A 0.450-kg ice puck, moving east with a speed of 5.34 m/s , has a head-on collision with a 0.990-kg puck initially at rest. Assume that the collision is perfectly elastic.1. What is the speed of the 0.450-kg puck after the collision?2. What is the direction of the velocity of the 0.450-kg puck after the collision?3. What ise the speed of the 0.990-kg puck after the collision?4. What is the direction of the velocity of the 0.990-kg puck after the collision?

Answers

Answer 1
Answer:

(1) -2.0025m/s.

(2) It moves west.

(3) 3.3375m/s.

(4) It moves east.

Explanation:

Assuming the collision occurred in an isolated system. This means that total momentum of the system of pucks is conserved. Since no external forces are acting on these pucks, the momentum of the pucks before collision is equal to the momentum of the pucks after collision. i.e

(p₁)₀ + (p₂)₀ = (p₁)₁ + (p₂)₁              -------------(i)

Where;

(p₁)₀ = momentum of the 0.45kg puck before collision

(p₂)₀ = momentum of the 0.990kg puck before collision

(p₁)₁ = momentum of the 0.45kg puck after collision

(p₂)₁ = momentum of the 0.990kg puck after collision

But;

(p₁)₀ = m₁ u₁  

[m₁ = mass of the 0.45kg, u₁ = speed of the 0.45kg before collision]

(p₂)₀ = m₂u₂

[m₂ = mass of the 0.990kg, u₂ = speed of the 0.990kg before collision]

(p₁)₁ = m₁v₁

[m₁ = mass of the 0.45kg, v₁ = speed of the 0.45kg after collision]

(p₂)₁ = m₂v₂

[m₂ = mass of the 0.990kg, v₂ = speed of the 0.990kg after collision]

Equation (i) then becomes;

m₁ u₁ + m₂u₂ = m₁v₁ + m₂v₂    ----------------(ii)

From the question:

m₁ = 0.450kg

u₁ = +5.34m/s   [Taking east direction as positive]

m₂ = 0.990kg

u₂ = 0m/s    [since the second puck is initially at rest]

Substitute these values into equation (ii)

(0.450 x 5.34) + (0.990 x 0) = 0.45 v₁ + 0.990 v₂

2.403 + 0 = 0.45 v₁ + 0.990 v₂

2.403 = 0.45 v₁ + 0.990 v₂           ------------------(iii)

Also, since the collision is perfectly elastic, it means that the kinetic energy is conserved. i.e the total kinetic energy before collision is equal to the total kinetic energy after collision.

=> [tex]\frac{1}{2}[/tex]m₁ u²₁ + [tex]\frac{1}{2}[/tex]m₂u²₂ = [tex]\frac{1}{2}[/tex]m₁v²₁ + [tex]\frac{1}{2}[/tex]m₂v²₂

Substitute the necessary values into the above equation:

[[tex]\frac{1}{2}[/tex] x 0.45 x 5.34²] + [0] = [[tex]\frac{1}{2}[/tex] x 0.45 x v²₁] + [[tex]\frac{1}{2}[/tex] x 0.990 x v²₂]

[6.41601] = [0.225 x v²₁] + [0.495 x v²₂]  ------------------(iv)

Now let's solve equations (iii) and (iv) simultaneously

2.403 = 0.45 v₁ + 0.990 v₂

6.41601 = 0.225 x v²₁ + 0.495 x v²₂

let

v₁ = x

v₂ = y

2.403 = 0.45 x + 0.990 y                  ------------(5)

6.41601 = 0.225 x² + 0.495 y²         -------------(6)

From equation (5), make x subject of the formula

2.403 = 0.45x + 0.990y

0.45x = 2.403 - 0.990y         [divide through by 0.45]

x = 5.34 - 2.2y    ----------------(m)

Substitute x into equation (6)

6.41601 = 0.225 (5.34 - 2.2y)² + 0.495 y²     [expand bracket]

6.41601 = 0.225 [28.5156 - 23.496y + 4.84y²] + 0.495 y²   [remove bracket]

6.41601 = 6.41601 - 5.2866y + 1.089y² + 0.495 y²

1.584y² - 5.2866y = 0

y(1.584y - 5.2866) = 0

y = 0       or       1.584y - 5.2866 = 0

y = 0       or       1.584y = 5.2866

y = 0       or       y = 3.3375

Since y = v₂ cannot be zero because the puck will definitely move after collision, the second value of y = 3.3375 is considered.

Substitute this value into equation (m)

x = 5.34 - 2.2y

x = 5.34 - 2.2(3.3375)

x = 5.34 - 7.3425

x = -2.0025

Therefore,

x = v₁ = -2.0025m/s

y = v₂ = 3.3375m/s

(1) From the analyses above, the speed of the 0.450kg puck after collision is -2.0025m/s.

(2) Since the speed is negative, it shows that the 0.45kg puck moves opposite the direction at which it was moving before collision. It moves west.

(3) The speed of the 0.990kg puck after collision is 3.3375m/s.

(4) Since the speed is positive, it shows that the 0.990kg puck moves east. Remember that east has been taking as the positive direction.


Related Questions

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s.A violinist is tuning her instrument to concert A (440 Hz). She plays the note while listening to an electronically generated tone of exactly that frequency and hears a beat of frequency 6.00 Hz, which increases to 7.00 Hz when she tightens her violin string slightly.
(a) What was the frequency of the note played by her violin when she heard the 3-Hz beats?
(b) To get her violin perfectly tuned to concert A, should she tighten or loosen her string from what it was when she heard the 3-Hz beats?

Answers

Answer:

A) 443 Hz

B) She has to loosen the string

Explanation:

A) Given;

Beat frequency;f_beat = 3 Hz

Frequency of electronically generated tone; f_e = 440 Hz

We know that formula for beat frequency is given by;

f_beat = |f1 - f2|

Now, applying it to this question, we have;

f_beat = f_v - f_e

Where f_v is frequency of the note played by the violinist

Thus, plugging in the relevant values;

3 = f_v - 440

f_v = 3 + 440

f_v = 443 Hz

B) In the concept of wave travelling in a string, the frequency is directly proportional to the square root of the force acting on the string.

Now, for the violinist to get her violin perfectly tuned to concert A from what it was when she heard the 3-Hz beats, the beat frequency will have to be zero. Which means it has to decrease by 3 Hz. For it to decrease, it means that the force applied has to decrease as we have seen that frequency is directly proportional to the square root of the force acting on the string.

Thus, she would have to loosen the string.

Two identical guitar strings are prepared such that they have the same length ( 0.62 m ) and are under the same amount of tension. The first string is plucked at one location, primarily exciting the second harmonic. The other string is plucked in a different location, primarily exciting the third harmonic. The resulting sounds give rise to a beat frequency of 324 Hz . What is the wave propagation speed on the guitar strings?

Answers

Answer:

The speed is 401.76m/s

Explanation:

To find the fundamental frequency we use

3rd.harmonic - 2nd.harmonic = 324Hz

And we know that

f=v/2L

So v=f2L

v= 324 x 2 x 0.62= 401.76m/s

On planet Q the standard unit of volume is called guppi. Space travelers from Earth have determined that one liter = 38.2 guppies. How many guppies are in 150 liters?

Answers

Answer:

5730 guppies

Explanation:

1 liter= 38.2 guppies

150 liters= 150×38.2

On planet Q the standard unit of volume is called guppi. Space travelers from Earth have determined that one liter = 38.2 guppies. Guppies in 150 liters is 5730 guppies.

What are units of volume?

Volume, which is measured in cubic units, is the 3-dimensional space occupied by matter or encircled by a surface. The cubic meter (m³), a derived unit, is the SI unit of volume. A addition new nickname for the cubic (dm³) is a liter (L).

Given that in the question on planet Q the standard unit of volume is called guppi, the conversion of different volume unit,

1 liter = 38.2 guppies, so

150 litre =  5730 guppies

Guppies in the given 150 liters of volume is 5730 guppies.

To learn more about volume refer to the link:

brainly.com/question/13338592

 #SPJ2

Discuss the characteristics of the waves in the visible portion of the electromagnetic spectrum

Answers

Explanation:

The electromagnetic spectrum consists of series of waves that are arranged from longest wavelength to shortest.

In the electromagnetic spectrum, between infrared and ultraviolet, visible light falls.The frequency of visible light is of the order of 10¹⁴ Hz.Its wavelength is about 740 nm.Visible light spectrum is visible to humans. It consists of colors like violet, blue, green, yellow etc.

A car accelerates from 14.0m/s to 21m/s in 6.0s. What was its acceleration?
How far did it travel in this time? Assume constant acceleration.

Answers

Answer:

Acceleration (a) = 1.167 m/s²

Distance travel = 105 m (Approx)

Explanation:

Given:

Initial speed (u) = 14 m/s

Final speed (v) = 21 m/s

Time taken (t) = 6 sec

Find:

Acceleration

Distance travel

Computation:

v = u + at

21 = 14 + a(6)

7 = 6a

Acceleration (a) = 1.167 m/s²

S = ut + (1/2)(a)(t²)

S = (14)(6) + (1/2)(1.167)(6²)

S = 84 + 21

S = 105 m (Approx)

Distance travel = 105 m (Approx)

A cosmic-ray proton in interstellar space has an energy of 10.0 MeV and executes a circular orbit having a radius equal to that of Mercury’s orbit around the Sun (5.80 × 1010m). What is the magnetic field in that region of space?

Answers

Answer:

The value is [tex]B = 7.7106 *10^{-12 } \ T[/tex]

Explanation:

From the question we are told that

The energy is [tex]E = 10.0 MeV = 10.0 *10^{6} \ eV = 10.0 *10^{6} * 1.60*10^{-19} = 1.6 *10^{-12} \ J[/tex]

The  radius  is  [tex]R = 5.80 *10^{10} \ m[/tex]

Generally the magnetic field is mathematically represented as

       [tex]B = \frac{m * v }{q * R }[/tex]

Where m is the mass of  proton with value  [tex]m = 1.6*10^{-27} \ kg[/tex]

 v is the velocity of the proton which is mathematically deduced  from the formula for kinetic energy as

                   [tex]v = \sqrt{ \frac{E }{ 0.5 * m } }[/tex]

Here E is also equivalent to kinetic energy of the proton so

                   [tex]v = \sqrt{ \frac{1.6 *10^{-12} }{ 0.5 * 1.6*10^{-27} } }[/tex]

                 [tex]v = 4.47214 *10^{7} \ m/s[/tex]

So  

     [tex]B = \frac{1.60 *10^{-27} * 4.47214 *10^{7}}{ 1.60 *10^{-19} * 5.80*10^{10}}[/tex]

      [tex]B = 7.7106 *10^{-12 } \ T[/tex]

The magnetic field in the region has a magnitude of 7.7×[tex]10^{-12}[/tex] T.

Let's consider the magnetic field in the region be B. Now the proton in the given orbit is in equilibrium under the influence of centripetal force ([tex]F_{c}[/tex]) and magnetic force([tex]F_{b}[/tex]).

[tex]F_{c}[/tex] = [tex]F_{b}[/tex]

m[tex]v^{2}[/tex]/R = qvB

⇒ B = mv/qR

where, m is the mass of proton =  1.6 × [tex]10^{-27}[/tex] kg

            q is the charge of proton = 1.6 × [tex]10^{-19}[/tex] C

            R is the radius of the orbit = 5.8 × [tex]10^{10}[/tex] m

now we can calculate the velocity of the proton, v, from the energy of the proton given in the question that is E = 10 MeV.

1 MeV = 1.6 × [tex]10^{-19}[/tex] J

           v = [tex]\sqrt{\frac{2E}{m} }[/tex]

               =[tex]\sqrt{\frac{2*10*10^{6}* 1.6*10^{-19} }{1.6*10^{-27} } }[/tex]

            v = 4.47 × [tex]10^{7}[/tex] m/s

now,

        B = mv/qR

           = [tex]\frac{1.6*10^{-27}*4.47*10^{7} }{1.6*10^{-19}*5.8*10^{10} }[/tex]

       B = 7.7 ×[tex]10^{-12}[/tex] T

Learn more about magnetic force :

https://brainly.com/question/25746114

Monochromatic light with wavelength 588 nm is incident on a slit with width 0.0351 mm. The distance from the slit to a screen is 2.7 m. Consider a point on the screen 1.3 cm from the central maximum. Calculate (a) θ for that point, (b) α, and (c) the ratio of the intensity at that point to the intensity at the central maximum.

Answers

Answer:

Explanation:

A. Using

Sinစ= y/ L = 0.013/2.7= 0.00481

စ=0.28°

B.here we use

Alpha= πsinစa/lambda

= π x (0.0351)sin(0.28)/588E-9m

= 9.1*10^-2rad

C.we use

I(စ)/Im= (sin alpha/alpha) ²

So

{= (sin0.091/0.091)²

= 3*10^-4

light of a wavelength 600 nm shines on a soap bubble film. For what soap film thickness will destructive interference occur

Answers

Answer:

The minimum thickness of the soap bubble for destructive interference to occur is 225.56 nm.

Explanation:

Given;

wavelength of light, λ = 600 nm

The minimum thickness of the soap bubble for destructive interference to occur is given by;

[tex]t = \frac{\lambda/n}{2}\\\\t = \frac{\lambda}{2n}[/tex]

where;

n is refractive index of soap film = 1.33

[tex]t = \frac{\lambda}{2n} \\\\t = \frac{600*10^{-9}}{2(1.33)}\\\\t = 2.2556 *10^{-7} \ m\\\\t = 225.56 *10^{-9} \ m\\\\t = 225.56 \ nm[/tex]

Therefore, the minimum thickness of the soap bubble for destructive interference to occur is 225.56 nm.

A 0.16 pF parallel - plate capacitor is charged to 10 V. Then the battery is disconnected from the capacitor. When 1.00 107 electrons are now placed on the negative plate of the capacitor, the voltage between the plates changes by

Answers

Answer:

The  value is   [tex]V_d  = 5 \ V[/tex]

Explanation:

From the question we are told that

    The  capacitance of the capacitor is [tex]C =  0.16 \  pF  =  0.16*10^{-12} \ F[/tex]

     The  voltage is  [tex]V  =  10\ V[/tex]

    The number of electrons present on the negative plate is  [tex]N_e  =  1.00 *10^{7} \  electrons[/tex]

     Generally the charge on the positive plate at 10 Volt is mathematically represented as

            [tex]Q_a =  C *  V[/tex]

     =>   [tex]Q_a  =  0.16*10^{-12} *  10[/tex]

     =>    [tex]Q_a  =  0.16*10^{-11} \  C[/tex]

Now the charge  on the plate when the electron where placed is evaluated as

    [tex]Q_b  =  Q_a  + ( N_e * e)[/tex]

Where e  is the charge on each electron with a value  [tex]e =  1.60 *10^{-19} \ C[/tex]

     =>    [tex]Q_b  =  0.16*10^{-11} + (1.0*10^7 *  1.60*10^{-19})[/tex]

      =>  [tex]Q_b  = 3.2 *10^{-12 } \  C[/tex]

Generally the voltage between the two plate is evaluated as

       [tex]V_d  =  \frac{ Q_b  -  Q_a  }{ 2 * C }[/tex]

=>     [tex]V_d  = \frac{ 3.2*10^{-12} -  0.16*10^{-11}}{2 *  0.16 *10^{-12}}[/tex]

=>      [tex]V_d  = 5 \ V[/tex]

     

 

Is the sinusoidal pattern on a string longer or shorter when there is a greater propagation velocity?

Answers

Answer:

We know that

Velocity= wavelength x frequency

And f= 1/2π√T/u

So we also know that

The frequency of nth harmonic standing wave is

fn=nv/2L

So from the relationships above

We can see that as propagational velocity increases, the frequency increases and the wavelength decreases. So finally we can say that sinusoidal pattern on a string is shorter when there is a greater propagation velocity

A truck travels due east for a distance of 1.9 km, turns around and goes due west for 9.3 km, and finally turns around again and travels 3.4 km due east.
What is the total distance that the truck travels? What are the magnitude and direction of the truck's displacement?

Answers

Answer:

The total distance of the truck is 14.6 km

The magnitude and direction of truck's disaplacement is 4 km east

Explanation:

Distance is the total path covered by the trcuk regardless of the direction since distance is described by magnitude only.

The total distance traveled by the truck is given by;

Total Distance = 1.9 km + 9.3 km + 3.4 km

Total Distance = 14.6 km

Displacement is the change in postion at a time interval. In dispalcement direction is considered.

let eastward motion be positive

let westward motion be negative

Displacement = 1.9km east - 9.3 km west + 3.4 km east

Displacement = -4 km east

Therefore, the magnitude and direction of dispalcement = 4 km east

The period of a simple pendulum is 8 s. If the same pendulum experiment is repeated on a plant of g = 1/4 gE, the new period is

Answers

Answer:

16 secs.

Explanation:

Data obtained from the question include the following:

Period on earth (T) = 8 secs

Next period (Tn) =?

Next, we shall determine the length of the string. This can be obtained as follow:

Period on earth (T) = 8 secs

Acceleration due to gravity on earth (gE) = 9.8 m/s²

Pi (π) = 3.14

Length of string (L) =?

T = 2π√(L/gE)

8 = 2 × 3.14 ×√(L/9.8)

8 = 6.28 × √(L/9.8)

Divide both side by 6.28

8/6.28 = √(L/9.8)

Take the square of both side

(8/6.28)² = L/9.8

Cross multiply

L = (8/6.28)² × 9.8

L = 15.9 m

Therefore the length of string is 15.9 m

Next, we shall determine the new period of the pendulum as follow:

Length (L) = constant = 15.9 m

Pi (π) = 3.14

Acceleration due to gravity on earth (gE) = 9.8 m/s²

Acceleration due to gravity on the planet (g) = 1/4 gE = 1/4 × 9.8 m/s²

= 2.45 m/s²

New period (Tn) =?

Tn = 2π√(L/g)

Tn = 2 × 3.14 × √(15.9/2.45)

Tn = 6.28 × √(15.9/2.45)

Tn = 16 secs.

Therefore, the new period of the pendulum is 16 secs.

Below is a car. The car's engine is providing a driving force of 10N. The parachute is creating air resistance of 7N. What is going to happen to the car? Explain.​

Answers

Answer:

The car is going to slow down to 3N.

Explanation:You have a force going 10N to the left and if you put an opposing force of 7N you get 3N of total force.

100 hectometer =_______ nanometers

Answers

10^12
Hope this is helpful

Answer:

Hey!

Your answer is 10^12 nanometers!

Explanation:

AKA

1,000,000,000,000  nm

HOPE THIS HELPS!!

When discharging a fire extinguisher, it is important to aim: a. On top of the flames b. At the top of the flames c. Anywhere d. At the base of the fire

Answers

Answer:

d. At the base of the fire

Explanation:

Most fire extinguisher utilize the non-combustible nature of carbon-dioxide . carbon-dioxide is what is used in fire extinguishers, because it does not support burning. Also, carbon-dioxide is denser than air, allowing it to sink down. Aiming at the bottom of the flame allows the carbon-dioxide to sink down on the base of the flame, covering the base of the fire, and cutting off its supply of oxygen which is vital for combustion.

dolphin swims due east for 1.90 km, then swims 7.20 km in the direction south of west. What are the magnitude and direction of the vector that will take the dolphin back to its starting point? (Enter the magnitude in km and the direction in degrees north of east.) Hint

Answers

Answer:

Explanation:

We shall represent all the displacement in terms of vector . We shall consider east as i , and north as j . south of west direction will be represented by - i - j .

displacement of 1.9 km due east

D₁ = 1.9 i

vector representing south of west = - i - j

unit vector = - i - j / √ 2

7.2 km south of west = 7.2 ( - i - j ) / √ 2

D₂ = - 5.09 ( i + j )

Total displacement

= D₁ + D₂

= 1.9 i  - 5.09 ( I + J )

D = - 3.19 i - 5.09 j .

magnitude of D = √ ( 3.19² + 5.09² )

= 6 km .

Direction of D

Tanθ = 5.09 / 3.19 = 1.59

θ = 58°

So direction will be 58° south of west .

To reach the starting point , he shall have to go in opposite direction .

So he shall have to go in the direction of north of east at angle 58° by a displacement of 6 km .

A passenger train travels 295 miles in the same amount of time it takes a freight train to travel 225 miles. The rate of the passenger train is 14 mph greater than the rate of the freight train. Find the rate of each train.

Answers

Answer:

9.3

Explanatin:

Answer:

4.5

Explanation:

A small uniform disk and a small uniform sphere are released simultaneously at the top of a high inclined plane, and they roll down without slipping. Which one will reach the bottom first?

Answers

Answer:

the sphere

Explanation:

From the given information,

A free flow body diagrammatic expression  for the small uniform disk and a small uniform sphere which are released simultaneously at the top of a high inclined plane  can be seen in the image attached below.

From the diagram;

The Normal force mgsinθ - Friction force F  = mass m × acceleration a

Meanwhile; the frictional force

[tex]F = \dfrac{I \alpha }{R}[/tex]

where

[tex]\alpha = \dfrac{a}{R}[/tex]  in a rolling motion

Then;

[tex]F = \dfrac{I a }{R^2}[/tex]

The Normal force mgsinθ - F  =  m ×  a     can be re-written as:

[tex]\mathtt{mg sin \ \theta- \dfrac{Ia}{R^2} = ma}[/tex]

making a the subject of the formula, we have:

[tex]a = (\dfrac{mg \ sin \theta}{m + \dfrac{I}{R^2}})[/tex]

Similarly;

I = mk²  in which k is the radius of gyration

replacing I = mk² into the above equation , we have:

[tex]a = (\dfrac{mg \ sin \theta}{m + \dfrac{mk^2}{R^2}})[/tex]

where;

the uniform disk [tex]\dfrac{k^2}{R^2 }= \dfrac{1}{2}[/tex]  

the uniform  sphere [tex]\dfrac{k^2}{R^2 }= \dfrac{2}{5}[/tex]

[tex]a = \dfrac{2}{3} \ g sin \theta \ for \ the \ uniform \ disk[/tex]

[tex]a = \dfrac{5}{7} \ g sin \theta \ for \ the \ uniform \ sphere[/tex]

We can now see that the uniform sphere is greater than the disk as such the sphere will reach the bottom first.

A sprinter can accelerate with constant acceleration for 4.00 s before reaching top speed. He can run the 100 m dash in 10 s. What is his speed as he crosses the finish line?

Answers

Answer:

12.5 m/s

Explanation:

If the avg speed is 10 m/s we assume that the speed is constant and not fluctuating

If VT (Velocity*Time) is equal to speed at the end of the 4 seconds of accel. it is the top speed at the finish line

since the speed is 0 at the beginning of races, we will use V0= 0 m/s

So the avg velocity for those 4 seconds is (VT+V0) /2 = or 1/2 VT (for 40% of the race)

The avg velocity for the last 6 seconds is, of course, VT (for 60% of the race

hence

(1/2VT)(.4) + VT(.6) = 10 m/s

VT = 10/.8 = 12.5 m/s

An object is continuously changing its velocity by the same rate. What is true
about this object?
a. The object has decreasing acceleration.
b. The object has increasing acceleration.
c. The object has zero acceleration.
d. The object has constant acceleration.

Answers

Answer:

d. The object has constant acceleration.

Explanation:

Acceleration is the rate of change of velocity.  So if the rate of change is the same at all times, then the acceleration is constant.

Scientist start by asking a question about something what is the initial explanation to that question called

Answers

A hypothesis. It is the first step to the scientific method.

Suppose that an object in free fall were somehow equipped with a speedometer. By how much would its speed readings increase with each second of fall? Express your answer to two significant figures and include the appropriate units.

Answers

Gravity is 9.8m/s^2
For every second the object travels 9.8m
When the object is at two seconds the distance travelled is gravity x time --> 9.8 x 2 = 19.6m/s^2
Finding the increase is just subtracting
19.6-9.8=9.8m/s^2

At what distance will a 80 W lightbulb have the same apparent brightness as a 120 W bulb viewed from a distance of 40 m

Answers

Answer:

32.6mm

Explanation:

Using area of a sphere(bulb) = 4πr²

So A is proportional to radius²

So the Energy will be proportional to r²

But 120/80 = 1.5 is the energy factor so

Using

1.5/d² = 1/r²

1.5/40²= 1/r^2

r = √( 40²/ 1.5)

r = 32.6m

What is the primary reason that astronomers suspect that some jovian moons were captured into their current orbits

Answers

Answer:

The summary including its circumstances in question is outlined in the illustrating segment below.

Explanation:

An astronomer seems to be an astronomical specialist who performs his information about the research problem or area elsewhere in Earth's view.A few other moons include "backward" orbits (particularly in comparison to the movement or rotation of another planet), or are favorably committed to the equator of their continent.

An object is at rest on a table. The gravitational force is 10 N down, and the normal force is 10 N up. What is the net force?
Fnormal
Fgravity
10 N up
OON
10 N down
20 N up

Answers

Answer:

0N

Explanation:

When trying to figure out opposing forces, you have to add them together,

10 + -10 = 0,

so your answer is 0N

Answer:the answer is 0 N

A star has a declination of approximately 45°. Which of these statements is correct about the star?

A) It is 45° above the celestial equator.
B) It is 45° below the celestial equator.
C) It is 45° to the left of the celestial equator.
D) It is 45° to the right of the celestial equator.

Answers

The star is approximately 45° above the celestial equator.  (A)

"Declination" on the celestial coordinate system is equivalent to latitude on the terrestrial one.

Why is the trajectory of the positively charged particle curving down, to the bottom of the screen, when moving in the magnetic field?

Answers

Answer:

Because the force F is always at 90 degrees to the velocity of the particle.

Explanation:

The factors that determine the path of a particle in a uniform magnetic field depend on

The magnetic flux density

The charge on the particle

The speed of the particle

Why is the trajectory of the positively charged particle curving down, to the bottom of the screen, when moving in the magnetic field?

Because the force F is always at 90 degrees to the velocity of the particle.

This can be illustrated by the equation

F = BQVSINØ

The magnetic force is always acting perpendicular to the particle velocity and its direction can be found using the left hand rule.

Marcus is teaching a lesson on energy transformations to his class. He cracks a glow stick to teach this practice and states that the energy in the glow stick is transformed. Assuming that the glowstick is a closed system and all of the energy gets transfomred, explain how the chemical energy of the glow stick compares to the light energy after he cracks it.

Answers

Answer:

this occurs because the energy of the constituents is greater than the energy of the material formed and this excess is what we can appreciate in the form of ligh

Explanation:

When the light stick breaks, it allows substances to mix, produced by a chemical reaction, which releases light into the environment.

The energy released is the energy energy of the excess when the transformation takes place, this occurs because the energy of the constituents is greater than the energy of the material formed and this excess is what we can appreciate in the form of light,

What are different ways to describe speed and velocity

Answers

Speed is a constant direction of where you go, constant motion, and velocity is the displacement of a objects direction, plus the direction the object is traveling to.

A sign is hung between two cables. If the sign weighs 350 N, what is the tension (in N) in each cable?

Answers

Answer:

Tension in each cable = 175 N

Explanation:

Given:

Number of cables = 2

Weight of sign board(T) = 350 N

Find:

Tension in each cable

Computation:

We know that,

T1 = T2

So,

T1 + T2 = T

T1 + T2 = 350

T1 + T1 = 350

T1 = 175

Tension in each cable = 175 N

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