"When the ball leaves the ramp at Point B, students measure the horizontal" distance traveled. They repeat the experiment five times, being careful to release the ball from the same starting Point A and find the average horizontal distance traveled to be 2.0 m. One student suggests they use a stopwatch to find the time the ball is in the air whereas another student suggests they use a meter stick. Given these materials, describe the procedure students should follow to minimize error and calculate the speed of the ball as it leaves the ramp.

Answer:

B or C

Explanation:

i don't if I'm right or not

Answer:

ANSWER : 2000 / 40 = 50 m / s

Answer:

We know that the gravitational acceleration in the surface of the Earth can be written as:

g = G*M/r^2

Where:

M = mass of the Earth

r = radius of the Earth.

G = gravitational constant.

The weight of an object of mass m, is written as:

W = m*g = m*(G*M/r^2)

Now, if we move our object to a place that has a mass equal to 1/6 times the mass of the Earth, and 1/3 the radius of the earth.

The gravitational acceleration on this planet is written as:

g' = G*(M/6)/(r/3)^2 = (1/6)*(G*M)/(r^2/9) = (9/6)*(G*M/r^2) = (3/2)*g

then the weight on this planet is:

W' = m*g' = m*(3/2)*g = (3/2)*(m*g)

and m*g was the weight on Earth, then:

W' = (3/2)*(m*g) = (3/2)*W

The new weight is 3/2 times the weight on Earth.

Answer:

[tex]1.28 \times 10 ^{-3} m^3/s[/tex]

Explanation:

Given that the radius of the pipe, r =0.0112 m

So, the area of the cross-section, [tex]a= \pi (0.0112)^2 = 3.941\times 10^{-4} m^2[/tex]

Speed of water in the pipe, v=3.25 m/s

Volume flow rate =[tex]av= 3.941\times 10^{-4} \times 3.25 = 1.28\times 10 ^{-3} m^3/s[/tex]

Hence, the volume flow rate is [tex]1.28 \times 10 ^{-3} m^3/s[/tex]

Answer:

For Acellus the answer is 0.00128

Explanation:

Trust me

Answer:

1. 57.99V

2. 37.5Hz

3. 0.7072A

4. 82 ohms

5. 5.18x10^-5F

Explanation:

In answer to this question, we have the Standard equation of AC emf to be

V = V0 x sin ωt

We have

V0 = 82V,

ω = 75π

1.

RMS Voltage =

V0/√2 = 82 /√2

= 82/1.414

= 57.99V

2.

ω = 2π* f

75π = 2πf

Frequency,f = 75π/2π

= 235.5/6.28

= 37.5 Hz

3.

RMS current

= Imax/√2

= 1.00/1.414

= 0.7072A

4.

Reactance

= Vrms/ Irms

= 57.99/0.7072

= 81.999

= 82.0 Ω

5.

Reactance = 1/ ω x C

Reactance = 82

ω = 75π

We put these values into the equation above and make c the subject of the formula

C = 1/82.0 x 75π

C = 1/ 82.0 x 75 x 3.14

C = 1/19311

Capacitance = 5.18x10^-5F

Answer:

v = 37.8 m/s

Explanation:

       [tex]F_{cent} = F_{g} - F_{n} (1)[/tex]

       [tex]v_{max} = \sqrt{g*r} =\sqrt{9.8 m/s2*146 m} = 37.8 m/s (5)[/tex]

Answer:

[tex]P=18,933.3Pa=18.9kPa[/tex]

Explanation:

Hello!

In this case, since we can compute the volume of the brick as shown below:

[tex]V=0.1m*0.1m*0.2m=0.002m^3[/tex]

Next, we can compute the mass of the brick given its density:

[tex]\rho =m/V\\\\m=V*\rho\\\\m=0.002m^3*19,300kg/m^3\\\\m=38.6kg[/tex]

Now, since the force exerted on the table corresponds to the weight of the brick, we use the gravity to obtain:

[tex]W=38.6kg*9.81m/s^2=378.7N[/tex]

Finally, since the surface of the brick in contact with the table corresponds to the 0.1x0.2 area (length and width), the area on which the weight force is exerted is:

[tex]A=0.1m*0.2m=0.02m^2[/tex]

Therefore, the pressure is:

[tex]P=\frac{F}{A}=\frac{W}{A}=\frac{378.7N}{0.02m^2}\\\\P=18,933.3Pa=18.9kPa[/tex]

Best regards!

Answer:

BMI is a simple indicator of weight for height and can't differentiate between muscle mass and fat mass. So BMI tends to overestimate the health risk for adults with a high muscle mass, such as some athletes, and underestimate the risk for adults with a low muscle mass, as can occur with sedentary lifestyles.

Explanation:

Hope it is helpful...

Answer:

The answer is 5.0

Explanation:

Hope that helped :D

Answer:

adbce i think

Explanation:

Answer:

vf = v₁/3 + 2v₂/3

Explanation:

Using the law of conservation of linear momentum,

momentum before impact = momentum after impact

So, Mv₁ + 2Mv₂ = 3Mv (since the railroad cars combine) where v₁ = initial velocity of first railroad car, v₂ = initial velocity of the other two coupled railroad cars, and vf = final velocity of the three railroad cars after impact.

Mv₁ + 2Mv₂ = 3Mvf

dividing through by 3M, we have

v₁/3 + 2v₂/3 = vf

vf = v₁/3 + 2v₂/3

Answer:

The answer is below

Explanation:

The equation for a linear line graph is given by:

y = mx + b, where y and x are variables, m is the slope of the graph and b is the y intercept (that is value of y when x is zero).

The slope (m) of a line passing through two points [tex](x_1,y_1)\ and\ (x_2,y_2)[/tex] is given by:

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

A) The first line passes through the point (0, 0) and (10, 60). It is represented as (time, velocity). Hence the slope is:

[tex]m=\frac{60 -0}{10-0}=6\ m/min^2[/tex]

B) The second line passes through the point (10, 60) and (15, 60). Hence the slope is:

[tex]m=\frac{60 -60}{15-10}=0\ m/min^2[/tex]

C) The third line passes through the point (15, 60) and (40, -40). Hence the slope is:

[tex]m=\frac{-40 -60}{40-15}=-4\ m/min^2[/tex]

D) The third line passes through the point (40, -40) and (55, 0). Hence the slope is:

[tex]m=\frac{0 -(-40)}{55-40}=2.67\ m/min^2[/tex]

Answer:

I believe the answer is sea floor spreading

Answer:

a) [tex]a=33.73mm/s^{2}[/tex]

b) mg>N

c) [tex]\%_{change}=0.343\%[/tex]

d) [tex]a=24.07mm/s^{2}[/tex]

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

[tex]a_{c}=\omega ^{2}r[/tex]

where:

[tex]\omega=\frac{2\pi}{T}[/tex]

we know the period of rotation of the earth is about 24 hours, so:

[tex]T=24hr*\frac{3600s}{1hr}=86400s[/tex]

so we can now find the angular speed:

[tex]\omega=\frac{2\pi}{86400s}[/tex]

[tex]\omega=72.72x10^{-6} rad/s^{2}[/tex]

So the centripetal acceleration will be:

[tex]a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)[/tex]

which yields:

[tex]a_{c}=33.73mm/s^{2}[/tex]

b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

[tex]\sum F=0[/tex]

so we get that:

[tex]N-mg+ma_{c} = 0[/tex]

and solve for the normal force:

[tex]N=mg-ma_{c}[/tex]

In this case, we can clearly see that:

[tex]mg>mg-ma_{c}[/tex]

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

[tex]mg=(60kg)(9.81m/s^{2})=588.6N[/tex]

and let's calculate the normal force:

[tex]N=m(g-a_{c})[/tex]

[tex]N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})[/tex]

N=586.58N

so now we can calculate the percentage change:

[tex]\%_{change} = \frac{mg-N}{mg}x100\%[/tex]

so we get:

[tex]\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%[/tex]

[tex]\%_{change}=0.343\%[/tex]

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

[tex]cos \theta = \frac{AS}{h}[/tex]

In this case:

[tex]cos \theta = \frac{r}{R_{E}}[/tex]

so we can solve for r, so we get:

[tex]r= R_{E}cos \theta[/tex]

in this case we'll use the average radius of earch which is 6,371 km, so we get:

[tex]r = (6371x10^{3}m)cos (44.4^{o})[/tex]

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

[tex]a=\omega ^{2}r[/tex]

[tex]a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m[/tex]

[tex]a=24.07 mm/s^{2}[/tex]

In this exercise we have to use the knowledge of mechanics to solve the magnitude of the acceleration and the acceleration of gravity, in this way we find that:

a)[tex]a=33.73 mm/s^2[/tex]

b)[tex]mg>N[/tex]

c) [tex]Change=0.343%[/tex]

d) [tex]a=24.07 mm/s^2[/tex]

Then calculating from the information given in the text;

a)In this case, we know the centripetal acceleration is given by the following formula:

[tex]a_c=w^2r\\w=\frac{2\pi}{T} \\T=24*\frac{3600}{60} = 86400 s[/tex]

so we can now find the angular speed:

[tex]w=\frac{2\pi}{86400} \\w=72.72*10^{-6} rad/s^2[/tex]

So the centripetal acceleration will be:

[tex]a_c=(72.72*10^{-6}rad/s^2)^2(6478*10^3m)\\a_c=33.73mm/s^2[/tex]

b)So we can do a sum of forces in equilibrium:

[tex]\sum F=o\\N-mg+ma_c=0\\N-mg+ma_c=0\\N=mg-ma_c\\mg>mg-ma_c\\mg>N[/tex]

This exist cause the centripetal increasing speed happen pulling united states of america upwards, that will create the size of the normal force tinier than the result or goods created of the bulk times the increasing speed of importance.

c) So let's calculate our weight and normal force:

[tex]mg=(60)(9.81)=588.6N\\N=m(g-a_c)\\N=(60)(9.81-33.73*10^{-3})\\N=586.58 N[/tex]

So now we can calculate the percentage change:

[tex]\%change= \frac{mg-N}{mg}*100\% \\=\frac{588.6-586.58}{588.6}*100\% \\=0.343\%[/tex]

d) There, we can see that the radius can be found by using the cos function:

[tex]cos\theta=\Delta S/h\\cos\theta=r/R_E\\r=R_E cos\theta\\r=(6371*10^3)cos(44.4)\\r=4,551.91 km[/tex]

And now we can calculate the acceleration at that point:

[tex]a=w^2r\\a=(72.72*10^{-6})^2(4,551.91*10^3)\\a=24.07 mm/s^2[/tex]

See more about acceleration at brainly.com/question/2437624

Answer:

88.14 m/s

Explanation:

From the question given above, the following data were obtained:

Radius (r) = length of blade = 80 m

Revolution (rev) = 1

Time (t) = 5.7 s

Velocity (v) =?

The velocity of the blade can be obtained by using the following formula:

v = (rev × 2πr) / t

NOTE: Pi (π) = 3.14

v = (1 × 2 × 3.14 × 80) / 5.7

v = 502.4 / 5.7

v = 88.14 m/s

Therefore, the velocity of the blade is 88.14 m/s

Given that,

6/7 and 24/28

To find,

The above number are in proportion or not.

Solution,

First number is 6/7

Other number is 24/28

Multiply numerator and denominator by 4 to the first number.

[tex]\dfrac{6}{7}=\dfrac{6}{7}\times \dfrac{4}{4}\\\\=\dfrac{6\times 4}{7\times 4}\\\\=\dfrac{24}{28}[/tex]

So, 6/7 and 24/28 are in proportion.

Fraction A is proportional to fraction B because the value of fraction A is equal to fraction B.

Given the following data:

In this exercise, you're required to determine whether or not the two fractions are proportionate i.e whether fraction A is proportional to fraction B.

Two fractions are considered to be proportionate when the value of fraction A is equal to fraction B.

We would write fraction B in its simplest form by dividing both the numerator and denominator by their highest common factor (HCF).

HCF of 24 and 28 = 4.

Dividing by 4, we have:

[tex]Fraction\;B =\frac{24}{28}=\frac{6}{7}[/tex]

In conclusion, fraction A is proportional to fraction B because the value of fraction A is equal to fraction B.

Read more: https://brainly.com/question/18676621

Answer:

375 J

Explanation:

Work done is the product of force and distance moved by the object.

W=F*d

Given that ;

Force= F= 250 N

Distance= d =1.5 m

W= 250 * 1.5 =375 J

Answer:

9800 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 5 kg

Height (h) = 200 m

Potential energy (PE) =?

Potential energy (PE) is the energy stored in an object by virtue of its location. Mathematically, it is expressed as:

PE = mgh

Where:

PE => is the potential energy.

m => is the mass of the object.

g => is the acceleration due to gravity.

h => is the height to which the object is located.

With the above formula, we can obtain the potential energy of the ball at its original height as follow:

Mass (m) = 5 kg

Height (h) = 200 m

Acceleration due to gravity (g) = 9.8 m/s²

Potential energy (PE) =?

PE = mgh

PE = 5 × 9.8 × 200

PE = 9800 J

Therefore, the potential energy of the ball at its original height is 9800 J

Answer:

All the answers are solved and explained below.

Explanation:

Note: This questions lacks the initial and most necessary data to answer these following questions. I have found a related question. I will be considering that question to carry out the answers.

Question: A car with a mass of 1000 kg is at rest at a spotlight. when the light turns green, it is pushed by a net force of 2000 N for 10 s. (This was the information missing in this question).

Data Given:

m = 1000 kg

F = 2000N

t = 10s

Q1 Solution:

Acceleration = a = ?

F = ma

a = F/m

a = 2000/ 1000

a = 2 [tex]m/s^{2}[/tex]

Q2: Solution:

Change in velocity = Δv = ?

acceleration = change in velocity / time

a = Δv/t

Δv = axt

Δv = 2 x 10

Δv = 20 m/s

Q3: Solution:

Impulse = I = ?

Impulse = Force x time

I = 2000 x 10

I = 20000 Ns

Q4: Solution:

Change in Momentum = Δp = ?

Δp = mΔv

Δp = 1000 x 20

Δp = 20000 Kgm/s

Q5: Solution:

Final velocity of the car at the end of 10 seconds = vf = ?

Δp = m x Δv

Δp = m x (vf-vi)

Δp = 1000 x (vf - 0 )

20000 = 1000 x vf

vf = 20000/1000

vf = 20 m/s

Q6: Solution:

Change in momentum the car experiences as it continues at this velocity?

Δp = ?

Δp = mΔv

Δp = m x (0)

Δp = 0

Q7: Solution:

Impulse = Change in momentum

Impulse = Δp

Implulse = 0

Q8: Solution:

Change in momentum = Δp = mΔv

Δp = m(vf-vi)

Δp = 1000 x (0-20)

Δp = -20000 kgm/s

Q9: Solution:

Impulse = Δp

Impulse = -20000 Ns

Q10: Solution:

Impulse = ?

Impulse = F x t

F = impulse/t

F = -20000/4s

F = -5000 N

Q11: Solution:

F = ma

a = ?

a = F/m

a = -5000/1000

a = -5[tex]m/s^{2}[/tex]

Explanation:

Helium is the type of atom, with 2 protons in the nucleus.

He-3 (which is the proper form), is an isotope of helium that has an atomic number of 3 (so 2 protons and 1 electron).

The most abundant helium isotope is He-4, just an extra fact.

Answer:

D

Explanation:

Answer:

[tex]magnetite[/tex]

[tex]hope \: helps...[/tex]

Answer:

shivering raises your body temperature

Answer:

The speed and direction of the tangled players is 2.89 m/s towards right.

Explanation:

Given;

mass of the player heading right, m₁ = 40 kg

initial velocity of the player heading right, u₁ = 9.0 m/s

mass of the player heading left, m₂ = 50 kg

initial velocity of the player heading left, u₂ = -2.0 m/s

let the speed of the tangled players = v

Apply the law of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(40 x 9) + (50 x - 2) = v(40 + 50)

360 - 100 = v(90)

260 = v(90)

v = 260 / 90

v = 2.89 m/s (towards right, since it is in position direction)

Therefore, the speed and direction of the tangled players is 2.89 m/s towards right.

Answer:

spring 1 has less potential energy

Explanation:

since spring 2 is clampped down further it will have a greater effect but since sping 1 is barely clamped down it wont show as much energy