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3 years ago

An object with a initial velocity of 0m/s accelerates at a rate of 4m/s how much time did it take to go 88m?

Physics

1 answer:

Citrus2011 [14]3 years ago

3 0

Answer:

6.633s

Explanation:

For this equation we use the following kinematic equation:

^D = Vi(t)+(1/2)(a)(t)^2

88m-0m = 0m/s(t)+(0.5)(4m/s^2)(t)^2

88m = 2m/s^2(t^2)

t^2 = 44s

t = sqrt(44)

t = 6.633s

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4 years ago

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A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.

babunello [35]

26.10 N is the vertical component of the force.

Rx represents the Horizontal component of force

Ry represents The Vertical component of force

According to the given diagram

Rx - Tcosθ = 0

Rx = Tcosθ

And,

Ry + Tsinθ = mg

Ry = mg - Tsinθ

The horizontal component of force =The Vertical component of force

Rx = Ry

Tcosθ = mg - Tsinθ

T(cosθ + sinθ) = 29 × 9.8 = 284.2 N

T√2 cosθ = 284.2 N

T × √2 ×0.544 = 284.2 N

T × 0.769 = 284.2 N

T = 370 N (app)

So,

Ry = 284.2 - 370 (sin 57°)

= 284.2 - 310.3 = -26.10 N

Hence, 26.10 N is the vertical component of the force exerted.

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7 0

2 years ago

Which set of arrows represents the induced magnetic force? A B C

vazorg [7]

The answer would be b because b has a set of arrows represents the induced magnetic force

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4 years ago

You are pushing an 80.0 N wheelbarrow as shown in (Figure 1). You lift upward on the handle of the wheelbarrow so that the only

AleksandrR [38]

Answer:

Wl = 1740 N

Explanation:

maximum lift weight unaided = force exerted (F) = 650 N

length of the wheelbarrow (L) = 1.4 m

weight of the wheelbarrow (w) = 80 N

distance of center of gravity of the wheel barrow from the wheel = 0.5 m

distance of center of gravity of the load from the wheel = 0.5 m

find the weight of the load (Wl)

from the diagram attached we can see that there is going to be a rotation about the axis A. let clockwise rotation be positive

ΣT = (F x 1.4) - ((Wl x 0.5) + (w x 0.5) = 0

(F x 1.4) = ((Wl x 0.5) + (w x 0.5)

Wl =

Wl = 1740 N

6 0

2 years ago

An object of mass 300 g, moving with an initial velocity of 5.00i-3.20j m/s, collides with an sticks to an object of mass 400 g,

Alexus [3.1K]

Answer:

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

Explanation:

Mass of object 1 , m₁ = 300 g = 0.3 kg

Mass of object 2 , m₂ = 400 g = 0.4 kg

Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s

Initial velocity of object 2 , v₂ = 3.00j m/s

Mass of composite = 0.7 kg

We need to find final velocity of composite.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s

Final momentum = 0.7 x v = 0.7v kgm/s

Comparing

1.5 i + 0.24 j = 0.7v

v = 2.14 i + 0.34 j

Magnitude of velocity

$v=\sqrt{2.14^2+0.34^2}=2.17m/s$

Direction,

$\theta =tan^{-1}\left ( \frac{0.34}{2.14}\right )=9.03^0$

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

7 0

3 years ago