Pls help I will give brainlist and don’t give me a link I can’t open them

A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 20.1 N; when it is completely i

Paha777 [63]

Answer:

A) $V = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}$

B) $d = 4181.49 kg/m^{3} = 4.18 g/cm^{3}$

Explanation:

A) Using the Archimedes' force we can find the weight of water displaced:

$W_{d} = W_{a} - W_{w}$

Where:

$W_{a}$ : is the weight of the block in the air = 20.1 N

$W_{w}$ : is the weight of the block in the water = 15.3 N

$W_{d} = W_{a} - W_{w} = 20.1 N - 15.3 N = 4.8 N$

Now, the mass of the water displaced is:

$m = \frac{W_{d}}{g} = \frac{4.8 N}{9.81 m/s^{2}} = 0.49 kg$

The volume of the block can be found using the mass of water displaced and the density of the water:

$V = \frac{m}{d} = \frac{0.49 kg}{997 kg/m^{3}} = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}$

B) The density of the block can be found as follows:

$d = \frac{W_{a}}{g*V} = \frac{20.1 N}{9.81 m/s^{2}*4.92 \cdot 10^{-4} m^{3}} = 4181.49 kg/m^{3} = 4.18 g/cm^{3}$

I hope it helps you!

6 0

3 years ago

A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on

MAXImum [283]

Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

$Ec=\frac{1}{2} *m*v^{2}$

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

$W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}$

$W=\frac{1}{2} *m*(v2^{2}-v1^{2})$

In this case:

W=?
m= 2,145 kg
v2= 12 $\frac{m}{s}$

v1= 25 $\frac{m}{s}$

Replacing:

$W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})$

W= -515,872.5 J

The work required is -515,872.5 J

3 0

3 years ago

The pointer of an analog meter is connected to a

dangina [55]

C. coil suspended by bearings.
but im not 100% sure

8 0

3 years ago

Read 2 more answers

Would a pair of scissors with really long blades and a short handle be practical? Why or why not?

Lana71 [14]

Yes, for balance.hope this helped.

6 0

3 years ago

Please I need help with this :(

Ann [662]

three charged particals are located at the corners of an equil triangle shown in the figure showing let (q 2.20 Uc) and L 0.650

3 0

2 years ago