Answer: 100
Explanation: 20 *5=100
Hope this helps!!Brainlist?plz
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.
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
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.
A horizontal pole is attached to the side of a building. There is a pivot P at the wall and a chain is connected from the end of the pole to a point higher up the wall. There is a tension force F in the chain. What is the moment of the force F about the pivot P?
Answer:
Fscos63
Explanation:
Given that a horizontal pole is attached to the side of a building. There is a pivot P at the wall and a chain is connected from the end of the pole to a point higher up the wall. There is a tension force F in the chain. What is the moment of the force F about the pivot P?
Taking the moment from the pivot point P, that means the moment at point p = 0
Then, if we consider the weight mg of the pole, according to the principle of equilibrium : sum of the upward forces equal to the sum of the downward forces.
Therefore, mg = Fsinø ....... (1)
Also, taking moment at point P
Let the length of the pole = s
The length of the weight of the pole = 1/2 S
Fscosø = mgs/2
The distance s will cancel out
2Fcosø = mg ...... (3)
Substitute mg in equation 1 into equation 3
2fcosø = fsinø
F will cancel out
Tanø = 2
Ø = tan^-1(2)
Ø = 63.4 degree
The moment of force F about pivot point P will be
Moment = force × distance
Moment = Fcos63 × S
Moment = Fscos63
A person runs 15.0 km north then turns around and runs 10.0 Km south. what is his dstance
Answer:
25km
Explanation:
The person runs 15km Northward
Turns around and runs 10km southward
The distance is the length of path covered by the person running.
This is given as:
Distance = Distance North + Distance South
Distance = 15km + 10km = 25km
Answer:
25km
Explanation:
cant explain but ik
Please help ASAP please ASAP
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
can someone please help me it’s 15 points of my major grade..
a.
b.
c.
e.
f.
g.
character limit thing
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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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]
The middle one please it’s due in 40 min
Answer:
3240000000 Joules
Explanation:
What distance is required for a train
to stop if its intial Velocity is 23 m/s
and its deceleration is 0.25m/s (Assume the train decelerates at a constant rate.)
Explanation:
what is time in this question
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.
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).
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
IS IT B!! if not pls helpp!
Answer:
I believe you are correct but we just started this unit
Explanation:
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.
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
What are five facts about magnets? Please help me
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
I NEED ANSWER ASAP!!!
At which point(s) will acceleration occur shown in the image???
Answer:
Gravity is an ever present force, and therefore acceleration is guaranteed to happen at every single one of those points (and in fact, everywhere in the universe).
On top of that, friction will be present in all four spots (friction with the rails, with the air, with the axles, etc.), and friction is a perfectly acceptable force that will cause acceleration, slowing the roller coaster down.
So the correct answer is every single point, regardless of what answer the teacher expects.
The object will be moving faster if the acceleration and velocity are pointing in the same direction. The object will also slow down if the acceleration is pointing in the opposite direction as the velocity.
What role of acceleration in the motion of object?When an object's speed, direction of motion, or both change, it accelerates. Even while it may appear to be virtually immediate in some circumstances, such as when a golf ball is struck by a club or during car collisions, changes in an object's speed are always continuous.
Since gravity is a constant force, acceleration will unavoidably occur at each of those locations and throughout the whole universe.
Therefore, In addition, there will be friction at all four locations—friction with the axles, the air, the rails, etc.—and friction is a completely normal force that will accelerate the roller coaster, slowing it down.
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A 10.0kg box sits at rest on ramp that has an incline of 20.0° from the horizontal. What is the force of gravity acting on the box?
A. -98.0N
B. 98.0N
C. -92.1N
D. -33.5N
Answer:
98.0N
Explanation:
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 ..
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.
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
If there is a difference in air pressure between two locations, what happens?
Answer:
High in the atmosphere, air pressure decreases. ... A low pressure system has lower pressure at its center than the areas around it. Winds blow towards the low pressure, and the air rises in the atmosphere where they meet. As the air rises, the water vapor within it condenses, forming clouds and often precipitation.
Explanation:
Henry designs an experiment to find out why plants without flowers tend to be wind-pollinated and not animal-pollinated. He takes samples from a variety of plants and studies them in the field. He finds that the evidence supports his hypothesis. Which of these was the hypothesis Henry was testing? Choose the correct answer. Nonflowering plants make less pollen than flowering plants. Nonflowering plants lack the structures to attract animal pollinators. Nonflowering plants grow close to the ground and capture fewer pollinators. Nonflowering plants grow close to the ground because they have no pollen grains.
Answer:
option 2
Explanation:
you are very welcom
Which of these pairs of atoms are isotpoes? (Physical Science) Pair A Pair B Pair C # protons 6 8 5 2 12 12 # neutrons 8 8 5 3 13 14
Answer:
I guess that the atoms are:
Protons: 6 8 5 2 12 12
Neutrons: 8 8 5 3 13 14
Now, two atoms are isotopes if they share the same number of protons (so both atoms are the same element) but they have a different number of neutrons.
From the given options, the only two that have the same number of protons but a different number of neutrons are:
Protons 12, neutrons 13
and
Protons 12, neutrons 14.
These two are isiotopes.
A ‘can-chiller’ is used to make a can of drink colder. The initial temperature of the liquid in the can was 25.0 °C. The can-chiller decreased the temperature of the liquid to 20.0 °C. The amount of energy transferred from the liquid was 6930 J. The mass of liquid in the can was 0.330 kg.
Calculate the specific heat capacity of the liquid. Give the unit.
Please help :)
Answer:
4200 J/°C/kg
Explanation:
The formula for heat transfer is given by :
Q= m*c*ΔT where;
Q= heat transferred = 6930 J
m=mass of the liquid = 0.330 kg
c= specific heat capacity=?
ΔT = 25-20 = 5.0°C
Applying the values in the formula as;
Q= m*c*ΔT
6930 = 0.330 * c * 5
6930 = 1.65 c
6930/1.65 = c
4200 = c
c= 4200 J/°C/kg
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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