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3 years ago

F a forklift raises a 76-kg load a distance of 2.5 m, about how much work has it done?

1 answer:

6 0

W= FxD. Since the weight is not put into force you’re going to convert it by multiplying it by 9.8 (gravity). 76kg x 9.8 = 7448. Then multiply that by the distance (2.5). Your answer is 1862.

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What is threshold of hearing?

gavmur [86]

Explanation:

<em>The absolute threshold of hearing is the minimum sound level of a pure tone that an average human ear with normal hearing can hear with no other sound present. The absolute threshold relates to the sound that can just be heard by the organism.</em>

5 0

3 years ago

A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,

n200080 [17]

Answer:

a. $t = \frac{v_{0} +/- \sqrt{v_{0} ^{2} - gD} }{g}$ b. D = v₀²/2g

Explanation:

Here is the complete question

A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance

Find the time at which the two balls collide.

Express your answer in terms of the variables D ,v₀ , and appropriate constants..

t = ?!

Part B

Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Express your answer in terms of the variables v₀ and g .

D =?!

Solution

The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.

The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.

The two balls collide when their vertical distances are equal. That is H₁ = H₂

So, D - v₀t + gt²/2 = v₀t - gt²/2

Collecting like terms

D + gt²/2 + gt²/2 = v₀t + v₀t

D +gt² = 2v₀t

gt² - 2v₀t + D = 0.

Using the quadratic formula,

$t = \frac{-(-2v_{0} ) +/- \sqrt{(-2v_{0} )^{2} - 4 X g XD} }{2g} = \frac{2v_{0} +/- \sqrt{4v_{0} ^{2} - 4gD} }{2g} = \frac{v_{0} +/- \sqrt{v_{0} ^{2} - gD} }{g}$

B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.

0 = v₀² - 2gs

s = v₀²/2g.

Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,

0 = v₀ - gt₀

t₀ = v₀/g

Also H = s₁ + s where s₁ = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.

So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g

6 0

3 years ago

Which change will always result in an increase in the gravitational force between two objects?

Anika [276]

The change that will always result in an increase in the gravitational force between two objects is increasing the masses of the objects and decreasing the distance between the objects.

8 0

3 years ago

Read 2 more answers

3. why is the sum of the maximum voltages across each element in a series r l c circuit usually greater than the maximum applied

Llana [10]

The sum of the maximum voltages across each element in a series RLC circuit is usually greater than the maximum applied voltage because voltages are added by vector addition.

<h3>What is the Kichoff's loop rule?</h3>

Kirchhoff's loop rule states that the algebraic sum of potential differences, as well as the voltage supplied by the voltage sources and resistances, in any loop must be equal to zero.

In a series RLCcircuit, the voltages are not added by scalar addition but by vector addition.

Kirchhoff's loop rule is not violated since the voltages across different elements in the circuit are not at their maximum values.

Therefore, the sum of the maximum voltages across each element in a series RLC circuit is usually greater than the maximum applied voltage because voltages are added by vector addition.

Learn more about Kichoff's loop rule at: https://brainly.in/question/35360816

#SPJ1

8 0

2 years ago

1. A kangaroo hops 84 m to the east in 7 seconds.

DENIUS [597]

Explanation:

Given parameters:

Distance hopped = 84m

Displacement = 84m due east

Time = 7s

Unknown:

Speed of kangaroo = ?

Velocity of kangaroo = ?

Solution:

To solve this problem,

Speed = $\frac{distance}{time }$ = $\frac{84}{7}$ = 12m/s

Velocity = $\frac{displacement}{time}$ = $\frac{84}{7}$ = 12m/s due east

3 0

3 years ago