Reading from left to right, describe the spacing between the dots in lines 1, 2, and 3.

If the mass of a material is 87 grams and the volume of the material is 14 cm3, what would the density of the material be?

tresset_1 [31]

Answer:

6.214g/cm³

Explanation:

The question is on density of a material

Density=mass/volume

Given, mass=87grams and volume= 14 cm³ density=?

Density=m/v 87/14 =6.214g/cm³

8 0

2 years ago

A tug boat pulls a ship with a constant net horizontal force of 5.00•10*3 N and causes the ship to move through a harbor. How mu

Murljashka [212]

The work done on the ship is $1.5\cdot 10^7 J$

Explanation:

The work done by a force on an object is given by:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

In this problem, we have:

$F=5.00\cdot 10^3 N$ (force acting on the ship)

d = 3.00 km = 3000 m (displacement of the ship)

$\theta=0^{\circ}$ (because the force is horizontal, and the displacement is horizontal as well)

Therefore, the work done on the ship is

$W=(5.00\cdot 10^3)(3000)(cos 0^{\circ})=1.5\cdot 10^7 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

8 0

2 years ago

Suppose the earth is shaped as a sphere with radius 4,0004,000 miles and suppose it rotates once every 24 hours. How many miles

Alenkinab [10]

Answer:

1047 miles

Explanation:

The radius of the Earth is

$r = 4000$ (miles)

So its circumference, which is the total length of the equator, is given by

$L=2\pi r= 2\pi(4000)=25133 mi$

Now we know that the Earth rotates once every 24 hours. So the distance through which the equator moves in one hour is equal to its total length divided by the number of hours, 24:

$L' = \frac{25133 mi}{24h}=1047 mi$

8 0

3 years ago

A car travels 40 miles in 30 minutes.

lukranit [14]

Answer:

(a)Average velocity ,v =128.74 Km/hr

(b)Kinetic Energy , K=958546.875 Joule

(c)Distance, s=268.8m

(d)Acceleration, a= - 2.38 $m/s^2$

<u>Explanation</u>:

<u>Given</u>:

Distance travelled = 40 miles

Time taken = 30 minutes.

(A) The average velocity in kilometres/hour

Converting 40 miles into km ,

we know that,

1 mile = 1.60934

40 miles = 40 x 1.60934

so 40 miles = 64.3738 Km

similarly converting 30 minutes into hours

1 minute = $\frac{1}{60}hours$

30 minute = $\frac{30}{60}hours$

30 minute = $\frac{1}{2}hours$

Now

Average velocity = $\frac{Speed}{time}$

Substituting Values,

Average velocity = $\frac{64.3738}{\frac[1}{2}}$

Average velocity = $64.3738 \times 2$

Average velocity =128.74 Km/hr

(B) If the car weighs 1.5 tons, what is its If the car weighs 1.5 tons, what is its kinetic energy in joules (Note: you will need to convert your velocity to m/s)? in joules (Note: you will need to convert your velocity to m/s)?

Converting 1.5 tons into kg we get

1 ton = 1000 kg

so 1.5 ton =1500 kg

converting velocity to m/s

$128.74 \times \frac{5}{18}$

=>35.75 m/s----------------------------------------------------------(1)

kinetic energy K= $\frac{1}{2}mv^2$

Substituting the values,

K= $\frac{1}{2}1500(35.75)^2$

K= $\frac{1}{2}1500(1278.06)$

K= $\frac{1500 \times (1278.06)}{2}$

K= $\frac{1917093.75}{2}$

K=958546.875 Joule---------------------------------------------(2)

(c)When the driver applies the brake, it takes 15 seconds to stop. How far does the car travel (in meters) while stopping

Lets use Distance formula,

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

Substituting the known values,

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

$s= (37.75)(15)+\frac{1}{2}a(15)^2$

$s=566.25+\frac{1}{2}a(225)$

$s=566.25+\frac{(225a)}{2}$ -------------------------------------(3)

(D) What is the average acceleration of the car (in m/s2) during braking?

Using the formula

v=u +at

re arranging the formula we get,

$a = \frac{v - u}{t}$

Substituting the values

$a = \frac{0 - 35.75}{15}$

$a = \frac{- 35.75}{15}$

a= - 2.38 $m/s^2$ ----------------------------------------(4)

Now substituting 4 in 3 we get

$s=566.25+\frac{(225( - 2.38)}{2}$

$s=566.25+\frac{-535.5}{2}$

$s=536.25-267.75$

s=268.8m--------------------------------------------------------------(5)

4 0

3 years ago

An elevator car weighs 5500 N. If the car accelerates upwards at a rate of 4.0 m/s2, what is the tension in the support cable li

nadya68 [22]

Answer:

Explanation:

The equation for this, since we are talking about weight on an elevator, is Newton's 2nd Law adjusted to fit our needs:

$F_n=ma+w$ where the Normal Force needed to lift that elevator car is the tension. So the equation then becomes

T = ma + w where T is the tension in the cable to lift the elevator, m is the mass of the elevator (which we have to solve for), a is the acceleration of the elevator (positive since it's going up), and w is the weight of the elevator (which we have as 5500 N). Solving first for mass:

w = mg and

5500 =- m(10) so

m = 550 kg. Now we have what we need to solve for the tension:

T = 550(4.0) + 5500 and

T = 2200 + 5500 so

T = 7700 N

4 0

3 years ago