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Harrizon [31]
2 years ago
13

the dimensions of a cuboidal block are 2.5 m * 2m * 1.2m its weight 900 n what is the minimum presseure exerted by the block on

the surface of its support.?​
Physics
2 answers:
aleksandr82 [10.1K]2 years ago
6 0

Answer:

your answer is here 180

Explanation:

900/2.5 * 2 = 180

daser333 [38]2 years ago
6 0

Answer:

The answer is 180 Pa.

Explanation:

<h3><u>Given</u>;</h3>

Here, weight = thrust = 900 N,

dimensions are 2.5 m × 2 m × 1.2 m

<h3><u>To </u><u>Find</u>;</h3>
  • Pressure exerted by the block on the surface of its support.
<h3><u>Formula</u>;</h3>
  • Pressure = Thrust ÷ Area

We know that, Pressure is inversely proportional to area. For the minimum pressure the area must be more.

So, if the block is resting with 2.5 m × 2 m surface on the support, the pressure will be minimum.

Pressure = 900 N ÷ 2.5 × 2.0 m² = 180 Pa

Thus, The pressure exerted by the block on the support is 180 Pa.

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A disk of mass M and radius R rotates at angular velocity ω0. Another disk of mass M and radius r is dropped on top of the rotat
AleksandrR [38]

Answer:

\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}

Explanation:

As we know that there is no external torque on the system of two disc

then the angular momentum of the system will remains conserved

So we will have

L_i = L_f

now we have

L_i = (\frac{1}{2}MR^2)\omega_o

also we have

L_f = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega

now from above equation we have

(\frac{1}{2}MR^2)\omega_o  = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega

now we have

\omega = \frac{MR^2\omega_o}{(MR^2 + Mr^2)}

\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}

6 0
3 years ago
Quasars are thought to be the nuclei of active galaxies in the early stages of their formation. A typical quasar radiates energy
uranmaximum [27]

Answer:

1.7531 smu/yn

Explanation:

check the attached file below for answer and explanation.

6 0
3 years ago
TRUE OR FLASE
BaLLatris [955]

Answer:

true for first and false for second

Explanation:

5 0
2 years ago
A block of aluminium of density 7900 kg/m3 has a volume of 250 cm. Find the<br>mass of the block.​
pashok25 [27]

Answer:

mass of block=2.7 gm

Explanation:

concept: Density=mass/volume

given:ρ=2700 kg/m^3 and v=250 cm^3 (in cm^3 not in m^3)

=> v= convert cm^3 to m^3

there fore= 1 cm^3=1 cm*1 cm*1 cm

i.e 1 cm^3=1/100*1/00*1/100 m^3 => 1 cm^3=1/1000000 m^3

ρ=m/v

=>mass=ρ*volume

=>mass=2700*1/10^-6

=>mass=2.7*10^-3 kg =>2.7 gm

8 0
2 years ago
A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The
lina2011 [118]

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s  

Explanation:

1) We can find the speed of the object by conservation of energy:

E_{i} = E_{f}

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2}                              

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}    

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2}      

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s      

 

3) When the object is at the equilibrium position, the speed of the object is:

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}    

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2}      

v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s

I hope it helps you!                                                                                        

8 0
3 years ago
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