When did the object have the highest average
speed? *
5 points
Time (s)
30
Distance (m)
90
240
60
120
580
180
1080
O Between 0 and 30 seconds
Between 30 and 60 seconds
Between 60 and 120 seconds
Between 120 and 180 seconds

Answer:

The height at point of release is 10.20 m

Explanation:

Given:

Spring constant : K= 5 x 10 to the 3rd power n/m

compression x = 0.10 m

Mass of block m= 0.250 kg

Here spring potential energy converted into potential energy,

mgh = 1/2 kx to the 2 power

For finding at what height it rise,

0.250 x 9.8 x h = 1/2 x 5 x 10 to the 3 power x (0.10)to the 2 power) - ( g= 9.8 m/8 to the 2 power

h= 10.20

Therefore, the height at point of release is 10.20 m

Answer: Gravitational Forces

Explanation:

Answer:

F = Changing momentum / time

Explanation:

change in momentum = final momentum - initial momentum

Answer: 25%

Explanation:

From the question, we are informed that the temperature of source and sink of cannot engine are 400K

and 300K respectively.

The efficiency will be calculated as:

= 1 - t/T

= 1 - 300/400

= 1 - 3/4

= 1/4

= 25%

Answer:

In picture one because string telephones are best heard when there is more tension on the string

Explanation:

Answer: 15000 J

Explanation:

Given ;

Temperature change, dθ = 10

Mass of substance, m = 5 kg

Specific heat capacity, C = 300 j/kg°C

Quantity of heat, Q required is obtained using the relation ;

Q = mCdθ

Q = 5kg * 300 j/kg°C * 10°C

Q = 5 * 300 J * 10

Q = 15000 J

Answer:

The equivalent resistance is 2 ohms.

Explanation:

let the first resistance = R₁ = 10 ohm

let the second resistance = R₂ = 10 ohm

let the third resistance = R₃ = 10 ohm

let the fourth resistance = R₄ = 10 ohm

let the fifth resistance = R₃ = 10 ohm

The equivalent resistance is calculated as;

[tex]\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5} \\\\\frac{1}{R_T} = \frac{(R_2R_3R_4R_5)+(R_1R_3R_4R_5)+(R_1R_2R_4R_5) +(R_1R_2R_3R_5)+(R_1R_2R_3R_4)}{R_1R_2R_3R_4R_5} \\\\R_T = \frac{R_1R_2R_3R_4R_5}{(R_2R_3R_4R_5)+(R_1R_3R_4R_5)+(R_1R_2R_4R_5) +(R_1R_2R_3R_5)+(R_1R_2R_3R_4)} \\\\R_T = \frac{(10^5)}{(10^4)+(10^4)+(10^4)+(10^4)+(10^4)} \\\\R_T = \frac{10^5}{5(10^4)} \\\\R_T = \frac{10}{5} \\\\R_T = 2 \ ohms[/tex]

Therefore, the equivalent resistance is 2 ohms.

Answer:

gap junctions,tight junctions,and desmosomes

Answer:

[tex]\mathrm{a.\:}123\:\mathrm{kg\cdot m/s},\\\mathrm{b.\:}95\:\mathrm{kg\cdot m/s},\\\mathrm{c.\:}28\:\mathrm{kg\cdot m/s}[/tex]

Explanation:

The momentum of an object is given by:

[tex]p=mv[/tex], where [tex]m[/tex] is the mass of the object and [tex]v[/tex] is the velocity of the object.

a)

The mass of the child and bike together is [tex]19+5.6=24.6[/tex] kilograms. Since they're moving at a velocity of 5.0 m/s, their momentum is:

[tex]p=24.6\cdot 5=\fbox{$123\:\mathrm{kg\cdot m/s}$}[/tex].

b)

The mass of the child is given as 19 kg. Since the child is on the bike moving at 5.0 m/s, it's implied the child is as well. Therefore, the momentum of the child is:

[tex]p=19\cdot 5=\fbox{$95\:\mathrm{kg\cdot m/s}$}[/tex].

c) The mass of the bike is given as 5.6 kg and it is moving at 5.0 m/s. Therefore, the momentum of the bike is:

[tex]p=5.6\cdot 5=\fbox{$28\:\mathrm{kg\cdot m/s}$}[/tex]

Answer:

a = 7.6\ mph/s

Explanation:

Motion With Constant Acceleration

It's a type of motion in which the velocity of an object changes uniformly in time.

The equation that describes the change of velocities is:

[tex]v_f=v_o+at[/tex]

Where:

a   = acceleration

vo = initial speed

vf  = final speed

t    = time

Solving the equation [for a:

[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]

The car accelerates from vo=0 to vf=60 mph in t=7.9 s, thus the acceleration is:

[tex]\displaystyle a=\frac{60 \ mph-0}{7.9}[/tex]

a = 7.6\ mph/s

Answer:

By teh way is isn't it question of law or science and the picture is of what electric light or not I have read it so I was asking isn't is question of science

Answer:

This is an example of Inelastic colission

Explanation:

Step one:

given:

mass of moose m1 = 620 kg

mass of train m2= 10,000kg

Initial velocity of moose  u1= 0 m/s

Initial velocity of train v1 = 10m/s

combined velocity of the system is given as v

Applying the conservation of momentum equation we have

m1u1+ m2u1= (m1+m2)V

substitutting we have

620*0+10000*10= (620+10000)V

100000= 10620V

divide both sides by 10620

V = 100000/10620

V=9.41m/s

The velocity of the moose after impact is 9.41m/s

Answer:

0J

Explanation:

work done= force x distance in direction of force.

since distance=0, work done=0

Answer:

3 m³ will occupy more volume.

Explanation:

We need to tell which occupies more volume : 3m³ or 100 ft³.

To compare, we first make the units same.

We know that,

1 m = 3.28 feet

1 m³ = 35.31 foot³

3 m³ = 105.94 foot³

Now, we can compare 105.94 foot³ and 100 ft³. It can be clear that 3 m³ will occupies more volume.

Answer:

Power of 5 watts is used

Explanation:

Mechanical Work and Power

Mechanical work is the amount of energy transferred by a force.

Being F the force vector and s the displacement vector, the work is calculated as:

[tex]W=\vec F\cdot \vec s[/tex]

If both the force and displacement are parallel, then we can use the equivalent scalar formula:

W=F.s

Power is the amount of energy converted per unit of time. The SI unit of power is the watt, equal to one joule per second.

The power can be calculated as:

[tex]\displaystyle P=\frac {W}{t}[/tex]

Where W is the work and t is the time.

It's required to calculate the power used in t=30 seconds when W=150 Joules of work are completed. Substitute in the formula:

[tex]\displaystyle P=\frac {150}{30}[/tex]

P = 5 Watt

Power of 5 watts is used

Answer:

Explanation:

while it is still falling it is turned into kinetic energy. While on the floor it turns onto potential energy. Remember energy cannot be created or the destroyed so the other ones are automatically out.

Answer:

50km i think but i am not very sure

Answer:

[tex]Position:(3.38m ,2.53m)[/tex]

[tex]Velocity=(1.72m/s,1.22m/s)[/tex]

Explanation:

From the question we are told that

Mass of particle 1  

 [tex]M_1=2.0kg[/tex]

Co-ordinate of particle 1 (2.0m,6.0m)

Velocity of Particle 1  

 [tex]V_1= (3.1 m / s , 2.6 m / s )[/tex]

Mass of particle 2  

[tex]M_2=4.5kg[/tex]

Co-ordinate of particle 2 (4.0m,1.0m)

Velocity of Particle 1  

[tex]V_2=( 1.1 m / s , 0.6 m / s )[/tex]

Generally the Position is mathematically given as  

[tex]X =\frac{ ((M_1 * Cx_1) + (M_2 *Cx_2)}{(M_1 + M_2)}[/tex]

 [tex]X=\frac{2*2+4.5*4}{2+4.5}[/tex]

[tex]X=3.38[/tex]

[tex]Y=\frac{(M_1* Cy_1 + M_2* Cy_2)}{(M_1 + M_2)}[/tex]

[tex]Y =\frac{(2 * 6 + 4.5 * 1)}{(2 + 4.5)}[/tex]

[tex]Y= 2.53 m[/tex]

Therefore the position is given as

[tex]Position:(3.38m ,2.53m)[/tex]

 Solving for Velocity

Generally the velocity of the system is mathematically Given as

[tex]V_x =\frac{ (M_1 * vx_1 + M_2 * Vx_2)}{(M_1 + M_2)}[/tex]

[tex]V_x=\frac{ (2 * 3.1 + 4.5 *1.1)}{(2 + 4.5)}[/tex]

[tex]V_x=1.72m/s[/tex]

For Y

[tex]V_y =\frac{(M_1* vy_1 + M_2* Vy_2)}{(M_1 + M_2)}[/tex]

[tex]V_y=\frac{ (2 * 2.6) + (4.5*0.6)}{(2 + 4.5)}[/tex]

[tex]V_y=1.22m/s[/tex]

Therefore Velocity

[tex]V=(1.72m/s,1.22m/s)[/tex]

Answer:

yo they deleted my answer. The answer is 0N

Explanation:

so when two forces pull on an object from opposite sides with the same force (in this case its 20N), then the object is in equilibrium at 0N.

So its clear that there is one person on the the opposite side.

SOOO  generally: (left or down) would be considered negative in an equation. And the other person  (right or up) would be considered positive. So if both forces are the same numbers on opposite sides then the answer is 0 (if you add both of them).

0 is the number of equilibrium.

OK, so the equation would be -20N + 20N and then badda bing badda boom viola, the answer: 0N

thanks for coming to my TED talk. I hope they don't delete this answer.

Answer:

5 x 10⁻⁷N

Explanation:

Given parameters:

Mass of object 1  = 100kg

Mass of object 2 = 300kg

Distance  = 2m

Unknown:

Force of gravitational attraction between the objects  = ?

Solution:

Newton's law of universal gravitation states that the gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

From Newton's law of universal gravitation we derive an expression:

       Fg  =  [tex]\frac{Gm_{1} m_{2} }{r^{2} }[/tex]

G is the universal gravitation constant = 6.67 x 10⁻¹¹

m is the mass

r is the distance between the bodies

Now insert the parameters and solve;

    Fg  = 6.67 x 10⁻¹¹ x [tex]\frac{100 x 300}{2^{2} }[/tex]    = 5 x 10⁻⁷N

Answer:

the relative error is similar in all of them, which is why they all have the same precision.

Explanation:

For this question we must be careful since there are several important concepts to take into account:

* the absolute error that is directly related to the precision or error in the mean given by the instrument

* the relative error to take into account the precision of the instrument and the magnitude of the measurement

The precisions of the different measuring instruments are

metric conta           ± 0.1 cm

micrometric screw ±0.001 cm

vernier                    ± 0.005 cm or 0.001 cm

goniometers           ± 1º

I tilt meters              ± 1º

Mobile applications varies a lot, but on the order of  ±0.01cm

We can see that the tape measure, the micrometric screw and the vernier are the only ones that give us a direct measurement, in the others some calculation must be made to obtain the distance reading, for which the error must also be propagated for the calculation.

We must also take into account that the vernier and micrometric screw are for short measurements only a few centimeters, the meters allow a medium to about 20 m, for measurements of more distance the other instruments are needed

If we only take into account the absolute error, the device with a smaller error is the most accurate, but this is not very correct.

The precision must be related to the magnitude of the measurement carried out, that is why the error or relative uncertainty was defined.

Let's take an example with the tape measure.

If we measure a distance of 1 cm the relative error is ±0.1, for a distance of 10 cm the relative error is ±0.01 which is very good

if we use a vernier to measure 2 cm the error is ±0.0025

if we use an inclinometer or a goniometer to measure a distance of 100 m in error it is of the order of ±0.09 m

With a cell phone it depends on the form of measurement, but all programs involve the measurement of time of a pulse of light, and assume a constant speed of light regardless of the refractive index of the medium that changes this speed. In principle this could be a very precise method, but you must know the calculation procedure and the approximations used.

As we can see, to give a correct answer we must use the relative error in this case the instruments that use the optical measurement method should be the most accurate, but the software for the calculation can involve large approximations.

Of the other instruments, the relative error is similar in all of them, which is why they all have the same precision.

Answer:

Angular velocity of merry-go-round is πk - 1 at t= T

Explanation:

From the question it is given that

[tex]\theta(t) = \pi k(t+k_e-\frac{t}{k} )[/tex] ..........................(1)

since mathematically, angular velocity is defined as

[tex]\omega(t) = \frac{d\theta(t)}{dt}[/tex] ........................(2)

on substituing the value of θ(t) from equation 1 in equation (2) we get

[tex]\omega(t) = \frac{d\theta(t)}{dt}[/tex] = [tex]\frac{d\pi k (t + k_e - \frac{t}{k} )}{dt}[/tex]   ............................(3)

on differentiating equation (3) with respect to time we get

ω(t) = πk(1 -[tex]\frac{1}{k}[/tex]) = πk - 1 angular velocity of merry-go-round

Therefore, angular velocity of merry-go-round is πk - 1 at t= T

The angular velocity of the merry-go-round at t=T is πk - 1. The pace at which angular displacement changes is defined as angular velocity.

The rate of change of angular displacement is defined as angular velocity.

It is stated as follows:

ω = θ t

The equation for the angular displacement is given as;

[tex]\theta (t) = \pi k (t+k_e-\frac{t}{k} )[/tex]

The angular velocity is found by the differentiation of the angular displacement.

[tex]\rm \omega(t) = \frac{d(\theta)}{dt} \\\\ \rm \omega(t) = \frac{d (\pi k (t+k_e-\frac{t}{k} )}{dt} \\\\ \rm \omega(t) = \pi k-1[/tex]

Hence the angular velocity of the merry-go-round at t=T is πk - 1.

To learn more about the angular speed refer to the link;

https://brainly.com/question/9684874

Answer:

  v_{f} = -0.693 m / s

Explanation:

The acceleration of the runner can be obtained from Newton's second law

       F = m a

        a = F / m

the bold are vectors, therefore the acceleration throughout the journey varies as the force has variations.

For the part of finding the velocity of the body we can use the relationship between the momentum and the variation of the momentum

            I = Δp

           ∫ F Δt = m [tex]v_{f}[/tex] - m v₀

           int F dt = m (v_{f}-v₀)

1) To find the change in velocity we must find the area under the curve of the graph, this can be done analytically if we know the functional of the curve or approximate it by intervals

a) between 0 <t <0.20 s

v) between 0.20 <t <0.30 s

a reasonable curve shape can be a Gaussian.

2) If we do not have the form of the cure, we can perform a graphical integration to find the area under the curve, we can do this by dividing the curve into small rectangles, finding the area of ​​each one and adding them.

3) Another even more approximate way is to create an average force in each interval and find the area of ​​this force, the average force is the average value of the force in the interval, let's use this method in the exercise

a) first interval 0 <t <0.20 Average force [tex]F_{mean}[/tex] = 300 N

    area = F_{mean} Δt

    area = I = 300 0.20

    I = 60 N s

the speed change is

    I = m Δv

    Δv = I / m

    Δv = 60/65

    Δv = 0.923 m / s

If we assume that the runner starts from rest, his final velocity is v = 0.923 m / s in the direction of the force.

b) second interval 0.2 <t <0.30s average force F_mean = 150 N

    area = I = 150 (0.30 - 0.20)

    I = 15 N s

the speed change is

    Δv = 15/65

    Δv = 0.23 m / s

Note that in this case the initial speed is not zero and since the two impulses are in the opposite direction the speed decreases

        [tex]v_{f}[/tex]= -0.923 + 0.23

        v_{f} = -0.693 m / s

Answer:

[tex]v_2 = 6.406 / cos25.29[/tex]

Explanation:

From the question we are told that

Speed of snow ball 7.6m/s

Speed of chunk 9.3m/s at [tex]19 \textdegree[/tex]

Generally the equation for the conservation  of momentum is mathematically given as

Let the snow ball be b

and the chunk b/2

According to conservation of momentum we have

        [tex]b_u = (b /2) v_1 cos\theta_1 + (b / 2) v_2cos\theta_2[/tex]

        [tex]v_2 cos\theta_2 = 2 * 7.6 -9.3 * cos_19[/tex]

        [tex]v_2 cos\theta_2 = 15.2 -8.793 = 6.406[/tex]    

Therefore

          [tex](b/2)v_1 sin\theta_1 = (b/2) v_2sin\theta_2[/tex]

          [tex]v_2 sin\theta_2 = 9.3 * sin_19[/tex]

          [tex]v_2 sin\theta_2 = 3.027[/tex]

Mathematically From  above equations

        [tex]tan\theta_2 = 0.4726[/tex]

        [tex]\theta_2 = 25.29 \textdegree[/tex]

Therefore the speed and direction of second chunk

        [tex]v_2 = 6.406 / cos25.29[/tex]

        [tex]v2 = 7.0855m/s[/tex]

Answer:

Looks like  streak color.

Explanation:

Answer:

184 feets

Explanation:

Given the data:

time (sec) __ velocity (ft/sec)

0 __________30

1 __________ 54

2 __________56

3 __________34

4 __________ 8

5 __________ 2

6 __________22

Using left end approximation:

(0,1) ___ f(0) = 30

(1,2) ___ f(1) = 54

(2,3) ___f(2) = 56

(3,4) ___f(3) = 34

(4,5) ___f(4) = 8

(5,6) __ f(5) = 2

Hence, the Total distance traveled during the 6 second interval is:

Change ; dT = 1

1 * (30 + 54 + 56 + 34 + 8 + 2) = 184

Answer:

65N

Explanation:

Answer:

65 N

Explanation:

i just did the assignment on edge 2021