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3 years ago

14

a force is applied to a block causing it to accelerate along a horizontal, frictionless surfaces. the energy gained by the block

is equal to the

1 answer:

GaryK [48]3 years ago

3 0

Energy given to the block

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Numerical Problems:

dangina [55]

Displacement=1200m
Time=4min=4(60)=240s

$\boxed{\sf Velocity=\dfrac{Displacement}{Time}}$

$\\ \sf\longmapsto Velocity=\dfrac{1200}{240}$

$\\ \sf\longmapsto Velocity=5m/s$

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3 years ago

At what water temperature will additional heat energy need to be added before the temperature will change again?

Katyanochek1 [597]

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Hope this helps! (If correct, please rank as brainliest answer) :)

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3 years ago

According to Kepler's harmonic law, Neptune would have a _________ orbit around the Sun than Mercury.

chubhunter [2.5K]

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3 years ago

Read 2 more answers

To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.7 kg and length 3.0 m

Zanzabum

Answer:

$T_{1}$ = 14.88 N

Explanation:

Let's begin by listing out the given variables:

M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,

g = 9.8 m/s²

At equilibrium, the sum of all external torque acting on an object equals zero

τ(net) = 0

Taking moment about $T_{1}$ we have:

(M + m) g * 0.5L - $T_{2}$ (L - d) = 0

⇒ $T_{2}$ = [(M + m) g * 0.5L] ÷ (L - d)

$T_{2}$ = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)

$T_{2}$ = 59.535 ÷ 2.4

$T_{2}$ = 24.80625 N ≈ 24.81 N

Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N

Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N

Using sum of equilibrium in the vertical direction, we have:

$T_{1}$ + $T_{2}$ = W + w ------- Eqn 1

Substituting T2, W & w into the Eqn 1

$T_{1}$ + 24.81 = 26.46 + 13.23

$T_{1}$ = <u>14.88</u> N

8 0

3 years ago

Calculate the mass of an ice block of volume 12m³. the density of the ice is 920kg/m³.

beks73 [17]

Mass = density • volume

so (12)(920)(kg/m^3•m^3)

11040 kg

4 0

3 years ago