at what speed does the kg ball move ?

Answer:

A

Explanation:

Answer:

D. Conduction

Explanation:

Convection heat transfer occurs in fluids, while conduction heat transfer occurs in solids.

Answer:

I think you are trying to ask 'what happens to the average speed (of humans or animals.) if the distance decreased and time stayed the same (the time such as what hour it is or what second?)' If that is the case then the average speed will be more since the distance decreased. And the time staying the same might affect it I am not fully sure so I will not say anything about that. I hope this helps!

Answer:

1.31×10¯⁶ N

Explanation:

From the question given above, the following data were obtained:

Mass of John (M₁) = 81 Kg

Mass of Mike (M₂) = 93 Kg

Distance apart (r) = 0.620 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force (F) =?

The gravitational force between the two students, John and Mike, can be obtained as follow:

F = GM₁M₂ / r²

F = 6.67×10¯¹¹ × 81 × 93 / 0.62²

F = 6.67×10¯¹¹ × 7533 / 0.3844

F = 1.31×10¯⁶ N

Therefore, the gravitational force between the two students, John and Mike, is 1.31×10¯⁶ N

Answer:

Let Vx = 35 m/s     speed of swimmer across river

     Vy = 22 m/s     speed of river dowstream

V = (Vx^2 + Vy^2)^1/2 = 41.3 m/s      net speed of swimmer

tan theta = Vy / Vx = .629    theta = 32.2 deg

(Theta would be zero if speed of river was zero)\

Answer:

Time period of horizontal Oscillation  = T = 2[tex]\pi[/tex][tex]\sqrt{\frac{m}{k} }[/tex]

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Explanation:

Solution:

As we know that:

F = Kx

Here,

K = Spring constant

x = displacement.

First, they are displacing it with 6 cm from its rest position for which Time period of the oscillation is T = 2 seconds.

But next, they want to know the effect on the time period of the oscillation if the displacement x is doubled from 6cm to 12 cm.

First of all, let us see the equation of the time period of the oscillation.

We need to check, if time period does depend on the displacement or not.

As we know,

Time period of horizontal Oscillation  = T = 2[tex]\pi[/tex][tex]\sqrt{\frac{m}{k} }[/tex]

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Since, K is the constant for a particular spring, we need to change the mass of the cart to change the time period.

Hence the Time period will remain same.

The time period will remain the same in both conditions. The time period of horizontal Oscillation will be [tex]2 \pi \sqrt{\frac{m}{k} }[/tex].

The period is the amount of time it takes for a particle to perform one full oscillation. T is the symbol for it. Taking the reciprocal of the frequency yields the frequency of the oscillation.

From Hooke's law;

F = Kx

Where,

K is the spring constant

x is the  displacement

First, they move it 6 cm from its rest position, with a T = 2 second oscillation period.

However, they want to know what influence doubling the displacement x from 6 cm to 12 cm has on the oscillation's time period.

Let's start by looking at the oscillation's time period equation. We need to see if the time period is affected by the shift.

The time period of the horizontal oscillation is given by;

[tex]\rm T = 2 \pi \sqrt{\frac{m}{k} }[/tex]

As the equation shows, the period of oscillation is determined by the mass and the spring constant. On the displacement, no.

We must modify the mass of the cart to change the time period since K is the constant for each spring.

Hence the time period will not change.

To learn more about the time period of oscillation refer to the link;

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Answer:

112.63km/hr

Explanation:

The given dimension is :

         70mph

We are to convert this to km/hr

         1 mile  = 1.609km

so;

     70mph x 1.609    = 112.63km/hr

So,

  The solution is 112.63km/hr

Answer:

7

Explanation:

We are given:

    w(x) = 3x   -   7

     w(x)  = 14

The problem here entails us to solve for x;

To solve for x; equate the two expressions:

          3x  - 7  = 14

           3x  = 14 + 7

           3x  = 21  

             x = 7

So the value of x  = 7

Answer:

Both balloons experiences same buoyant force.

Explanation:

Buoyant force is defined as the upward force that is exerted on an object which is wholly or partly immersed in a fluid. We can also define it as the upward force exerted by any fluid upon a body immersed in it. We may also refer to this buoyant force as the upthrust.

This buoyant force is the push of air on the balloon and it is independent of the contents of the balloons. Hence, both balloons experiences same buoyant force.

Answer:

(a) 34.4°

(b) 49.4 N

Explanation:

(a) From the diagram,

Amgle between the x axis can be calculated as,

cosΘ = adjacent/hypoteneous

cosΘ = 72.1/87.4

cosΘ = 0.8249

Θ = cos⁻¹(0.8249)

Θ = 34.4°.

Hence the angle between the x axis is 34.4°

(b) To find the component along the y axis we make use of pythagoras theorem.

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

Where a = 87.4 N, b = 72.1 N, c = y.

Substitute  these values into equation 1

87.4² = 72.1² + y²

y² = 87.4²-72.1²

y² = 2440.35

y = √(2440.35)

y = 49.4 N

Answer:

08 and 09 with be 67

Explanation:

i know this because it was the vase major

Answer:

The weight of the girl is 250 N

Explanation:

Static Equilibrium

Static equilibrium occurs when an object is at rest, i.e., neither rotating nor translating.

In the static rotational equilibrium, the total torque is zero with respect to any rotational axis.

The torque applied by a force F perpendicular to a displacement X with respect to a reference rotating point is:

T = F*X

The seesaw will be in rotational equilibrium if the torque applied by the boy of mass m1=40 Kg at x1=2 m from the pivot is equal to the torque applied by the girl of unknown mass m2 at x2=3.2 m from the pivot.

The force applied by both children is their weight:

[tex]F_1 = W_1 = m_1g[/tex]

[tex]F_2 = W_2 = m_2g[/tex]

It must be satisfied:

[tex]m_1gx_1=m_2gx_2[/tex]

Simplifying:

[tex]m_1x_1=m_2x_2[/tex]

Solving for m2:

[tex]\displaystyle m_2=\frac{m_1x_1}{x_2}[/tex]

[tex]\displaystyle m_2=\frac{40*2}{3.2}[/tex]

[tex]m_2=25\ kg[/tex]

Her weight is:

[tex]\mathbf{W_2=25*10 = 250\ N}[/tex]

The weight of the girl is 250 N

Answer:

no

Explanation:

it is faster at the equator

Answer:

Uh, I could be wrong but doesn’t it mean that the wave and particle are reacting together to make light? I think it’s something like that... I hope this helps!

The correct answer is 5 km/h

Explanation:

The speed at which the duck travels can be found by using the equation that is given (speed= distance/ time). The first step to do this is to replace distance and time using the values given. Here is the process:

speed = distance / time

speed = 10 km / 2 hours

Now, solve this equation

speed = 10 km/ 2 hours

10 / 2 or 10 divided 2 = 5

Finally, use the units, in this case, the correct units are km/h

Answer:

14.6m/s

Explanation:

Given parameters:

Mass of the student  = 62kg

Initial velocity  = 0m/s

Height of slide = 10.6m

g  = 10m/s²

Unknown:

Speed at the bottom of the slide = ?

Solution:

The speed at the bottom of the slide is the final velocity;

             v ² = u²  + 2gh

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

h is the height

             v² = 0²  +  2x 10 x 10.6

             v²  = 212

              v  =  14.6m/s

Answer:

[tex]0.15\: \mathrm{Hz}[/tex]

Explanation:

The frequency is of an object is given by [tex]f=\frac{1}{T}[/tex], where [tex]T[/tex] is the orbital period of the object.

Since the racecar makes 24 revolutions around a circular track in 162 seconds, it will take the racecar [tex]\frac{162}{24}=6.75\:\mathrm{s}[/tex] per revolution.

Therefore, the frequency of the racecar is [tex]\frac{1}{6.75}=\fbox{$0.15\:\mathrm{Hz}$}[/tex] (two significant figures).

The radius of the track is irrelevant in this problem.

Answer:

.

Explanation:

F = kx so k = 800/((10-5)/100) = 16000 N/m

W = 1/2 kx^2 = 1/2 * 16000 * .05^2 = 20 J.

(sorry if it's wrong)

Answer:

Explanation:

Maximum value of force will be possible when both the sphere will have same charge . In that case charge on each sphere = Q / 2 =.5Q

F( max ) = k .5Q x .5Q / R²

=.25kQ² /R²

For the second case

F = k q ( Q-q)/  R²

F = .25kQ² /3R²

.25kQ² /3R² =  k q ( Q-q)/  R²

.25 Q² = 3qQ - 3q²

3q² - 3qQ + .25 Q² = 0

q =

Answer:

most likely be included in the analysis section of a lab report

Explanation:

Answer: Surface water is replaced by the water cycle.

Explanation: The water cycle is a cycle that describes the movement of water.

Answer:

it is water cycle when it's one of the process occurs and process is precipation this replaces surface water

Answer:

9 Nm

Explanation:

The formula for torque is;

τ = lever arm * force applied

τ = 30/100 * 30

τ = 9 Nm

The torque on a bolt will be "9 Nm".

Given values are:

Force applied,

Lever arm,

As we know the formula,

→ [tex]\tau = Lever \ arm\times force \ applied[/tex]

By substituting the values, we get

→    [tex]= \frac{30}{100}\times 30[/tex]

→    [tex]= 9 \ Nm[/tex]

Thus the above response is correct.    

Learn more:

https://brainly.com/question/20372081

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

a)

       [tex]\tau = I * \alpha (1)[/tex]

       [tex]\alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)[/tex]

        [tex]\omega_{f}^{2} - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)[/tex]

       [tex]0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)[/tex]

       [tex]\omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)[/tex]

        [tex]v = \omega * r (8)[/tex]

       [tex]v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)[/tex]

b)    

       [tex]a_{t} = \alpha * r (9)[/tex]

       [tex]a_{t} = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)[/tex]

       [tex]a_{c} = \omega^{2} * r (11)[/tex]

       [tex]a_{c} = \omega^{2} * r = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)[/tex]

       [tex]a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)[/tex]

Answer:

1960 joule

Explanation:

Answer:

The solution to this question can be defined as follows:

Explanation:

In point (a):

[tex]v_i= 100 \ mV\\\\ i_I = 100 \mu \ A\\\\ v_O = 10 \ V\\\\ R_L = 100 \ \Omega \\\\i_L = \frac{V_0}{R_L} = \frac{10}{100} = 100 \ MA \\\\A_v = \frac{V_0}{V_i} = \frac{10}{100 \times 10^{-3}} =100 \\\\A_v(db) = 20 \lag (100) =40 \ db \\\\ A_i= \frac{i_L}{i_i} = \frac{100 \times 10^{-3}}{100 \times 10^{-6}} =1000 \\\\A_i(db) = 20 \lag (100) =60 \ db \\\\[/tex]

[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_L}{v_i i_i} = \frac{ 10(100 \times 10^{-3})}{100 \times 10^{-6} \times 100 \times 10^{-3}} =100000\\\\ A_p(db) =10 \log (100000) =50 \ db \\\\[/tex]

In point (b):

[tex]v_i = 10 \mu V\\\\ i_i = 100 \ nA \\\\ v_O =1 \ V \\\\ R_L= 10 \ k \Omega \\\\i_0 = \frac{V_0}{R_L} = \frac{1}{10 \ K} = 100 \ \muA \\\\A_v = \frac{V_0}{V_i} = \frac{10}{10 \times 10^{-6}} =100000 \\\\A_v(db) = 20 \lag (100000) =100 \ db \\\\ A_i= \frac{i_0}{i_i} = \frac{100 \times 10^{-6}}{100 \times 10^{-9}} =1000 \\\\A_i(db) = 20 \lag (1000) =60 \ db \\\\[/tex]

[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_0}{v_i i_i} = \frac{ 1 \times 100 \times 10^{-6})}{10 \times 10^{-6} \times 100 \times 10^{-9}} =100000000\\\\A_p(db) =10 \log (100000000) =80 \ db \\\\[/tex]

In point (C):

[tex]v_i =1\ V\\\\ i_I = 1 \ mA\\\\ v_O =5\ V \\\\ R_L = 10 \ \Omega \\\\i_0 = \frac{V_0}{R_L} = \frac{5}{10 } = 0.5 \ A \\\\A_v = \frac{V_0}{V_i} = \frac{5}{1} =5 \\\\A_v(db) = 20 \log 5 =13.97 \ db = 14 \db \\\\ A_i= \frac{i_0}{i_i} = \frac{0.5}{1\times 10^{-3}} =500 \\\\A_i(db) = 20 \log (500) =53.97 \ db = 54 \db \\\\[/tex]

[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_0}{v_i i_i} = \frac{ 5 \times 0.5 }{1 \times 1 \times 10^{-3}} =2500\\\\A_p(db) =10 \log (2500) = 33.97 \ db = 34 \db\\\\[/tex]

Answer:

hmmm thats too hard for me.

Explanation:

Answer:

  v = 1.4 m / s

Explanation:

To solve this problem we use the law of conservation of momentum, for this we define a system formed by the two blocks in such a way that the force during the collision have been internal

initial instant. Just before the crash

           p₀ = M v₁ - m v₂

final instant. Right after the crash

          p_f = (M + m) v

the moment is preserved

          p₀ = p_f

          M v₁ - m v₂ = (M + m) v

           v = [tex]\frac{1}{M+m}[/tex]  (M v₁ - mv₂)

let's calculate

           v = [tex]\frac{1}{4+1}[/tex]  (4 2 - 1 1)

           v = 1.4 m / s

in the same direction as the largest block (M)

The approximate speed of block X after the collision is 1.4 m/s.

Given information:

Block X and block Y travel toward each other along a horizontal surface with block X traveling in the positive direction.

Block X has a mass of [tex]m_1=[/tex] 4 kg and a speed of [tex]u_1=[/tex] 2 m/s.

Block Y has a mass of [tex]m_2=[/tex] 1 kg and a speed of [tex]u_2=[/tex] -1 m/s.

The collision between them is perfectly inelastic. So, the final velocity of both the objects will be the same.

The momentum of the system will be conserved. Let v be the final velocity of both the blocks.

The final velocity v can be calculated as,

[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\4\times 2-1\times1=(4+1)v\\5v=7\\v=1.4\rm\;m/s[/tex]

Therefore, the approximate speed of block X after the collision is 1.4 m/s.

For more details about collision, refer to the link:

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