Ask question

Ask question

All categories

3 years ago

11

You're using a wedge to split a log. you are hitting the wedge with a large hammer to drive it into the log. it takes a force of

1900 n to push the wedge into the wood. if the wedge moves 0.23 m into the log, how much work have you done on the wedge?

2 answers:

KIM [24]3 years ago

6 0

The work performed on an object is the force multiplied by the distance it is moved, provided the movement is parallel to the force. Since that is the case here, we can get the work by W=Fd=1900N x 0.23m = 437J. This energy is used to split the wood.

RoseWind [281]3 years ago

5 0

Answer: 437 J

Work done can be defined as the product of force and displacement. It is the work required to move an object to certain distance using a certain force.

$work=force\times displacement$

It is given that:

$Force=1900 N$

$Displacement=0.23 m$

Then, $work=1900N\times 0.23m = 437 N-m=437J$

Hence, the work done on the wedge to drive it into log is 437 J

You might be interested in

An electric grinder uses a grinding wheel

luda_lava [24]

(1500 rev/min)(min / 60 s) / (3.0 s) = 8.33 rev/s²

<span>(B) </span>
<span>(1/2)(8.33 rev/s²)(3.0 s)² = 37.5 rev </span>

<span>(C) </span>
<span>(1500 rev/min)(min / 60 s)[2π(0.12 m) / rev] = 18.8 m/s</span>

3 0

3 years ago

Read 2 more answers

Potassium ions (K+) move across a 7.0 -mm- thick cell membrane from the inside to the outside. The potential inside the cell is

Reil [10]

Explanation:

Relation between potential energy and charge is as follows.

U = qV

or, $\Delta U = q \times \Delta V$

= $1.6 \times 10^{-19} \times 70 \times 10^{-3}$

= $112 \times 10^{-22}$ J

or, = $1.12 \times 10^{20} J$

Therefore, we can conclude that change in the electrical potential energy $\Delta U$ is $1.12 \times 10^{20} J$ .

7 0

3 years ago

A body is projected from the ground at an angle of 30° with the horizontal at an initial speed of 128 ft/s. Ignoring air frictio

Elanso [62]

Answer

given,

v = 128 ft/s

angle made with horizontal = 30°

now,

horizontal component of velocity

vx = v cos θ = 128 x cos 30° = 110.85 ft/s

vertical component of velocity

vy = v sin θ = 128 x sin 30° = 64 m/s

time taken to strike the ground

using equation of motion

v = u + at

0 =-64 -32 x t

t = 2 s

total time of flight is equal to

T = 2 t = 2 x 2 = 4 s

b) maximum height

using equation of motion

v² = u² + 2 a h

0 = 64² - 2 x 32 x h

64 h = 64²

h = 64 ft

c) range

R = v_x × time of flight

R = 110.85 × 4

R = 443.4 ft

4 0

3 years ago

A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar

denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of $\beta$

Using formula of $\beta$

$\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}$

Put the value into the formula

$\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}$

$\beta=8.86\times10^{9}$

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

$T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}$

Put the value into the formula

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}$

$T=5.45\times10^{-4}$

(b). We need to calculate the tunnel probability for width 1.00 nm

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}$

$T=7.74\times10^{-8}$

Hence, The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

6 0

3 years ago

a rocket with a mass of 4kg accelerates up from the ground at a rate of 20ft m/s^2, what is the drag force acting on the rocket

Maru [420]

The drag force acting on the rocket is 80N.

<h3>Give an explanation of drag force?</h3>

The divergence in velocity between the fluid and the item, also known as drag, exerts a force on it. Between the liquid and the solid object, there should be motion. Drag is absent in the absence of motion.

The air molecules are more compressed (pushed together) on the surfaces that are facing the front while being more dispersed (spread out) on the surfaces facing the back. Turbulent flow, which occurs when air layers split from the surface and start to swirl, is what causes this.

The drag force acting on the rocket F = ma

Given,

m = 4kg, a = 20ftm/s²

Substituting m and a values in the above formula,

The drag force acting on the rocket F = 4×20

The drag force acting on the rocket F = 80N.

To know more about drag force visit:

brainly.com/question/15144984

#SPJ4

8 0

1 year ago