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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

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If it takes 100 N to get a 10 kg object to accelerate at 10m/s/s, how much force will it take to get a 20 kg object to accelerat

yan [13]

Answer: C) 200 N

Explanation:

The force $F$ is defined as:

$F=m.a$

Where:

$m=20 kg$ is the mass of the object

$a=10 m/s^{2}$ is the acceleration

Then:

$F=(20 kg)(10 m/s^{2})$

Finally:

$F=200 N$

Hence, the correct option is C.

7 0

3 years ago

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),

yuradex [85]

Answer:

$v=3.564\ m.s^{-1}$

$\Delta v =2.16\ m.s^{-1}$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

<u>Using the equation of motion:</u>

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3$

$v_J=5.94\ m.s^{-1}$

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30\times 5.94+0=(30+20)v$

$v=3.564\ m.s^{-1}$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

<u>Now we find the force along the slide due to the body weight:</u>

$F=m_J.g.sin\theta$

$F=30\times 9.8\times \frac{1.8}{3}$

$F=176.4\ N$

<em><u>Hence the net force along the slide:</u></em>

$F_R=71.4\ N$

<em>Now the acceleration of John:</em>

$a_j=\frac{F_R}{m_J}$

$a_j=\frac{71.4}{30}$

$a_j=2.38\ m.s^{-2}$

<u>Now the new velocity:</u>

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2\times 2.38\times 3$

$v_J_n=3.78\ m.s^{-1}$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^{-1}$

8 0

3 years ago

1. Predict whether the energy required to remove an electron from magnesium and potassium would be more or less than that requir

Inga [223]

In both cases less energy is required

But comparetively Mg require more energy than K

Let's see the electron configuration of Both

[Mg]=1s²2s²2p⁶3s²=[Ne]3s²
[K]=1s²2s²2p⁶3s²3p⁶4s¹=[Ar]4s¹

K has only one valence electron so very less ionization enthalpy so less energy required

Mg has 2 so more IE hence more energy required

8 0

1 year ago

A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0o above the horizontal towar

Artist 52 [7]

Answer:

a) v₀ = 32.64 m / s , b) x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

x = v₀ₓ t

y = $v_{oy}$ t - ½ g t²

We look for the components of speed with trigonometry

sin 43 = v_{oy} / v₀

cos 43 = v₀ₓ / v₀

v_{oy} = v₀ sin 43

v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

t = x / v₀ cos 43

y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

y = x tan 43 - ½ g x² / v₀² cos² 43

1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

v₀ = √ (35280 / 33.11)

v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

y = $v_{oy}$ t - ½ g t²

.y = v₀ sin43 t - ½ g t²

25 = 32.64 sin 43 t - ½ 9.8 t²

4.9 t² - 22.26 t + 25 = 0

t² - 4.54 t + 5.10 = 0

We solve the second degree equation

t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

t = (4.54 ± 0.46) / 2

t₁ = 2.50 s

t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

x = v₀ₓ t

x = v₀ cos 43 t

x = 32.64 cos 43 2.50

x = 59.68 m

8 0

3 years ago

A student carried out an experiment adding different weights to a spring and recording the results. Look at the table of results

MAXImum [283]

Answer:

0.25 m.

Explanation:

We'll begin by calculating the spring constant of the spring.

From the diagram, we shall used any of the weight with the corresponding extention to determine the spring constant. This is illustrated below:

Force (F) = 0.1 N

Extention (e) = 0.125 m

Spring constant (K) =?

F = Ke

0.1 = K x 0.125

Divide both side by 0.125

K = 0.1/0.125

K = 0.8 N/m

Therefore, the force constant, K of spring is 0.8 N/m

Now, we can obtain the number in gap 1 in the diagram above as follow:

Force (F) = 0.2 N

Spring constant (K) = 0.8 N/m

Extention (e) =..?

F = Ke

0.2 = 0.8 x e

Divide both side by 0.8

e = 0.2/0.8

e = 0.25 m

Therefore, the number that will complete gap 1is 0.25 m.

5 0

2 years ago