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3 years ago

9

A block weighing 10 newtons is resting on a plane inclined 30° to the horizontal. What is the magnitude of the normal force acti

ng on the block? Round your answer to the nearest whole number.

1 answer:

aleksley [76]3 years ago

7 0

Magnitude of normal force acting on the block is 7 N

Explanation:

10N = 1.02kg

Mass of the block = m = 1.02 kg

Angle of incline Θ = 30°

Normal force acting on the block = N

From the free body diagram,

N = mgCos Θ

N = (1.02)(9.81)Cos(30)

N = 8.66 N

Rounding off to nearest whole number,

N = 7 N

Magnitude of normal force acting on the block = 7 N

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In a nuclear fusion reaction two 2H atoms are combined to produce one 4He.

Mrrafil [7]

Answer:

a)= 0.025602u

b) = 23.848MeV

c) N = 1.546 × 10¹³

Explanation:

The reaction is

²₁H + ²₁H ⇄ ⁴₂H + Q

a) The mass difference is

Δm = 2m(²₁H) - m (⁴₂H)

= 2(2.014102u) - 4.002602u

= 0.025602u

b) Use the Einstein mass energy relation ship

The enegy release is the mass difference times 931.5MeV/U

E = (0.025602) (931.5)

= 23.848MeV

c)

the number of reaction need per seconds is

N = Q/E

= 59W/ 23.848MeV

$= \frac{59}{(23.848 \times 10^6 )(1.6 \times 10^1^9) } \\\\= 1.546 \times 10^1^3$

N = 1.546 × 10¹³

5 0

3 years ago

When a 5.0-kilogram cart moving with a speed of 2.8 meters per second on a horizontal surface collides with a 2.0 kilogram cart

mixer [17]

Here in all such collision type question we can use momentum conservation as we can see that there is no external force on this system

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

as we know that

$m_1 = 5 kg$

$m_2 = 2 kg$

$v_{1i} = 2.8 m/s$

$v_{2i} = 0 m/s$

now from above equation we have

$5(2.8) + 2(0) = m_1v+ m_2v$

$14 = (5+ 2) v$

$v = 2 m/s$

so the speed of combined system is 2 m/s

8 0

3 years ago

Plz help I will give brainly to the correct answer

Tresset [83]

Answer:

D. x-rays

Explanation:

Lower frequency: Radio waves, microwaves and infrared have lower frequency than visible light. Shorter wavelength: Ultraviolet, x-rays and gamma rays have a shorter wavelength than visible light.

4 0

3 years ago

The vacuum pressure of a condenser is given to be 80 kpa. if the atmospheric pressure is 98 kpa, what is the gage pressure and a

dimaraw [331]

The absolute pressure is given by the equation,

$P_{abs}=P_{atm}-P_{vac}$

Here, $P_{abs}$ is absolute pressure, $P_{atm}$ is atmospheric pressure and $P_{vac}$ is vacuum pressure.

Therefore,

$P_{abs}=98 kPa-80 kPa=18kPa$

The gage pressure is given by the equation,

$P_{gage}=P_{abs}-P_{atm}$ .

Thus,

$P_{gage}=18kPa-98 kPa=-80 kPa$ .

In kn/m^2,

The absolute pressure,

$P_{abs}=18kPa(\frac{1kN/m^2}{kPa}) =18\ kN/m^2$

The gage pressure,

$P_{gage}=-80kPa(\frac{1kN/m^2}{kPa}) =-80\ kN/m^2$ .

In lbf/in2

The absolute pressure,

$P_{abs}=18\ kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=2.6\ lbf/in^2$

The gage pressure,

$P_{gage}=-80kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=-11.6\ lbf/in^2$

In psi,

The absolute pressure,

$P_{abs}=18\ kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa})=2.610\ psi$ .

The gage pressure,

$P_{gage}=-80kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa} )=-11.6030\ psi$

In mm Hg

The absolute pressure,

$P_{abs}=18kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})= 135\ mm\ of\ Hg$

The gage pressure,

$P_{gage}=-80kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})=-600\ mm\ of\ Hg$

3 0

3 years ago

From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum up

NemiM [27]

Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

K.E = (0.5)(mv₀²)

where,

K.E = initial kinetic energy of projectile = 1430 J

m = mass of projectile = 0.41 kg

v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ = (83.5 m/s) Cos(40.5°)

<u>v₀ₓ = 63.5 m/s</u>

y- component = v₀y = v₀Sinθ = (83.5 m/s) Sin(40.5°)

<u>v₀y = 54.2 m/s</u>

7 0

2 years ago