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Furkat [3]
3 years ago
9

1)A chocolate bar measures 10 cm long by 2 cm wide and is 2 cm thick

Physics
1 answer:
Rudiy273 years ago
3 0
A) volume = length*width*height
so the volume = 10*2*2
= 40cm^2
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Force between two charges is given by

F =\frac{kq_1q_2}{r^2}

F =\frac{9*10^9* 1* 10^{-6}* 1 * 10^{-6}}{4^2}

F = 5.625 * 10^{-4} N

Now in order to find the acceleration of each mass

we can use

F = ma

5.625 * 10^{-4} = 1 * a

a= 5.625 * 10^{-4} m/s^2

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A grandfather clock works by swinging a pendulum back and forth with a
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For example, a wave with a time period of 2 seconds has a frequency of

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2 years ago
Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after
adoni [48]

Answer: v = 1.19 * 10^{6} m/s

Explanation: q = magnitude of electronic charge = 1.609 * 10^{-19} c

mass of an electronic charge = 9.10 * 10^{-31} kg

V= potential difference = 4V

v = velocity of electron

by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.

kinetic energy = \frac{mv^{2} }{2},  potential energy = qV

hence, \frac{mv^{2} }{2} = qV

\frac{9.10 *10^{-31} * v^{2}  }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31}  * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s

7 0
3 years ago
A loader sack of total mass
vampirchik [111]

Question: A loader sack of total mass

is l000 grams falls down from

the floor of a lorry 200 cm high

Calculate the workdone by the

gravity of the load.​

Answer:

19.6 Joules

Explanation:

Applying

W = mgh........................ Equation 1

Where W = Workdone by gravity on the load, m = mass of the loader sack, h = height, g = acceleration due to gravity

From the question,

Given: m = 1000 grams = (1000/1000) kilogram = 1 kg, h = 200 cm = 2 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

W = (1×2×9.8)

W = 19.6 Joules

Hence the work done by gravity on the load is 19.6 Joules

8 0
3 years ago
A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the
olga2289 [7]

Answer:

a) F = 30 N, b)   I = 12 N s , c)  I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

     p₀ = m v

The speed can be found by kinematics

     v = v₀ - g t

     v = 0 - 10 0.4

     v = -4.0 m / s

Final after division

     pf = m₁ v₁f + m₂ v₂f

    p₀ = pf

    M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

     F = mg

     F = 3 10 = 30 N

b) The expression for momentum is

     I = Ft

Before the explosion the only force that acts is the weight

    I = mg t

    I = 3 10 0.4

    I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

    M v = m₁ v₁f + m₂ v₂f

    v₂f = (M v - m₁ v₁f) / m₂

    v₂f = (3 (-4) - 1 6) / 2

   v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

    I = F t = Dp

   F t = pf –po)

   F t= [m₁ v₁f + m₂ v₂f]

   

   I = [1 6 + 2 (-9) -0]

   I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

 ΔI =If – I₀  

ΔI = - 12 – (-12)

ΔI = -0 N s

3 0
3 years ago
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