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kirill [66]
3 years ago
5

Subtract 14/3 from 5

m5" alt="subtact14 \3 from5" align="absmiddle" class="latex-formula">
​

Physics
1 answer:
WITCHER [35]3 years ago
3 0
The answer is 1/3
Steps:

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Which type of force can be used to kick a soccer ball?
Airida [17]

Answer:

The answer is Friction.

Hope this helps :)

Please mark me brainliest :)

6 0
2 years ago
Read 2 more answers
A still ball of mass 0.514kg is fastened to a cord 68.7cm long and is released when the cord is horizontal. At the bottom of its
Alex787 [66]

Answer:

1.21 m/s

Explanation:

From the law of conservation of energy,

U₁ + K₁ + E₁ = U₂ + K₂ + E₂

where U₁ = initial potential energy of system =initial potential energy of still ball = mgh where m = mass of still ball = 0.514 kg, g = acceleration due to gravity = 9.8 m/s² and h = height = length of cord = 68.7 cm = 0.687 m.

K₁ = initial kinetic energy of system = 0

E₁ = initial internal energy of system = unknown and

U₂ = final potential energy of system = 0

K₁ = final kinetic energy of system = final kinetic energy of ball + steel block = 1/2(m + M)v² where m = mass of still ball, M = mass of steel block = 2.63 kg and v = speed of still ball + steel block

E₁ = final internal energy of system = unknown

So,

U₁ + K₁ + E₁ = U₂ + K₂ + E₂

mgh + 0 + E₁ = 0 + 1/2(m + M)v² + E₂

mgh = 1/2(m + M)v² + (E₂ - E₁)

Given that (E₂ - E₁) = change in internal energy = ΔE = 1/2ΔK where ΔK = change in kinetic energy. So, ΔE = 1/2ΔK = 1/2(K₂ - K₁) = K₂/2 = 1/2(m + M)v²/2 = (m + M)v²/4

Thus, mgh = 1/2(m + M)v² + (E₂ - E₁)

mgh = 1/2(m + M)v² + (m + M)v²/4

mgh = 3(m + M)v²/4

So, making v subject of the formula, we have

v² = 4mgh/3(m + M)

taking square root of both sides, we have

v = √[4mgh/3(m + M)]

Substituting the values of the variables into the equation, we have

v = √[4 × 0.514 kg × 9.8 m/s² × 0.687 m/{3(0.514 kg + 2.63 kg)}]

v = √[13.8422/{3(3.144 kg)}]

v = √[13.8422 kgm/s²/{9.432 kg)}]

v = √(1.4676 m²/s²)

v = 1.21 m/s

6 0
3 years ago
A coin placed on the cover of a book just begins to move when the cover makes an angle of 38 degrees with the horizontal. What i
RSB [31]

Answer:

m g sin theta = force of object along incline due to gravity

N μ = frictional of incline on object where N is the normal force

N = m g cos theta     force perpendicular to incline

m g sin theta = N μ = μ m g cos theta

μ  = tan theta = tan 38 = .78

6 0
2 years ago
A free falling object has the velocity time graph shown. What is the objects displacement between 0.0 and 6.0s
zloy xaker [14]

For free fall motion the displacement can be found by graphically as well as by kinematics equation

Here acceleration of object is constant as it fall due to gravity so we can use

d = v_i * t + \frac{1}{2}at^2

here if body starts with zero initial speed then we can say

d = 0 + \frac{1}{2}*9.8*t^2

here we need to find the displacement from t = 0 to t = 6s

so we can say

d = \frac{1}{2}*9.8*6^2

d = 176.4 m

so the displacement will be 176.4 m

in order to find the displacement from the graph of velocity and time we need to find the area under the graph for given time interval that will also give us same displacement for given period of time.

6 0
4 years ago
Read 2 more answers
PLEASE HELP ME WITH THIS ONE QUESTION
yaroslaw [1]

Answer:

Energy =  1.5032*10^(-10) Joules

Explanation:

By Einstein's relativistic energy equation, we know that the energy of a given particle is given by:

Energy = rest energy + kinetic energy.

            = m*c^2  + (γ - 1)*mc^2

Where γ depends on the velocity of the particle.

But if the proton is at rest, then the kinetic energy is zero, and γ = 1

Then the energy is just given by:

Energy = m*c^2

Where we know that:

mass of a proton = 1.67*10^(-27) kg

speed of light = c = 2.9979*10^(8) m/s

Replacing these in the energy equation, we get:

Energy = ( 1.6726*10^(-27) kg)*( 2.9979*10^(8) m/s)^2

Energy = 1.5032*10^(-10) kg*m^2/s^2

Energy =  1.5032*10^(-10) J

5 0
3 years ago
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