Ask question

Ask question

All categories

3 years ago

7

A large crate is pushed across the floor with an effort of 45 Newtons. The box is pushed a distance of 3.5 meters. How much work

is done?

1 answer:

I am Lyosha [343]3 years ago

3 0

Answer:

$\boxed {\boxed {\sf 157.5 \ Joules }}$

Explanation:

Work is equal to the product of force and distance.

$W=F*d$

The box is being pushed with an effort (force) of 45 Newtons and the distance is 3.5 meters.

$F= 45 \ N \\d= 3.5 \ m$

Substitute the values into the formula.

$W= 45 \ N * 3.5 \ m$

Multiply.

$W= 157.5 \ N*m$

1 Newton meter is equal to 1 Joule
Our answer of 157.5 N*m equals 157.5 J

$W= 157.5 \ J$

<u>157.5 Joules </u> of work are done on the crate.

You might be interested in

In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so

svetlana [45]

Answer:

(a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

Explanation:

Given that,

Radius of the circle = 0.681 m

Angular acceleration = 67.7 rad/s²

Angular speed =18.6 rad/s

We need to calculate the centripetal acceleration of the ball

Using formula of centripetal acceleration

$a_{c}=\omega^2\times r$

Put the value into the formula

$a_{c}=(18.6)^2\times0.681$

$a_{c}=235.5\ m/s^2$

We need to calculate the tangential acceleration of the ball

Using formula of tangential acceleration

$a_{t}=r\alpha$

Put the value into the formula

$a_{t}=0.681\times67.7$

$a_{t}=46.104\ m/s^2$

(a). We need to calculate the magnitude of the total acceleration of the ball

Using formula of total acceleration

$a=\sqrt{a_{c}^2+a_{t}^2}$

Put the value into the formula

$a=\sqrt{(235.5)^2+(46.104)^2}$

$a=239.97\ m/s^2$

(b). We need to calculate the angle of the total acceleration relative to the radial direction

Using formula of the direction

$\theat=\tan^{-1}(\dfrac{a_{t}}{a_{c}})$

Put the value into the formula

$\theta=\tan^{-1}(\dfrac{46.104}{235.5})$

$\theta=11.0^{\circ}$

Hence, (a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

5 0

3 years ago

Risks are only in a persons mind would know from facts to back them up

lyudmila [28]

Risks are only in a persons

8 0

3 years ago

Why do we need to learn about median and mode?

siniylev [52]

We need to learn about median and mode because they are both important in statistical data and statistical data is what shows us important information about the world around us.

3 0

3 years ago

2. Keith took a cube and measured the mass and volume of a

galben [10]

Answer:

<h2>6 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 36 g

volume = 6 cm³

We have

$density = \frac{36}{6} = 6 \\$

We have the final answer as

<h3>6 g/cm³</h3>

Hope this helps you

7 0

2 years ago

At a certain harbor, the tides cause the ocean surface to rise and fall a distance d (from highest level to lowest level) in sim

OleMash [197]

Answer:

1.869 hours

Explanation:

$T$ = Time period = 11.1 h

Angular frequency is given by

$\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{11.1}\\\Rightarrow \omega=0.566\ rad/h$

The distance moved from highest to lowest level is given by

$d=2x_{m}\\\Rightarrow x_m=\frac{d}{2}\\\Rightarrow x_m=0.5d$

At the ocean surface $x_m=0.25d$

$x=x_mcos(\omega t+\phi)$

$\phi$ = Phase constant = 0 as clock is started at $x_0=x_m$

$0.25d=0.5dcos(0.56t)\\\Rightarrow cos(0.56t)=\frac{0.25}{0.5}\\\Rightarrow 0.56t=cos^{-1}0.5\\\Rightarrow t=\frac{1.047}{0.56}\\\Rightarrow t=1.869\ h$

The time taken for the water to fall the distance is 1.869 hours

7 0

3 years ago