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2 years ago

2. If you exert a force of 10.0 N to lift a box a distance of 0.9 m, how much work have you done?

Physics

2 answers:

KonstantinChe [14]2 years ago

6 0

force of 10.0 N

distance of 0.9 m
w=f*d
w=10*0.9
=9.0 j

hichkok12 [17]2 years ago

5 0

Answer :

(2) The work done is, 9.0 J

(3) The power of a machine measures its rate of doing work.

(4) The power used, 700 W

(5) A 750-W motor might also be rated as a 1-horsepower motor.

Solution for part (2) :

Formula used : $W=F\times d$

where,

W = work done = ?

F = applied force = 10 N

d = displacement = 0.9 m

Now put all the given values in the above formula, we get the work done.

$W=(10N)\times (0.9m)}=9J$

Therefore, the work done is, 9.0 J

Explanation for part (3) :

Power : it measures the rate of doing work by time.

Formula for power is, $Power=\frac{Work}{time}$

Solution for part (4) :

Formula used : $P=\frac{W}{t}=\frac{F\times d}{t}$

where,

W = work done

F = applied force = 700 N

d = displacement = 6 m

t = time taken = 6 s

Now put all the given values in the above formula, we get the work done.

$P=\frac{700N\times 6m}{6s}=700W$

Therefore, the power used, 700 W

Solution for part (5) :

As, 746 watt is equal to 1 horsepower

So, 750 watt is equal to $\frac{750}{746}=1.005=1\text{ horsepower}$

Therefore, a 750-W motor might also be rated as a 1-horsepower motor.

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Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

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This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

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