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nirvana33 [79]
3 years ago
15

A person wants to use a lever to lift a dumpster weighing 4200 N.4200 N. The lever is arranged so that it is approximately horiz

ontal and makes contact below the dumpster's center of mass at a distance of 0.30 m0.30 m from the lever's pivot. About how much torque will the dumpster exert on the lever?
Physics
1 answer:
natta225 [31]3 years ago
4 0

Answer:

1260 Nm

Explanation:

As one end of the lever makes contact at the point right below the dumpster center of mass and has a distance of 0.3 m to the lever pivot point, plus the fact that the level is also approximately horizontal, we can safely assume that the dumpster gravity is directly perpendicular to the level.

This would make the torque caused by the dumpster gravity on the level is

T = LF = 4200*0.3 = 1260 Nm

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Andrei [34K]

Answer:

Pressure = ρgh

pressure (p) is measured in pascals (Pa)

density (ρ) is measured in kilograms per metre cubed (kg/m3)

The fore of gravitational field strength (g) is measured in N/kg or m/s 2

height of column (h) is measured in metres (m)

Answer = 235,200 Pa

Explanation:

Pressure = ρgh

Pressure = 1,000 x 9.8 x 24

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When a +0.00235 C charge
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3 years ago
A car traveling at 14 m/s accelerates at 3.5 m/s² for 5 seconds. How much distance does it travel during that time?​
Rainbow [258]

Answer: 113.75

Explanation:

You know

acceleration = a = 3.5 m/s²

time = t = 5 seconds

initial velocity = u = 14 m/s

Unknown is distance = s = ?

Use equation: s = ut + \frac{1}{2} at²

Substitute all the known values inside the equation:

s = (14*5) + 0.5 * 3.5 * 5²

s = 70 + 43.75 = 113.75 m

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3 0
3 years ago
A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w
trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

KE = PE

\frac{1}{2} mv ^2 = \frac{1}{2} k A^2

Here,

m = Mass

v = Velocity

k = Spring constant

A = Amplitude

Rearranging to find the Amplitude we have,

A = \sqrt{\frac{mv^2}{k}}

Replacing,

A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}

A = 0.0744m

(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

T = 2\pi \sqrt{\frac{m}{k}}

Replacing,

T = 2\pi \sqrt{\frac{0.750}{13}}

T= 1.509s

Now the velocity is described as,

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v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}

We have all the values, then replacing,

v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}

v = 0.2049m/s

7 0
3 years ago
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