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2 years ago

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How much work would a child do while puling a 12-kg wagon a distance of 3m with a 22 N force directed 30 degrees with respect to

the horizontal? (A) 82J (B) 52J (C) 109J (D) 95J

1 answer:

Alika [10]2 years ago

6 0

Answer:

The work done will be 57.15 J

Explanation:

Given that,

Mass = 12 kg

Distance = 3 m

Force = 22 N

Angle = 30°

We need to calculate the work done

The work done is defined as,

$W = Fd\cos\theta$

Where, F = force

d = displacement

Put the value into the formula

$W=22\times3\times\cos30^{\circ}$

$W=22\times3\times\dfrac{\sqrt{3}}{2}$

$W = 57.15\ J$

Hence, The work done will be 57.15 J

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What are TWO benefits of scientific using a diagram to model the water cycle? [PLS HELP ASAP]~pick 2 letters~

LenKa [72]

Answer:

A.) it can show changes that occur in many different parts of earth at the same time

D.) it can be used to show how the parts of the cycle relate to one another

Explanation:

When a scientific diagram is used as a model, it explains the phenomena described in the research.

Scientific models describing water cycle help students understand that a group of parts which relate can combine to form a larger system. The different parts of a water cycle such as evaporation and condensation all work together to facilitate continuity of the water cycle. Additionally, the models highlights the individual processes contributed by the stages of water cycle showing changes and resulting processes.

Option B and C are incorrect because the model diagrams show much details including individual processes contributed by each of the parts leading to functioning of the whole water cycle process.

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2 years ago

a bus is moving with the velociity of 36 km/hr . after seeing a boy at 20 m ahead on the road, the driver applies the brake and

Klio2033 [76]

Answer:

Assumption: the acceleration of this bus is constant while the brake was applied.

Acceleration of this bus: approximately $\left(-6.0\; \rm m \cdot s^{-2}\right)$ .

It took the bus approximately $1.7\;\rm s$ to come to a stop.

Explanation:

Quantities:

Displacement of the bus: $x = 10\; \rm m$ .
Initial velocity of the bus: $\displaystyle u = 36\; \rm km \cdot hr^{-1} = 36\; \rm km \cdot hr^{-1}\times \frac{1\; \rm m \cdot s^{-1}}{3.6\; \rm km\cdot hr^{-1}} = 10\; \rm m \cdot s^{-1}$ .
Final velocity of the bus: $v = 0\; \rm m\cdot s^{-1}$ because the bus has come to a stop.
Acceleration, $a$ : unknown, but assumed to be a constant.
Time taken, $t$ : unknown.

Consider the following SUVAT equation:

$\displaystyle x = \frac{1}{2}\, \left(a\, t^2\right) + u\, t$ .

On the other hand, assume that the acceleration of this bus is indeed constant. Given the initial and final velocity, the time it took for the bus to stop would be inversely proportional to the acceleration of this bus. That is:

$\displaystyle t = \frac{v - u}{a}$ .

Therefore, replace the quantity $t$ with the expression $\displaystyle \left(\frac{v - u}{a}\right)$ in that SUVAT equation:

$\displaystyle x = \frac{1}{2}\, \left(a\, \left(\frac{v -u}{a}\right)^2\right) + u\, \left(\frac{v - u}{a}\right)$ .

Simplify this equation:

$\begin{aligned}x &= \frac{1}{2}\, \left(a\, {\left(\frac{v -u}{a}\right)}^2\right) + u\, \left(\frac{v - u}{a}\right) \\ &= \frac{1}{2}\left(\frac{{(v - u)}^2}{a}\right) + \frac{u\, (v - u)}{a} =\frac{1}{a}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right)\end{aligned}$ .

Therefore, $\displaystyle a= \frac{1}{x}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right)$ .

In this question, the value of $x$ , $u$ , and $v$ are already known:

$x = 10\; \rm m$ .
$\displaystyle u =10\; \rm m \cdot s^{-1}$ .
$v = 0\; \rm m\cdot s^{-1}$ .

Substitute these quantities into this equation to find the value of $a$ :

$\begin{aligned} a &= \frac{1}{x}\, \left(\frac{{(v - u)}^2}{2} + u\, (v - u)\right) \\ &= \frac{1}{10\; \rm m}\times \left(\frac{{\left(0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}\right)}^2}{2} + \left(0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}\right)\times 10\; \rm m \cdot s^{-1}\right)\\ &\approx -6.0\; \rm m \cdot s^{-2}\end{aligned}$ .

(The value of acceleration $a$ is less than zero because the velocity of the bus was getting smaller.)

Substitute $a \approx -6.0\; \rm m \cdot s^{-2}$ (alongside $u = 10\; \rm m \cdot s^{-1}$ and $v = 0\; \rm m \cdot s^{-1}$ ) to estimate the time required for the bus to come to a stop:

$\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{0\; \rm m \cdot s^{-1} - 10\; \rm m \cdot s^{-1}}{-6.0\; \rm m \cdot s^{-2}} \approx 1.7\; \rm s\end{aligned}$ .

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2 years ago

a man is pulling a block with a mass of 6.2kg across a horizontal surface and accelerates the block at a rate of 0.50m/s.the coe

klasskru [66]

Mg = 6.2 x 9.81 = 60.822 This is also normal force. Coefficient of friction times normal force is the force due to friction: 60.822 x 0.24 = 14.6N F = MA so F(your force) - F(friction) = 6.2 x 0.5 = 3.1 Your answer is 3.1+ 14.6 I hope this is correct though I might be wrong.

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2 years ago

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The former soviet union launched the first artificial earth satellite. What is the name of this satelitte?

faltersainse [42]

Answer:

Sputnik I.

Explanation:

The Soviet Union (former) launched the first artificial earth satellite called Sputnik I on October 4, 1957.

It had a diameter of 58 cm and it weighed 83.6 kg. It had an orbital period of 5880 seconds around the earth. It orbited the 11 weeks (officially 3 weeks but then, its battery died and it orbited for 8 weeks before falling back into the earth's atmosphere).

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2 years ago

A cannon sends a projectile towards a target a distance 1420 m away. The initial velocity makes an angle 35◦ with the horizontal

ss7ja [257]

Answer:

$v_{o}=141.51m/s$

Explanation:

From the exercise we know the final x distance, the angle which the projectile is being released and acceleration of gravity

$x=1420m\\g=-9.8m/s^{2}$

From the equation of x-position we know that

$x=v_{ox}t=v_{o}cos(35)t$

Solving for $v_{o}$

$v_{o}=\frac{x}{tcos(35)} =\frac{1420m}{tcos(35)}$ (1)

Now, if we analyze the equation of y-position we got

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

At the end of the motion y=0

$0=v_{o}sin(35)t+\frac{1}{2}gt^{2}$

Knowing the equation for $v_{o}$ in (1)

$0=\frac{1420}{tcos(35)}tsin(35)-\frac{1}{2}(9.8)t^{2}$

$\frac{1}{2}(9.8)t^{2}=1420tan(35)$

Solving for t

$t=\sqrt{\frac{2(1420tan(35))}{9.8} } =14.25s$

Now, we can solve (1)

$v_{o}=\frac{1420m}{(14.25s)cos(35)}=141.51m/s$

6 0

2 years ago

Read 2 more answers