What are 3 things you could you do this week to help you connect better with kids in
your classes?

Answer:

b 1.39 m/s²

Explanation:

Given the following data;

Time = 12 seconds

Distance, S = 100 m

Since it's starting from rest, the initial velocity is equal to 0m/s.

To find the acceleration, we would use the second equation of motion;

[tex] S = ut + \frac {1}{2}at^{2}[/tex]

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

100 = 0(12) + ½*a*12²

100 = 0 + 72

100 = 72a

Acceleration, a = 100/72

Acceleration, a = 1.389 ≈ 1.39 m/s²

Answer:

grams

Explanation:

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car [tex]V_{c}[/tex] = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

[tex]V_{f}[/tex]  = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

so

[tex]V_{f}[/tex] =  [tex]V_{i}[/tex] + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁  = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁  will be

[tex]d_{c}[/tex] = [tex]V_{c}[/tex] × t₁

we substitute

[tex]d_{c}[/tex] = 22.352 × 7

[tex]d_{c}[/tex]  = 156.46 m

now distance between polive car and speeding car

Δd =  [tex]d_{c}[/tex] - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( [tex]V_{f}[/tex] - [tex]V_{c}[/tex] )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = [tex]V_{f}[/tex] × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b)  how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

Answer:

C: Equal to

Explanation:

In calculating workdone, it is pertinent to know that it doesn't depend on time. The only relationship between work and time is when we want to calcite power where workdone/time taken = power.

Now, even if it took 2T time to love the same distance, it just means lesser force was used but still the workdone doesn't change.

Thus, the workdone in the first case will be equal to the workdone in the second case.

Explanation:

The energy of the system before the collision must equal the energy after the collision.

After the collision the bullet and the block have a total mass of 18.41 kg and they move at a speed of 8.8 m/s. The kinetic energy after the collision is

[tex]\frac{18.41 kg (8.8 m/s)^2}{2} = 713 J[/tex]

Before the collision only the bullet has kinetic energy.

So we can now determine the speed of the bullet using

[tex]\frac{0.11kg (v^2)}{2} = 713 J\\v = 114 m/s[/tex]

Answer:

The time of motion is 0.64 s.

Explanation:

Given;

mass of the apple, m = 107 g

height of fall, h = 2 m

The velocity of the apple when it hits the ground is calculated from the law of conservation of energy;

[tex]P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\gh = \frac{1}{2} v^2\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8\times 2} \\\\v = 6.261 \ m / s[/tex]

The time of motion is calculated;

v = u + gt

6.261 = 0 + 9.8t

6.261 = 9.8t

t = 6.261 / 9.8

t = 0.64 s

Therefore, the time of motion is 0.64 s

The time taken for the apple to hit the ground is 0.64 s.

The time taken for the apple to hit the ground can be calculated using the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Solve for t.

Hence, The time taken for the apple to hit the ground is 0.64 s

Learn more about time taken here: https://brainly.com/question/4931057

Answer:

7 would be C, a cell.

Explanation:

Hi.

7 would be C, a cell.

A cell is the basic unit of structure and function in all living things.

If it is living, it is made of cells.

Hope this helps.

Answer:

7. Cell

8. Organelle

Explanation:

The vertical component of the salmon's velocity is 7.2 m/s x sin 29 = 3.49 m/s

If g = 9.81 m/s^2, the salmon takes

(3.49 m/s) / (9.81 m/s^2) = 0.356 s to reach the highest point of its trajectory.

It takes another 0.356 s to fall back into the water again.

So the salmon is out of the water for a total of 0.712 s.

In this time the salmon travels horizontally with a velocity of 7.2 m/s x cos 29 = 6.30 m/s

We can now calculate the horizontal distance tavelled by multiplying the horizontal velocity by the time spent out of water;

0.712 s x 6.30 m/s = 4.48 m

Answer:

the required speed/velocity of the stunt double is 13.633 km/h

Explanation:

Given the data in the question;

velocity of train V = 150 km/h

distance = length of train + distance between the tunnel and the end

= 2 km + 20 km = 22 km

first we calculate time t taken by the train to reach the tunnel;

t = distance / velocity

we substitute

t = 22 km / 250 km/h

t = 0.1467 hr

so the velocity of the of the stunt double will be;

velocity = distance / time

we substitute

velocity = 2 km / 0.1467 hr

velocity = 13.633 km/h

Therefore, the required speed/velocity of the stunt double is 13.633 km/h

Answer:

84kliometers

Explanation:

divide one hundred and sixty eight kilo meters by two hours

when the photosynthetic began to pump free oxygen into oceans, the ocean had enough oxygen to support the life of non-photosynthetic organisms. So, non-photosynthetic organisms came into being.

Answer:

its x and y component is 24.749m/s

Explanation:

Given

Speed of the dog = 35m/s

x component of the speed = xcos theta

y component of the speed = ycos theta

Given theta =45 degrees

x-component = 35cos45

x-component = 35(0.7071)

x-component = 24.749m/s

y-component = 35sin45

y-component = 35(0.7071)

y-component = 24.749m/s

Hence its x and y component is 24.749m/s

Answer:

Explanation:

The distance of searchlight will act as radius R and velocity of car may be supposed to be tangential velocity v  . We are required to calculate  angular velocity ω .

v = 80 ft /s

R = 3000 ft

ω = v / R

= 80  / 3000 = .027 rad / s

For angular acceleration the formula is

angular acceleration α = a / R

a is linear acceleration = 15 ft / s²

α = 15 / 3000 = .005 rad / s².

Answer:

8.5 hours

Explanation:

Answer:

in the acceleration process the quantity α and w must increase

the deceleration process the alpha quantity must constant  a direction opposite to the angular velocity

Explanation:

Acceleration and angular velocity are related to linear

           v = w xr

            a = αx r

The bold letters indicate vectors and the cross is a vector product, therefore if

we can see that the relationship between linear and angular variables is direct

therefore in the acceleration process the quantity α and w must increase as well as their linear counterparts

in the deceleration process the alpha quantity must constant as the linear acceleration and must have a direction opposite to the angular velocity

Answer:

Untrasound

Explanation:

Your welcome :)

Answer:

Water > Box of books > Stone > Ball

Explanation:

We'll begin by calculating the potential energy of each object. This can be obtained as follow:

For stone:

Mass (m) = 15 Kg

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

Height (h) = 3 m

Potential energy (PE) =?

PE = mgh

PE = 15 × 10 × 3

PE = 450 J

For water:

Mass (m) = 10 Kg

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

Height (h) = 9 m

Potential energy (PE) =?

PE = mgh

PE = 10 × 10 × 9

PE = 900 J

For ball:

Mass (m) = 1 Kg

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

Height (h) = 20 m

Potential energy (PE) =?

PE = mgh

PE = 1 × 10 × 20

PE = 200 J

For box of books:

Mass (m) = 25 Kg

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

Height (h) = 2 m

Potential energy (PE) =?

PE = mgh

PE = 25 × 10 × 2

PE = 500 J

Summary:

Object >>>>>>>> Potential energy

Stone >>>>>>>>> 450 J

Water >>>>>>>>> 900 J

Ball >>>>>>>>>>> 200 J

Box of books >>> 500 J

Arranging from greatest to least, we have:

Object >>>>>>>> Potential energy

Water >>>>>>>>> 900 J

Box of books >>> 500 J

Stone >>>>>>>>> 450 J

Ball >>>>>>>>>>> 200 J

Water > Box of books > Stone > Ball

Answer:

[tex]v=\dfrac{1200}{t}\ m/s[/tex]

Explanation:

Given that,

The car traveled a total of 1,200 meters during this test.

We need to find the average speed of the car. The average speed of the car is given by total distance covered divided by the time taken. So,

[tex]v=\dfrac{1200}{t}\ m/s[/tex]

But putting the value of t we can find the average speed of the car.

Answer:

a)  v = 4.37 10⁶ m / s,  speed is much less than c

b)     v = 2.01 10⁸  m / s,  this value is 67% of the speed of light, , for which relativistic corrections should be used

Explanation:

The bohr model for the hydrogen atom and dendroids is a classical model with a quantization of the angular momentum

let's start by using Newton's second law with the electric force

         F = m a

Coulomb's law electric force

         F = [tex]k \frac{q_1q_2}{r^2}[/tex]

in this case in an atom the number of protons is equal to the atomic number and there is only one electron

        q₁ = Ze

        q₂ = e

acceleration is centripetal

         a = v² / r

we substitute

        [tex]k \frac{Z e^2}{r^2} = m \frac{v^2}{r}[/tex]

        v² =  [tex]k \frac{Ze^2}{m r}[/tex]

quantization is imposed without justification in this model,

       L = p x r = n [tex]\hbar[/tex]

       \hbar= h /2π

if we consider circular orbits, the speed and position are perpendicular

       m v r = n \hbar

       r = [tex]\frac{n \hbar}{m v}[/tex]

we substitute

     v² = [tex]k \frac{Z e^2}{m} \frac{m v}{n \hbar}[/tex]

     v = [tex]k \frac{Z e^2 }{ n \hbar}[/tex]

let's apply this equation

     \hbar= h / 2π

     \hbar= 6.626 10-34 / 2π

     \hbar= 1.05456 10⁻³⁴ J s

a) He1 ion,  the atomic number of helium is 2

      v =  [tex]\frac{9 \ 10^9 \ 2 ( 1.6 \ 10^{-19})^2 }{n \ 1.0546 \ 10^{-34}}[/tex]

       v =4.3695 10⁶ / n m / s

the ground state occurs for N = 1

        v = 4.37 10⁶ m / s

the relationship of this value to the speed of light is

        v / c = 4.37 10⁶/3 10⁸

        v / c = 1.46 10⁻²

speed is much less than c

b) the uranium ion with atomic number Z = 92

       v = [tex]\frac{9 \ 10^9 \ 92 ( 1.6 \ 10^{-19})^2 }{n \ 1.054 \ 10^{-34} }[/tex]

       v = 2.01 10⁸  m / s

       v/c = [tex]\frac{2.01 \ 10^8 }{3 \ 10^8}[/tex]

       v/c =  0.67

this value is 67% of the speed of light, for atoms with a higher atomic number the effects are increasingly important, for which relativistic corrections should be used

Answer: 1. The law of consevation of energy sates that energy can neither be created nor destroyed. It can only be transformed or transfered from one form to another. The law of conservation of energy is found everywhere for example, Water falls from the sky, converting potential energy to kinetic energy.

2. Different forms of energy are related because energy cannot be created or destroyed. they can all be transformed into from one form to another.

Explanation:

Answer:

Speed of the melon = 0.25 m/s

we would normally don't see the melon moving due to friction with the resting surface.

Explanation:

We use conservation of momentum:

Pi = Pf

with Pi = 0.1 kg * 30 m/s = 3 kg m/s

and Pf = 0.1 kg * 20 m/s + 4.0 kg * V = 2 kg m/s + 4 * V

Then using the equality above, we solve for V (velocity of the melon)

3 kg m/s = 2 kg m/s + 4 V

1 kg m/s = 4 kg * V

Then V = 1 / 4  M/s = 0.25 m/s

So we would normally don't see the melon moving due to friction with the resting surface.

Answer:

I.72m/s²

II.8m/s²

Explanation:

acceleration equal velocity² divided by length

Answer:

the magnitude of the tension on the cable necessary to keep the system in equilibrium is 1060.9 N

Explanation:

Given that;

Length L = 10 m

mass of beam m_b = 150 kg; weight W_beam = 150×9.8

mass of sign m = 75 kg

distance of sign hung from the beam from the wall d = 2.50 m

angle ∅ = 60°

g = 9.8 m/s²

Now,

Torque acting at one end of the beam will be;

[tex]T_{net}[/tex] = Tsin∅ × L - mg(d)-W × (L/2)

for equilibrium, [tex]T_{net}[/tex] = 0

therefore, 0 = Tsin∅ × L - mg(d)-W × (L/2)

so we substitute

Tsin(60°) × 10 - 75×9.8(2.50) - 150 × 9.8× (10/2) = 0

Tsin(60°) × 10 - 1837.5 - 7350 = 0

Tsin(60°) × 10 - 9187.5 = 0

Tsin(60°) × 10 = 9187.5

divide both side by 10

Tsin(60°) = 918.75

T × 0.8660 = 918.75

T = 918.75 / 0.8660

T = 1060.9 N

Therefore, the magnitude of the tension on the cable necessary to keep the system in equilibrium is 1060.9 N

The magnitude of the tension on the cable that keep the system in equilibrium is 1060.9 N.

Torque acting at one end of the beam,  

= Tsin∅ × L - mg(d)-W × (L/2)

When equilibrium  = 0  

Tsin∅ × L - mg(d)-W × (L/2) = 0

Where,

L - Length  =  10 m

m - mass of sign bord= 75 kg

g- gravitational accelaration = 9.8 m/s²

W - weight of beam =  150×9.8 = 1470 kg

Put the values in the formula,

Tsin(60°) × 10 - 75×9.8(2.50) - 150 × 9.8× (10/2) = 0  

Tsin(60°) × 10 - 1837.5 - 7350 = 0  

Tsin(60°) × 10 - 9187.5 = 0  

Tsin(60°) × 10 = 9187.5    

Tsin(60°) = 918.75  

T × 0.8660 = 918.75  

T = 918.75 / 0.8660  

T = 1060.9 N

Therefore, the magnitude of the tension on the cable that keep the system in equilibrium is 1060.9 N.

To know more about magnitude of the tension,

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Answer:

Q = 1.2*10⁻¹² C

Explanation:

       [tex]C = \frac{Q}{V} (1)[/tex]

       [tex]C = \frac{\epsilon_{0}*\epsilon _{r} * A}{d} (2)[/tex]

       [tex]Q =\frac{\epsilon_{0} * \epsilon_{r} *A* V}{d} = \frac{(8.85*(4.7)^{2}*79.5)e-24 (F/m*m2*V)}{1.3e-8m} = 1.2e-12 C = 1.2 pC (3)[/tex]

Answer:

The balls velocity is 1 divided by 3

The velocity of the ball is 18.85 m/s.

The ball’s centripetal acceleration is 236.87 m/s².

The ball's centripetal force is  118.44 Newton.

Centripetal acceleration is a characteristic of an object's motion along a circular path. Centripetal acceleration applies to any item travelling in a circle with an acceleration vector pointing in the direction of the circle's center.

Given parameters:

length of the string: l = 1.5 meters.

Time interval = 60 seconds.

Total number of complete rotation = 120.

Hence, the velocity of the ball = 120×2π×1.5/60 m/s

= 18.85 m/s.

The ball’s centripetal acceleration = (velocity)²/ radius

= (18.85)²/1.5 m/s²

= 236.87 m/s²

The ball's centripetal force = mass × centripetal acceleration

= 0.5 × 236.87 Newton

= 118.44 Newton

Learn centripetal acceleration here:

https://brainly.com/question/14465119

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Answer:

B. in both directions until the temperature is equal in the water and the air

Explanation:

When a warm body is in contact with a cool body , there is exchange of heat energy in both sides until there is attainment of equilibrium temperature . At this temperature both the body attains equal temperature . Initially rate of heat radiated by warm body is more than that from cool body , but after attainment of equilibrium , the rate becomes equal to each other . This is called dynamic equilibrium .

Hence option B is correct .

Answer:

A. 2.63 s B. 12.38 m

Explanation:

A. For what time interval (in s) after the light turned green is the bicycle ahead of your car?

The time interval at which the bicycle is ahead of the car is the time it takes for the car to reach the bicycle's speed of 21.0 mi/h.

So, using v = u + at where u = initial speed of car = 0 mi/h, v = final speed of car = 21.0 mi/h, a = acceleration of car = 8.00 mi/h/s and t = time taken for acceleration.

So, v = u + at

t = (v - u)/a

substituting the values of the variables into the equation, we have

t = (21.0 mi/h - 0 mi/h)/8.00 mi/h/s

= 21.0 mi/h ÷ 8.00 mi/h/s

= 2.63 s

B. By what maximum distance does the bicycle lead the car?

To find this distance, we find the distance moved by both the car in this time of t = 2.63 s

So, using s = ut + 1/2at² where u = initial speed of car = 0 mi/h = 0 m/s, t = time = 2.63 s, a = acceleration of car = 8.00 mi/h/s = 8.00 × 1609 m/3600 s = 3.58 m/s/s = 3.58 m/s² and s = distance moved by car.

So, substituting the values of the variables into the equation, we have

s = ut + 1/2at²

s = 0 m/s × 2.63 s + 1/2 × 3.58 m/s² × (2.63 s)²

s = 0 m + 1/2 × 3.58 m/s² × (2.63 s)²

s =  1.79 m/s² × 6.9169 s²

s = 12.38 m

which is also the maximum distance with which the bicycle leads the car.