An airplane flies at 400 km/h in a 100km/h hurricane crosswind. Find the resultant speed of the plane.

What is the net work done on the 20kg block while it moves the 4 meters?

Vinil7 [7]

Answer:

The answer to your question is 784.8 J. None of your answer, did you forget some information?

Explanation:

Data

mass = 20 kg

distance = 4 m

work = ?

Formula

Work = force x distance

Force = mass x gravity

Process

1.- Calculate the weight of the block

Weight = 20 x 9.81

Weight = 196.2 N

2.- Calculate the work done

Work = 196.2 x 4

Work = 784.8 J

5 0

3 years ago

Find the net charge of a system consisting of 6.25×10^6 electrons and 7.75×10^6 protons. Express your answer using three signifi

jok3333 [9.3K]

Answer:

2.40 x 10⁻¹³ C

Explanation:

$n_{e}$ = number of electrons = 6.25 x 10⁶

$q_{e}$ = charge on electron = - 1.6 x 10⁻¹⁹ C

$n_{p}$ = number of protons = 7.75 x 10⁶

$q_{p}$ = charge on proton = 1.6 x 10⁻¹⁹ C

Net charge is given as

Q = $n_{e}$ $q_{e}$ + $n_{p}$ $q_{p}$

Q = (- 1.6 x 10⁻¹⁹) (6.25 x 10⁶) + (1.6 x 10⁻¹⁹) (7.75 x 10⁶)

Q = 2.40 x 10⁻¹³ C

6 0

3 years ago

A 1500 kg car traveling 5.0 m/s collides head on with a 3000 kg truck traveling

Setler [38]

m1=1500kg
m_2=3000kg
v_1=5m/s
v_2=7m/s

Using law of conservation of momentum

$\\ \sf\Rrightarrow m_1v_1-m_2v_2=(m1+m2)v_3$

$\\ \sf\Rrightarrow 1500(5)-3000(7)=(1500+3000)v_3$

$\\ \sf\Rrightarrow 7500-21000=4500v_3$

$\\ \sf\Rrightarrow -13500=4500v_3$

$\\ \sf\Rrightarrow v_3=-3m/s$

6 0

2 years ago

A converging lens of focal length 20 cm is placed in contact with, and to the left of, a diverging lens of focal length 30 cm. I

scZoUnD [109]

Answer:

Magnification will be equal to 3

Explanation:

We have given focal length of the converging lens $F_1=20cm$

Focal length of the diverging lens $F_2=30cm$

Object is placed 40 cm to the length of the converging lens d = 40 cm

Combination of the focal length will be equal to

$\frac{1}{F}=\frac{1}{F_1}+\frac{1}{F_2}$

$\frac{1}{F}=\frac{1}{20}+\frac{1}{-30}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}$

F = 60 cm

So combination of the focal length will be 60 cm

Magnification is given by

$M=\frac{F}{F-d}=\frac{60}{60-40}=3$

So magnification will be equal to 3

3 0

3 years ago

An electron is placed on a line connecting two fixed point charges of equal charge but the opposite sign. The distance between t

viktelen [127]

Answer:

a) F_net = 6.48 10⁻¹⁸ ( $\frac{1}{x^2} + \frac{1}{(0.300-x)^2}$ ), b) x = 0.15 m

Explanation:

a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

F = $k \frac{q_2q_2}{r^2}$

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron r₁ = x

right side -electron r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

F_net = k q Q ( $\frac{1}{r_1^2}+ \frac{1}{r_2^2}$ )

F _net = kqQ ( $\frac{1}{x^2} + \frac{1}{(d-x)^2}$ )

let's substitute the values

F_net = 9 10⁹ 1.6 10⁻¹⁹ 4.50 10⁻⁹ ( $\frac{1}{x^2} + \frac{1}{(0.30-x)^2}$ )

F_net = 6.48 10⁻¹⁸ ( $\frac{1}{x^2} + \frac{1}{(0.300-x)^2}$ )

now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values of the distance (x) and the net force are given

x (m) F (N)

0.05 27.0 10-16

0.10 8.10 10-16

0.15 5.76 10-16

0.20 8.10 10-16

0.25 27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m

4 0

3 years ago