QUCIK!! SOMEONE PLEASE HELP! I’LL MARK BRAINLIEST!!

Answer:

B

Explanation:

Motion is movement, the teacher's movement is motion

Answer:

We know the momentum after the collision MUST be equal to the momentum BEFORE the collision.  

Momentum is a VECTOR quantity having both magnitude and direction.  The first ball has momentum P =m*v = 2*4 = 8 at 90degrees.  The second ball has momentum P = 1*8 = 8 at -90 or 270 degrees.  They sum to zero when you perform vector addition.

Explanation:

Answer:

It is refraction

Explanation:

Well we know it has to be greater than 300,000 km/s since we can't see it.

We can't calculate it any closer than that using the given information.

Answer:

the force needed to give the truck the acceleration is 29,760 N.

Explanation:

Given;

mass of truck, m = 4800 kg

acceleration of the truck, a = 6.2 m/s²

The force needed to give the truck the acceleration is calculated as;

F = ma

F = 4800 x 6.2

F = 29,760 N

Therefore, the force needed to give the truck the acceleration is 29,760 N.

Answer:

Oscillation whose amplitude reduce with time are called damped oscillation. This happen because of the friction. In oscillation if its amplitude doesn't change with time then they are called Undamped oscillation

Damped and undamped vibration refer to two different types of vibrations. The main difference between damped and undamped vibration is that undamped vibration refer to vibrations where energy of the vibrating object does not get dissipated to surroundings over time, whereas damped vibration refers to vibrations where the vibrating object loses its energy to the surroundings.

Answer:

D

Explanation:

Answer:

jjnn ok jjjmkkmmkijnnkko

Explanation:

Effective communication with patients will enable one to know the needs of the patient better as well as reducing the barriers to understanding each other for both parties.

To be an effective communicator while educating patients, the person must:

Answer:

megaliter > kiloliter > liter >centiliter >mililiter > deciliter > nanoliter

Explanation:

plz mark brainlest

Answer:

d. About 112 meters.

Explanation:

From the question, John's displacement forms a right angle triangle as below.

Using Pythagoras theorem,

a² = b²+c²....................... Equation 1

Where a = John's displacement from the starting point, b = 100 m, c = 50 m

Substitite these values into equation 1

a² = 100²+50²

a² = 10000+2500

a² = 12500

a = √12500

a = 111.8 meters.

a = about 112 meters.

The right answer is d. About 112 meters.

Answer:

A. 240 J

B. 60 J

C. 90 J

D. 0 J

E. 50 J

Explanation:

A. Determination of the potential energy of the coconut while it is still in the tree

Mass (m) = 2 Kg

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

Height (h) = 12 m

Potential energy (PE) =.?

PE = mgh

PE = 2 × 10 × 12

PE = 240 J

B. Determination of the potential energy of the coconut when it hits the bystander on the head,

Mass (m) = 2 Kg

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

Height (h) = 3 m

Potential energy (PE) =.?

PE = mgh

PE = 2 × 10 × 3

PE = 60 J

C. Determination of the potential energy of the coconut when it bounces up to its maximum height,

Mass (m) = 2 Kg

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

Height (h) = 3 + 1.5 = 4.5 m

Potential energy (PE) =.?

PE = mgh

PE = 2 × 10 × 4.5

PE = 90 J

D. Determination of the potential energy of the coconut when it lands on the ground,

Mass (m) = 2 Kg

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

Height (h) = 0 m

Potential energy (PE) =.?

PE = mgh

PE = 2 × 10 × 0

PE = 0 J

E. Determination of the potential energy of the coconut when it rolls into a ground hole, and falls 2.50 m to the bottom of the hole.

Mass (m) = 2 Kg

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

Height (h) = 2.50 m

Potential energy (PE) =.?

PE = mgh

PE = 2 × 10 × 2.50

PE = 50 J

(a) The potential energy of the coconut relative to the ground while it is still in the tree is 235.2 J.

(b) The potential energy of the coconut relative to the ground when it hits the bystander on the head is 58.8 J.

(c) The potential energy of the coconut relative to the ground when it bounces up to its maximum height is 88.2 J.

(d) The potential energy of the coconut relative to the ground when it lands on the ground is 0 J.

(e) The potential energy of the coconut when it rolls into a groundhog hole, and falls 2.50 m to the bottom of the hole is 49 J.

The given parameters;

The potential energy of the coconut relative to the ground while it is still in the tree;

[tex]P.E = mgh\\\\P.E = 2 \times 9.8 \times 12\\\\P.E = 235.2 \ J[/tex]

The potential energy of the coconut relative to the ground when it hits the bystander on the head;

[tex]P.E = 2 \times 9.8 \times 3 \\\\P.E = 58.8 \ J[/tex]

The potential energy of the coconut relative to the ground when it bounces up to its maximum height;

[tex]P.E = 2 \times 9.8 (1.5 + 3)\\\\P.E = 88.2 \ J[/tex]

The potential energy of the coconut relative to the ground when it lands on the ground;

[tex]P.E = 2 \times 9.8 \times 0\\\\P.E = 0 \ J[/tex]

The potential energy of the coconut relative to the ground when it rolls into a groundhog hole, and falls 2.50 m to the bottom of the hole;

[tex]P.E = 2\times 9.8 \times 2.5 \\\\P.E = 49 \ J[/tex]

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Answer: The answer is D. KNO3

Explanation:

The graph shows that the KN03 going straight up from the temperature sign so you reversed that so that it will make it to 90°C to 30°C

To solve this we must be knowing each and every concept related to solubility. Therefore, the correct option is option D among all the given options.

The greatest amount of one material that may be dissolved in the other is referred to as its solubility. It is the most solute that may be dissolved into a solvent near equilibrium, resulting in a saturated solution.

When specific circumstances are satisfied, more solute can be dissolved further than the solubility limit point, resulting in a supersaturated solution. Adding extra solute after saturation or supersaturation does not enhance the concentration in the solution. Rather, the excess solute begins to precipitated out of solution. KNO[tex]_3[/tex] is the compound that would have the greatest percentage recovered after cooling a saturated solution of that compound from 90°C to 30°C.

Therefore, the correct option is option D.

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Answer:

25 m/s

Explanation:

200/8 = 25

Answer:

f

Explanation:

f

Answer:

hello your question is incomplete attached below is the complete and the required circuit diagrams

answer :

Ai) This proves that when the current across the resistor is doubled the value of the voltage across the resistor doubles as well

B) It is better to take more measurements in the range 20mA < I < 40mA because of the amount of temperature reached by the bulb and the change in resistance is affected by the temperature

hence At 0 mA current, there won't be any noticeable change

Explanation:

Ai) The voltage across the resistor will double when you double the current through the resistor

Given that : V = I*R.  

lets assume : I = 2 amperes , R = 3 ohms

V = 2*3 = 6 v

secondly lets assume double the value of  (I)   i.e. I = 4 amperes

hence : V = 4*3 = 12 volts

This proves that when the current across the resistor is doubled the value of the voltage across the resistor doubles as well

Aii) Showing the two data points from simulation

I1                    V1            I2= 2I1         V2=2V1       V1/ I1 =V2/I2

0.9*10^3     9 * 10^3     1.8*10^3       18*10^3          10 ohms

1.6 * 10^3    16 * 10^3    3.2*10^3     32*10^3         10 ohms

B) It is better to take more measurements in the range 20mA < I < 40mA because of the amount of temperature reached by the bulb and the change in resistance is affected by the temperature

hence At 0 mA current, there won't be any noticeable change

it's up in Gogle trust me

Answer:

Please see below as the answer is self explanatory.

Explanation:

This question is incomplete, the missing diagram is uploaded along this Answer below.

Answer:

a) the work done on the cart by the spring is 4.875 lb-ft

b) the work done on the cart by its weight is - 3.935 lb-ft

Explanation:

Given the data in the question;

(a) determine the work done on the cart by the spring

we calculate the work done on the cart by the spring as follows;

[tex]W_{spring}[/tex] = 1/2×k( [tex]x^{2} _{1}[/tex] - [tex]x^{2} _{2}[/tex] )

where k is spring constant ( 3 lb/in )

we substitute  

[tex]W_{spring}[/tex] = 1/2 × 3( (-8)² - (5)² )      

[tex]W_{spring}[/tex] = 1/2 × 3( 64 - 25 )

[tex]W_{spring}[/tex] = 1/2 × 3( 39 )

[tex]W_{spring}[/tex] = 58.5 lb-in

we convert to pound force-foot

[tex]W_{spring}[/tex] = 58.5 × 0.0833333 lb-ft

[tex]W_{spring}[/tex] = 4.875 lb-ft

Therefore, the work done on the cart by the spring is 4.875 lb-ft

b) the work done on the cart by its weight

work done by its weight;

[tex]W_{gravity}[/tex] = -mgsin∅( x₂ - x₁ )        

we substitute in of values from the image below;

[tex]W_{gravity}[/tex] = -14 × sin(15°)( 5 - (-8) )  

[tex]W_{gravity}[/tex] = -14 × 0.2588 × 13

[tex]W_{gravity}[/tex] = -47.1  lb-in

we convert to pound force-foot

[tex]W_{gravity}[/tex] = -47.1 × 0.0833333 lb-ft

[tex]W_{gravity}[/tex] = - 3.935 lb-ft

Therefore, the work done on the cart by its weight is - 3.935 lb-ft

a) the work done on the cart by the spring is 4.875 lb-ft.

b) the work done on the cart by its weight is - 3.935 lb-ft.

a. The work done on the cart by the spring is

= 1/2 × 3( (-8)² - (5)² )      

= 1/2 × 3( 64 - 25 )

= 1/2 × 3( 39 )

= 58.5 lb-in

Now we have to convert to pound force-foot

So,

= 58.5 × 0.0833333 lb-ft

= 4.875 lb-ft

b) Now

work done by its weight;

= -mgsin∅( x₂ - x₁ )        

So,

= -14 × sin(15°)( 5 - (-8) )  

= -14 × 0.2588 × 13

= -47.1  lb-in

Now we convert to pound force-foot

= -47.1 × 0.0833333 lb-ft

= - 3.935 lb-ft

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Answer:

a. d/2 mid-way between the charges.

b. d/2 mid-way between the charges.

Explanation:

(a) Find the location of all points, if any, where the electric potential is zero.

Since the charges are of equal magnitude and opposite charge and separated by a distance, d, the electric potential due to the +Q charge is V = kQ/x and that due to the -Q charge is V' = -kQ/(d - x) where x is the point of zero electric potential.

The potential is zero when  V + V' = 0, and this can only be midway between the charges. This is shown below

So, kQ/x + [-kQ/(d - x)] = 0

kQ/x - kQ/(d - x) = 0

kQ/x = kQ/(d - x)

1/x = 1/(d - x)

(d - x) = x

d = x + x

d = 2x

x = d/2 which is mid-way between the charges.

(b) Find the location of all points, if any, where the electric field is zero.

Since the charges are of equal magnitude and opposite charge and separated by a distance, d, the electric field due to the +Q charge is E = kQ/x² and that due to the -Q charge is E' = -kQ/(d - x)² where x is the point of zero electric field.

The electric field is zero when  E + E' = 0 and this can only be midway between the charges. This is shown below.

So, kQ/x² + [-kQ/(d - x)²] = 0

kQ/x² - kQ/(d - x)² = 0

kQ/x² = kQ/(d - x)²

1/x² = 1/(d - x)²

(d - x)² = x²

d - x = ± x

d = x ± x

d = x - x or x + x

d = 0 or 2x

d = 0 or d = 2x

Since d ≠ 0, d = 2x ⇒ x = d/2 which is midway between the charges.

Answer:

Density of liquid

Depth of liquid

Acceleration due to gravity

Answer:

v= 20.8 m/s

Explanation:

       [tex]v_{f} = v_{o} + a*t = v_{o} + g*t (1)[/tex]

        [tex]\Delta y = v_{o} *t + \frac{1}{2} * a* t^{2} (2)[/tex]

       [tex]v_{o} = \frac{(\Delta y- \frac{1}{2} *a*t^{2})}{t} = \frac{-22m+19.6m}{2.00s} = -1.2 m/s (4)[/tex]

       [tex]v_{f} = v_{o} + g*t = -1.2 m/s - (9.8m/s2*2.00s) = -20.8 m/s (5)[/tex]

Answer:

a) p(total) = <0.05, 0.1, 0.1 > kg m/s

b) p = <0.05, 0.1, 0.1 > kg.m/s

c) v_f = < 1, 2, 2 > m/s

Explanation:

a.)

Mass of each lump = 25 g = 0.025 kg

Velocity of lump 1 = < -2, 0, -7 > m/s

Momentum of lump 1 = Mass×Velocity

                                   = 0.025×< -2, 0, -7 >

                                   = < -0.05, 0, 0.175> kg m/s

Velocity of lump 2 = < 4, 4, -3 > m/s

Momentum of lump 2 = Mass×Velocity

                                    = 0.025×< 4, 4, -3 >

                                    = < 0.1, 0.1, -0.075> kg m/s

Total momentum before impact  =  < -0.05,  0,  0.175 > + < 0.1, 0.1, -0.075>

                                                      = < 0.05, 0.1, 0.1 > kg m/s

⇒p(total) = <0.05, 0.1, 0.1 > kg m/s

b)

As we know that,

By the law of conservation of linear momentum,

The total momentum will be the same before and after the collision.

⇒Momentum of the stuck together  after the collision = Total momentum of the lumps just before impact.

⇒ p = <0.05, 0.1, 0.1 > kg m/s

c)

Let the final velocity =  v_f

Total mass = 0.025 + 0.025 = 0.05 kg

As

Momentum = mass ×velocity

⇒ <0.05, 0.1, 0.1 > = 0.05 ×v_f

⇒ v_f = <0.05, 0.1, 0.1 > / 0.05

          = < 1, 2, 2 > m/s

⇒v_f = < 1, 2, 2 > m/s

The distance travelled by toy car B using the same fully charged battery as car A is 875 m

E = fd

E = 35 × 2000

E = 70000 J

E = fd

70000 = 80 × Distance of car B

Divide both sides by 80

Distance of car B = 70000 / 80

Distance of car B = 875 m

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Answer:

22m/s

Explanation:

Mass, m=60 kg

Force constant, k=1300N/m

Restoring force, Fx=6500 N

Average friction force, f=50 N

Length of barrel, l=5m

y=2.5 m

Initial velocity, u=0

[tex]F_x=kx[/tex]

Substitute the values

[tex]6500=1300x[/tex]

[tex]x=\frac{6500}{1300}=5[/tex]m

Work done due to friction force

[tex]W_f=fscos\theta[/tex]

We have [tex]\theta=180^{\circ}[/tex]

Substitute the values

[tex]W_f=50\times 5cos180^{\circ}[/tex]

[tex]W_f=-250J[/tex]

Initial kinetic energy, Ki=0

Initial gravitational energy, [tex]U_{grav,1}=0[/tex]\

Initial elastic potential energy

[tex]U_{el,1}=\frac{1}{2}kx^2=\frac{1}{2}(1300)(5^2)[/tex]

[tex]U_{el,1}=16250J[/tex]

Final elastic energy,[tex]U_{el,2}=0[/tex]

Final kinetic energy, [tex]K_f=\frac{1}{2}(60)v^2=30v^2[/tex]

Final gravitational energy, [tex]U_{grav,2}=mgh=60\times 9.8\times 2.5[/tex]

Final gravitational energy, [tex]U_{grav,2}=1470J[/tex]

Using work-energy theorem

[tex]K_i+U_{grav,1}+U_{el,1}+W_f=K_f+U_{grav,2}+U_{el,2}[/tex]

Substitute the values

[tex]0+0+16250-250=30v^2+1470+0[/tex]

[tex]16000-1470=30v^2[/tex]

[tex]14530=30v^2[/tex]

[tex]v^2=\frac{14530}{30}[/tex]

[tex]v=\sqrt{\frac{14530}{30}}[/tex]

[tex]v=22m/s[/tex]

Answer: custodian,

Explanation: they never saw any colored women in the division before

Answer:

awful

Explanation:

can you help me with my question Which of the themes of Hawthorne's "Dr. Heidegger's Experiment" is illustrated by this passage? Paragraph 36: The most singular effect of their gayety was an impulse to mock the infirmity and decrepitude of which they had so lately been the victims. A. Gayety produces an impulse to mock. B. It is a great release to look back on our problems happily C It is human nature to mock or make light of problems we have been delivered from