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3 years ago

Does salt or fresh water make a difference in floating of ships? plz answer ASAP will give brainliest to the best answer

Physics

2 answers:

ch4aika [34]3 years ago

5 0

Answer:

The density of salt water is more than that of density of fresh water, so so salt and fresh water makes a difference in floating of ships.

qaws [65]3 years ago

3 0

Answer:

Yes, salt or fresh water makes a difference in floating of ships.

Explanation:

Salt water is heavier than freshwater, the ships float higher in salt water than fresh water.

The density is higher, so less salt water will be displaced, and the ship will float higher.

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Starting from rest, a dragster travels a straight 1/4 mi racetrack in 7.10 s with constant acceleration. What is its velocity wh

Gennadij [26K]

268.6567 mph is its velocity when it crosses the finish line

d=(v1+v2 /2) x t

.25=(0+v2 /2) x 6.7/3600 hours

900=v2/2 x 6.7

v2=268.6567 mph as the speed with which the dragster crosses the finish

<h3>When acceleration is not zero, can speed remain constant?</h3>

The answer is that an accelerated motion can have a constant speed. Consider a particle travelling uniformly around a circle; it experiences acceleration since the motion's direction is changing, but it maintains a constant speed along the tangential axis throughout the motion.

Acceleration is the frequency of a change in velocity. Acceleration is a vector with magnitude and direction, much as velocity. For instance, if a car is moving in a straight path and speeding up, it is said to have forward (positive) acceleration, and if it is slowing down, it is said to have backward (negative) acceleration.

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brainly.com/question/24681896

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5 0

1 year ago

A roller coaster, traveling with an initial speed of 15 meters per second, decelerates uniformly at â7.0 meters per second2 to a

Harrizon [31]

<span>By algebra, d = [(v_f^2) - (v_i^2)]/2a. Thus, d = [(0^2)-(15^2)]/(2*-7) d = [0-(225)]/(-14) d = 225/14 d = 16.0714 m With 2 significant figures in the problem, the car travels 16 meters during deceleration.</span>

8 0

3 years ago

Consider an old-fashion bicycle with a small wheel of radius 0.17 m and a large wheel of radius 0.92 m. Suppose the rider starts

Gala2k [10]

Answer:

10259.6 m

Explanation:

We are given that

Radius of small wheel,r=0.17 m

Radius of large wheel,r'=0.92 m

Initial velocity,u=0

Time,t=2.7 minutes=162 s

1 min=60 s

Velocity,v=10m/s

Time,t'=13.7 minutes=822 s

Time,t''=4.1 minutes=246 s

$v=u+at$

Substitute the values

$10=0+162a=162a$

$a=\frac{10}{162}=0.0617m/s^2$

$s=ut+\frac{1}{2}at^2$

Substitute the values

$s=\frac{1}{2}(0.0617)(162)^2=809.6 m$

$s'=vt'=10\times 822=8220 m$

$a'=\frac{v}{t''}=\frac{10}{246}$

$s''=\frac{1}{2}a't''^2=\frac{1}{2}\times \frac{10}{246}(246)^2=1230 m$

Total distance traveled by rider=s+s'+s''=809.6+8220+1230=10259.6 m

6 0

3 years ago

Read 2 more answers

Calculate the power of convex lens of focal length 20cm

maxonik [38]

D = 1/f, where D is the power in diopters and f is the focal length in meters.

D=1/20

<u>D=0.05</u>

6 0

3 years ago

If the mass of an object is 44 kilograms and its velocity is 10 meters per second east, how much Kinetic Energy does it have?

olya-2409 [2.1K]

Given: Mass m = 44 Kg; Velocity v = 10 m/s

Required: Kinetic energy K.E = ?

Formula: K.E = 1/2 mv²

K.E 1/2 (44 Kg)(10 m/s)²

K.E = 2,200 Kg.m²/s²

K.E = 2,200 J Answer is A

8 0

3 years ago