All categories

1 year ago

Joe used a 750 watt 1/2" cordless drill to put together a bookcase. Calculate the work involved in this thirty minute process.

Physics

2 answers:

Nataliya [291]1 year ago

3 0

750W 30mins...750x30x60 joules ... 75000x18 ... about 1,500,000j

RoseWind [281]1 year ago

3 0

Answer:

1,350,000Joules

Explanation:

Work done by Joe in 30minutes can be calculated using the formula

Work done = power × time

Given power = 750watts

Time = 30minutes

Time = 30×60 = 1800secs

Workdone = 750×1800

Work done by Joe= 1,350,000Joules

You might be interested in

If a metal wire is 4m long and a force of 5000n causes it to stretch by 1mm, what is the strain?

barxatty [35]

Answer:

$2.5\cdot 10^{-4}$

Explanation:

The strain is defined as the ratio of change of dimension of an object under a force:

$S=\frac{\Delta L}{L_0}$

where

$\Delta L$ is the change in length of the object

$L_0$ is the original length of the object

In this problem, we have $L_0 = 4 m$ and $\Delta L=1 mm=0.001 m$ , therefore the strain is

$S=\frac{\Delta L}{L_0}=\frac{0.001 m}{4 m}=2.5\cdot 10^{-4}$

5 0

2 years ago

16) A wheel of moment of inertia of 5.00 kg-m2 starts from rest and accelerates under a constant torque of 3.00 N-m for 8.00 s.

KiRa [710]

Answer:

57.6Joules

Explanation:

Rotational kinetic energy of a body can be determined using the expression

Rotational kinetic energy = 1/2Iω²where;

I is the moment of inertia around axis of rotation. = 5kgm/s²

ω is the angular velocity = ?

Note that torque (T) = I¶ where;

¶ is the angular acceleration.

I is the moment of inertia

¶ = T/I

¶ = 3.0/5.0

¶ = 0.6rad/s²

Angular acceleration (¶) = ∆ω/∆t

∆ω = ¶∆t

ω = 0.6×8

ω = 4.8rad/s

Therefore, rotational kinetic energy = 1/2×5×4.8²

= 5×4.8×2.4

= 57.6Joules

6 0

2 years ago

Read 2 more answers

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

DENIUS [597]

Answer:

$v_{oy}=11.29\ m/s$

$y=7.39\ m$

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle $\theta$ with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

Where $y_o$ is the initial height above the ground level, $v_{oy}$ is the vertical component of the initial velocity and t is the time

The y-component of the speed is

$v_y=v_{oy}-gt$

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of $v_o$

The object will reach the maximum height when $v_y=0$ . It allows us to compute the time to reach that point

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum heigh is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Solving for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Replacing the known values

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

5 0

2 years ago

An individual white LED (light-emitting diode) has an efficiency of 20% and uses 1.0 W of electric power. a. How many LEDs must

Neporo4naja [7]

Answer:

8, 8 W

Explanation:

The useful power of 1 Light Emitting Diode is

$0.2\times 1=0.2\ W$

Total power required is 1.6 W

Number of Light Emitting Diodes would be

$n=\dfrac{1.6}{0.2}\\\Rightarrow n=8$

The number of Light Emitting Diodes is 8.

Power would be

$P=8\times 1=8\ W$

The power that is required to run the Light Emitting Diodes is 8 W

7 0

2 years ago

Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic

Nuetrik [128]

Answer:

(a) A = $3.90 \AA$

(b) $A = 4.50 \AA$

(d) $A = 9.02 \AA$

Solution:

As per the question:

Radius of atom, r = 1.95 $\AA = 1.95\times 10^{- 10} m$

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = $2\times 1.95 = 3.90 \AA$

(b) For body centered cubic lattice:

$A = \frac{4}{\sqrt{3}}r$

$A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA$

$A = 2{\sqrt{2}}r$

$A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA$

(d) For diamond lattice:

$A = 2\times \frac{4}{\sqrt{3}}r$

$A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA$

6 0

1 year ago