A rolling ball moves from x1 = 8.0 cm to x2 = -4.1 cm during the time from t1 = 2.9 s to t2 = 6.0 s .

Answer:

Explanation:

Given;

initial velocity of the ball, u = 23 m/s

time of motion, t = 2 s

The position of the ball after 2 s is given by;

h = ut - ¹/₂gt²

h = (23 x 2) - ¹/₂ x 9.8 x 2²

h = 46 - 19.6

h = 26.4 m

The velocity of the ball after 2 s is given by;

v² = u² + 2(-g)h

v² = u² - 2gh

v² = 23² - (2 x 9.8 x 26.4)

v² = 529 - 517.44

v² = 11.56

v = √11.56

v = 3.4 m/s

Answer:

Explanation:

Step one:

given data

we are told that 1 large pizza can be cut into 8 even slices

Required

we want to find how many slices are there in 4 large pizzas

Step two:

so if 1 pizza has 8 slices

       4 pizza will have x

cross multiply we have

x= 8*4

x=32 slices

As the pushing force x increases, it would be opposed by the static frictional force. As x passes a certain threshold and overcomes the maximum static friction, the block will start moving and will require a smaller magnitude x to maintain opposition to the kinetic friction and keep the block moving at a constant speed. If x stays at the magnitude required to overcome static friction, the net force applied to the block will cause it to accelerate in the same direction.

Let w denote the weight of the block, n the magnitude of the normal force, x the magnitude of the pushing force, and f the magnitude of the frictional force.

The block is initially at rest, so the net force on the box in the horizontal and vertical directions is 0:

n + (-w) = 0

n = w = m g = (5.3 kg) (9.80 m/s²) = 51.94 N

The frictional force is proportional to the normal force, so that f = µ n where µ is the coefficient of static or kinetic friction. Before the block starts moving, the maximum static frictional force will be

f = 0.67 (51.94 N) ≈ 35 N

so for 0 < x < 35 N, the block remains at rest and 0 < f < 35 N as well.

The block starts moving as soon as x = 35 N, at which point f = 35 N.

At any point after the block starts moving, we have

f = 0.48 (51.94 N) ≈ 25 N

so that x = 25 N is the required force to keep the block moving at a constant speed.

As x  is increasing it will be opposed by a static frictional force and for the object to start moving and maintain its acceleration, the magnitude of x must exceed the magnitude of the static frictional force and kinetic frictional force

Given data :

mass of block at rest ( m ) = 5.3 kg

Coefficient of static friction ( μ_s ) =0.67

Coefficient of kinetic friction is ( μ_k ) = 0.48

Horizontal force applied to block = x  

First step : magnitude of normal force ( n ) when object is at rest

n = w            where w = m*g

n - w = 0

n - ( 5.3 * 9.81 ) = 0     ∴  n = 51.94 N

Second step : Required magnitude of x before the movement of object

F =  μ_s * n

F = 0.67 * 51.94  = 34.79 N  ≈ 35 N

∴ The object will start moving once F and x = 35 N

Final step : Magnitude of x  after object start moving

F = μ_k  * n

  = 0.48 * 51.94 = 24.93 N  ≈ 25 N

∴ object will continue to accelerate at a constant speed once F and x = 25N

Learn more : https://brainly.com/question/21444366

Answer:

v = 0.823 m/s

Explanation:

A man walks south at a speed of 2.00 m/s for 60.0 minutes.

The distance covered in South = 60 × 60 × 2 = 7200 m

He then turns around and walks north a distance 3000 m in 25.0 minutes.

As they moved in opposite direction, net displacement will be : 7200 - 3000 = 4200 m

Average velocity of the man = net displacement/time

[tex]v=\dfrac{4200\ m}{(60+25)\times 60}\\\\=0.823\ m/s[/tex]

So, the average velocity of the man is 0.823 m/s.

Answer:

For any given projector, the width of the image (W) relative to the throw distance (D) is know as the throw ratio D/W or distance over width. So for example, the most common projector throw ratio is 2.0. This means that for each foot of image width, the projector needs to be 2 feet away or D/W = 2/1 = 2.0.

Answer:

Acceleration = 18g

Explanation:

Given the following data;

Initial velocity, u = 26m/s

Final velocity, v = 0

Time = 0.15 secs

To find the acceleration;

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]

Substituting into the equation, we have;

[tex]a = \frac{0 - 26}{0.15}[/tex]

[tex]a = \frac{26}{0.15}[/tex]

Acceleration = 173.33m/s2

To express it in magnitude of g;

Acceleration = 173.33/9.8

Acceleration = 17.7 ≈ 18g

Acceleration = 18g

Answer:

220v

Explanation:

Sorry, the question is incomplete

Answer:

on the potential difference applied and on the resistance of the wire.

Explanation:

Ohms law state that the current through a conductor between two points is directly proportional to the potential difference across the two points. Imtroducing the comstant of proportionality, the resistance, one arrives at the usual athematical equation that describes this relationship: I = V/R.

Answer:

a) The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty in its perimeter is 0.8 centimeters.

Explanation:

a) From Geometry we remember that the perimeter of the rectangle ([tex]p[/tex]), measured in centimeters, is represented by the following formula:

[tex]p = 2\cdot (w+l)[/tex] (1)

Where:

[tex]w[/tex] - Width, measured in centimeters.

[tex]l[/tex] - Length, measured in centimeters.

If we know that [tex]w = 6.4\,cm[/tex] and [tex]l = 8.3\,cm[/tex], then the perimeter of the rectangle is:

[tex]p = 2\cdot (6.4\,cm+8.3\,cm)[/tex]

[tex]p = 29.4\,cm[/tex]

The perimeter of the rectangle is 29.4 centimeters.

b) The uncertainty of the perimeter ([tex]\Delta p[/tex]), measured in centimeters, is estimated by differences. That is:

[tex]\Delta p = 2\cdot (\Delta w + \Delta l)[/tex]  (2)

Where:

[tex]\Delta w[/tex] - Uncertainty in width, measured in centimeters.

[tex]\Delta l[/tex] - Uncertainty in length, measured in centimeters.

If we know that [tex]\Delta w = 0.2\,cm[/tex] and [tex]\Delta l = 0.2\,cm[/tex], then the uncertainty in perimeter is:

[tex]\Delta p = 2\cdot (0.2\,cm+0.2\,cm)[/tex]

[tex]\Delta p = 0.8\,cm[/tex]

The uncertainty in its perimeter is 0.8 centimeters.

Answer:

p₂ / p₁ = 2 (v₁ / v₂)

Explanation:

The moment is a very useful concept, since it is one of the quantities that is conserved during shocks and explosions, for which it had to be redefined to be consistent with special relativity,

         p = m v / √[1+ (v/c)² ]

for the case of speeds much lower than the speed of light this expression is close to

         p = m v

In this exercise they indicate that the moment of the second particle is twice the moment of the first, when their velocities are small

        p₂ = 2 p₁

       p₂/p₁ = 2

in consecuense

       m v₂ = 2 m v₁

       v₂ = 2 v₁

consider particles of equal mass.

By the time their speeds increase they enter the relativistic regime

        p₂ = mv₂ /√(1 + v₂² /c²)

        p₁ = m v₁ /√(1 + v₁² / c²)

let's look for the relationship between these two moments

       p₂ / p₁ = mv₂ / mv₁   [√ (1+ v₁² / c²) /√ (1 + v₂² / c²)

from the initial statement

      p₂ / p₁ = 2 √(c² + v₁²) / (c² + v₂²)

we take c from the root

      p₂ / p₁ = 2 √ [(1+ v₁²) / (1 + v₂²)]

this is the exact result, to have an approximate shape suppose that the velocities are much greater than 1

      p₂ / p₁ = 2 √ [v₁² / v₂²] = 2 √ [(v₁ / v₂)²]

      p₂ / p₁ = 2 (v₁ / v₂)

we see the value of the moment depends on the speed of the particles

Answer:

B: lodestone

Explanation:

Each magnet has its magnetic poles, north (N) and south (S). Diversified ones are attracted and reptiles of the same name are repelled, similarly to charges, so it was considered possible to separate the magnet at the north and south poles.

Magnetic properties can be lost if the magnet is exposed to high temperatures if it falls or due to some mechanical shocks.

Answer:

The displacement of the car after 6s is 43.2 m

Explanation:

Given;

velocity of the car, v = 12 m/s

acceleration of the car, a = -1.6 m/s² (backward acceleration)

time of motion, t = 6 s

The displacement of the car after 6s is given by the following kinematic equation;

d = ut + ¹/₂at²

d = (12 x 6) + ¹/₂(-1.6)(6)²

d = 72 - 28.8

d = 43.2 m

Therefore, the displacement of the car after 6s is 43.2 m

Answer:Final volume after pressure is applied=4,292cm3

Explanation:

Using the bulk modulus formulae

We have that The bulk modulus of waTer is given as  

K =-V dP/dV

Where  K, the bulk modulus of water = 2.15 x 10^9N/m^2

2.15 x 10^9N/m^2= - 4,300 x  4 × 106N/m2 / dV

dV = - 4,300 x  4 × 10^6N/m^2/ 2.15 x 10^9N/m^2

dV (change in volume)= -8.000cm^3

Final volume after pressure is applied,

V= V+ dV

V= 4300cm3 + (-8.000cm3)

=4300cm3 - 8.000cm3

Final Volume, V =4,292cm3

Answer:82. Since you have a distance and a force, then the easiest principle to use is energy, i.e. work.

The work done by friction is F * d. This work cancels out the kinetic energy of the stone (1/2)mv^2

Fd = (1/2)mv^2

F = (1/2)mv^2/d.

Plug in m = 20 kg, v = 3 m/sec, d = 40 m.

83. With more mass, the kinetic energy is higher now. The work needed is higher. W = F * d and F is the same.

Explanation:Hope I helped :)

Answer:

B, which is why ice skaters often keep their arms close to their body when doing spins and jumps to minimize resistance.

Answer:

The final velocity of the car is 26.65 m/s.

Explanation:

Given;

acceleration of the racecar, a = 6.5 m/s²

initial velocity of the car, u = 0

time of motion, t = 4.1 s

The final velocity of the car is given by;

v = u + at

where;

v is the final velocity of the car

suvstitute the givens

v = 0 + (6.5)(4.1)

v = 26.65 m/s.

Therefore, the final velocity of the car is 26.65 m/s.

Answer:

10J

Explanation:

In this question we have the following information

The charge of the particle is q = 5 C

The equipotenetial level is V1 = 5.5 v

and also the

equipotenetial level is V2 = 3.5 v

So we calculate the

work done W=q x (v1-v2)

workdone = 5 x (5.5-3.5)

= 5x2

=10 J

Workdone = 10 J

So we conclude that the workdone on a particle with these information is 10j

Answer:

476.35 km

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 11000 km/hr

Final velocity (v) = 0 km/hr (at maximum height)

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

Maximum height (h) = ?

Next, we shall convert 9.8 m/s² to km/hr². This is illustrated below:

1 m/s² = 12960 km/hr²

Therefore,

9.8 m/s² = 9.8 m/s² × 12960 km/hr² / 1 m/s²

9.8 m/s² = 127008 km/hr²

Thus, 9.8 m/s² is equivalent to 127008 km/h²

Finally, we shall determine the maximum height reached by the projectile.

This is illustrated below:

Initial velocity (u) = 11000 km/hr

Final velocity (v) = 0 km/hr (at maximum height)

Acceleration due to gravity (g) = 127008 km/hr²

Maximum height (h) = ?

v² = u² – 2gh (since the projectile is going against gravity)

0² = 11000² – (2 × 127008 × h)

0 = 121×10⁶ – 254016h

Collect like terms

0 – 121×10⁶ = – 254016h

– 121×10⁶ = – 254016h

Divide both side by – 254016

h = – 121×10⁶ / – 254016

h = 476.35 km

Thus, the maximum height reached by the projectile is 476.35 km

Answer:1 because

Explanation: it’s pointing to the earth and gravity

Pulls things down to earth

Answer:

Answer is (b) Mercury, venus and Mars.

Explanation:

i think b is correct!!

;-) :-) :-) :-)

Answer:

2.9 cm

Explanation:

Assuming that the rear wheel has a radius of 0.330 m

Given that

r(a) = 12 cm -> 0.12 m

w(a) = 0.6 rev/s -> 3.77 rad/s

v = 5 m/s

r(w) = 0.330 m

The speed on any point on the rim at the sprocket in the front is

v(a) = w(a).r(a) = 3.77 * 0.12 = 0.4524 m/s

Also,

v(a) = speed at any point on the chain

v(b) = speed at any point on the rim of the rear sprocket

v(a) = v(b)

where v(b) = w(b).r(b)

Recall that the speed at any point on the rear wheel is v, where

v = w(b).r(w)

5 = w(b) * 0.330

w(b) = 5/0.330

w(b) = 15.15 rad/s

On substituting this in the equation, we have

v(b) = w(b).r(b).

Remember also, that v(a) = v(b), so

0.4524 = 15.15 * r(b)

r(b) = 0.4524 / 15.15

r(b) = 0.029 m -> 2.9 cm

Therefore, the radius of the rear sprocket needed is 2.9 cm

Answer:

distance i think

Explanation: