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2 years ago

11

PLEASE HELPP!!! I NEED ANSWERS!!!

1 answer:

const2013 [10]2 years ago

7 0

Answer:

0.5kg

Explanation:

Using formula f=ma,

6N=mx12m/s^2

6N/(12m/s^2)=m

0.5kg=m

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What intitial action is the reason behind beginning a scientific investigation

Nataly [62]

To come to a conclusion would be the initial action that would be the reason behind beginning a scientific investigation.

When you want to prove your point and come to a conclusion, you would start an investigation. Your conclusion helps you form a ending statement.

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3 years ago

An oscillating object takes 0.10 s to complete one cycle; that is, its period is 0.10 s. what is its frequency f? express your a

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2 years ago

An alarm clock has a resistance of 14,000 ohms and is plugged into a 120-volt outlet. How much power does the clock use?

yan [13]

Power used by the clock=1.03 W

Explanation:

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8 0

3 years ago

Read 2 more answers

irga5000 [103]

Answer:

a

Explanation:

Because I searched it up

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2 years ago

Read 2 more answers

The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a

grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

$\sin\theta=\dfrac{3}{5}$

The horizontal component is

$T_{x}=T\cos\theta$

The vertical component is

$T_{y}=T\sin\theta$

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

$T\sin\theta\times 4=380+1200$

$T=\dfrac{1580\times5}{3\times 4}$

$T=658.33\ N$

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

$F_{x}=T\cos\theta$

Put the value into the formula

$F_{x}=658.33\times\dfrac{4}{5}$

$F_{x}=526.66\ N$

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0

3 years ago