At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball's acceleration is 950 m/s2 and the vertical or y component of its acceleration is 750 m/s2. The ball's mass is 0.35 kg. What is the magnitude of the net force acting on the soccer ball at this instant?

Answer:

by using it's buoyant or floating effect by Archimedes.

the buoyant force act on the astronauts body and make he/ she feels like in low gravity.

the buoyant force equation is

F = Density of liquid x earth gravitational field x volume of astronauts body and suit.

the Weight of astronauts in the pools will be less than in the land or air.

Weight in water = weight in air/land - buoyant force

so the astronauts will feel like in the outer space with low gravity.

Answer:

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

Explanation:

Given;

wheel rotates from rest with constant angular acceleration.

Initial angular speed v = 0

Time t = 2.50

Distance x = 8 rev

Applying equation of motion;

x = vt +0.5at^2 ........1

Since v = 0

x = 0.5at^2

making a the subject of formula;

a = x/0.5t^2 = 2x/t^2

a = angular acceleration

t = time taken

x = angular distance

Substituting the values;

a = 2(8)/2.5^2

a = 2.56 rev/s^2

velocity at t = 2.50

v1 = a×t = 2.56×2.50 = 6.4 rev/s

Through the next 5 second;

t2 = 5 seconds

a2 = 2.56 rev/s^2

v2 = 6.4 rev/s

From equation 1;

x = vt +0.5at^2

Substituting the values;

x2 = 6.4(5) + 0.5×2.56×5^2

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

Answer:  Its b, The only problem with this is is there supposed to be a picture?

Explanation: NASA has used there fancy gadgets to figure this out but if there was a picture, this answer could be different.

Answer:

a. 42N

b. 11.8m/s

c. 1.69s

d. 160N

Explanation:

a)  The tension of the rope is 130.66 N.

b) The speed of the bucket while strike the water = 4.64 m/s.

c) The time of fall is  = 4.303 second.

d)  While the bucket is falling, what is the force exerted on the cylinder by the axle is  130.66 N.

Mass of the water bucket; M = 15.0 kg

Mass of the cylinder; m =  12.0 kg

Height of the bucket; h = 10.0 m.

They are connected by a rope and a pivots.

So, acceleration of them is same and let it be a.

So equation of motion of both of them be:

Mg - T = Ma

and, T - mg = ma

Hence, a = g(M-m)/(M+m)

= 9.8(15-12)/(15+12)

= 1.08 m/s²

And, T = m(g+a)

= 12.0(9.8+1.08)

= 130.66 N.

a) so tension of the rope is 130.66 N.

b) speed of the bucket while strike the water = √2ah =√(2×1.08×10.0) m/s = 4.64 m/s.

c) The time of fall is = √2h/a = √(2×10/1.08) second = 4.303 second.

d) While the bucket is falling, what is the force exerted on the cylinder by the axle is tension of the rope, that is, 130.66 N.

Learn more about speed here:

https://brainly.com/question/28224010

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Answer:

T₁ = 378.33 k = 105.33°C

Explanation:

From Fourier's Law of heat conduction, we know that:

Q = - KAΔT/t

where,

Q = Heat Transfer Rate = 1400 W

K = Thermal Conductivity of Material (Aluminum) = 237 W/m.k

A =Surface Area through which heat transfer is taking place=circular bottom

A = π(radius)² = π(0.15 m)² =  0.0707 m²

ΔT = Difference in Temperature of both sides of surface = T₂ - T₁

T₁ = Temperature of outer surface = ?

T₂ = Temperature of inner surface = 105°C + 273 = 378 k

ΔT = 388 k - T₁

t = thickness of the surface (Bottom of Pot) = 0.4 cm = 0.004 m

Therefore,

1400 W = - (237 W/m.k)(0.0707 m²)(378 k - T₁)/0.004 m

(1400 W)/(4188.14 W/k) = - (378 k - T₁)

T₁ = 0.33 k + 378 k

T₁ = 378.33 k = 105.33°C

Answer:

a = 17 m / s²

Explanation:

For this experiment that the astronauts are carrying out, the value of the relation of gravity is cosecant, therefore we can use the kinematic relations

         v² = v₀² + 2a y

They indicate the initial speed 1.2 m / s the final speed 4.3 m / s and the distance remembered 0.5 m

we clear

        a = (v² - v₀²) / 2y

we calculate

       a = (4.3² -1.2²) / 2 0.5

       a = 17 m / s²

this is the gravity of the new planet

Answer:

Overall charge still remains zero on conductor until touched by charged rod.

Explanation:

Here, we want to know what has happened to the overall charge on the conductor.

Since the conductor is neutral, the overall charge on the conductor must remain zero because positive charge is induced on close end to rod then equal and negaitve charge is induced on far end to rod.

Thus, overall charge still remains zero on conductor until touched by charged rod.

Answer:

If the acceleration is constant, the movements equations are:

a(t) = A.

for the velocity we can integrate over time:

v(t) = A*t + v0

where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:

[tex]\int\limits^{10}_0 {A*t + v0} \, dt = ((A/2)10^2 + v0*10) = (A*50 + v0*10)[/tex]

Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.

Answer:

27J

Explanation:

From conservation of Thermal energy, the total internal energy is the total sum of energy supplied or taken from the system plus work done for or on the system.

Now the change in internal energy would be the sum of the received energy substended in the gas plus the work done by the system which is workdone that it will sustend in pushing the lid. This is expressed mathematically as;

U = Q + (F×d);

U- change in internal energy

Q is the energy received by the system and is positive when energy is received by the system.

Fxd is the workdone and is positive since the gas pushes up the lid- the system does work.

U=12+(3×5)= 27J

Answer:

Approximately [tex]0.608[/tex] (assuming that [tex]g = 9.81\; \rm N\cdot kg^{-1}[/tex].)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

This answer will approach this question in two steps:

For simplicity, let [tex]a_{T}[/tex] represent the tangential acceleration ([tex]1.90\; \rm m \cdot s^{-2}[/tex]) of this car.

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to [tex]90^\circ[/tex] or [tex]\displaystyle \frac{\pi}{2}[/tex] radians.

The angular acceleration of this car can be found as [tex]\displaystyle \alpha = \frac{a_{T}}{r}[/tex]. ([tex]a_T[/tex] is the tangential acceleration of the car, and [tex]r[/tex] is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity ([tex]u[/tex] and [tex]v[/tex]) to (tangential) acceleration [tex]a_{T}[/tex] and displacement [tex]x[/tex]:

[tex]v^2 - u^2 = 2\, a_{T}\cdot x[/tex].

The idea is to solve for the final angular velocity using the angular analogy of that equation:

[tex]\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta[/tex].

In this equation, [tex]\theta[/tex] represents angular displacement. For this motion in particular:

Solve this equation for [tex]\omega(\text{final})[/tex] in terms of [tex]a_T[/tex] and [tex]r[/tex]:

[tex]\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}[/tex].

Let [tex]m[/tex] represent the mass of this car. The centripetal force at this moment would be:

[tex]\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}[/tex].

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, [tex]m\, g[/tex].

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let [tex]\mu_s[/tex] denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

[tex]F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g[/tex].

The size of this force should be equal to that of the centripetal force when the car is about to skid:

[tex]\mu_s\, m\, g = \pi\, m\, a_{T}[/tex].

Solve this equation for [tex]\mu_s[/tex]:

[tex]\mu_s = \displaystyle \frac{\pi\, a_T}{g}[/tex].

Indeed, the expression for [tex]\mu_s[/tex] does not include any unknown letter. Let [tex]g = 9.81\; \rm N\cdot kg^{-1}[/tex]. Evaluate this expression for [tex]a_T = 1.90\;\rm m \cdot s^{-2}[/tex]:

[tex]\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608[/tex].

(Three significant figures.)

Answer:

A burning candle. (chemical energy into energy of heat and light, i.e. thermal and wave)

Explanation:

Answer:

Explanation:

The Law of Energy Conservation states that K1 + U1 = K2 + U2

m= 72.0 kg

h= 3.90 m

a)

K1 + U1 = K2 + U2

0 + mgh = 1/2mvf^2 + 0

mass cancels out so gh=1/2vf^2

(9.8 m/s^2)(3.9 m)=(.5)(vf^2)

vf= 8.74 m/s

b)

1/2mv^2 + mgh = 1/2mv^2 + 0

mass cancels again

(.5)(2.9^2 m/s) + (9.8 m/s^2)(3.9 m) = (.5)(vf^2)

vf= 9.21 m/s

c)

This would be the same as the past problem as the velocity gets squared so direction along the axis doesn't matter. Thus, vf= 9.21 m/s

Answer:

A and B is positive charge

C_negative

Explanation:

because when an ebonite is rubbed with fur produce negative charge due to law of electrostatic like charge repel and unlike attract

Answer:

To solve this problem we just need to graph two forces with same direction, pointing to different sides with intensities of 40 N and 60 N.

The image attached shows these forces.

Notice that the vectors are parallel, that's because they have the same direction, but they point to different sides, and their magnitudes have a difference of 20 N.

Answer:

D

Explanation:

Insoluble impurities would not change the constituent of the substance. Soluble would for example salt water takes longer time for the water to become vapour when subjected to the same temperature that normal water.

Wind would affect, the more windy the tendency for particles of the liquid to be moved into the atmosphere.

With an increase in surface area, the evaporation rate increase . Take a clue from water placed on the ground and exposed to the atmosphere and that same quantity of water is placed in a cup. That on the floor would evaporate faster.

Similarly the higher the temperature a substance is subjected to the easier is it's rate of evaporation. Take for instance water in a cup placed in the sun and that same placed in a room with mild temperatures than that of the sun.With time that in the sun decreases in volume faster than that in the room.

Answer:

32mi

Explanation:

If 1lb contains 3,500 Cal

It means the number of hours required to burn 3500cal would be;

3500/331 = 10.57hours

But a brisk walk is 3.0 mi/h,

It means a distance of 3.0 × 10.57 mi would be covered = 31.71 miles

32miles{ approximated to the nearest whole}

Note Distance = speed × time

Answer:

The power is  [tex]P_p = 4.6 \ W[/tex]

Explanation:

From the question we are told that

   The power delivered is  [tex]P_{s} = 1.15 \ W[/tex]

   Let it resistance be denoted as R

    The resistors are connected in series so the equivalent resistance is  

     [tex]R_{eqv} = R+ R = 2 R[/tex]

Considering when it is connected in series    

Generally power is mathematically represented as

     [tex]P_s = V * I[/tex]

Here I is the current which is mathematically represented as

       [tex]I = \frac{V}{2R}[/tex]

The power becomes

     [tex]P_s = V * \frac{V}{2R}[/tex]

     [tex]P_s = \frac{V^2}{2R}[/tex]

substituting value

    [tex]1.15 = \frac{V^2}{2R}[/tex]

Considering when resistance is connected in parallel

  The equivalent resistance becomes

    [tex]R_{eqv} = \frac{R}{2}[/tex]

So The current  becomes

       [tex]I = \frac{V}{\frac{R}{2} } = \frac{2V}{R}[/tex]

And the power becomes

     [tex]P_p = V * \frac{2V}{R} = \frac{2V^2}{R} = \frac{4 V^2}{2 R} = 4 * P_s[/tex]

 substituting values

     [tex]P_p = 4 * 1.15[/tex]

     [tex]P_p = 4.6 \ W[/tex]

Answer:

  c.  in the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.

Explanation:

First law: things keep doing what they are doing, unless force is applied.

Complete Question

The complete question is shown on the first uploaded image

Answer:

The mass is    [tex]M =1.43 *10^{32} \ kg[/tex]

Explanation:

From the  question we are told that

       The mass of the stars are [tex]m_1 = m_2 =M[/tex]

        The orbital speed of each star is  [tex]v_s = 240 \ km/s =240000 \ m/s[/tex]

         The orbital period is [tex]T = 12.5 \ days = 12.5 * 2 4 * 60 *60 = 1080000\ s[/tex]

The centripetal force acting on these stars is mathematically represented as

      [tex]F_c = \frac{Mv^2}{r}[/tex]

The gravitational force acting on these stars is mathematically represented as

      [tex]F_g = \frac{GM^2 }{d^2}[/tex]

So  [tex]F_c = F_g[/tex]

=>        [tex]\frac{mv^2}{r} = \frac{Gm_1 * m_2 }{d^2}[/tex]

=>      [tex]\frac{v^2}{r} = \frac{GM}{(2r)^2}[/tex]

=>      [tex]\frac{v^2}{r} = \frac{GM}{4r^2}[/tex]

=>    [tex]M = \frac{v^2*4r}{G}[/tex]

The distance traveled by each sun in one cycle is mathematically represented as

     [tex]D = v * T[/tex]

      [tex]D = 240000 * 1080000[/tex]

      [tex]D = 2.592*10^{11} \ m[/tex]

Now this can also be represented as

      [tex]D = 2 \pi r[/tex]

Therefore

                  [tex]2 \pi r= 2.592*10^{11} \ m[/tex]

=>   [tex]r= \frac{2.592*10^{11}}{2 \pi }[/tex]

=>    [tex]r= 4.124 *10^{10} \ m[/tex]

So  

       [tex]M = \frac{v^2*4r}{G}[/tex]

=>    [tex]M = \frac{(240000)^2*4*(4.124*10^{10})}{6.67*10^{-11}}[/tex]

=>    [tex]M =1.43 *10^{32} \ kg[/tex]

Answer:

Explanation:

Using the law of conservation of momentum which states that 'the sum of momentum of bodies before collision is equal to their sum after collision. The bodies will move together with a common velocity after collision.

Momentum = Mass * Velocity

Before collision;

Momentum of receiver m1u1= 0 kgm/s (since the receiver is standing still)

Momentum of the tackler

m2u2 = 2.60*122 = 317.2 kgm/s

where m2 and u2 are the mass and velocity of the tacker respectively.

Sum of momentum before collision = 0+317.2 = 317.2 kgm/s

After collision

Momentum of the bodies = (m1+m2)v

v = their common velocity

m1 = mass of the receiver

Momentum of the bodies = (122+m1)(1.30)

Momentum of the bodies = 158.6+1.30m1

According to the law above;

317.2 = 158.6+1.30m1

317.2-158.6 = 1.30m1

158.6 = 1.30m1

m1 = 158.6/1.30

m1 = 122kg

The mas of the receiver is 122kg

Answer:

  a = - e E / m

a = - 1,758 10¹¹ E

Explanation:

For this exercise we can use Newton's second law

        F = m a

where the force is electric

 the forces given by the product of the charge by the electric field

         F = q E

in this case tell us that the charge is the charge of the electron

         q = -e = - 1.6 10⁻¹⁹ C

we substitute

        - e E = m a

          a = - e E / m

we calculate

           a = - 1.6 10⁻¹⁹ / 9.1 10⁻³¹ E

           a = - 1,758 10¹¹ E

The negative sign indicates that the acceleration is in the opposite direction to the electric field

Answer:

The final speed of the ball is 9 m/s.

Explanation:

We have,

A steel ball is dropped from a diving platform. It is required to find the velocity of the ball 0.9 seconds after its released. It will move under the action of gravity. Using equation of motion to find it as :

[tex]v=u+at[/tex]

u = 0 (at rest), a = g

[tex]v=gt\\\\v=10\times 0.9\\\\v=9\ m/s[/tex]

So, the final speed of the ball is 9 m/s.

Complete question is;

Assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charge the battery with power rating, 12 V, 50 Ampere-minutes

Answer:

Amount of water required to charge the battery = 7.35 m³

Explanation:

The formula for Potential energy of the water at that height = mgh

Where;

m = mass of the water

g = acceleration due to gravity = 9.8 m/s²

H = height of water = 50 cm = 0.5 m

We know that in density, m = ρV

Where;

ρ = density of water = 1000 kg/m³

V = volume of water

So, potential energy is now given as;

Potential energy = ρVgH = 1000 × V × 9.8 × 0.5 = (4900V) J

Now, formula for energy of the battery is given as;

E = qV

We are given;

q = 50 A.min = 50 × 60 = 3,000 C

V = 12 V

Thus;

qV = 3,000 × 12 = 36,000 J

E = 36,000 J

At a 100% conversion rate, the energy of the water totally powers the battery.

Thus;

(4900V) = (36,000)

4900V = 36,000

V = 36,000/4900

V = 7.35 m³

Answer:

a.[tex]5.66ms^{-2}[/tex]

b.55 m

Explanation:

We are given that

Mass ,m=0.9 kg

Length of string,l=50 cm=[tex]\frac{50}{100}=0.50 m[/tex]

1 m=100 cm

[tex]\theta=30^{\circ}[/tex]

R=55 m

a.Centripetal acceleration

[tex]a_c=gtan\theta[/tex]

[tex]a_c=9.8tan30^{\circ}[/tex]

[tex]a_c=5.66 m/s^2[/tex]

Hence, the radial acceleration of the bus=[tex]5.66m/s^2[/tex]

b. Radius of curve,R=55 m

Answer:

0.981 rad/sec^2

Explanation:

mass that pulls on string = 5 kg

weight due to mass = 5 x 9.81 = 49.05 N

radius of rod = 0.1 m

torque produced by this force on the rod = force x radius

torque = 49.05 x 0.1 = 4.905 N-m

mass of disk = 125 kg

radius of disk = 0.2 m

moment of inertia of the disk I = m[tex]r^{2}[/tex]

I = 125 x [tex]0.2^{2}[/tex] = 5 kg-m^2

from the equation, T = Iα

where T is torque

I is moment of inertia

α is angular acceleration

imputing values,

4.905 = 5α

α = 4.905/5 = 0.981 rad/sec^2

Answer:

Explanation:

The power rating of the battery isn't provided. But let us assume that it is one of the common batteries with ratings of 12 V and 50 A.h

Potential energy possessed by water at that height = mgh

m = mass of the water = ρV

ρ = density of water = 1000 kg/m³

V = volume of water = ?

g = acceleration due to gravity = 9.8 m/s²

h = height of water = 50 cm = 0.5 m

Potential energy = ρVgh = 1000 × V × 9.8 × 0.5 = (4900V) J

Energy of the battery = qV

q = 50 A.h = 50 × 3600 = 180,000 C

V = 12 V

qV = 180,000 × 12 = 2,160,000 J

Energy = 2,160,000 J

At a 100% conversion rate, the energy of the water totally powers the battery

(4900V) = (2,160,000)

4900V = 2,160,000

V = (2,160,000/4900)

V = 440.82 m³

Hence, with our assumed power ratings for the battery (12 V and 50 A.h), 440.82 m³ of water at the given height of 50 cm would power the battery.

Incase the power ratings of the battery in the complete question is different, this solution provides you with how to obtain the correct answer, given any battery power rating.

Hope this Helps!!!

Answer:

a = -0.8 m/s²

Here, negative sign indicates that the acceleration has opposite direction to the direction of motion.

Explanation:

First we find the angular acceleration of the ball from the following formula:

α = (ωf - ωi)/t

where,

α = angular acceleration = ?

ωf = final angular velocity = 7 rad/s

ωi = initial angular velocity = 13 rad/s

t = Time taken = 15 s

Therefore,

α = (7 rad/s - 13 rad/s)/15 s

α = - 0.4 rad/s

negative sign shows that acceleration is in opposite direction to the direction of motion.

Now, for the linear acceleration, we use the formula:

a = rα

where,

a = linear acceleration = ?

r = radius of circular path = length of rope = 2 m

therefore,

a = (2 m)(- 0.4 rad/s²)

a = -0.8 m/s²

Here, negative sign indicates that the acceleration has opposite direction to the direction of motion.

Answer: [tex]2m/s^2[/tex]

Explanation:

[tex]Formula: F=ma[/tex]

Where;

F = force

m = mass

a = acceleration

Solve for a;

[tex]a=\frac{F}{m}[/tex]

[tex]a=\frac{30N}{15kg}\\ a=2m/s^2[/tex]

Answer:

Answer D is correct

Explanation:

[tex]pressure = \frac{force}{area} \\ = \frac{320}{0.5} \\ = 640[/tex]

hope this helps

brainliest appreciated

good luck! have a nice day!

Answer:

29.16 J

Explanation:

From Hook's law,

W = 1/2(ke²)..................... Equation 1

Where W = work done, k = Spring constant, e = extension.

Given: W = 9 J, e = 0.5 m.

Substitute into equation 1

9 = 1/2(k×0.5²)

Solve for k

k = 18/0.5²

k = 72 N/m.

The work done required to stretch the spring by additional 0.4 m is

W = 1/2(72)(0.4+0.5)²

W = 36(0.9²)

W = 29.16 J.