A ball rolls at a speed of 4 m/s off a flat surface which is a height 1.3m above the ground. how long (in seconds) is the ball in the air before it hits the ground

Answer:

Explanation:

The forces acting along the playground are:

Kinetic friction force Fk and

Wsintheta (weight along the plane)

Since sintheta = opp/hyp

Sintheta = 4/7

Taking the sum of forces along the playground

\sumFx = max

Wsintheta - Fk = max

Since the speed is constant, ax = 0, the equation becomes

Wsin theta - Fk = 0

Fk = Wsintheta

From the question

W = mg

W = 27×9.81

W = 264.87N

Fk = 264.87 × 4/7

Fk = 1059.48/7

Fk = 151.35N

Hence the magnitude of the kinetic friction force acting on the child is 151.35N

The dog starts from rest and has a speed of 2.10 m/s at the bottom. The height of the slope will be 0.225 m

Friction is a type of force that resists or prevents the relative motion of two physical objects when their surfaces come in contact.

The friction force prevents any two surfaces of objects from easily sliding over each other or slipping across one another. It depends upon the force applied to the object.

As there is almost zero friction between the total potential energy height is equal to the kinetic energy

m*g*h = 1/2* m*v²

Where m is the mass of the dog

g is the acceleration due to gravity,h is the height of the slope

v is the velocity of the sliding dog

g*h = 1/2* v²

9.81*h = 0.5* 2.1²

           =0.225 m

The height of the slope would be 0.225 m.        

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Complete Question

The complete question is shown on the first uploaded image

Answer:

a

  [tex]F = 20 \ kg[/tex]

b

 [tex]F = 18 \ kg[/tex]

c

 [tex]F = 10 \ kg[/tex]

Explanation:

From the diagram we are told that

   The first mass is  [tex]m_1 = 10 \ kg[/tex]

    The second mass is  [tex]m_2 = 3 \ kg[/tex]

    The third mass is  [tex]m_3 = 5 \ kg[/tex]

     The fourth mass is  [tex]m_4 = 2 \ kg[/tex]

Gnerally the total mass accelerated by right force is mathematically evaluated as

        [tex]F = m_1 + m_2 +m_3 + m_4[/tex]

        [tex]F = 10 + 3 +5 + 2[/tex]

      [tex]F = 20 \ kg[/tex]

Gnerally the total mass accelerated by cord 3  is mathematically evaluated as

        [tex]F = m_1 + m_2 +m_3[/tex]

        [tex]F = 10 + 3 +5[/tex]

      [tex]F = 18 \ kg[/tex]

Gnerally the total mass accelerated by cord 3  is mathematically evaluated as

        [tex]F = m_1[/tex]

        [tex]F = 10 \ kg[/tex]

Answer:

The graphs are attached

Explanation:

We are told that he started at rest at rest and travelled down the hill reaching a velocity of 20 m/s after 13 seconds.

Acceleration is gotten from;

v = u + at

a = (v - u)/t

a = (20 - 0)/13

a = 20/13 m/s² or 1.54 m/s²

Distance in this period is gotten from;

v² = u² + 2as

s = (v² - u²)/2s

s = (20² - 0²)/(2 × 20/13)

s = 400/(40/13)

s = 130 m

We are told that after reaching the bottom of the hill, he travelled at a constant velocity of 20 m/s for the next 7 seconds.

At constant velocity, acceleration is 0.

Thus,distance in this period is;

s1 = vt = 20 × 7 = 140 m

I've attached the graphs

Answer:

A. 38 nm

B. 22 nm

Explanation:

This question has given us m, mass of the ball as 3.5kg

M = The mass of arms = 4kg

R = length of arms as 70cm

70 cm = 0.7m

0.7m/2 = 0.35m

rmg + Rmg

g = 9.8m/s

= 0.7(3.5)(9.8)+(0.35)(4)(9.8)

= 24.01 + 13.72

= 37.72

~38

B.

55 degrees below horizontal

Cos 55⁰ = 0.5736

= 0.7(3.5)(9.8)(0.5736)+(0.35)(4)(9.8)(0.5736)

= 13.772136+7.869792

= 21.642

~22Nm

The magnitude of the torque about his shoulder, if he held his arm straight is 21.64Nm

The formula for calculating the magnitude of the torque about his arm is expressed as:

[tex]\tau = (F_1r + F_2R) cos\theta[/tex]

F1 and F2 are the forces

[tex]\theta[/tex] is the given angle

F1 = 3.5 * 9.8 = 34.3 N

r = 70cm = 0.7m

F2 = 4.0 * 9.8 = 39.2N

R = 0.7/2 = 0.35m

Substitute the given values into the formula:

[tex]\tau = (34.3(0.7) + (39.2)(0.35)) cos55^0\\\tau = (24.01+13.72)cos55\\\tau = 37.73cos55\\\tau=21.64Nm[/tex]

Hence the magnitude of the torque about his shoulder, if he held his arm straight is 21.64Nm

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Answer: that one is wrong i did it on flocabulary its false

Explanation:

Since electromagnetic radiations do not require a material medium for their propagation and can travel through vacuum, the statement that electromagnetic radiation cannot travel through a vacuum is False, option B.

!An electromagnetic radiation is a radiation which results from.the interaction of the electric and magnetic fields.

Electromagnetic radiations do not require a material medium for their propagation, and hence can travel through vacuum.

An example of electromagnetic radiation is radio waves.

Hence, the statement that electromagnetic radiation cannot travel through a vacuum is False, option B.

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Answer:

A force vector is a representation of a force that has both magnitude and direction.

Answer:

10 m/h^2

Explanation:

a=v/t

a=60/6

a=10 m/h^2

Answer:

Hello your question has some missing part and diagram attached below is the missing part and diagram

Determine the x component of force at point C in the pipe assembly. Neglect the weight of the pipe. Take F1 = { 348 i - 427 j} lb and F2 = {-307 j + 122 k} lb.

Determine the y component of the force at point C in the pipe assembly.

Determine the x component of the force at point C in the pipe assembly.

Determine the x component of the moment at point C in the pipe assembly.

Determine the y component of the moment at point C in the pipe assembly.

Determine the z component of the moment at point C in the pipe assembly.

answer: attached below is the detailed solution

Explanation:

Attached below is the detailed solution

Q= Q1 +Q2 +Q3+Q4+ Q5

3066 J

Answer:

Thermal expansion

Explanation:

One of the effects of heat is that it causes expansion of materials.

When the copper ball and aluminium plate are both heated, the both materials expand. When they expand, the copper ball can is found to fit more easily into the aluminium plate after both metals have become wider than they were at room temperature.

The coefficient of thermal expansion for aluminium is 35% greater than that of copper. This is the reason why the copper ball can fit into the aluminum plate more easily at higher temperature. The aluminum has expanded 35% more than copper making it easier and leaving a lot of space for the copper to fit into the aluminum plate.

Answer:

a = 2.85714 m/s^2

Explanation:

Fnet=ma

eaqual and oppisate forces means that the 100N pushed on the ball comes back to the child.

100 = 35*a

a= 100/35

Answer:

4R

Explanation:

The centripetal acceleration a is given by a = v²/r where V = speed of object and r = radius of circle. Now, it is given that V = speed of car = V and radius of circle = R, so its centripetal acceleration a = V²/R.

Now, if the speed is doubled to V' = 2V with the same centripetal acceleration, we find the new radius of the curve R' from

a = V'²/R'

R' = V'²/a

R' = (2V)² ÷ V²/R

R' = 4V²R/V²

R' = 4R

So, the new radius is 4 times the initial radius.

Answer: Solubility; decreasing

Explanation:

Answer:2.6

Explanation:

Answer:

The answer is "[tex]\bold{+9.0 \ \frac{m}{s^2}}[/tex]"

Explanation:

Its rate (in respect of time) was its derivative of its specified function x, as well as the speed is the derivative of the speed.  Therefore, in Newton's second legislation [tex]a=2c- 3(2.0)(2.0)t[/tex] is used:  

[tex]F=(2.0 \ kg)a=40c -24t[/tex] (understood by SI units).  

At [tex]t=3.0s, the \ F=-36 \ N[/tex] is told.  

Therefore, the solution of c can be used to solve [tex]F=-36\ N=4.0c-24(3.0)[/tex]. This results in [tex]c=+9.0 \ \frac{m}{s^2}[/tex].

Answer:

Explanation:

a) The total distance traveled by the balloon = (100 + 60) m

                                                          = 160 metes

b) speed = [tex]\frac{distance covered}{time taken}[/tex]

The average speed for the first flight = [tex]\frac{100}{10}[/tex]

                                                    = 10 m/s

c) The speed during its next flight = [tex]\frac{60}{6}[/tex]

                                                    = 10 m/s

d) Average speed for the entire trip = [tex]\frac{(10 + 10)}{2}[/tex]

                                                    = 10 m/s

e) Total displacement = [tex]\sqrt{(100)^{2} + (60)^{2} }[/tex]

                                   = [tex]\sqrt{13600}[/tex]

                                   = 116.62 m

f) velocity = [tex]\frac{displacement}{time}[/tex]

Average velocity during the first 10 seconds = [tex]\frac{100}{10}[/tex]

                              = 10 m/s

g) Average velocity during the next 6 seconds = [tex]\frac{60}{6}[/tex]

                               = 10 m/s

h) Average velocity for the entire trip = [tex]\sqrt{(10)^{2} + (10)^{2} }[/tex]

                                                              = [tex]\sqrt{200}[/tex]

                                                              = 14.14 m/s

Answer:

0.71 N

Explanation:

The inertia of a body, I, is the product of the force applied on it and the time which it acts. Also, it is equal to the change in  the momentum, P, of a body.

So that:

I = Ft

I = ΔP

Ft = m(v - u)

Where F is the force, t is the time, m is the mass of object, v is the final velocity and u is the initial velocity.

Given that: m = 4.33 kg, u = -2.74 m/s, v = 6.35 m/s and t = 55.33 s

F x 55.33 = 4.33(6.35 - (-2.74))

F x 55.33 = 4.33(6.35 + 2.74)

55.33 F = 4.33 x 9.09

55.33 F = 39.3597

F = [tex]\frac{39.3597}{55.33}[/tex]

  = 0.711363

The average force on the ball by the player is 0.71 N.

Answer:

d = 21450 m

Explanation:

Given that,

The speed of a bullet, v = 330 m/s

We need to find the distance covered by the bullet in 65 s.

We know that,

Speed of an object = distance covered divided by time taken.

It means,

d = vt

d = 330 m/s × 65 s

d = 21450 m

So, the bullet will cover a distance of 21450 m.

Answer:

12.6 km/hr

explanation:

25.2 x  [tex]\frac{1}{2}[/tex]

Answer:

The speed of the mouse is 10 m/s

Explanation:

Given;

speed of the cat, u = 8 m/s

let the mass of the cat = m

then, mass of the mouse, = m / 10

let the speed of the mouse = v

Let the kinetic energy of the mouse = K.E

When the cat speeds up by 2 m/s, its new speed = 10 m/s,

then its kinetic energy is given as;

¹/₂m(10)² = 10K.E

[tex]50 m = 10[\frac{1}{2} *\frac{m}{10}.v^2][/tex]

50 = ¹/₂v²

v² = 100

v = √100

v = 10 m/s

Therefore, the speed of the mouse is 10 m/s

Answer:

V₂₋₁  = 50 [miles/h]

Explanation:

In order to solve this problem we must use the concept of relative velocities.

Where:

V1 = 50 [miles/h]

V2 = 75 [miles/h]

V₂₋₁ = Relative velocity among the two cars.

V₂₋₁  = 75 - 25 = 50 [miles/h]

That is, the car that goes at 50 miles per hour, sees how the second car approaches at 50 miles per hour.

Answer:

Shake the bottle while the chemicals are reacting.

Explanation:

Answer:

C

Explanation:

If you take the balloon off, the product will leak.

Given :

If 2000 ft³ of air is crossing an evaporator coil and is cooled from 80°F to 60°F.

To Find :

The volume of air in ft³ exiting the evaporator coil.

Solution :

We know, relation between initial and final temperature and volume is given by :

[tex]\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}[/tex]

Here, temperature is in Kelvin .

We know, Kelvin temperature is given by :

[tex]K=(F-32)\times \dfrac{5}{9} + 273[/tex]

Putting F = 60° F,

[tex]K=(60-32)\times \dfrac{5}{9} + 273\\\\K = 288.56\ K[/tex]

Putting F = 80° F,

[tex]K=(80-32)\times \dfrac{5}{9} + 273\\\\K = 299.67\ K[/tex]

Putting all values, we get :

[tex]\dfrac{V_1}{2000}=\dfrac{60}{80}\\\\V_1=\dfrac{60\times 2000}{80}\ ft^3\\\\V_1=1500\ ft^3[/tex]

Hence, this is the required solution.

Answer:

The magnitude of the force F on a positive charge q in the case that the velocity v and the magnetic field B are perpendicular is [tex]F = q\cdot v\cdot B[/tex], measured in newtons.

Explanation:

From classical theory on Magnetism, the vectorial form of the magnetic force on a particle is given by:

[tex]\vec F = q\cdot \vec v \times \vec B[/tex] (Eq. 1)

Where:

[tex]\vec F[/tex] - Magnetic force, measured in newtons.

[tex]q[/tex] - Electric charge, measured in coulombs.

[tex]\vec v[/tex] - Velocity of the particle, measured in meters per second.

[tex]\vec B[/tex] - Magnetic field, measured in tesla.

By definition of cross product, we get that magnitude of magnetic force on a positive charge [tex]q[/tex] is:

[tex]F = q\cdot v\cdot B \cdot \sin \theta[/tex] (Eq. 2)

Where:

[tex]v[/tex] - Speed of the particle, measured in meters per second.

[tex]B[/tex] - Magnitude of the magnetic field, measured in tesla.

[tex]\theta[/tex] - Angle between the velocity of the particle and magnetic field, measured in sexagesimal degrees.

If velocity and magnetic field are perpendicular, then (Eq. 2) is reduced into this form: ([tex]\theta = 90^{\circ}[/tex])

[tex]F = q\cdot v\cdot B[/tex]

Answer:

x₂ =6*x₁

Explanation:

       [tex]\frac{1}{2}*k*x_{1}^{2} = \frac{1}{2}*m*v^{2} (1)[/tex]

        [tex]\frac{1}{2}*k*x_{2} ^{2} = \frac{1}{2} * 4*m*(3*v)^{2} =\frac{1}{2}*4*m*9*v^{2} = \frac{1}{2}* 36 m*v^{2} (2)[/tex]

        [tex]x_{2} ^{2} = 36 * x_{1} ^{2}[/tex]

Answer:

c it is free fall when an object force act on the force of gravity