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2 years ago

The value of the normal force exerted on a 2,000 gram block on a table

Physics

1 answer:

kobusy [5.1K]2 years ago

4 0

Answer:

N = 19.6 N

Explanation:

Given that,

Mass of a block, m = 2000 g

1 kg = 1000 g

It means, 2000 g = 2kg

We need to find the value of normal force on the block on a table. Normal force is balanced by the weight of the block as follows :

N = mg, g is acceleration due to gravity

N = 2 kg × 9.8 m/s²

N = 19.6 N

So, the normal force acting on the block is 19.6 N.

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what is required to cause a body of mass 500g to accelerate uniformly from rest across a smooth horizontal surface so that it wi

Y_Kistochka [10]

A constant force of 1.25N

Explanation:

This is a kinematics problem.

First find acceleration,

then use F = ma to find force.

Given:

mass = .5kg

delta x = 20m

t = 4s

a = ?

Friction = 0

From the kinematics equations:

delta x = Vi + (1/2)at^2

Plug in terms that are given:

20m = 0 + (1/2)a(4^2)

(2*20m)/(16s^2) = a

40m/(16s^2)= 2.5m/s^2

Now use F = ma to find force exerted on object.

F = (0.5kg)*(2.5m/s^2)

F = 1.25N

8 0

2 years ago

The term 'balanced diet' best means a person is getting which of the following? O A fewer nutrients and vitamins than it needs.

Sidana [21]

Answer:

A Balanced diet is the receiving of the right amount of nutrients and vitamins that the body needs for proper functioning

3 0

2 years ago

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In the United States, household electric power is provided at a frequency of 60 HzHz, so electromagnetic radiation at that frequ

grigory [225]

Answer:

the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

Explanation:

Given the data in the question;

To determine the maximum intensity of an electromagnetic wave, we use the formula;

$I$ = $\frac{1}{2}$ ε₀cE $_{max$ ²

where ε₀ is permittivity of free space ( 8.85 × 10⁻¹² C²/N.m² )

c is the speed of light ( 3 × 10⁸ m/s )

E $_{max$ is the maximum magnitude of the electric field

first we calculate the maximum magnitude of the electric field ( E $_{max$ )

E $_{max$ = 350/f kV/m

given that frequency of 60 Hz, we substitute

E $_{max$ = 350/60 kV/m

E $_{max$ = 5.83333 kV/m

E $_{max$ = 5.83333 kV/m × ( $\frac{1000 V/m}{1 kV/m}$ )

E $_{max$ = 5833.33 N/C

so we substitute all our values into the formula for intensity of an electromagnetic wave;

$I$ = $\frac{1}{2}$ ε₀cE $_{max$ ²

$I$ = $\frac{1}{2}$ × ( 8.85 × 10⁻¹² C²/N.m² ) × ( 3 × 10⁸ m/s ) × ( 5833.33 N/C )²

$I$ = 45 × 10³ W/m²

$I$ = 45 × 10³ W/m² × ( $\frac{1 kW/m^2}{10^3W/m^2}$ )

$I$ = 45 kW/m²

Therefore, the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

7 0

2 years ago

Two identical cylindrical vessels with their bases at the same level each contain a liquid of density 1.23 g/cm3. The area of ea

motikmotik

Explanation:

Work done by gravity is given by the formula,

W = $\rho A (h_{1} - h)g (h - h_{2})$ ......... (1)

It is known that when levels are same then height of the liquid is as follows.

h = $\frac{h_{1} + h_{2}}{2}$ ......... (2)

Putting value of equation (2) in equation (1) the overall formula will be as follows.

W = $\frac{1}{4} \rho gA(h_{1} - h_{2})^{2})$

= $\frac{1}{4} \times 1.23 g/cm^{3} \times 9.80 m/s^{2} \times 3.89 \times 10^{-4} m^{2}(1.76 m - 0.993 m)^{2})$

= 0.689 J

Thus, we can conclude that the work done by the gravitational force in equalizing the levels when the two vessels are connected is 0.689 J.

3 0

3 years ago

Which piece of equipment would give the MOST accurate measurement of 45 mL of a liquid? A) A 100 mL graduated cylinder B) A 50 m

Ludmilka [50]

b is the answer i think

3 0

3 years ago

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