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2 years ago

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A net force of 24 N is acting on a 4.0-kg object. Find the acceleration in m/s2.

1 answer:

Inessa [10]2 years ago

6 0

Hi there!

We can use Newton's Second Law:
$\Sigma F = ma$

ΣF = Net force (N)
m = mass (kg)
a = acceleration (m/s²)

We can rearrange the equation to solve for the acceleration.

$a = \frac{\Sigma F}{m}\\\\a = \frac{24}{4} = \boxed{6 \frac{m}{s^2}}$

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A 65 kg box is lifted by a person pulling a rope a distance of 15 meters straight up at a constant speed. How much Power is requ

Y_Kistochka [10]

Answer:

Power is 1061.67W

Explanation:

Power=force×distance/time

Power=65×9.8×15/9 assuming gravity=9.8m/s²

Power=3185/3=1061.67W

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3 years ago

Which of the following processes is not regarded as contributing to the creation of greenhouse gases? deforestation creating ele

DiKsa [7]

Creating electricity from wind is not regarded as a process contributing to the creation of greenhouse gases. Meanwhile, the processes such as deforestation, the creation of electricity from coal <span>and the use of fertilizers </span><span>are greatly contributing to the making of greenhouse gases.</span>

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3 years ago

Read 2 more answers

The current supplied by a battery as a function of time is ) -(0.64 A)e-/(6.0 hr). What is the total number of electrons transpo

Paha777 [63]

Answer:

The total number of electrons is $8.6\times10^{22}$

(3) is correct option.

Explanation:

Given that,

The current equation is

$I=0.64 e^{\dfrac{-t}{6.0 hr}}$

We know that,

The formula of charge

$q=\int_{0}^{\infty}{I dt}$

$q=\int_{0}^{\infty}{0.64 e^{\dfrac{-t}{6.0 hr}}dt$

$q=0.64\int_{0}^{\infty}{e^{\dfrac{-t}{6.0 hr}}dt$

$q=0.64\int_{0}^{\infty}{e^{\dfrac{-t}{21600}}dt$

$q=0.64(21600e^{\dfrac{-t}{21600}})_{0}^{\infty}$

$q=0.64(0-21600)$

$q=0.64\times21600$

$q=13824\ C$

We need to calculate the number of electron

Using formula of charge

$q=ne$

$n=\dfrac{q}{e}$

$n=\dfrac{13824}{1.6\times10^{-19}}$

$n=8.6\times10^{22}$

Hence, The total number of electrons is $8.6\times10^{22}$

7 0

3 years ago

A 0.171-kilogram hockey puck is struck by a hockey stick. If the force from the hockey stick is 30.2 Newtons, but the puck also

Komok [63]

Answer:

a = 156.14 m/s^2

Explanation:

Using the laws of newton:

∑F = ma

where ∑F is the sumatory of forces, m the mass and a the aceleration.

so:

F - $f_k$ = ma

where F is the force from the hockey stick and $f_k$ is the force of friction.

Replacing values, we get:

(30.2N) - (3.5N)= (0.171kg)a

Finally, solving for a:

a = 156.14 m/s^2

4 0

3 years ago

A 1.0 kg block is placed against an ideal spring with spring constant 800 N/m and initially compressed 0.20 m. The spring and bl

densk [106]

Answer:

the release will be at 3.266 m distance

Explanation:

mass = 1 Kg

spring constant (k) = 800 N/m

initial compression = 0.20 m

θ = 30⁰

$U= \dfrac{1}{2}kx^2\\U= \dfrac{1}{2}800\times 0.2^2\\U= 16 J$

$U=mgh\\h=\dfrac{U}{mg}\\h=\dfrac{16}{1 \times 9.8}\\h = 1.633m$

$d=\dfrac{h}{sin \theta}\\d=\dfrac{1.633}{sin 30}\\\\d= 3.266m$

hence the release will be at 3.266 m distance.

5 0

3 years ago