All categories

1 year ago

Dźwig podniósł kontener o masie m = 80 kg na wysokość h = 10 m. Pierwsze 5 m kontener przebył z przy-

Physics

1 answer:

Nady [450]1 year ago

7 0

Answer:

a) W_total = 8240 J , b) W₁ / W₂ = 1.1

Explanation:

In this exercise you are asked to calculate the work that is defined by

W = F. dy

As the container is rising and the force is vertical the scalar product is reduced to the algebraic product.

W = F dy = F Δy

let's apply this formula to our case

a) Let's use Newton's second law to calculate the force in the first y = 5 m

F - W = m a

W = mg

F = m (a + g)

F = 80 (1 + 9.8)

F = 864 N

The work of this force we will call it W1

We look for the force for the final 5 m, since the speed is constant the force must be equal to the weight (a = 0)

F₂ - W = 0

F₂ = W

F₂ = 80 9.8

F₂ = 784 N

The work of this fura we will call them W2

The total work is

W_total = W₁ + W₂

W_total = (F + F₂) y

W_total = (864 + 784) 5

W_total = 8240 J

b) To find the relationship between work with relate (W1) and work with constant speed (W2), let's use

W₁ / W₂ = F y / F₂ y

W₁ / W₂ = 864/784

W₁ / W₂ = 1.1

You might be interested in

The net force on a boat causes it to accelerate at 1.55 m/s2. The mass of the boat is 215 kg. The same net force causes another

jekas [21]

Answer:

2666 kg

0.11567 m/s²

Explanation:

m = Mass of boat

a = Acceleration of boat

From Newton's second law

Force

$F=ma\\\Rightarrow F=215\times 1.55\\\Rightarrow F=333.25\ N$

Force on the first boat is 333.25 N

$F=ma\\\Rightarrow m=\frac{F}{a}\\\Rightarrow m=\frac{333.25}{0.125}\\\Rightarrow m=2666\ kg$

Hence, mass of the second boat is 2666 kg

Combined mass = 2666+215 = 2881 kg

$F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{333.25}{2881}\\\Rightarrow a=0.11567\ m/s^2$

The acceleration on the combined mass is 0.11567 m/s²

6 0

1 year ago

Object A of mass M is moving east at speed v. It collides with object B of mass 2M that was initially at rest. The motion of the

Firlakuza [10]

Answer:

$v_B=\frac{v}{3}$

Explanation:

Given that:

mass of object A, $m_A=M$

mass of object B, $m_B=2M$

speed of object A, $v_A=v$

So, according to the conservation of momentum, the momentum before collision is equal to the momentum after conservation.

$m_A.v+m_B\times 0=m_A\times v_A +m_B\times v_B$

$M\times v+0 = M\times \frac{v}{3}+2M\times v_B$

$2M\times v_B= \frac{2M\times v}{3}$

$v_B=\frac{v}{3}$

3 0

1 year ago

A split highway has a number of lanes for traffic. For traffic going in one direction, the radius for the inside of the curve is

Arte-miy333 [17]

Cuál es tu pregunta

7 0

1 year ago

An isolated charged point particle produces an electric field with magnitude E at a point 2 m away. At a point 1 m from the part

guajiro [1.7K]

Explanation:

The electric field at a distance r from the charged particle is given by :

$E=\dfrac{kq}{r^2}$

k is electrostatic constant

if r = 2 m, electric field is given by :

$E_1=\dfrac{kq}{(2)^2}\\\\=\dfrac{kq}{4}\ .....(1)$

If r = 1 m, electric field is given by :

$E_2=\dfrac{kq}{r_2^2}\\\\=\dfrac{kq}{1}\ ....(2)$

Dividing equation (1) and (2) we get :

$\dfrac{E_1}{E_2}=\dfrac{\dfrac{kq}{4}}{kq}\\\\\dfrac{E_1}{E_2}=\dfrac{1}{4}\\\\E_2=4\times E_1$

So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.

4 0

1 year ago

Let’s consider tunneling of an electron outside of a potential well. The formula for the transmission coefficient is T \simeq e^

ioda

Answer:

L' = 1.231L

Explanation:

The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:

$T \approx e^{-2CL}$

L: width of the barrier

C: constant that includes particle energy and barrier height

You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let's say, L', the value of the transmission coefficient is T'=0.025.

To find the new value of the L' you can write down both situation for T and T', as in the following:

$0.050=e^{-2CL}\ \ \ \ (1)\\\\0.025=e^{-2CL'}\ \ \ \ (2)$

Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):

$ln(0.050)=ln(e^{-2CL})=-2CL\ \ \ \ (3)\\\\ln(0.025)=ln(e^{-2CL'})=-2CL'\ \ \ \ (4)$

Next, you divide the equation (3) into (4), and finally, you solve for L':

$\frac{ln(0.050)}{ln(0.025)}=\frac{-2CL}{-2CL'}=\frac{L}{L'}\\\\0.812=\frac{L}{L'}\\\\L'=\frac{L}{0.812}=1.231L$

hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L' is 1.231 L

8 0

1 year ago