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Svetllana [295]
3 years ago
12

An object is hanging from a rope. When it is held in air the tension in the rope is 8.86 N, and when it is submerged in water th

e tension is 7.84 N. What is the density of the the object
Physics
1 answer:
SSSSS [86.1K]3 years ago
8 0

Answer:

8684.2 kg/m³

Explanation:

Tension in the rope as a result of the weight = 8.86 N

Tension in the rope when submerge in water = 7.84

upthrust = 8.86 - 7.84 =1.02 N = mass of water displaced × acceleration due to gravity

Mass of water displaced = 1.02 / 9.81 = 0.104 kg

density of water = mass of water / volume of water

make volume subject of the formula

volume of water displaced = mass / density ( 1000) = 0.104 / 1000 = 0.000104 m³

volume of the object = volume of water displaced

density of the object = mass of the object / volume of the object = (8.86 / 9.81) / 0.000104 = 0.9032 / 0.000104 = 8684.2 kg/m³

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4 0
4 years ago
The engine of a locomotive exerts a constant force of 6.8 105 N to accelerate a train to 80 km/h. Determine the time (in min) ta
Setler [38]

Answer:

t = 6 minutes

Explanation:

Given that,

Force,F=6.8\times 10^5\ N

Initial speed of the train, u = 0

Final speed of the train, v = 80 km/h = 22.22 m/s

The mass of the train, m=1.1\times 10^7\ kg

We need to find the time taken by the train to come to rest. We know that,

F = ma

F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.1\times10^7\times (22.22-0)}{6.8\times 10^5}\\\\t=359.44\ s

or

t = 6 minutes (approx)

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3 0
3 years ago
A plank 2.00 cm thick and 13.0 cm wide is firmly attached to the railing of a ship by clamps so that the rest of the board exten
charle [14.2K]

Answer:

9.93 MPa

Explanation:

Given:

 - mass of the man = 68.4 kg

 - Deflection dx = 5.2 cm

 - thickness of plank t = 2.0 cm

 - width of plank w = 13.0 cm

 - Length subtended L = 2.0 m

Find:

Shear Modulus of Elasticity S :

                 S = shear stress / shear strain

                        Shear stress = F / A

                        Shear stress = 68.4*9.81 / 0.02*0.13

                        Shear stress = 258078.4615 Pa

                        Shear strain = dx / L

                        Shear Strain = 0.052 / 2

                        Shear Strain = 0.026  

Hence,

                  S = 258078.4615 / 0.026

                  S = 9.93 MPa

5 0
3 years ago
A wheel initially rotating at 12 rad/s decelerates uniformly to rest in 0.4 s. If the wheel has a rotational inertia of 0.5 kg.m
Firdavs [7]

Answer:

(B) 15 N.m

Explanation:

The deceleration of the wheel is first found by using the following formula:

\alpha = \frac{\omega_f - \omega_i}{t}

where,

α = angular acceleration = ?

ωf = final angular speed = 0 rad/s

ωi = initial angular speed = 12 rad/s

t = time = 0.4 s

Therefore,

\alpha = \frac{0\ rad/s - 12\ rad/s}{0.4\ s}\\\\\alpha = -30\ rad/s^2

here, the negative sign shows deceleration.

Now, we find the torque responsible for this deceleration:

\tau = I\alpha

where,

τ = torque = ?

I = rotational inertia = 0.5 kg.m²

Therefore,

\tau = (0.5\ kg.m^2)(30\ rad/s^2)\\

τ = 15 N.m

Therefore, the correct answer is:

<u>(B) 15 N.m</u>

8 0
3 years ago
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