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NNADVOKAT [17]
3 years ago
5

A piece of wood that floats on water has a mass of 0.0175 kg. A lead weight is tied to the wood, and the apparent mass with the

wood in air and the lead weight submerged in water is 0.0765 kg. The apparent mass with the wood and the lead weight both submerged in water is 0.0452 kg. What is the specific gravity of the wood
Physics
1 answer:
crimeas [40]3 years ago
5 0

Answer:

Specific gravity is 0.56

Explanation:

We know that

mass of water displaced by the wood is, m1( apparent mass when wood in air and lead is submerged in water) - m2(the apparent mass when wood and lead both are submerged in water)

= 0.0765 - 0.0452 = 0.0313 Kg

So the specific gravity of the wood is, = mass of wood / mass of water displaced by the wood

= 0.0175/0.0313

=0.56

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Which event is an example of condensation?
ale4655 [162]

Answer: D

If the fog disappears when the Sun comes out, then this is an example of condensation because:

the Sun actually dries up the fog, and it makes it into higher clouds.

Hope this helps you!

3 0
3 years ago
A pendulum on earth swings with angular frequency ω. On an unknown planet, it swings with angular frequency ω/ 4. The accelerati
Afina-wow [57]

Answer:

g / 16

Explanation:

T = 2π \sqrt{\frac{l}{g} }

angular frequency ω = 2π /T

= \sqrt{\frac{g}{l} }

ω₁ /ω₂ = \sqrt{\frac{g_1}{g_2} }

Putting the values

ω₁ = ω ,     ω₂ = ω / 4

ω₁ /ω₂ = 4

4 =  \sqrt{\frac{g}{g_2} }

g₂ = g / 16

option d is correct.

6 0
3 years ago
An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an
kirill [66]

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁=  m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

v_{f1} = \sqrt{2*a_{1}*d_{1}  }  Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁  :Newton's second law

-Ff₁+Wx₁ = ma₁   , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁      Equation (2)

∑Fy₁=0   : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁  =  m*g*cos30°

We replace   N₁  =  m*g*cos30 and  Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁   :  We divide by m

-μ₁*g*cos30°+g*sin30° = a₁  

g*(-μ₁*cos30°+sin30°) = a₁  

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

v_{f1} = \sqrt{2*3.2*3}  }

v_{f1} =\sqrt{2*3.2*3}

v_{f1} = 4.38 m/s

Rough surface  kinematics

vf₂²=v₀₂²+2*a₂*d₂   v₀₂=vf₁=4.38 m/s

0   =4.38²+2*a₂*d₂  Equation (3)

Rough surface  kinetics (μ= 0.3)

∑Fx₂=ma₂  :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂   Equation (4)

∑Fy₂= 0  :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂   We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0   =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= -  6.79*3 = 20.4 N*m

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

6 0
4 years ago
Which of these is a chemical property of a substance?
My name is Ann [436]
Reactivity is a chemical property of substance
5 0
3 years ago
A child is riding a merry-go-round that has an instantaneous angular speed of 12 rpm. If a constant friction torque of 12.5 Nm i
sammy [17]

Answer:

-0.25 rad/s^2

Explanation:

The equivalent of Newton's second law for rotational motions is:

\tau = I \alpha

where

\tau is the net torque applied to the object

I is the moment of inertia

\alpha is the angular acceleration

In this problem we have:

\tau = -12.5 Nm (net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)

I=50.0 kg m^2 is the moment of inertia

Solving for \alpha, we find the angular acceleration:

\alpha = \frac{\tau}{I}=\frac{-12.5 Nm}{50.0 kg m^2}=-0.25 rad/s^2

3 0
3 years ago
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