A bicyclist travels 30 km in 2 hours east. Calculate its velocity?

Answers

Answer 1

Answer:

Its velocity would be

15 km per hour

or

0.00416667 per second


Related Questions

Which scientist is credited with having the greatest contribution to early microscopy and was the first to observe and describe single-celled organisms?

Answers

Answer:

Antonie van Leeuwenhoek

Explanation:

Market researchers were interested in the relationship between the number of pieces in a brick-building set and the
cost of the set. Information was collected from a survey and was used to obtain the regression equation ý =
0.08x + 1.20, where x represents the number of pieces in a set and ŷ is the predicted price in dollars) of a set.
What is the predicted price of a set that has 500 pieces?
$40
$41.20
$600
$6,235

Answers

Plug in 500 for x and solve. 500(0.08) = 40

Then do 40 + 1.2 = 41.20

Answer B

For a horizontally projectile object if the initial velocity doubled, the Vertical displacement (angle y) will be :

Answers

Answer:

Unchanged

Explanation:

horizontal projection always has initial vertical velocity equal to zero.

gravity works pretty much the same at any point on earth and will accelerate any falling object at the same rate traveling the same vertical distance in the same time. Neglecting air resistance of course.

By the way, Δy is read "delta y" not "angle y" which would be "∠ y"

delta represents a change in something, in this case the y coordinate.

50 Points Help Asap !!

Briefly describe V-T graph For every point please if its increasing if its flat or if its positive or negative help fast

Answers

#A

It's flat as no change can be seen on the graph of A part

#B

Its increasing as the line goes up in B part.

#C.

It's flat as no change can be seen on the graph of C part

#D

It's decreasing as the graph goes down in D part.

#E

It's decreasing as the graph goes down in E part

#F

It's decreasing as the graph goes down in F part.

#G

It's increasing as the graph goes up in G part.
G partt hope helpsss

what is the approximate distance from the surface of the earth center 2900km 700km 50000km 6400km

Answers

Answer:

6400km is the closest

Explanation:

Which force, in real life, will have the least effect on a bowling ball rolling down a lane toward bowling pins?

A) magnetism

B) air resistance

C) gravity

D) friction

Answers

Answer:

Its Friction

Explanation:

the pins are not floating and they are not a magnet and not involved with air

The force, in real life, that will have the least effect on a bowling ball rolling down a lane toward bowling pins is magnetism. The correct option is A.

What is magnetism?

Magnetism is basically the force which indeed magnets exert when they attract or even repel one another. The movement of electric charges resulting in magnetism.

Every substance is composed of tiny units referred to as atoms. Each atom contains electrons, which are charged particles.

To increase stability, the pins themselves have a low center of gravity. Because they are spherical in shape, they can roll and strike other pins in a variety of directions.

The force acting on the bowling ball is friction and air resistance. The friction force is equal to the friction coefficient multiplied by the normal force, and thus mass times acceleration.

Thus, the correct option is A.

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What is non examples of enlarged

Answers

reduction

deflation

these are examples of the opposite of enlarged to make something smaller is really the key thing here

A meter stick is attached to one end of a rigid rod with negligible mass of length l = 0.302 m. The other end of the light rod is suspended from a pivot point, as shown in the figure below. The entire system is pulled to a small angle and released from rest. It then begins to oscillate. A meter stick hung from a rod of length l. The rod is attached to the ceiling. The rod and meter stick extend downward in a straight line making a small angle with the vertical. (a) What is the period of oscillation of the system (in s)? (Round your answer to at least three decimal places.)

Answers

The period of oscillation of the system nearest to three decimal places

= 1.092 seconds

The period of an oscillation occurring in a system is the time taken to complete one cycle.

The formula that is used to calculate the period of oscillation (T) is

                = 2π√[tex]\frac{l}{g}[/tex]

But,

π = 3.14159 (constant)

g= 10m/s² (acceleration due to gravity)

l = 0.302 m

Therefore T = 2 × 3.14159 × √[tex]\frac{0.302}{10}[/tex]

                    = 6.28318 x √0.0302

                    = 6.28318 x 0.17378

                    = 1.09189s

                    = 1.092 seconds ( to the nearest three decimal places)

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In a car crash, how are force, mass, acceleration, and velocity related?

Answers

Answer:

the answer is simple but click on the brainiest and thank me.

Explanation:

F=force, m=mass, a=acceleration, v=velocity

v=displcement/time

a= v/t

F= m×a

F= m× (v/t)

If you do 72 J of work in 1.2 seconds, how much power is produced

Answers

Answer:

Explanation:

Power is the rate of doing work

P = 72 / 1.2 = 60 Watts

1
2) Mighty Mouse can lift seven mammoths with a total mass of 525 kg. Using the acceleration of
gravity, find the total WEIGHT that Mighty Mouse is lifting. (answer: 5250 N)

Answers

Answer:

5250N

Explanation:

Weight= Mass × acceleration due to gravity

= 525kg × 10m/s^2

= 5250N

Am object of mass in a circular path of radius 100metres with the speed of 10metres/second.calculate the acceleration towards the center​

Answers

Answer:

is 6

Explanation:

h

When an elastic object is changed from its original shape:
A:Energy is released
B:Work is done
C:It is ruined
D:It makes a twanging sound

Answers

Answer:

deformation : elastic deformation is reversed when the force is removed. inelastic deformation is not fully reversed when the force is removed – there is a permanent change in shape.

Explanation:

tysm

Vesta is a minor planet (asteroid) that takes 3.63 years to orbit the Sun.
Calculate the average sun -Vesta distance

Answers

Using Kepler's third law, the average sun -Vesta distance is 2.36 AU.

According to Kepler's laws, the square of the period of revolution of planets are proportional to the cube of their average distances from the sun. Hence, we can write; [tex]T^{2} =r^{3}[/tex]

Where;

T = period of the planet

r = average distance of the planet

When;

T =  3.63 years

r = [tex]\sqrt[3]{T^2}[/tex]

r = [tex]\sqrt[3]{(3.63)^2}[/tex]

r = 2.36 AU

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Before the collision, the toy train is travelling at 0.5 m/s. The train and the stationary
truck both have a mass of 75 g.
Calculate the momentum of the toy train before the collision?

Answers

Answer:

Explanation:

p = mv = 0.075(0.5) = 0.0375 kg•m/s

Option B.
Consider a setup in which two springs are attached to a mass in parallel.
Convince yourself that in this setup, the compression of each spring must be the same. Using
this fact, derive the effective spring constant for springs in parallel
This is asking, "ll1 replace the two springs by a single imaginary spring, what would its spring
constant be such that the force stays the same?" Your answer should only depend on k, and k

Answers

Answer:

it would be...

Explanation:

A Scooter has a mass of 250 kg. A constant force is exerted on it for 6.0 s. During the time the force is exerted, the scooter increases its speed from 6.00 m/s to 280 m/s. What is the magnitude of the force exerted on the scooter

Answers

Answer:

917 N

Explanation:

917 N, this is your answer!!

Glad to help.

 what is the difference between repelling and attracting

Answers

Answer:

Attracting means pulling toward you and repelling means pushing away

Explanation:

Answer: Repelling is when something will not connect with another object. The force will cause a repel between the two objects. Attracting is when something is attracted or being pulled to another object.

Explanation: Hope this helps!

A girl of mass m1=60.0 kilograms springs from a trampoline with an initial upward velocity of vi=8.00 meters per second. At height h=2.00 meters above the trampoline, the girl grabs a box of mass m2=15.0 kilograms. (Figure 1)

For this problem, use g=9.80 meters per second per second for the magnitude of the acceleration due to gravity.

What is the speed vbefore of the girl immediately before she grabs the box?
Express your answer numerically in meters per second.

What is the speed vafter of the girl immediately after she grabs the box?
Express your answer numerically in meters per second.

What is the maximum height hmax that the girl (with box) reaches? Measure hmax with respect to the top of the trampoline.

Answers

The conservation of momentum and energy allows to shorten the results for the movement of the girl on the trampoline holding the box are:

     a) the girl's speed is v = 4.98 m / s

     b) The speed of the girl + box system is: v_f = 0.996 m / s

     c) the maximum height is: y = 2.05 m

 

Kinematics studies the movement of bodies, looking for relationships between the position, velocity and acceleration of bodies.

The momentum is defined by the product of mass and the velocity, when a system is isolated the momentum is conserved.

The mechanical energy is the sum of the kinetic energy plus the potential energies, when there is no friction in the system the mechanical energy is conserved.

Let's solve this exercise in parts:

a) Let's use kinematics to find the speed of the girl before she grabs the box

              v² = v₀² - 2 g y₁

              v² = 8² - 2 9.8 2.00

              v = R 24.8 = 4.98 m / s

b) Let's use momentum conservation for when the speed of the girl and the box together. Let's write the moment in two moments.

Initial instant. Just before you grab the box.

              p₀ = M v + 0

Final moment. Right after taking the box

             [tex]p_f[/tex]  = (m + M) [tex]v_f[/tex]

         

In system this form by the girl and the box therefore it is an isolated system and the momentum is conserved.

           [tex]p_o = p_f[/tex]  

           mv = (m + M) [tex]v_f[/tex]  

           [tex]v_f = \frac{m}{m+M} \ v[/tex]

Let's calculate

           [tex]v_f = \frac{15}{15+ 60} \ 4.98[/tex]  

           [tex]v_f[/tex]  = 0.996 m / s

c) Now we use conservation of energy after the girl has the box.

Starting point. When the girl has the box

           Em₀ = K + U

           Em₀ = ½ (m + M) v² + (m + M) g y₁

Final point. At the highest point of the trajectory

          [tex]Em_f[/tex]  = U

         [tex]Em_f[/tex] = (m + M) g y₁

As there is no friction, the energy is conserved.

           [tex]Em_o = Em_f[/tex]  

          ½ (m + M) v² + (m + M) g y₁ = (m + M) g y

          y = [tex]\frac{v^2}{2g} + y_1[/tex]  

Let's calculate

           y = [tex]\frac{0.996^2}{2 \ 9.8} + 2.0[/tex]

           y = 2.05 m

In conclusion using the conservation of momentum and energy we can shorten the results for the movement of the girl on the trampoline holding the box are:

     a) the girl's speed is v = 4.98 m / s

     b) The speed of the girl + box system is: v_f = 0.996 m / s

     c) the maximum height is: y = 2.05 m

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Answer:

Vbefore = 4.98 m/s

Vafter = 3.98 m/s

Hmax = 2.81 m

A 15 g bullet is fired at 630 m/s into a 4.9 kg block that sits at the edge of a 75-cm-high table. The bullet embeds itself in the block and carries it off the table.

How far from the point directly below the table's edge does the block land?

Answers

The horizontal distance traveled below the table's edge is 29.4 cm.

The given parameters;

mass of the bullet, = 15 g = 0.015 kgspeed of the bullet, = 630 m/smass of the block, = 4.9 kg

The final velocity of the bullet-block system is calculated as follows;

[tex]m_1 u_1 + m_2u_1 = v(m_1 + m_2)\\\\0.015(630) + 4.9(0) = v(0.015 + 4.9)\\\\9.45 = 4.915 v\\\\v = \frac{9.45}{4.915} \\\\v = 1.923 \ m/s[/tex]

The time for the bullet-block system to reach the ground from the table is calculated as follows;

[tex]h = v_0t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\2h = gt^2\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 0.75}{9.8} }\\\\t = 0.153 \ s[/tex]

The horizontal distance traveled below the table's edge is calculated as follows;

[tex]X = vt\\\\X = 1.923 \times 0.153\\\\X = 0.294 \ m\\\\X = 29.4 \ cm[/tex]

Thus, the horizontal distance traveled below the table's edge is 29.4 cm.

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An object moves 45m due East and then 15m due West, the total distance is

Answers

Answer:30m east

Explanation: 45m-15m=30m, which cancels out the one to the west. All that is left is 30m east.

Is this a test/quiz btw?

Alex is x years old . June is 7years older than Alex . their total combined ages is 29 years . find June,s age . show all work algebraically

Answers

Answer:

18 years

Explanation:

Given,

Let Alex be = x years

Then June will be = (7 + x) years

We know that,

Their total combined age is 29 years

Therefore,

By the problem,

=> x + (7 + x) = 29

=> 2x = 29 - 7

=> 2x = 22

=> x = 22 ÷ 2

=> x = 11

So,

Required age of June is = (7 + x) years

= (7 + 11) years

= 18 years (Ans)

Which of Newton’s Laws of motion help explain the following scenario?
A paddle wheel boat pushes on the water and the water pushes back causing the boat to move

Answers

Answer:

Newton #3

Explanation:

For every action there is an equal and opposite reaction.

Acceleration of a car that speeds from 4.3 m/s to 12.9 m/s in 2 seconds

Answers

Explanation:

let v1 = 4.3 m/s

v2 = 12.9 m/s

t = 2 seconds

v2 = v1 + at

12.9 = 4.3 + a×2

2a = 12.9 - 4.3 = 8.6

a = 8.6/2

a = 4.3 m/s^2

A proton is released in a uniform electric field and it experiences an electric force of 2.18x10^-14N toward south .What are the magnitude and direction of the electric field

Answers

Answer:

F = E q

E = 2.18E-14 / 1.6E-19 = 1.36E5 N/C

The direction of the field will be the same as force on the proton - southerly

1.25 is closer to 1.04 or not ?
plz heelp plz

Answers

Answer:

No, it is closer to 1.30

Explanation:

A driver entering the outskirts of a city takes her foot off the accelerator so that the car slows down from 90 km/h to 50 km/h in 10.0s. Find the car’s average acceleration

Answers

Answer:

Explanation:

a = (vf - vi) / t

a = (50 - 90) / 10.0

a = -4 km/h/s(1000 m/km / 3600 s/h)

a = - 1.11 m/s²

Colette launches an air rocket in the upward, positive direction. It launches with an initial velocity of 25.5 m/s. It accelerates in the downward, negative direction at a rate of 9.81 m/s2. After 3.5 seconds, what is the magnitude of the rocket's displacement?

Answers

Answer:

Give me some hint please

What on earth is equal to 9.8m/s/s

Answers

Answer:

Acceleration due to gravity

A rocket ship has several engines and thrusters. While the Solid Rocket Booster (SRB) and main engines only work together during the first 2 minutes of flight, the main engines operate for a total of 8.5 minutes after the launch. Once the SRBs are released, the main engines alone accelerate the rocket from about 1341 m/s to 7600 m/s.

What is the acceleration of the SRB and main engine during the first 2.0 minutes of flight?

A. 52 m/s2
B. 13 m/s2
C. 9.8 m/s2
D. 11 m/s2

Answers

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

A rocket ship has several engines and thrusters. We can divide its initial movement into 2 parts:

From t = 0 min to t = 2.0 min, the SRB and the main engines act together and the speed goes from 0 m/s (rest) to 1341 m/s.From t = 2.0 min to t = 8.5 min, the main engines alone accelerate the ship form 1341 m/s to 7600 m/s.

We want to know the acceleration in the first part (first 2.0 minutes). We need to consider that:

The speed increases from 0 m/s to 1341 m/s.The time elpased is 2.0 min.1 min = 60 s.

The acceleration of the ship during the first 2.0 minutes is:

[tex]a = \frac{\Delta v }{t} ) \frac{(1341m/s-0m/s)}{2.0min} \times \frac{1min}{60s} = 11 m/s^{2}[/tex]

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

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