2. A nutty professor leaves a history classroom and walks 10. meters north to a drinking fountain. Then, in a daze, she turns and walks 30. meters south to an art classroom. (draw vectors to scale to show your work)
(a) What is the professor’s total displacement from the classroom to the art classroom?
(b) What was the total distance she traveled?

Answer:

Vertebrate zoology

Explanation:

Have a great day!

Time taken by the skydiver to land on the ground is 4.57 seconds

The applied force, F = 700 N

The mass of the skydiver, m = 80 kg

The velocity, v = 40 m/s

To calculate the time taken by the skydiver to fall to the ground will be calculated using the formula

[tex]F=\frac{mv}{t}[/tex]

Substitute F = 700, m = 80, and v = 40 to solve for t

[tex]700=\frac{80(40)}{t}\\\\700t=3200\\\\t=\frac{3200}{700}\\\\t=4.57 s[/tex]

Time taken by the skydiver to land on the ground is 4.57 seconds

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Answer:

folk dancing is old music and ball room is young dancing

Answer:

a=Head

b=Tip

c=South pole of the magnet

d=North pole of the magnet

Gravitational Force

Explanation:-

The magma flows through the gap between the rocks which are stacked up on top of one another.

The gravity attracts the magma to do that.

Also the rocks are stacked because of the gravitational force between them

Hence Option D is correct

Hope it helps

Answer:

pls mark me as brainliest

Explanation:

last one

Answer:

373.15 Kelvin

Explanation:

Boiling Point of water

= 100 degree Celsius

= (100 + 273.15) Kelvin

= 373.15 Kelvin

Answer:

y-component = 45.75N

x-component = 0N

Explanation:

[tex]\text{Given that,}\\\\\text{Mass}~ m = 68 ~\text{g} = 68 \times 10^{-3}~ \text{kg}\\\\\text{Volume}~ V = 80 ~ \text{cm}^3 = 80 \times 10^{-6}~ \text{m}^3 = 8\times 10^{-5}~ \text{m}^3\\\\\\\text{Density} ~\rho = \dfrac mV = \dfrac{68\times 10^{-3}}{8 \times 10^{-5}} = 8.5 \times 10^2 = 850 ~ \text{kg}~\text{m}^{-3}[/tex]

Answer:

I hope this helps:)

Explanation:

good luck:)

Answer:

60n would be the needed force

The force needed to push a box weighing 30 N up a ramp that has an ideal mechanical advantage of 3 is equal to 10 N.

The mechanical advantage can be described as the ratio of the input force to the output force. The mechanical advantage of any machine can be determined by the ratio of the forces involved to do the work.

The ratio of the resistance force to the effort is called the actual mechanical advantage which will be comparatively less. The efficiency of a machine is always determined by equating the ratio of its output to its input.

The efficiency of the machine is equal to the ratio of the actual mechanical advantage (M.A.) and theoretical mechanical advantage. Mechanical advantage can be defined as the force produced by a machine to the force applied to it.

Given the load = 30 N and the ideal mechanical advantage = 3

Mechanical advantage = Load/ Effort

Input force or effort = Load/ M.A.

Force = 30/3

Input Force = 10 N

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#SPJ2

Answer:

This appears to be the formula for the electrical force between 2 objects.

You need this constant to calculate the electrical force between 2 charged objects.

Note that k = (1 / 4 * π ∈0)

∈0 is called the permittivity constant and ∈0 = 8.85E-12 farad/m

k = ( 1 / (4 * 3.28 * 8.85E-12) = 8.99E9

This makes F reasonably easy to remember as

F = 9E9 Q1 * Q2 / R^2 Newtons for the electrical force between Q! and Q2 separated by R

Answer:

Below!

Explanation:

Referring to the picture, we can conclude that the picture J will have the most effect on ocean tides.

Whichever model is similar to model J in the picture will be your answer.

Hoped this helped.

the answer is c, inherited traits are controlled by genes, which are found in chromosomes

did you end up getting the answer- im struggling

Answer:

I would think human error could cause differences like not pressing the stop watch on time.

Explanation:

cut the square and make it a triangle

Answer:

KE = 225000000 J

Explanation:

Answer:

10.954. from equation. KE=1/2m(v^2)

2KE=m(v^2) 6000=50(v^2) (v^2)=6000/50

(v^2)=120. square root for all v=10.954 m/s

The galaxy is 7200 km/s moving away from us.

A Hubble is a large telescope in space.

The question can be solved using the formula below

⇒ Formula

⇒ Where:

⇒ Given:

⇒ Substitute these values into equation 1

Hence, the galaxy is 7200 km/s moving away from us.

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Answer:

I just put a random number for the work

Explanation:

I hope it is ok.

Answer:

d. distance travelled

Explanation:

answer

Hey there! I'll try to provide you with my best answer.

Answer: In a collision, the velocity change is always computed by subtracting the initial velocity value from the final velocity value. If an object is moving in one direction before a collision and rebounds or somehow changes direction, then its velocity after the collision has the opposite direction as before.

1. • Here, force of gravity on the block = 20 N.

• Therefore, the normal force will also be the same, i.e., 20 N [According to Newton's Third Law, on every action, there is an equal and opposite reaction]

• The coefficient

[tex] u_{k} = 0.4[/tex]

• Force of friction =

[tex]u_{k} \times \: normal \: \: \: force \\ = 0.4 \times 20N \\ = 8N[/tex]

• Hence, the force of sliding friction between the block and the ground is 8 N.

• So, it is option c. 8 N

2. The answer is option d. continue in the same direction with no change in speed.

We know, force = mass × acceleration. When force is 0, then acceleration will also be 0 since mass cannot be 0. So, there will be no change in speed.

3. It is option b. force that is required to give a one kilogram object the acceleration of 1 m/s^2.

Newton is the SI unit of force. As mentioned earlier, force = mass × acceleration. The SI unit of mass and acceleration is Kg and m/s^2 respectively.

So, 1 N = 1 Kg × 1 m/s^2.

4. It is d. not zero.

Acceleration is the change in speed. So, if the force is zero, then acceleration will not occur.

5. Force = 2 N

Acceleration of the object A = 2 m/s^2.

Acceleration of the object B = 1 m/s^2.

Therefore, mass of the object A = 2 N ÷ 2 m/s^2 = 1 Kg

And, mass of the object B = 2 N ÷ 1 m/s^2 = 2 Kg

So, the mass of object B is greater than that of object A.

Hence, the answer is option c. Object B has more mass.

Hope you could get an idea from here.

Doubt clarification - use comment section.

By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s

Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:

mass m =  0.170 kg

initial speed u = 6 m/s.

Distance covered s = 61 m

To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V

To do this, let us first calculate the kinetic energy at which the ball move.

K.E = 1/2m[tex]U^{2}[/tex]

K.E = 1/2 x 0.17 x [tex]6^{2}[/tex]

K.E = 3.06 J

The work done on the ball is equal to the kinetic energy. That is,

W = K.E

But work done = Force x distance

F x S = K.E

F x 61 = 3.06

F = 3.06/61

F = 0.05 N

From here, we can calculate the acceleration of the ball from Newton second law

F = ma

0.05 = 0.17a

a = 0.05/0.17

a = 0.3 m/[tex]s^{2}[/tex]

To calculate the final velocity, let us use third equation of motion.

[tex]V^{2}[/tex] = [tex]U^{2}[/tex] + 2as

[tex]V^{2}[/tex]  = [tex]6^{2}[/tex] + 2 x 0.3 x 61

[tex]V^{2}[/tex] = 36 + 36

[tex]V^{2}[/tex] = 72

V = [tex]\sqrt{72}[/tex]

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

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Answer:

The effects of radiation depend on the type, energy, and location of the radiation source, and the length of exposure.

Answer:

3m

Explanation:

I can't make a drawing right know, but it's all around vectors. First of all the ball travels down 5 meters, so we have a vector pointing down, then it is reflected and travels 2 meters up. You have to sum the vectors so you obtain that the distance traveled is: 5-2 meters = 3m

Answer:

0.1g to 0.0000001g hope it helps uu

Answer:

a 6.5 m

Explanation:

[tex]f = kx = > x = \frac{f}{k} = \frac{78}{12} = 6.5 m[/tex]

The answer is that the resultant force is the vector sum of the component forces. In your simple example, where the component forces (F1 and F2) are at right angles to one another, as illustrated below, then the resultant force (r) is given by:

[tex]r= \sqrt{(f_{1}) {}^{2} +( f_{2})^{2} + 2f_{1}f_{2}. cos \: Θ, } \\ where \\ \: Θ, = angle \: between \: f_{1} \: and \: f_{2} \\ r = \sqrt{(2) {}^{2} + (5) {}^{2} + 2 \times 2 \times 5 \times cos90 } \\ \sqrt{4 + 25} = \sqrt{29} = 5.385 \\ nearest \: hundred = 5.39[/tex]

Answer: 3N to the right

Explanation: 2N- 5N= 3N