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Anuta_ua [19.1K]

2 years ago

14

PLEASE HELP 15 POINTS!!!!

1 answer:

goldfiish [28.3K]2 years ago

8 0

Answer:

15 degrees Celsius

Explanation:

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Frequency is measured in units called?

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What would the velocity of an object be if it was on a 1.75m

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Answer:

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2 years ago

You place a 55.0 kg box on a track that makes an angle of 28.0 degrees with the horizontal. The coefficient of static friction b

Radda [10]

Answer:

$\theta=34 \textdegree$

Explanation:

From the question we are told that:

Mass $m=55kg$

Angle $\theta =28.0$

Coefficient of static friction $\alpha =0.680$

Generally, the equation for Newtons second Law is mathematically given by

For

$\sum_y=0$

$N=mgcos \theta$

for

$\sum_x=0$

$F_{s}=mgsin\theta$

Where

$F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta$

$F_{s}=0.68*55*9.8*cos 28$

$F_{s}=323.62N$

Therefore

$\alpha mgcos \theta=mg sin \theta$

$\theta=tan^{-1}(0.68)$

$\theta=34 \textdegree$

6 0

2 years ago

A horse of mass 242 kg pulls a cart of mass 224 kg. The acceleration of gravity is 9.8 m/s 2 . What is the largest acceleration

Rufina [12.5K]

To solve this problem it is necessary to apply the concepts related to Newton's second Law and the force of friction. According to Newton, the Force is defined as

F = ma

Where,

m= Mass

a = Acceleration

At the same time the frictional force can be defined as,

$F_f = \mu N$

Where,

$\mu =$ Frictional coefficient

N = Normal force (mass*gravity)

Our values are given as,

$m_h = 242 kg\\m_c = 224 kg\\\mu = 0.894\\$

By condition of Balance the friction force must be equal to the total net force, that is to say

$F_{net} = F_f$

$m_{total}a = \mu m_hg$

$(m_h+m_c)a = \mu*m_h*g$

Re-arrange to find acceleration,

$a= \frac{\mu*m_h*g}{(m_h+m_c)}$

$a = \frac{0.894*242*9.8}{(242+224)}$

$a = 4.54 m/s^2$

Therefore the acceleration the horse can give is $4.54m/s^2$

3 0

2 years ago