Ask question

Ask question

All categories

strojnjashka [21]

3 years ago

8

A boy is using a rope to pull a sled to the left on a horizontal surface. The rope is parallel to the surface. What are the dire

ctions of the_______. (a) tension force and the (b) friction force on the sled

1 answer:

trasher [3.6K]3 years ago

3 0

Answer:

Explanation:

Check attachment

You might be interested in

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

juin [17]

Answer:

Explanation:

Given

For first case

launch angle $\theta =45^{\circ}C$

at highest point $h=150 m/s$

$150=u\cos 45$

$u=\frac{150}{\cos 45}=212.132 m/s$

For second case

$\theta _2=37^{\circ}C$

at highest Point velocity is $u\cos \theta _2$

$=212.132\times \cos 37$

$=169.41 m/s$

as there is no acceleration in x direction therefore horizontal velocity is same

7 0

3 years ago

Read 2 more answers

The earth rotates every 86,160 seconds. What is the tangential speed (in m/s) at Livermore (Latitude 37.6819° measured up from e

Lena [83]

Answer:

The tangential speed at Livermore is approximately 284.001 meters per second.

Explanation:

Let suppose that the Earth rotates at constant speed, the tangential speed ( $v$ ), measured in meters per second, at Livermore (37.6819º N, 121º W) is determined by the following expression:

$v = \left(\frac{2\pi}{\Delta t}\right)\cdot R \cdot \sin \phi$ (1)

Where:

$\Delta t$ - Rotation time, measured in seconds.

$R$ - Radius of the Earth, measured in meters.

$\phi$ - Latitude of the city above the Equator, measured in sexagesimal degrees.

If we know that $\Delta t = 86160\,s$ , $R = 6.371\times 10^{6}\,m$ and $\phi = 37.6819^{\circ}$ , then the tangential speed at Livermore is:

$v = \left(\frac{2\pi}{86160\,s} \right)\cdot (6.371\times 10^{6}\,m)\cdot \sin 37.6819^{\circ}$

$v\approx 284.001\,\frac{m}{s}$

The tangential speed at Livermore is approximately 284.001 meters per second.

4 0

3 years ago

A ship 1200m off shore fires a gun. how long after the gun is fired will it be heard on the shore?

ryzh [129]

Answer:

We know that the speed of sound is 343 m/s in air

we are also given the distance of the boat from the shore

From the provided data, we can easily find the time taken by the sound to reach the shore using the second equation of motion

s = ut + 1/2 at²

since the acceleration of sound is 0:

s = ut + 1/2 (0)t²

s = ut <em>(here, u is the speed of sound , s is the distance travelled and t is the time taken)</em>

Replacing the variables in the equation with the values we know

1200 = 343 * t

t = 1200 / 343

t = 3.5 seconds (approx)

Therefore, the sound of the gun will be heard at the shore, 3.5 seconds after being fired

6 0

3 years ago

This is a Science question pls help give 15 coins for answer

Vlad1618 [11]

Answer:

<u>Conservation</u>: using less water

<u>Xeriscaping</u>: replanting your yard with plants that do not require great amounts of water

<u>Desalination</u>: process of removing salt from water so that it can be used for consumption

<u>Water Budget</u>: finite amount of usable water available

<u>Potable</u>: water that is safe to use a drink

7 0

3 years ago

Read 2 more answers

How much work is done in holding a 15 N sack of potatoes while waiting in line at the

marta [7]

Answer: 2700 Joules

Explanation:

Given variables are:

Workdone (w) = ?

Weight of sack (w) = 15N

Time taken (t) = 3 minutes

The standard unit of time is seconds, so convert 3 minutes to seconds

(If 1 minute = 60 seconds

3 minutes = 3 x 60 = 180 seconds)

Recall that work is done is when force is applied on an object over a distance

i.e Workdone = force x distance.

(However, the force acting on the sack of potatoes is weight)

Hence, workdone = weight x time

w = 15N x 180 seconds

w = 2700 J

thus, 2700 joules of work was done.

4 0

3 years ago