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Studentka2010 [4]
2 years ago
6

Un trineo de 8.00kg se mueve en línea recta sobre una superficie horizontal sin fricción.En cierto punto su rapidez es de 4.00m/

s 2.50m más adelante es de 6.00m/s
Physics
1 answer:
galben [10]2 years ago
4 0

Answer:

An 8 kg sled mass moves in a straight line on a surface without horizontal friction at a certain point its speed is 4 m / s 2.50 m later its speed is 6 m / s using the work and energy theorem, determine the force acting on the sled assuming it is constant and acts in the direction of the sled movement

The force acting on the 8 kg sled has a value of 32 N.

Explanation:

We have initially look for the work that is applied to the sled, and this will be nothing more than the difference in kinetic energy from the end point to the initial, then:

ΔEc = Ec₂ - Ec₁

ΔEc = 0.5 * m * [V₂² - V₁²]

ΔEc = 0.5* 8 kg * [(6m / s) ² - (4m / s) ²]

ΔEc = 80 J

So, the work is 80 J, now we know that work is a relation distance and force , such that:

W = F * d

F = W / d

F = (80 J) / (2.50 m)

F = 32 N

Therefore, the force acting on the sled is 32 N.

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The flagpole is held vertical by two ropes. The first of these ropes has a tension in it of 100 N and is at an angle of 60° to t
KatRina [158]

Answer:

T₂ = 123.9 N,  θ = 66.2º

Explanation:

To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.

The tension T1 = 100 N, we create a reference frame centered on the pole

X axis

       T₁ₓ - T_{2x} = 0

        T_{2x}= T₁ₓ

Y axis y

      T_{1y} + T_{2y} - 200N = 0

      T_{2y} = 200 -T_{1y}

let's use trigonometry to find the component of the stresses

         sin 60 = T_{1y} / T₁

         cos 60 = t₁ₓ / T₁

         T_{1y} = T₁ sin 60

         T1x = T₁ cos 60

         T_{1y}y = 100 sin 60 = 86.6 N

         T₁ₓ = 100 cos 60 = 50 N

for voltage 2 it is done in the same way

         T_{2y} = T₂ sin θ

         T₂ₓ = T₂ cos θ

we substitute

         

           T₂ sin θ= 200 - 86.6 = 113.4

           T₂ cos θ = 50              (1)

to solve the system we divide the two equations

           tan θ = 113.4 / 50

           θ = tan⁻¹ 2,268

           θ = 66.2º

we caption in equation 1

           T₂ cos 66.2 = 50

           T₂ = 50 / cos 66.2

           T₂ = 123.9 N

8 0
2 years ago
11. A 6.0‐m wire with a mass of 50 g, is under tension. A transverse wave, for which the frequency is 810 Hz, the wavelength is
Irina-Kira [14]

Answer:

Time will be 19 ms so option (a) is correct option

Explanation:

We have given that mass of wire m = 50 gram = 0.5 kg

Frequency f = 810 Hz

Wavelength = 0.4 m

Velocity is given by

v=wavelength\times frequency=810\times 0.4=324m/sec

Amplitude is given as d = 6 m

So time t=\frac{distance}{velocity}=\frac{6}{324}=0.01851=19ms

So option (a) is correct option

4 0
3 years ago
A amusement park moves riders in a circle at a rate of 6.0m/s if the radius is 9.0 meters what is the acceleration of the ride
oksano4ka [1.4K]

Centripetal acceleration is (speed-squared) / (radius)

CA = (6 m/s)² / (9 m)

CA = (36 m²/s²) / (9 m)

CA = (36/9) (m²/m·s²)

<em>Centripetal acceleration = 4 m/s²</em>

4 0
3 years ago
How many total oxygen atoms are in the compound Molybdenum (V) Dichromate?
alekssr [168]

Answer:

1.2646\times10^{25}\ atoms

Explanation:

-The chemical formula for Molybdenum (V) Dichromate is Mo(Cr_2O_7)_3

-There are 21 moles of oxygen per one mole of Molybdenum (V) Dichromate

-We apply Avogadro's constant to find the number of atoms of oxygen:

Avogadro's \ Constant=6.022\times 10^{23} \ mol_1\\\\No\ of \ Atoms=Moles\times Avogadro's \ Constant\\\\=21\times 6.022\times 10^{23} \\\\=1.2646\times10^{25}\ atoms

Hence, there are 1.2646\times10^{25} \ atoms

3 0
3 years ago
The gravitational force between two asteroids is 6. 2 × 108 N. Asteroid Y has three times the mass of asteroid Z. If the distanc
Sliva [168]

Answer:Let m =  mass of asteroid y.Because asteroid y has three times the mass of asteroid z, the mass of asteroid z is m/3.Given:F = 6.2x10⁸ Nd = 2100 km = 2.1x10⁶ mNote thatG = 6.67408x10⁻¹¹ m³/(kg-s²)The gravitational force between the asteroids isF = (G*m*(m/3))/d² = (Gm²)/(3d²)orm² = (3Fd²)/G     = [(3*(6.2x10⁸ N)*(2.1x10⁶ m)²]/(6.67408x10⁻¹¹ m³/(kg-s²))    = 1.229x10³² kg²m = 1.1086x10¹⁶ kg = 1.1x10¹⁶ kg (approx)Answer: 1.1x10¹⁶ kg

Explanation:

7 0
2 years ago
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