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almond37 [142]
3 years ago
7

1. What is the weight of a 12 kg dog on the moon? (acceleration of gravity is 1.63 m/s2)

Physics
1 answer:
Drupady [299]3 years ago
3 0
3 is the answe bc it is awesome and cool
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A cylindrical storage tank has a radius of 1.35 m. When filled to a height of 3.45 m, it holds 14,014 kg of a liquid industrial
Kruka [31]

Answer:

709.93 kg/m³

Explanation:

Density: This can be defined as the ratio of the mass of a body to the volume of that body. The S.I unit of density is kg/m³

From the question above,

D = m/v.............................. Equation 1

Where D = density, m = mass, v = volume.

Note: The volume of the liquid is equal to the volume of the height occupied by the liquid in the container

Since the tank is cylindrical,

v = πr²h........................ Equation 2

Where r = radius of the the tank, h = height of the liquid in the tank

Substitute equation 2 into equation 1

D = m/(πr²h)............... Equation 3

Given: m = 14014 kg, r = 1.35 m, h = 3.45 m, π = 3.14

Substitute into equation 3

D = 14014/(3.14×1.35²×3.45)

D = 14014/19.74

D = 709.93 kg/m³

6 0
3 years ago
I will give more points <br>for this thank you I u helped<br>me ​
scZoUnD [109]

Answer:

weight = mass×gravity

  • 15×9.8
  • 147N

<h2>stay safe healthy and happy.</h2>
6 0
3 years ago
The rotational kinetic energy term is often called the kinetic energy in the center of mass, while the translational kinetic ene
Nataliya [291]

Question in proper order

The rotational kinetic energy term is often called the <em>kinetic energy </em><em>in</em> the center of mass, while the translational kinetic energy term is called the <em>kinetic energy </em><em>of</em> the center of mass.

You found that the total kinetic energy is the sum of the kinetic energy in the center of mass plus the kinetic energy of the center of mass. A similar decomposition exists for angular and linear momentum. There are also related decompositions that work for systems of masses, not just rigid bodies like a dumbbell.  

It is important to understand the applicability of the formula  

Ktot=Kr+Kt

Which of the following conditions are necessary for the formula to be valid?

a. The velocity vector v  must be perpendicular to the axis of rotation

b.The velocity vector v  must be perpendicular or parallel to the axis of rotation

c. The moment of inertial must be taken about an axis through the center of mass

Answer:

Option c

Explanation:

K_{total} = K_{rotational}+K_{translational}

The first two conditions are untrue, this is because, you can have rotation in any direction and translation in any direction of any collection of masses. Rotational and translational velocities of masses do not depend on each other

The last statement is true because by definition, the moment of inertia, which is a measure of reluctance, is usually taken about a reference point which is the center of mass

4 0
3 years ago
Read 2 more answers
A dentist’s drill starts from rest. After 3.20 s of constant angu-lar acceleration, it turns at a rate of 2.51 3 104 re v/m i n.
Gekata [30.6K]

Answer:

ΔTita = 4205.6 rad

Explanation:

w_{i} means initial angular velocity, which is 0 rev/min

w_{f} means final angular velocity, which is 2.513*10^{4}rev/min

t means time t= 3.20 s

one revolution is equivalent to 2πrad so the final angular velocity is:

w_{f} = (2π/60) *2.513*10^{4} rad/s

w_{f}= 2628.5 rad/s

so the angular acceleration, α will be:

α = 2628.5 rad/s / 3.20 s

a = 821.40 rad/s^{2}

so the rotational motion about a fixed axis is:

w^{2} _{f} =w^{2} _{i} + 2αΔTita    where ΔTita is the angle in radians

so now find the ΔTita the subject of the formula

ΔTita = \frac{w^{2} _{f}-w^{2} _{i}  }{2a}

ΔTita = ((2628.5)^{2} - (0 rev/min)^{2}) / 2* 821.40

ΔTita = 4205.6 rad

7 0
4 years ago
Read 2 more answers
If the period of a pendulum were 2.4 s on the moon, which has an acceleration due to gravity of 1.67 m/s^2, then how long would
guajiro [1.7K]
<h2>Reffer the attachment </h2>

Mark as brainlist ❤️❤️

5 0
3 years ago
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