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3 years ago

A 50kg block is at rest on the ground.

Physics

1 answer:

yKpoI14uk [10]3 years ago

7 0

Answer:

(b) 490 N downwards

Explanation:

(b) Formula : w= mg

m is the mass of the object

g is a gravitational constant (9.8)

w= mg

= 50×9.8

= 490 N downwards

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The atmospheric pressure on the top of the Engineering Sciences Building (ESB) is 97.6 kPa, while that in Room G39-ESB (ground f

Stells [14]

Answer:

Δ h = 52.78 m

Explanation:

given,

Atmospheric pressure at the top of building = 97.6 kPa

Atmospheric pressure at the bottom of building = 98.2 kPa

Density of air = 1.16 kg/m³

acceleration due to gravity, g = 9.8 m/s²

height of the building = ?

We know,

Δ P = ρ g Δ h

(98.2-97.6) x 10³ = 1.16 x 9.8 x Δ h

11.368 Δ h = 600

Δ h = 52.78 m

Hence, the height of the building is equal to 52.78 m.

6 0

3 years ago

What does basketballs have inside of it

FrozenT [24]

Hi there!

Great question!

Basketballs have air inside them. A special pump is used to insert the air. That's why you can lift the basketballs off the ground easily. If it was a solid, though, you'd hardly be able to lift the ball up! Basketballs can float, too, because anything with air inside can float. If it were solid, it would sink in the water easily.

Hope this helps! :D

4 0

3 years ago

The total energy of a block—spring system is 0.18 J. The amplitude is 14.0 cm and the maximum speed is 1.25 m/s. Find: (a) the m

algol13

a) The mass is 0.23 kg

b) The spring constant is 1.25 N/m

c) The frequency is 1.42 Hz

d) The speed of the block is 1.08 m/s

Explanation:

We can find the mass of the block by applying the law of conservation of energy: in fact, the total mechanical energy of the system (which is sum of elastic potential energy, PE, and kinetic energy, KE) is constant:

$E=PE+KE=const.$

The potential energy is given by

$PE=\frac{1}{2}kx^2$

where k is the spring constant and x is the displacement. When the block is crossing the position of equilibrium, x = 0, so all the energy is kinetic energy:

$E=KE_{max}=\frac{1}{2}mv_{max}^2$ (1)

where

m is the mass of the block

$v_{max}=1.25 m/s$ is the maximum speed

We also know that the total energy is

$E=0.18 J$

Re-arranging eq.(1), we can find the mass:

$m=\frac{2E}{v_{max}^2}=\frac{2(0.18)}{(1.25)^2}=0.23 kg$

The maximum speed in a spring-mass system is also given by

$v_{max} =\sqrt{\frac{k}{m}} A$

where

k is the spring constant

m is the mass

A is the amplitude

Here we have:

$v_{max}=1.25 m/s$ is the maximum speed

m = 0.23 kg is the mass

A = 14.0 cm = 0.14 m is the amplitude

Solving for k, we find the spring constant

$k=\frac{v_{max}^2}{A^2}m=\frac{1.25^2}{0.14^2}(0.23)=18.3 N/m$

The frequency in a spring-mass system is given by

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k is the spring constant

m is the mass

In this problem, we have:

k = 18.3 N/m is the spring constant (found in part b)

m = 0.23 kg is the mass (found in part a)

Substituting and solving for f, we find the frequency of the system:

$f=\frac{1}{2\pi}\sqrt{\frac{18.3}{0.23}}=1.42 Hz$

We can solve this part by using the law of conservation of energy; in fact, we have

$E=PE+KE=\frac{1}{2}kx^2 + \frac{1}{2}mv^2$

Where v is the speed of the system when the displacement is equal to x.

We know that the total energy of the system is

E = 0.18 J

Also we know that

k = 18.3 N/m is the spring constant

m = 0.23 kg is the mass

Substituting

x = 7.00 cm = 0.07 m

We can solve the equation to find the corresponding speed v:

$v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.18)-(18.3)(0.07)^2}{0.23}}=1.08 m/s$

#LearnwithBrainly

3 0

3 years ago

An inductor is connected to the terminals of a battery that has an emf of 16.0 V and negligible internal resistance. The current

balu736 [363]

Answer:

(a) The resistance R of the inductor is 2480.62 Ω

(b) The inductance L of the inductor is 1.67 H

Explanation:

Given;

emf of the battery, V = 16.0 V

current at 0.940 ms = 4.86 mA

after a long time, the current becomes 6.45 mA = maximum current

Part (a) The resistance R of the inductor

$R = \frac{V}{I_{max}} = \frac{16}{6.45*10^{-3}} = 2480.62 \ ohms$

Part (b) the inductance L of the inductor

$\frac{Rt}{L} = -ln(1-\frac{I}I_{max}})\\\\L = \frac{Rt}{-ln(1-\frac{I}I_{max})}}$

where;

L is the inductance

R is the resistance of the inductor

t is time

$L = \frac{Rt}{-ln(1-\frac{I}I_{max})}} = \frac{2480.62*0.94*10^{-3}}{-ln(1-\frac{4.86}{6.45})} \\\\L =\frac{2.3318}{1.4004} = 1.67 \ H$

Therefore, the inductance is 1.67 H

5 0

3 years ago

a lost puppy of 5 kg is 25 m away from his owner of 90 kg. What is the gravitional attraction between the two

Fudgin [204]

The gravitational attraction 50 m

5 0

3 years ago

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