ion always has more electrons than protons
When can a high speed velocity cause damage?'
Answer:
50 Mph.
Explanation:
According to the National Severe Storms Laboratory, winds can really begin to cause damage when they reach 50 mph. But here’s what happens before and after they reach that threshold, according to the Beaufort Wind Scale (showing estimated wind speeds): - at 19 to 24 mph, smaller trees begin to sway.
it serves as the fuctional unit of the nervous system
Answer:
i would say the neuron
how to make measurements of length, volume and time?
Answer:
The volume of a regular object can be calculated by multiplying its length by its width by its height. Since each of those is a linear measurement, we say that units of volume are derived from units of length.
Explanation:
Answer:
length×Width×Height
Explanation:
Length×Width×Height is the formula for volume
3. A bee flies forward at 4.9 for 33 s, lands on a flower and stays there for 7s, then flies back along its previous route at 1.9 for 39 s. What is the average speed of the bee
during the entire time?
O 1.701
O 3.001
O 4.466
O 3.801
None of these is correct.
Answer:
None of these is correct.
Explanation:
The average speed can be derived from the sum of the total distance traveled and the total time taken.
Total distance = 4.9 + 1.9 = 6.8
Total time taken = 33 + 39 = 72
So;
Average speed = [tex]\frac{total distance}{total time}[/tex] = [tex]\frac{6.8}{72}[/tex] = 0.014
None of the answer choices given is correct.
whats the result of rounding 52.8015 into five significant figures
Answer:
52.802
Explanation:
"Significant figures" in Mathematics refer to the digits that give accuracy to the value of a measurement. There are specific rules when it comes to determining the significant figures. For example, all non-zero digits are considered significant and zeroes located in-between non-zero numbers are significant. In the number given above, the digit "0" is located between "8" and "1," therefore, it is significant. All the digits above are significant.
The problem is only asking for "five" significant figures. We can do this by counting from the left to the right. By this means, we know that the number will be rounded off to the nearest thousandths, which is "1." The number after 1 is 5, which means that 1 digit will be added to number 1, thus, making the digit into "2." The last digit (5) will then be removed.
Explanation:
five significant of 52.8015=52.801 ..
How would I solve this? It's Newton's 2nd law
Answer:
5.65 m/s²
Explanation:
We'll begin by calculating the mass of PJ when in San Diego (i.e Earth). This can be obtained as follow:
Weight of PJ on Earth (Wₑ) = 545 N
Acceleration due to gravity (g) on Earth (gₑ) = 10 m/s²
Mass of PJ on Earth (mₑ) =.?
Wₑ = mₑ × gₑ
545 = mₑ × 10
Divide both side by 10
mₑ = 545 / 10
mₑ = 54.5 Kg
Thus, the mass of PJ on San Diego (i.e Earth) is 54.5 Kg
Finally, we shall determine the acceleration due to gravity of planet Koja. This can be obtained as follow:
Weight of PJ on Koja (Wₖ) = 308 N
Mass of PJ on Koja (mₖ) = mass of PJ on Earth (mₑ) because mass is constant irrespective of location.
Mass of PJ on Earth (mₑ) = 54.5 Kg
Mass of PJ on Koja (mₖ) = 54.5 Kg
Acceleration due to gravity of on Koja (gₖ) =?
Wₖ = mₖ × gₖ
308 = 54.5 × gₖ
Divide both side by 54.5
gₖ = 308 / 54.5
gₖ = 5.65 m/s²
Thus, the acceleration due to gravity on planet Koja is 5.65 m/s²
An empty airplane with a mass of 200,000 kg must have a speed of 82 m/s to achieve takeoff. Once it is fully loaded, the airplane has a mass of 350,000 kg. It has 3200 m of runway. a) How much force is needed to get the full airplane safely in the air?
b) How much runway would the empty airplane use if its engines generated the same force?
Answer:
a) A force of 367718.75 newtons is needed to get the full airplane safely in the air.
b) The empty airplane would need a runway of 1828.571 meters.
Explanation:
a) This problem can be solved by using the Work-Energy Theorem, which states that work needed by the airplane to get minimum speed is equal to its change in translational kinetic energy, both measured in joules. The resulting formula is presented below:
[tex]F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2})[/tex] (1)
Where:
[tex]F[/tex] - Minimum net force, measured in newtons.
[tex]\Delta s[/tex] - Runway length, measured in meters.
[tex]m[/tex] - Mass of the airplane, measured in kilograms.
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds of the airplane, measured in meters per second.
If we know that [tex]m = 350000\,kg[/tex], [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v_{f} = 82\,\frac{m}{s}[/tex] and [tex]\Delta s = 3200\,m[/tex], then the minimum net force needed by the airplane to get itself safely in the air:
[tex]F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}[/tex]
[tex]F = \frac{(350000\,kg)\cdot \left[\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (3200\,m)}[/tex]
[tex]F = 367718.75\,N[/tex]
A force of 367718.75 newtons is needed to get the full airplane safely in the air.
b) If we know that [tex]m = 200000\,kg[/tex], [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v_{f} = 82\,\frac{m}{s}[/tex] and [tex]F = 367718.75\,N[/tex], then the length of the runway is:
[tex]\Delta s = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot F}[/tex]
[tex]\Delta s = \frac{(200000\,kg)\cdot \left[\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (367718.75\,N)}[/tex]
[tex]\Delta s = 1828.571\,m[/tex]
The empty airplane would need a runway of 1828.571 meters.
I was having trouble with this problem, and problems like it: A 3.2 kg pelican, with a 1.73 kg fish in its mouth, is flying 1.52 m/s at a height of 40 m when the fish wiggles free and fall back toward the ocean. How fast is the fish moving when it hits the water?
Answer:
28.1 m/s
Explanation:
[tex]u_x[/tex] = Initial velocity of the fish = 1.52 m/s
y = Height of the bird = 40 m
[tex]a_y[/tex] = Acceleration in y axis = [tex]9.81\ \text{m/s}^2[/tex]
[tex]u_y[/tex] = Initial velocity in y axis = 0
[tex]y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow 40=0+\dfrac{1}{2}\times 9.81t^2\\\Rightarrow t=\sqrt{\dfrac{40\times 2}{9.81}}\\\Rightarrow t=2.86\ \text{s}[/tex]
[tex]v_y=u_y+a_yt\\\Rightarrow v_y=0+9.81\times 2.86\\\Rightarrow v_y=28.057\ \text{m/s}[/tex]
The final velocity in x direction will remain the same as the initial velocity as there is no acceleration in the x direction [tex]u_x=v_x=1.52\ \text{m/s}[/tex]
Resultant velocity is given by
[tex]v=\sqrt{v_x^2+v_y^2}\\\Rightarrow v=\sqrt{1.52^2+28.057^2}\\\Rightarrow v=28.1\ \text{m/s}[/tex]
The fish is moving at a velocity of 28.1 m/s when it hits the water.
What are five facts about magnets? Please help me
A vector
of magnitude 5 units and another vector ū of magnitude 3 units point in directions
differing by 60°. Find out (a) the scalar product of the two vectors and (b) the vector product of the two vectors
Answer:
7.5 units
13 units
Explanation:
[tex]|v|=5\ \text{units}[/tex]
[tex]|u|=3\ \text{units}[/tex]
[tex]\theta[/tex] = Angle between the vectors = [tex]60^{\circ}[/tex]
Scalar product is given by
[tex]u\cdot v=|u||v|\cos\theta\\ =3\cdot 5\cdot \cos60^{\circ}\\ =7.5\ \text{units}[/tex]
The scalar product of the vectors is 7.5 units.
Vector product is given by
[tex]u\times v=|u||v|\sin\theta\\ =3\times 5\sin60^{\circ}\\ =13\ \text{units}[/tex]
The vector product of the vectors is 13 units.
An object, 5 cm high, is placed on the principal axis of a diverging lens of focal length 20 cm. The object is 30 cm from the lens.
Use a scaled diagram to locate the image formed by the lens.
Answer:
The answer is 70 cm
Explanation:
If you add All the numbers together, you receive an 55 cm then you add 15 because the points on the diagram also count.
what volume of alcohol will have the same mass as 4.2m^3 of petrol? (density of alcohol 0.4kg/m^3 and petrol is 0.3kg/m^3)
Answer:
3.15m³
Explanation:
To solve this problem, let us first find the mass of the petrol from the given dimension.
Mass = density x volume
Volume of petrol = 4.2m³
Density of petrol = 0.3kgm⁻³
Mass of petrol = 4.2 x 0.3 = 1.26kg
So;
We can now find the volume of the alcohol
Volume of alcohol = [tex]\frac{mass}{density}[/tex]
Mass of alcohol = 1.26kg
Density of alcohol = 0.4kgm⁻³
Volume of alcohol = [tex]\frac{1.26}{0.4}[/tex] = 3.15m³
A 5.75 g bullet is fired with a velocity of 1.50 x 102 m/s toward a stationary solid block resting on a frictionless surface. The bullet embeds but the block does not move. 1. What is the change in momentum of the bullet if it embeds in the block? 2. What is the change in momentum of the bullet if it bounces off the block in the opposite direction with a speed of 100 m/s?
Answer:
1. -0.863 kgm/s 2. -1.438 kgm/s
Explanation:
1. What is the change in momentum of the bullet if it embeds in the block?
Since the block does not move, the velocity of the bullet after hitting the block , v is zero. That is v = 0 m/s
Now, the momentum change of the bullet ΔP = m(v - u) where m = mass of block = 5.75 g = 5.75 × 10⁻³ kg, u = initial velocity of bullet = 1.50 × 10² m/s and v = final velocity of bullet after hitting the block = 0 m/s (since it embeds in the block and the block does not move).
So, ΔP = m(v - u)
= 5.75 × 10⁻³ kg(0 m/s - 1.50 × 10² m/s)
= 5.75 × 10⁻³ kg(- 1.50 × 10² m/s)
= -8.625 × 10⁻¹ kgm/s
= -0.8625 kgm/s
≅ -0.863 kgm/s
2. What is the change in momentum of the bullet if it bounces off the block in the opposite direction with a speed of 100 m/s?
If it bounces off the block in the opposite direction with a speed of 100 m/s, then its final velocity is v = -100 m/s.
So, our momentum change ΔP' = m(v - u) where m = mass of block = 5.75 g = 5.75 × 10⁻³ kg, u = initial velocity of bullet = 1.50 × 10² m/s and v = final velocity of bullet after hitting the block = -100 m/s = -1 × 10² m/s
So, ΔP = m(v - u)
= 5.75 × 10⁻³ kg(-1 × 10² m/s - 1.50 × 10² m/s)
= 5.75 × 10⁻³ kg(-2.50 × 10² m/s)
= -14.375 × 10⁻¹ kgm/s
= -1.4375 kgm/s
≅ -1.438 kgm/s
Which change of state takes place when a gas loses energy?
A. condensation
B. evaporation
C. molting
D. solidification
Answer:
Condensation (((((((((((((
A bucket is filled partly with water such that its combined mass is 2.5 kg. It is tied to a rope and whirled in a circle with a radius of 1.4 m. The speed at the top of the circle is 4.0 m/s and the speed at the bottom of the circle is 6.0 m/s. Determine the magnitude of the net force acting on the bucket at the bottom of the circle.
Answer:
1. Simply τ = m x g x r = 54kg x 9.8m/s² x 0.050m = 26 N·m
2. The bucket creates a torque
τ = 75kg x 9.8m/s² x 0.075m = 55 N·m,
so we must create the same torque with the handle.
55 N·m = F x 0.25m
F = 220 N
Explanation:
Hope this is helpful
A 4.00 kg block is pushed along the ceiling with a constant applied force of 85.0 N that acts at an angle of 55.0 degrees with the horizontal. The block accelerates to the right at 6.00 m/s2. Determine the coefficient of kinetic friction between block and ceiling.
Answer:
0.35
Explanation:
According to Newton's second law;
\sum Fx = ma
Fm - Ff =ma
Fm is the moving force = Wsin theta
Fm = 4(9.8)sin55
Fm = 32.1N
Ff is the frictional force = nmgcos theta
Ff = n(4)(9.8)cos55
Ff = 22.48n
Acceleration a = 6.0m/s²
Substitute the given values into the formula and get the coefficient of friction
32.11-23.48n = 4(6)
32.11-24= 23.48n
8.11 = 23.48
n = 8.11/23.48
n = 0.35
Hence the coefficient of friction is 0.35
Please help ASAP please ASAP
The eagles suck, so do the giants and the jets and jaguars are irrelevant
anyone wanna argue
Naw ur pretty accurate, heck collage is the only football worth watching most the time. Hook'um horns!
When fireworks explode, sound and light are produced. These are examples of(1 point)
macroscopic inputs.
macroscopic outputs.
microscopic inputs.
microscopic outputs.
Answer: macroscopic outputs
Explanation:
When fireworks explode, sound and light are produced. These are examples of macroscopic outputs. Because, explosion from fireworks is an exothermic process which releases massive heat energy to the surroundings.
What is exothermic reaction?Exothermic reaction are those which evolve heat energy to the surroundings. The change in enthalpy of the reaction is negative here. Whereas, in an endothermic reaction energy is absorbed by the reactants.
Exothermic reactions sometimes results in massive explosion. The heat energy released to the surroundings from the fire works is macroscopic level.
The small scale process or quantity that cannot be measured using normal scales are called microscopic units. Therefore, the sound, light, and heat from the explosion all are macroscopic outputs.
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A 50Kg girl jumps off a 5-meter-high diving board. What is her kinetic energy right before she
hits the water?
A. 0 J
B. 25 J
C. 1225 J
D. 2450 J
Answer:
D is the correct answer
Explanation:
Ek=m*g*h=50*9.8*5=2450
The kinetic energy right before she hits the water is 2450J. So, the correct option is D.
What is Kinetic energy?Kinetic energy is defined as the energy that is due to the motion of an object. If we want to accelerate an object, we must apply a force, by applying a force we need to do work. After the work is done, energy has been transferred to the object, and the object will continue to move with a new constant speed.
A 50Kg girl jumps off a 5-meter-high diving board.
We need to find the kinetic energy of the girl before she enters the water which means that the kinetic energy becomes equal to the potential energy such that,
P.E.=K.E. = mgh
where, m=mass of the object
g= acceleration due to gravity [tex](9.8m/s^2)[/tex]
h= height
So, K.E= 50* 9.8*5 = 2450 J
Thus, the kinetic energy right before she hits the water is 2450J. So, the correct option is D.
Learn more about Kinetic energy, here:
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Puck 1 is moving 10 m/s to the left and puck 2 is moving 8 m/s to the right. They have the same mass, m.
a. What is the total momentum of the system before the collision? (Answer in terms of the mass, m.) (0.5 points)
b. What is the total momentum of the system after the collision? (Answer in terms of the mass, m.) (0.5 points)
c. Write puck 1's velocity after the collision in component form. (1 point)
d. What is the y-component of puck 2's velocity after the collision? (1 point)
e. What is the x-component of puck 2's velocity after the collision? (1 point)
f. At what angle does puck 2 move after the collision? Determine the angle and draw it on the diagram. (1 point)
g. What is the magnitude of puck 2's velocity after the collision? (1 point)
Answer:
(a) the total momentum of the system before the collision = -2m kg.m/s.
(b) the total momentum of the system after the collision = -2m kg.m/s.
(c) puck 1's velocity after the collision in component form = (5.44 i, 2.54 j)
Explanation:
Given;
mass of Puck 1 , = m
mass of Puck 2, = m (since they have the same mass m)
initial velocity of Puck 1, u₁ = 10 m/s to the left
initial velocity of Puck 2, u₂ = 8 m/s to the right
Let the rightward direction be positive direction
Let the leftward direction be negative direction
(a) the total momentum of the system before the collision;
P₁ = (initial momentum of Pluck 1) + (initial momentum of Pluck 2)
P₁ = (-mu₁) + mu₂
P₁ = mu₂ - mu₁
P₁ = m(u₂ - u₁)
P₁ = m(8 - 10)
P₁ = -2m kg.m/s
(b) the total momentum of the system after the collision;
Based on the principle of conservation of linear momentum, the total momentum before collision is equal to the total momentum after collision.
Thus, the total momentum of the system after the collision is -2m kg.m/s.
(c) puck 1's velocity after the collision in component form
[tex]v = (v_x, v_y)\\\\v = (vcos \theta , vsin \theta)\\\\v = (6cos 25^0 , 6sin25^0)\\\\v = (5.44i, 2.54j)m/s[/tex]
A ball falls from a tower
a) The two forces acting on the tennis ball are equal and opposite. What is the resultant force
?acting on the ball
Answer:
Zero
Explanation:
The resultant force acting on the ball would be zero.
Since only two forces were acting on the tennis ball and these forces negate and cancel each other in magnitude, the resultant effect on the tennis ball would be zero.
Assuming that one of the forces is 5N and acting from the positive side and the other force is also 5N but acting from the negative side.
Resultant = -5 + 5 = 0 N
The middle one please it’s due in 40 min
Answer:
3240000000 Joules
Explanation:
An astronaut sitting on the launch pad on Earth's surface is 6,400 kilometers from Earth's center and weighs 400 newtons. Calculate her weight when she reaches an altitude of 6,400 kilometers above the surface of Earth.
Answer:
weight at height = 100 N .
Explanation:
The problem relates to variation of weight due to change in height .
Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .
At the surface :
Applying Newton's law of gravitation
mg₀ = G Mm / R²
At height h from centre
mg₁ = G Mm /h²
Given mg₀ = 400 N
400 = G Mm / R²
400 = G Mm / (6400 x 10³ )²
G Mm = 400 x (6400 x 10³ )²
At height h from centre
mg₁ = 400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²
= 400 / 4
= 100 N .
weight at height = 100 N
Appliances connected so that they form a single pathway for
charges to flow are connected in a(n)
A. series circuit.
B. parallel circuit.
C. off circuit.
D. open circuit
Appliances connected so that they form a single pathway for charges to flow are connected in a(n)
Answer:A. Series circuit
#CARRYONLEARNING #STUDYWELLAppliances connected so that they create a single pathway for charges to flow are connected in a series circuit. Therefore, option (A) is correct.
What is the resistance of resistors connected in series?In a series combination of appliances, they are connected end-to-end. Consider two resistors, R₁ and R₂ which are connected in a series combination then their effective resistance can be given by:
Total Resistance of the series circuit, R = R₁ + R₂
In a series combination, the current flows through one appliance and then through another appliance. The same current flows through each appliance in one direction. The total voltage of the series circuit is equal to the sum of all the voltage drops across all appliances.
A potential difference of the series circuit, V = V₁ + V₂
Therefore, when appliances are connected in a series circuit they form a single pathway for charges to flow.
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Please what is the work done by a man who is pulling a box of 45kg of mass by means of rope which makes angle of 45 degrees ?
Answer:
No work is done since no distance is given
Explanation:
Since no distance is given, the force is not doing any work
No work is done by the man since we do not know the distance or displacement.
Work is only said to be done when the force applied on an object moves it through a particular distance.
Work done = Force x distance.
Since no distance is given in this problem, we can as well assume that the force applied is doing no work on the object.
Fig. 2.1 shows a train
Fig. 2.1
The total mass of the train and its passengers is 750000kg. The train is travelling at a speed of 84m/s.
The driver applies the brakes and the train takes 80s to slow down to a speed of 42m/s.
(a) Calculate the impulse applied to the train as it slows down,
impulse =
[3]
(b) Calculate the average resultant force applied to the train as it slows down,
force =
(2)
Answer:
[tex]\mathrm{(a)\:}32,000,000\:\mathrm{Ns},\\\mathrm{(b)\:}390,000\:\mathrm{N}[/tex]
Explanation:
The impulse-momentum theorem states the impulse on an object is equal to the change in momentum of that object. Momentum is given by [tex]p=mv[/tex]. Since mass is constant, the train's change in momentum is:
[tex]\Delta p=m\Delta v=750,000\cdot42=31,500,000=\fbox{$32,000,000\:\mathrm{Ns}$}[/tex](two significant figures).
Impulse is also given as [tex]\Delta p = F\Delta t[/tex], where [tex]F[/tex] is the average force applied and [tex]\Delta t[/tex] is change in time. Since [tex]t[/tex] is given as [tex]80\mathrm{s}[/tex], we have the following equation:
[tex]F\Delta t=\Delta p\\\\F=\frac{\Delta p}{\Delta t},\\\\F=\frac{31,500,000}{80},\\\\F=393,750=\fbox{$390,000\:\mathrm{N}$}[/tex](two significant figures).
can someone please help me it’s 15 points of my major grade..
a.
b.
c.
e.
f.
g.
character limit thing
The phases of the moon are caused by which of the following
A. Earths rotation around its axis
B. Earths orbit around the sun
C. The moons orbit around the earth
D. Earths shadow falling on the moon
Answer: C: The moons orbit around the earth.
Explanation:
A jet aircraft is traveling at 262 m/s in horizontal flight. The engine takes in air at a rate of 85.9 kg/s and burns fuel at a rate of 3.92 kg/s. The exhaust gases are ejected at 921 m/s relative to the aircraft. Find the thrust of the jet engine. Answer in units of N Find the delivered power. Answer in units of W.
Answer:
[tex]F_T=60132.52N[/tex]
[tex]P=15814852.76W[/tex]
Explanation:
From the question we are told that
Velocity of aircraft [tex]V=263m/s[/tex]
Engine air intake rate [tex]\triangle M_a=85.9kg/s[/tex]
Fuel burn rate [tex]\triangle M_f =3.92kg/s[/tex]
Velocity of exhaust gas [tex]V_e =921m/s[/tex]
Generally the Mass change rate of Rocket is mathematically given by
[tex]\triangle M = \triangle M_a+\triangle M_f[/tex]
[tex]\triangle M= 85.9+3.92[/tex]
[tex]\triangle M=89.82kg/s[/tex]
Generally the Trust of the rocket is given mathematically by
[tex]F_T=(\triangle M *V_e)-(dM_a/dt)*(V)[/tex]
[tex]F_T=(89.82 *921)-(85.9)*(263)[/tex]
[tex]F_T=60132.52N[/tex]
Generally the Rocket's delivered power is mathematically given by
Delivered power P
[tex]P=V*F_T[/tex]
[tex]P=263*60132.52N[/tex]
[tex]P=15814852.76W.[/tex]
