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3 years ago

Lab report 1.04 hypothesis

1 answer:

5 0

Answer:a supposition or proposed explanation made on the basis of limited evidence as a starting point for further investigation.

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Which statement about covalent bonds is true

g100num [7]

In covalent bonds the atoms share electrons.

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A 15 kg box sits still on a rough inclined plane (no math required here).

nexus9112 [7]

Answer:

Explanation:

The force of friction equals the sine component of the force due to gravity

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3 years ago

Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30∘ above the

Pavlova-9 [17]

Answer:1.71 m/s

Explanation:

Given

mass of Susan $m=12 kg$

Inclination $\theta =30^{\circ}$

Tension $T=29 N$

coefficient of Friction $\mu =0.18$

Resolving Forces Along x axis

$F_x=T\cos \theta -f_r$

where $f_r=friction\ Force$

$F_y=mg-N-T\sin \theta$

since there is no movement in Y direction therefore

$N=mg-T\sin \theta$

and $f_r=\mu N$

Thus $F_x=T\cos \theta -\mu N$

$F_x=29\cos (30)-\0.18\times (12\times 9.8-29\sin (30))$

$F_x=25.114-18.558$

$F_x=6.556 N$

Work done by applied Force is equal to change to kinetic Energy

$F_x\cdot x=\frac{1}{2}\cdot mv_f^2-\frac{1}{2}\cdot mv_i^2$

$6.556\times 2.7=\frac{1}{2}\cdot 12\times v_f^2$

$v_f^2=\frac{6.556\times 2.7\times 2}{12}$

$v_f^2=2.95$

$v_f=1.717 m/s$

8 0

3 years ago

what is the kinetic energy of an object that has mass of 30 kilograms and move with a velocity of 20 m/s

kotegsom [21]

Kinetic Energy = 1/2 * m * v² 1/2 * 30 * 20² 1/2 * 30 * 400 12000/2 6000 J.

7 0

3 years ago

What is the magnitude of the total acceleration of point A after 2 seconds? The bar starts from rest and has a constant angular

monitta

Answer:

a_total = 2 √ (α² + w⁴) , a_total = 2,236 m

Explanation:

The total acceleration of a body, if we use the Pythagorean theorem is

a_total² = a_T²2 + $a_{c}$ ²

where

the centripetal acceleration is

a_{c} = v² / r = w r²

tangential acceleration

a_T = dv / dt

angular and linear acceleration are related

a_T = α r

we substitute in the first equation

a_total = √ [(α r)² + (w r² )²]

a_total = 2 √ (α² + w⁴)

Let's find the angular velocity for t = 2 s if we start from rest wo = 0

w = w₀ + α t

w = 0 + 1.0 2

w = 2.0rad / s

we substitute

a_total = r √(1² + 2²) = r √5

a_total = r 2,236

In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m

a_total = 2,236 m

7 0

3 years ago