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maks197457 [2]
3 years ago
5

A 50 g mass is freely hanging from a horizontal meter stick at a distance of 99 cm from the pivot. Calculate the weight force W

that the mass exerts on the meter stick? ·Calculate the torque that the weight force exerts about the pivot? 4. What does the pivot exert on the meter stick?
Physics
1 answer:
Neko [114]3 years ago
6 0

Answer:

W = 0.49 N

τ = 0.4851 Nm

Force

Explanation:

The weight force can be found as:

W = mg

W = (0.05 kg)(9.8 m/s²)

<u>W = 0.49 N</u>

The torque about the pivot can be found as:

τ = W*d

where,

τ = torque

d = distance between weight and pivot = 99 cm = 0.99 m

Therefore,

τ = (0.49 N)(0.99 m)

<u>τ = 0.4851 Nm</u>

The pivot exerts a  <u>FORCE </u>on the meter stick because the pivot applies force normally over the stick and has a zero distance from stick.

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A baseball pitcher throws a ball across home plate. The ball travels 13.40m in 0.357s Determine the
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I hope this helps! :)

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2 years ago
Which process is responsible for changing the composition of rock?
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3 years ago
Read 2 more answers
You illuminate the grating in a spectrometer at normal incidence θi=0° with a beam of light that has a wavelength of 6562.8 Å. T
monitta

Answer:

a) θ₁ = 23.14 ° , b) θ₂ = 51.81 °

Explanation:

An address network is described by the expression

     d sin θ = m λ

Where is the distance between lines, λ is the wavelength and m is the order of the spectrum

The distance between one lines, we can find used a rule of proportions

     d = 1/600

     d = 1.67 10⁻³ mm

    d = 1-67 10⁻³ m

Let's calculate the angle

    sin θ = m λ / d

    θ  = sin⁻¹ (m λ / d)

First order

    θ₁ = sin⁻¹ (1 6.5628 10⁻⁷ / 1.67 10⁻⁶)

    θ₁ = sin⁻¹ (3.93 10⁻¹)

    θ₁ = 23.14 °

Second order

     θ₂ = sin⁻¹ (2 6.5628 10⁻⁷ / 1.67 10⁻⁶)

     θ₂ = sin⁻¹ (0.786)

     θ₂ = 51.81 °

3 0
3 years ago
A body is thrown up with a velocity of 78.4 m per second.How high will it rise and how much time it will take to return to its p
a_sh-v [17]

Answer:

The maximum height reached by the body is 313.6 m

The time to return to its point of projection is 8 s.

Explanation:

Given;

initial velocity of the body, u = 78.4 m/s

at maximum height (h) the final velocity of the body (v) = 0

The following equation is applied to determine the maximum height reached by the body;

v² = u² - 2gh

0 = u² - 2gh

2gh = u²

h = u²/2g

h = (78.4²) / (2 x 9.8)

h = 313.6 m

The time to return to its point of projection is calculated as follows;

at maximum height, the final velocity becomes the initial velocity = 0

h = v + ¹/₂gt²

h = 0 + ¹/₂gt²

h =  ¹/₂gt²

2h = gt²

t² = 2h/g

t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 313.6}{9.8} }\\\\t = 8 \ s

4 0
3 years ago
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