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3 years ago

A 7800 W lift took 10 seconds to move a 5000 N weight. How far did the lift move this weight?

Physics

1 answer:

Eduardwww [97]3 years ago

7 0

Answer:

Both are moving at 30 km/h, so their speed is the same. ... enough fuel for the trip/how long it will take. 4 Weight is a force, and so is a vector. ... c At 10 seconds David's displacement is.

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An object starts at 11 m/s with an acceleration of 12 m/s? What is its velocity after 4.5 seconds?

marshall27 [118]

Answer:

65 m/s

Explanation:

v=v0+at <=> v = 11 + 12 t ∧ t = 4.5 s <=> v = 11 + 12×4.5 <=> v = 65 m/s

5 0

3 years ago

Children are sled riding on a hill One little girl pulls her sled back up the hill and does 379.5 joules of work while pulling i

Serggg [28]

Answer:

2.2N

Explanation:

Given parameters:

Work done = 379.5J

Height = 173m

Unknown:

Amount of force exerted on the sled = ?

Solution:

The amount of force she exerted on the sled is the same as her weight.

Work done is the force applied to move a body through a distance.

Work done = mgh

m is the mass

g is the acceleration due to gravity

h is the height

mg = weight;

Work done = weight x h

379.5 = weight x 173

weight = $\frac{379.5}{173}$ = 2.2N

4 0

3 years ago

What concerns people in regards to<br> the federal government?

inysia [295]

Answer:

Mulitple Changes.

Explanation:

Most people want something to be done, and never change again. But some others do, since they have there own beliefs. Alot of people are concerned about tax rates, taxes, nationial polices, own beliefs. People have there own beliefs and when someone is elected into the federal Government, they have there own Beliefs (Mostly Republican Or Democrate). People go against on what they dont think and go with what they think. At the end whoever sways the people on who to believe and vote for, Is elected.

3 0

3 years ago

A rock is being twirled in a circle on the end of a string. The string provides the centripetal force needed to keep the ball mo

KonstantinChe [14]

Answer:

Explanation:

The force of tension exerted by the string on the rock acts as centripetal force, so its direction is always towards the centre of the circle.

However, the direction of motion of the rock is always tangential to the circle: this means that the force is always perpendicular to the direction of motion of the rock.

As we know, the work done by a force on an object is

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the force and the displacement

In this situation, F and d are perpendicular, so $\theta=90^{\circ}$ , therefore $cos \theta = 0$ and the work done is zero:

$W=0$

4 0

3 years ago

6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri

mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a = - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0

3 years ago