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vladimir1956 [14]

2 years ago

15

During a 980. miles cruise, a ship travelled 590 miles in 7 hours first. Then the ship changed its speed for the rest of the tri

p. If 13 hours were spent on the whole cruise, what was the speed for the rest of the trip? mph

2 answers:

vfiekz [6]2 years ago

7 0

Explanation:

The average speed of a modern cruise ship is roughly 20 knots (23 miles per hour), with maximum speeds reaching about 30 knots (34.5 miles per hour).

nekit [7.7K]2 years ago

7 0

980-590 = 390 miles remaining
13-7 = 6 hours remaining
Miles/hours gives miles per hour for rest of trip therefore 390/6 = 65mph.

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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

How do the discovery of gravity and the invention of electronic satellites most likely relate to the processes of scientific inv

Leya [2.2K]

Answer:

c

Explanation:

because i said so

5 0

3 years ago

Two long parallel wires are separated by 6.0 mm. The current in one of the wires is twice the other current. If the magnitude of

lions [1.4K]

Answer:

Explanation:

Magnitude of force per unit length of wire on each of wires

= μ₀ x 2 i₁ x i₂ / 4π r where i₁ and i₂ are current in the two wires , r is distance between the two and μ₀ is permeability .

Putting the values ,

force per unit length = 10⁻⁷ x 2 x i x 2i / ( 6 x 10⁻³ )

= .67 i² x 10⁻⁴

force on 3 m length

= 3 x .67 x 10⁻⁴ i²

Given ,

8 x 10⁻⁶ = 3 x .67 x 10⁻⁴ i²

i² = 3.98 x 10⁻²

i = 1.995 x 10⁻¹

= .1995

= 0.2 A approx .

2 i = .4 A Ans .

6 0

3 years ago

The spin cycle of a clothes washer extracts the water in clothing by greatly increasing the water's apparent weight so that it i

andreyandreev [35.5K]

The apparent weight of a 1.1 g drop of water is 4.24084 N.

<h3>What is Apparent Weight?</h3>

According to physics, an object's perceived weight is a characteristic that describes how heavy it is. When the force of gravity acting on an object is not counterbalanced by a force of equal but opposite normality, the apparent weight of the object will differ from the actual weight of the thing.
By definition, an object's weight is equal to the strength of the gravitational force pulling on it. It follows that even a "weightless" astronaut in low Earth orbit, with an apparent weight of zero, has almost the same weight that he would have if he were standing on the ground; this is because the gravitational pull of low Earth orbit and the ground are nearly equal.

Solution:

N = Speed of rotation = 1250 rpm

D = Diameter = 45 cm

r = Radius = 22.5 cm

M = Mass of drop = 1.1 g

Angular speed of the water = $\omega = \frac{2\pi N}{60}$

$\omega = \frac{2\pi \times 1250}{60}$

$\omega = 130.89 rad/s$

Apparent weight is given by

$W _a = M\omega^{2}R$

$W_a = 1.1 \times 10^-^3\times (130.89)^2\times 0.225$

$W_a$ = 4.24084 N

Know more about Apparent weight brainly.com/question/14323035

#SPJ4

Question:

The spin cycle of a clothes washer extracts the water in clothing by greatly increasing the water's apparent weight so that it is efficiently squeezed through the clothes and out the holes in the drum. In a top loader's spin cycle, the 45-cm-diameter drum spins at 1250 rpm around a vertical axis. What is the apparent weight of a 1.1 g drop of water?

7 0

1 year ago

The vertical force f acts downward at A on the two membered frames. Determine the magnitude of the two components of F directed

galben [10]

Answer:

The magnitude will be "353.5 N". A further solution is given below.

Explanation:

The given values is:

F = 500 N

According to the question,

In ΔABC,

⇒ $\angle BCA = (90-30)$

⇒ $=60^{\circ}$

then,

⇒ $\angle BAC=(180-45-60)$

⇒ $=75^{\circ}$

Now,

The corresponding angle will be:

⇒ $\angle FAC=60^{\circ}$

⇒ $\angle FAB=70+60$

⇒ $=135^{\circ}$

Aspect of F across the AC arm will be:

= $F\times cos(60)$

On putting the values of F, we get

= $500\times (.5)$

= $200 \ Newton$

Component F along the AC (in magnitude) will be:

= $F\times cos(135)$

= $500\times (-.707)$

= $-353.5 \ N \$

4 0

2 years ago