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3 years ago

The net external force on a golf cart is 411 N north of tha cart has a total mass of 281 kg what is tha cart acceleration

Physics

1 answer:

Alisiya [41]3 years ago

8 0

1.46m/s²

Explanation:

Given parameters:

Mass of cart = 281kg

Net external force = 411N

Unknown:

Acceleration of the cart = ?

Solution:

According to newton's second law " the net force on a car is the product of its mass and acceleration";

Force = mass x acceleration

Since acceleration of the cart is unknown;

Acceleration = $\frac{force}{mass}$ = $\frac{411}{281}$

Acceleration = 1.46m/s²

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Answer:

F = G M m / R^2 gravitational force on planet of mass m.

None of these quantities change in the given hypothesis so

there will be no change in the orbit of mass m

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2 years ago

PLEASE PROVIDE AN EXPLANATION.<br><br> THANKS!!!

ziro4ka [17]

Answer:

(a) A = 0.0800 m, λ = 20.9 m, f = 11.9 Hz

(b) 250 m/s

(d) Positive x-direction

(e) 6.00 m/s

(f) 0.0365 m

Explanation:

(a) The standard form of the wave is:

y = A cos ((2πf) t ± (2π/λ) x)

where A is the amplitude, f is the frequency, and λ is the wavelength.

If the x term has a positive coefficient, the wave moves to the left.

If the x term has a negative coefficient, the wave moves to the right.

Therefore:

A = 0.0800 m

2π/λ = 0.300 m⁻¹

λ = 20.9 m

2πf = 75.0 rad/s

f = 11.9 Hz

(b) Velocity is wavelength times frequency.

v = λf

v = (20.9 m) (11.9 Hz)

v = 250 m/s

T = v²ρ

where ρ is the mass per unit length.

T = (250 m/s)² (0.0200 kg/m)

T = 1250 N

(d) The x term has a negative coefficient, so the wave moves to the right (positive x-direction).

(e) The maximum transverse speed is Aω.

(0.0800 m) (75.0 rad/s)

6.00 m/s

(f) Plug in the values and find y.

y = (0.0800 m) cos((75.0 rad/s) (2.00 s) − (0.300 m⁻¹) (1.00 m))

y = 0.0365 m

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2 years ago

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A particle moves along line segments from the origin to the points (1, 0, 0), (1, 5, 1), (0, 5, 1), and back to the origin under

Kaylis [27]

Answer:

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Explanation:

Since work done W = ∫F.dr and F(x, y, z)= z²i + 4xyj + 5y²k and dr = dxi + dyj + dzk

F.dr = (z²i + 4xyj + 5y²k).(dxi + dyj + dzk) = z²dx + 4xydy + 5y²dz

W = ∫F.dr = ∫z²dx + 4xydy + 5y²dz = z²x + 2xy² + 5y²z

We now evaluate the work done for the different regions

W₁ = work done from (0,0,0) to (1,0,0)

W₁ = {z²x + 2xy² + 5y²z}₀₀₀¹⁰⁰ = 0²(1) + 2(1)(0)² + 5(0)²(0) - [(0)²(0) + 2(0)(0)² + 5(0)²(0)] = 0 - 0 = 0 J

W₂ = work done from (1,0,0) to (1,5,1)

W₂ = {z²x + 2xy² + 5y²z}₁₀₀¹⁵¹ = (1)²(1) + 2(1)(5)² + 5(5)²(1) - [0²(1) + 2(1)(0)² + 5(0)²(0)] = 1 + 50 + 125 - 0 = 176 J

W₃ = work done from (1,5,1) to (0,5,1)

W₃ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁵¹ = 1²(0) + 2(0)(5)² + 5(5)²(1) - [(1)²(1) + 2(1)(5)² + 5(5)²(1)] = 125 - (1 + 50 + 125) = 125 - 176 = -51 J

W₄ = work done from (0,5,1) to (0,0,0)

W₄ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁰⁰ = (0)²(0) + 2(0)(0)² + 5(0)²(0) - [1²(0) + 2(0)(5)² + 5(5)²(1)] = 0 - 125 = -125 J

The total work done W is thus

W = W₁ + W₂ + W₃ + W₄

W = 0 J + 176 J - 51 J - 125 J

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3 years ago

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V125BC [204]

Answer:

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Explanation:

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I = m (v_f - v₀)

take the ball's exit direction as positive, whereby the ball velocities

v₀ = -90mph, the final velocity v_f = + 54 m / s

Let's reduce the units to

I = 0.142 [54- (-40.23) ]

the SI system

v₀ = - 90 mph (1609.34 m / 1 mile) (1h / 3600 s = -40.23 m / s

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we calculate

I = 0.142 [54- (-40) ]

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I = ∫ F .dt

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3 years ago