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3 years ago

Billy pushes 15.0 kg block at a rate of 2.5 m/s/s eastward. What is the net force Billy applied to the block?

Physics

2 answers:

vekshin13 years ago

8 0

Answer:

,mjmhgjk,mnhbfgtyhyiuhyglufkcvlbvkch

Explanation:

sladkih [1.3K]3 years ago

5 0

I believe the answer would be 37.5

Since F=ma plug the numbers

F=15*2.5

15*2.5=37.5

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Perform the following mathematical operation and report answer to the correct number of significant figures 143.6 divided by 21.

Svetllana [295]

The answer to the correct number of significant figures is 6.774.

<h3>What is quotient?</h3>

When a number(big) divided smaller number, the answer obtained greater than zero is called a quotient.

Divide 143.6 ÷ 21.2

143.6/21.2 = 1436/212

=6.77358

The quotient is rounded to three significant figures after decimal

143.6 ÷ 21.2 = 6.774

Thus, the answer to the correct number of significant figures is 6.774

Learn more about quotient

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7 0

2 years ago

Evaluate ( 64800 ms ) 2 to three significant figures and express the answer in Si units. Express your answer using three signifi

Sergeu [11.5K]

Answer:

42.0×10² second²

Explanation:

Here, time is given in milisecond

(64800 ms)²

= 4199040000 ms²

The SI unit is seconds

1 second = 1000 milisecond

$1\ milisecond=\frac{1}{1000}\ second$

$\\\Rightarrow 1\ milisecond^2=\left(\frac{1}{1000}\right)^2\ second^2$

$4199040000\ ms^2=4199040000\times \left(\frac{1}{1000}\right)^2\ second^2=4199.04\ second^2$

42.0×10² second²

6 0

3 years ago

Which of the following is an example of a conductor?

nordsb [41]

Answer: B. a gold chain

7 0

3 years ago

One Newton is equivalent to<br> A. 1 kg/s2<br> B. 1 kg*m/s<br> C. 1 kg*m/s2<br> D. 1 kg/s

tiny-mole [99]

Answer:

Explanation:

7 0

2 years ago

An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e

White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

$x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)$

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

$\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}$

with this value you can compute the frequency:

$f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz$

the mass of the block is given by the formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg$

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

$\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s$

Finally, the amplitude is:

$x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m$

5 0

3 years ago