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3 years ago

The mass of an object is independent of its location. true or false

Physics

1 answer:

Arturiano [62]3 years ago

8 0

That's true. The mass of your body is like the money
in your pocket. It doesn't change, no matter where you
take it.

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A 100 kg box sits on an incline held at rest by a very thin rope attached to another mass hanging as shown. The force of frictio

MAVERICK [17]

If friction is acting along the plane upwards

then in this case we will have

For equilibrium of 100 kg box on inclined plane we have

$mgsin\theta = F_f + T$

also for other side of hanging mass we have

$T = Mg = 50(9.8) = 490 N$

now we have

$100(9.8)sin\theta = 100 + 490$

$980sin\theta = 590$

$sin\theta = 0.602$

$\theta = 37 degree$

In other case we can assume that friction will act along the plane downwards

so now in that case we will have

$mgsin\theta + F_f = T$

also we have

$T = Mg = 50(9.8) N$

now we have

$100(9.8)sin\theta + 100 = 50(9.8)$

$980sin\theta + 100 = 490$

$980 sin\theta = 490 - 100$

$sin\theta = 0.397$

$\theta = 23.45 degree$

<em>So the range of angle will be 23.45 degree to 37 degree</em>

6 0

3 years ago

Two blocks A and B with mA = 2.6 kg and mB = 0.81 kg are connected by a string of negligible mass. They rest on a frictionless h

vladimir1956 [14]

Answer:

Explanation:

Given

Mass of block A $(m_a)$ =2.6 kg

Mass of block B $(m_b)$ =0.81 kg

Force of 6.5 N is applied on Block A

Force on Block A is F, Tension

thus

$F-T=m_a\times a$ -----1

where a=acceleration of the system

For Block B

$T=m_b\times a$ -----2

$F=m_b\times a+m_a\times a$

$F=\left ( m_a+m_b\right )a$

$F=\left ( 2.6+0.81\right )a$

$a=\frac{6.5}{3.41}$

$a=1.96 m/s^2$

$T=0.81\times 1.96=1.54 N$

5 0

3 years ago

Which examples are a COMPLETE description of motion? Select ALL that apply.

SIZIF [17.4K]

All of them apply because they all are doing some thing to move

8 0

2 years ago

Spiral fracture of bone: Spiral fracture of bone occurs due to twisting of the limb, and is a very common skiing accident. The f

Pavlova-9 [17]

Answer:

principal stresses :б1 = 32.62mPa б2 = 31.38mPa

Max Shear stress : 16.31 mPa

Orientation of max principle plane = 44.43°

Orientation of minimum principal plane = 134.43°

Explanation:

Given data:

Torque = 50 N-m

weight = 80 kgs

half of weight is subjected to each leg

radius of bone = 10 mm = 0.010 m

<u>a) Determine the principal stresses and shear stress</u>

first calculate the max shear stress ( this will occur in the outermost element

= 16T / π*d^3 where : T = 50 , d = 0.020 m

hence max shear stress = 32 mPa

next determine compressive stress

= ( 40*g) / π/4*d^2 . where : d = 0.020 m , g = 9.81

hence compressive stress = 1.24 mPa

draw and calculate the radius of Mohr's circle

radius of Mohr's circle = 32.0060

Hence principal stresses = 32.0060 ± 0.62

б1 = 32.62mPa

б2 = 31.38mPa

attached below is the remaining part of the solution

3 0

3 years ago

2. What is the mass of an object that was accelerated at a rate of 1 m/s with a force of 2 N

e-lub [12.9K]

Answer:

2 kilograms

Explanation:

F = ma

2 = 1m = 2

3 0

2 years ago