Answer:
12.31 m/s
Explanation:
If we recall from the previous knowledge we had about speed,
we will know that:
speed = distance/ time.
As such:
The average speed of the rider bicycle is
average speed = total distance/ total time
Mathematically, it can be computed as:
[tex]v_{avg} = \dfrac{d+d}{\dfrac{d}{v_1}+ \dfrac{d}{v_2}}[/tex]
[tex]v_{avg} = \dfrac{2d}{\dfrac{d}{10 \ m/s}+ \dfrac{d}{16 \ m/s}}[/tex]
[tex]v_{avg} = \dfrac{2}{\dfrac{1}{10 \ m/s}+ \dfrac{1}{16 \ m/s}}[/tex]
[tex]v_{avg} = \dfrac{2}{\dfrac{13}{80 \ m/s}}[/tex]
[tex]\mathbf{v_{avg} =12.31 \ m/s}[/tex]
Consider a simple pendulum consisting of a massive bob suspended from a fixed point by a string. Let T denote the time (the period of the pendulum) that it takes the bob to complete one cycle of oscillation (the time it takes for the pendulum to swing back and forth one time). How does the period of the swing of the simple pendulum depend on the quantities that define the pendulum and the quantities that determine the motion
Answer:
The period of the swing depends on only the length of the string and not on the mass of the bob and the period of the pendulum depends on only the horizontal component of g.
Explanation:
The period of the swing depends on only the length of the string and not on the mass of the bob. Since the length of the string and the mass of the bob define the pendulum.
Also, the properties that define the motion are the component of the weight of the bob in the horizontal direction which determines the to and fro movement of the bob. So, the period of the pendulum depends on only the horizontal component of g.
So, T = 2π√(l/g) where l = length of pendulum and g = acceleration due to gravity.
I WILL GIVE YOU BRAINLIEST! PLEASE HELP
A daydreaming soccer player takes a 0.47 kg ball to the face experiencing an impact force of 1060.9 N. If the ball hit the
player's face with a speed of 14.5 m/s and bounces off in the opposite direction with the same speed, calculate the time of
impact. Note: The time of impact will be a fraction of a second so answer with at least 5 decimal places
Answer:
0.00547s
Explanation:
Step one:
given data
mass= 0.4kg
force= 1060.9N
velocity = 14.5m/s
Frome
Ft= mv
substitute
t= mv/F
t=0.4*14.5/1060.9
t=5.8/1060.9
t=0.00547s
Two spherical objects with a mass of 3.17 kg each are placed at a distance of 2.96 m apart. How many electrons need to leave each object so that the net force between them becomes zero
Answer:
1.704 × 10¹⁹ electrons
Explanation:
The gravitational force due to the identical masses with mass, m = 3.17 kg at a distance r = 2.96 m from each other is F = Gm²/r².
Since equal number of electrons have to leave both masses, we have a charge q on each mass acting to oppose each other at distance r.
So, the electrical force of repulsion is thus
F'= kq²/r²
For the net force to be zero, the gravitational force of attraction must balance the electrical force of repulsion.
So, F = F'
So, Gm²/r² = kq²/r²
Gm² = kq²
q² = Gm²/k
taking square root of both sides, we have
q = m√(G/k)
So, substituting the values of the variables into the equation, we have
q = m√(G/k)
= 3.17 kg√(6.67 × 10⁻¹¹ Nm²/kg²/9 × 10⁹ Nm²/C²)
= 3.17 kg√(0.741 × 10⁻²⁰ C²/kg²)
= 3.17 kg × 0.861 C/kg
= 2.73 C
Now, q = ne where n = number of electrons and e = electron charge = 1.602 × 10⁻¹⁹ C
n = q/e
= 2.73 C ÷ 1.602 × 10⁻¹⁹ C
= 1.704 × 10¹⁹ electrons
Small household electrical devices, such as vacuum cleaners, televisions, and floor lamps, each draw a different amount of current, but all require 120 volts to operate. Are the outlets in of a power-strip, then, wired in series or parallel
Answer:
the selected configuration is PARALLEL
Explanation:
There are two ways to connect electrical circuits.
In series in this type of circuit the equipment is connected to the same line, in this case the current of the circuit circulates through all the inputs is the same, but the voltage in each one is a part of the total voltage, this circuit has the problem that if an appliance is disconnected or damages the power and the entire circuit is interrupted
V_total = ∑ V_i
In a parallel circuit the equipment is connected in two two cables, for which the current is divided in each branch, the voltage in all the branches is the same, this type of circuit has the advantage that if one equipment is damaged or disconnects the others they can continue working
V_total = V
Therefore, in consequence of the above, the selected configuration is PARALLEL
Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s as it leaves the hose nozzle. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation of the hose until the water takes 3.00 to reach a building 41.0m away. You can ignore air resistance; assume that the end of the hose is at ground level.
Required:
a. Find the angle of elevation of the hose.
b. Find the speed in m/s of the water at the highest point in its trajectory.
c. Find the acceleration in m/s^2 of the water at the highest point in its trajectory.
d. How high above the ground in m does the water strike the building?
e. How fast is it moving in m/s just before it hits the building?
Answer:
a) θ = 58.3º
b) vfh = 13.7 m/s
c) g = -9.8 m/s2
d) h = 22.2 m
e) vfb = 15.5 m/s
Explanation:
a)
Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:vₓ₀ = v * cos θ (1)where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:[tex]x_{f} = v_{ox} * t = v_{o} * cos \theta * t (2)[/tex]
Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:[tex]cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)[/tex]
⇒θ = cos⁻¹ (0.526) = 58.3º (4)b)
At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)c)
At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)d)
Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:[tex]\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)[/tex]
Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)Replacing (7) in (6), we get:[tex]\Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)[/tex]
e)
When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.The horizontal component, since it keeps constant, is just v₀x:v₀ₓ = 13.7 m/sThe vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:[tex]v_{fy} = v_{oy} - g*t (9)[/tex]
Replacing by the time t (a given), g, and v₀y from (7), we can solve (9) as follows:[tex]v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s (10)[/tex]
Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:[tex]v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)[/tex]
If the force of gravity suddenly stopped acting on planets, they would
A.) spiral slowly towards the sun
B.) continue to orbit the sun
C.) move in straight lines tangent to thier orbits
D.) spiral slowly away from the sun
E.) fly straight away from the sun
What happens to the sum of the ball's kinetic energy and potential energy as the ball rolls from point A to point E? Assume there's no friction between the ball and the ground.
А. The sum decreases.
В. The sum increases.
C. The sum remains the same.
D. The sum always equals zero.
Answer:
C. The sum remains the same.
Explanation:
The sum of the kinetic and potential energy remains the same as the all rolls from point A to E.
We know this based on the law of conservation of energy that is in play within the system.
The law of conservation of energy states that "energy is neither created nor destroyed within a system but transformed from one form to another".
At the top of the potential energy is maximum As the ball rolls down, the potential energy is converted to kinetic energy. Potential energy is due to the position of a bodyKinetic energy is due to the the motion of the bodyLet A be the second to last digit and let B be the last two digits of your 8-digit student ID. Example: for 20245347, A = 4 and B = 47.A ball rolls off a table. The table top is 1.2 m above the floor and the ball lands 3.6 m from the base of the table. Determine the speed of the ball at the time it rolled over the edge of the table? Calculate the answer in m/s and rounded to three significant figures.
Answer:
7.35 m/s
Explanation:
Using y - y' = ut - 1/2gt², we find the time it takes the ball to fall from the 1.2 m table top and hit the floor.
y' = initial position of ball = 1.2 m, y = final position of ball = 0 m, u = initial vertical velocity of ball = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for ball to hit the ground.
So, substituting the values of the variables into the equation, we have
y - y' = ut - 1/2gt²
0 - 1.2 m = (0 m/s)t - 1/2(9.8 m/s²)t²
- 1.2 m = 0 - (4.9 m/s²)t²
- 1.2 m = - (4.9 m/s²)t²
t² = - 1.2 m/- (4.9 m/s²)
t² = 0.245 s²
t = √(0.245 s²)
t = 0.49 s
Since d = vt where d = horizontal distance ball moves = 3.6 m, v = horizontal velocity of ball = unknown and t = time it takes ball to land = 0.49 s.
So, d = vt
v = d/t
= 3.6 m/0.49 s
= 7.35 m/s
Since the initial velocity of the ball is 7.35 m/s since the initial vertical velocity is 0 m/s.
It is shown thus V = √(u² + v²)
= √(0² + v²)
= √(0 + v²)
= √v²
= v
= 7.35 m/s
A basketball of mass 0.23kg is thrown horizontally against a rigid vertical wall with a velocity of 20m/s. It rebounds with a velocity of 15m/s. Calculate the impulse of the force of the wall on the basketball.
Answer:
[tex]8.1\:\mathrm{Ns}[/tex]
Explanation:
The impulse-momentum theorem gives the impulse on an object to be equal to the change in momentum of that object. Since mass is maintained, the change in momentum of the basketball is:
[tex]\Delta p = m\Delta v[/tex], where [tex]m[/tex] is the mass of the basketball and [tex]\Delta v[/tex] is the change in velocity.
Since the basketball is changing direction, its total change in velocity is:
[tex]\Delta v = 20-(-15)=35\:\mathrm{m/s}[/tex].
Therefore, the basketball's change in momentum is:
[tex]\Delta p = m\Delta v = 0.23\cdot 35= 8.05=8.1\:\mathrm{kg\cdot m/s}[/tex].
Thus, the impulse on the basketball is [tex]\fbox{$8.1\:\mathrm{Ns}$}[/tex] (two significant figures).
A typical ceiling fan running at high speed has an airflow of about 1.85 ✕ 10^3 ft^3/min, meaning that about 1.85 ✕ 10^3 cubic feet of air move over the fan blades each minute.
Determine the fan's airflow in m^3/s.
Answer:
0.83 m³/s
Explanation:
The speed of the airflow is given as;
1.85 x 10³ ft³/min
Now we are to express this unit in m³/s
1ft = 0.3m
60s = 1 min
So;
1.85 x 10³ x ft³ x [tex]\frac{1}{min}[/tex] x [tex]\frac{(0.3m)^{3} }{ft^{3} }[/tex] x [tex]\frac{1min}{60s}[/tex]
= 0.83 m³/s
A beaker with water resting on a scale weighs 40 N. A block
suspended on a hanging spring weighs 20 N. The spring scale
reads 15 N when a block is fully submerged in the water. What is
the reading of a scale on which the beaker with water rests, while
the block is submerged in the water after detached from the
hanging spring?
A. 25 N B. 60 N C. 55 N D. 45 N
Answer:
D. 45 N
Explanation:
The weight of the block is 20 N, when the block is fully immerged in water, it weighs 15 N. Hence the loss of weight = 20 N - 15 N = 5 N.
The loss of weight is as a result of the buoyant force. The buoyant force is the upward force exerted by a fluid when an object is fully or partially immersed in a fluid.
The buoyant force of 5 N acts in the upward direction, the weight of the beaker that would be read by the scale when the beaker is immersed in water = 40 N + 5 N = 45 N
A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 34.5 N is applied tangent to the rim of the disk.
a) What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through .200 revolution?
b) What is the magnitude a of the resultant acceleration of a point on the rim of the disk after the disk has turned through .200 evolution?
Answer:
a) v = 1.01 m/s
b) a = 5.6 m/s²
Explanation:
a)
If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:[tex]\tau = I * \alpha (1)[/tex]
Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:τ = F*r (2)For a solid uniform disk, the rotational inertia regarding an axle passing through its center is just I = m*r²/2 (3).Replacing (2) and (3) in (1), we can solve for α, as follows:[tex]\alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)[/tex]
Since the angular acceleration is constant, we can use the following kinematic equation:[tex]\omega_{f}^{2} - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)[/tex]
Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:[tex]0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)[/tex]
Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:[tex]\omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)[/tex]
Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:[tex]v = \omega * r (8)[/tex]
where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of the disk, r becomes the radius of the disk, 0.200 m.Replacing this value and (7) in (8), we get:[tex]v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)[/tex]
b)
There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:[tex]a_{t} = \alpha * r (9)[/tex]
where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.Replacing this value and (4), in (9), we get:[tex]a_{t} = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)[/tex]
Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.The radial acceleration is just the centripetal acceleration, that can be expressed as follows:[tex]a_{c} = \omega^{2} * r (11)[/tex]
Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.Replacing this value and (7) in (11) we get:[tex]a_{c} = \omega^{2} * r = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)[/tex]
The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:[tex]a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)[/tex]
A 4 kg object moving to the left collides with and sticks to a 3 kg object moving to the right. Which of the following is true of the motion of the combined objects immediately after the collision?
a. They must be moving to the left.
b. They must be moving to the right.
c. They must be at rest.
d. The motion cannot be determined without knowing the speeds of the objects before the collision.
Answer:
D. The motion cannot be determined without knowing the speeds of the objects before the collision.
Explanation:
This question is tricky! We know the object moving to the left has a greater mass than the one moving to the right. We'd assume they would move to the left because the leftwards object has a greater mass, right?
Not. So. Fast.
We can solve for the objects' final velocity using the formula for momentum, m₁v₁ + m₂v₂ = (m₁ + m₂)v .
Now here's where the trap is sprung: we don't think about the equation. This shows that the final velocity of the objects and the direction depends on both the mass of the objects and their initial velocity.
Basically, what if the 3 kg object is moving at 1 m/s and the 4 kg object is moving at –0.5 m/s? The objects would move to the right after the collision!
Do we know the velocity of these objects? No, right?
That means we can't determine the direction of their motion unless we know their initial, pre-collision velocity. This question is tricky because we look at the 4 kg vs. 3 kg and automatically assume the 4 kg object would dictate the direction of motion. That's not true. It depends on velocity as well.
I hope this helps you! Have a great day!
please help thank you
Answer:
[tex]\theta \approx 59.036^{\circ}[/tex], [tex]T_{2} \approx 23.324\,N[/tex]
Explanation:
First we build the Free Body Diagram (please see first image for further details) associated with the mass, we notice that system consist of a three forces that form a right triangle (please see second image for further details): (i) The weight of the mass, (ii) two tensions.
The requested tension and angle can be found by the following trigonometrical and geometrical expressions:
[tex]\theta = \tan^{-1} \frac{W}{T_{2}}[/tex] (1)
[tex]T_{1} = \sqrt{W^{2}+T_{2}^{2}}[/tex] (2)
Where:
[tex]W[/tex] - Weight of the mass, measured in newtons.
[tex]T_{1}[/tex], [tex]T_{2}[/tex] - Tensions from the mass, measured in newtons.
If we know that [tex]W = 20\,N[/tex] and [tex]T_{2} = 12\,N[/tex], then the requested values are, respectively:
[tex]\theta = \tan^{-1} \frac{20\,N}{12\,N}[/tex]
[tex]\theta \approx 59.036^{\circ}[/tex]
[tex]T_{2} = \sqrt{(20\,N)^{2}+(12\,N)^{2}}[/tex]
[tex]T_{2} \approx 23.324\,N[/tex]
If 478 watts of power are used in 14 seconds,how much work was done
Answer:
6692J
Explanation:
Power is defined as the rate at which work is being done.
So,
Power = [tex]\frac{workdone}{time }[/tex]
Work done = Power x time
Given parameters:
Power = 478watts
Time = 14s
So;
Work done = 478 x 14 = 6692J
Cecily is inflating her bicycle tyre with the pump below. When she pushes the plunger down, it is doing work against the gas. This means that the plunger is transferring what to the gas particles?
The plunger is transferring energy to the gas particles.
What the plunger is transferring to the gas particles?When Cecily is applying pressure to the plunger, the gas inside is being compressed, and the speed of gas molecules increases.
When these air molecules enter the tire, adiabatically, the temperature of the gas rises.
It happens due to an increase in the average kinetic energy of the gas particles.
Thus, the gas particles receive energy from the plunger.
Learn more about the plunger, here:
https://brainly.com/question/4190018
Give an example of each element,compound, and mixture
Answer:An element cannot be broken down into any other substance. ... Compounds are substances made from atoms of different elements joined by chemical bonds. They can only be separated by a chemical reaction. Common examples are water (H2O), salt (sodium chloride, NaCl), methane (CH4).
Explanation:I hope this helps
Answer:
Mixture is a substance that is formed when two or more compounds or elements are mixed in any ratio. For example- sherbet, air. sand, etc.
Explanation:
An element cannot be broken down into any other substance. ... Compounds are substances made from atoms of different elements joined by chemical bonds. They can only be separated by a chemical reaction. Common examples are water (H2O), salt (sodium chloride, NaCl), methane (CH4).
;-; sorry hopes this helps a little
How many joules does your 1600W electric hair dryer transfer if it takes 1 minutes to dry your hair?
____ joules
Answer:
96,000joules!!!
Explanation:
Hope this helps u
Waves in the ocean are tearing apart the shoreline. Which of the following two Earth Systems are interacting with each other.
Answer:
the Indian Ocean on 26 December 2004. This event claimed 227,898 dead and missing from 14 countries. The difference in mortality rates between these tsunamis reflects, in part, the benefits of understanding how tsunami waves are generated and move, and educating citizens to make scientifically
sound and potentially life-saving decisions.
A tsunami is a series of rapidly propagating, shallow-water ocean waves that develops when a submarine earthquake, landslide, or volcanic eruption displaces a large volume of water. Powerful earthquakes, with magnitudes of 9 or greater, caused both the 2004 and 2011 tsunamis. The earthquakes resulted from the movement of large tectonic plates. The 11 March 2011 earthquake occurred at 32 km (20
mi.) deep in Earth’s crust about 130 km (81 mi.) east of the city of Sendai. This location is on the boundary between two tectonic plates—the Pacific plate to the east and North American plate to the west. This
boundary fractured, releasing energy that was transmitted through the rocks and elevated portions of the
ocean floor. This drastic movement transmitted energy to the overlying ocean water, which generated
tsunami waves that radiated outward. The waves washed over the nearby coastlines and were felt around
the globe within hours (Figure 1.1).
Explanation:
Answer:
I believe Geosphere (lithosphere) and Hydrosphere
Explanation:
I hope it's right if not please notify me.
How is the Moon thought to have formed
what happens if we add a lump of ice to a tumbler containing water? how does heat flow?
Answer:
The ice will melt
Heat flows from the water to the ice
Explanation:
If we add a lump of ice to a tumbler containing water, the ice lump will melt in the water.
This is a simple physical change of state from solid to liquid.
At the end both the solid ice and the original liquid maintains the liquid state.
The heat will flow from the ice to the water in the tumbler. Heat generally flows from a body at higher temperature to one with a lower temperature. The water has a higher temperature when compared to the ice. So, heat flows from the liquid water to the ice until thermal equilibrium is reached.In the figure, given ∆x=30cm, k=200N/cm, g=10m/s². Find the mass of the object
Answer:
600 Kg
Explanation:
From the question given above, the following data were obtained:
Extention (∆x) = 30 cm
Spring constant (K) = 200 N/cm Acceleration due to gravity (g) = 10 m/s²
Mass (m) of object =?
Next, we shall determine force exerted. This can be obtained as follow:
Extention (∆x) = 30 cm
Spring constant (K) = 200 N/cm
Force (F) =?
F = K∆x
F = 200 × 30
F = 6000 N
Finally, we shall determine the mass of the object. This can be obtained as follow:
Acceleration due to gravity (g) = 10 m/s²
Force (F) = 6000 N
Mass (m) of object =?
F = mg
6000 = m × 10
Divide both side by 10
m = 6000 / 10
m = 600 Kg
Thus, the mass of the object is 600 Kg
If an object has applied force of 20 N and a frictional force of 5 N what is the net force?
Answer:
Net force = 15 N
Explanation:
Given that,
Applied force on an object = 20 N
Frictional force = 5 N
We need to find the net force acting on the object.
Friction is an opposing force. It acts in the opposite direction.
Net force = Applied force - Frictional force
= 20 N - 5 N
= 15 N
Hence, the net force acting on the object is 15 N.
Starting from the front door of your ranch house, you walk 50.0 m due east to your windmill, and then you turn around and slowly walk 30.0 m west to a bench where you sit and watch the sunrise. It takes you 27.0 s to walk from your house to the windmill and then 47.0 s to walk from the windmill to the bench. For the entire trip from the front door to the bench, what are your :
a. average velocity
b. average speed
Answer:
Explanation:
Total displacement for entire trip = final position - initial position
= 50 m - 30 m = 20 m
Total time = 27 + 47 = 74 s
Average velocity = Total displacement / total time
= 20 / 74 = .27 m /s
Total distance covered in entire trip = 50 + 30 = 80 m
Total time = 74 s
Average speed = Total distance covered / total time
= 80 / 74 = 1.08 m /s .
Austin invested $11,000 in an account paying an interest rate of 5.7% compounded quarterly. Assuming no deposits or withdrawals are made, how much money, to the nearest dollar, would be in the account after 6 years?
Answer:
15448
Explanation:
A=11000(1.01425)^{24}
A=11000(1.01425)
24
Austin invested $11,000 in an account paying an interest rate of 5.7% compounded quarterly. Assuming no deposits or withdrawals are made, the money to the nearest dollar, would be in the account after 6 years is 15448.
What is Compound interest?The compound interest occurs when the interest is reinvested rather than paying it out. It's basically earning interest over interest.
The formula is:
Compound interest, [tex]A = P ( 1 +\frac{r}{n} )^{nt}[/tex]
Where:
A = final Amount
P = initial principal balance
r = interest rate
n = number of times interest applied per time period
t = number of time periods elapsed
Austin invested P=$11000 in an account with an interest rate of r=5.7% = 0.057 (decimal) during t=6 years compounded quarterly. Since there are 4 quarters in a year, n=4.
Thus, Substituting all the values in the given formula,
A = 11000 ( 1 + [tex]\frac{0.057}{4} )^{6*4}[/tex]
= 11000 × 1.4043662796
= 15448.0290
The money to the nearest dollar, would be in the account after 6 years is 15448.
Learn more about Compound interest,
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In the picture shown below A represents a characteristic of only geocentric model, B represents a characteristic common to both geocentric and heliocentric models, C represents a characteristic of only heliocentric model, and D represents a characteristic which the geocentric and heliocentric models do not have.
Under which label will the characteristic, "The sun and planets revolve around a central moon in the solar system" fall?
A
B
C
D
Kiara starts at 4, walks 6 blocks left and 2 blocks right. What is her displacement?
3.
What is a star? Why does the sun appear so big whereas the other stars do not?
Ans.
Plz help this is so confusing
The correct answer is 5 km/h
Explanation:
The speed at which the duck travels can be found by using the equation that is given (speed= distance/ time). The first step to do this is to replace distance and time using the values given. Here is the process:
speed = distance / time
speed = 10 km / 2 hours
Now, solve this equation
speed = 10 km/ 2 hours
10 / 2 or 10 divided 2 = 5
Finally, use the units, in this case, the correct units are km/h
Megan walks 1100\,\text m1100m1100, start text, m, end text to the left in 330\,\text s330s330, start text, s, end text. What was her average speed in \dfrac{\text m}{\text s} s m start fraction, start text, m, end text, divided by, start text, s, end text, end fraction?
Answer:
v = 3.34 m/s
Explanation:
Given that,
Distance, d = 1100 m
Time, t = 330 s
We need to find the average speed of the Megan. It is equal to the total distance divided by total time taken.
[tex]v=\dfrac{1100\ m}{330\ s}\\\\v=3.34\ m/s[/tex]
So, the average speed of Megan is 3.34 m/s.
Answer:
33.3
Explanation:
