Plz help

you push a sled of mass 15 kg across the snow with a force of 180 N for a distance of 2.5 m. There is no friction. if the sled s

Oxana [17]

<span>7.7 m/s First, determine the acceleration you subject the sled to. You have a mass of 15 kg being subjected to a force of 180 N, so 180 N / 15 kg = 180 (kg m)/s^2 / 15 kg = 12 m/s^2 Now determine how long you pushed it. For constant acceleration the equation is d = 0.5 A T^2 Substitute the known values getting, 2.5 m = 0.5 12 m/s^2 T^2 2.5 m = 6 m/s^2 T^2 Solve for T 2.5 m = 6 m/s^2 T^2 0.41667 s^2 = T^2 0.645497224 s = T Now to get the velocity, multiply the time by the acceleration, giving 0.645497224 s * 12 m/s^2 = 7.745966692 m/s After rounding to 2 significant figures, you get 7.7 m/s</span>

5 0

3 years ago

Please help don't for get to show work

emmasim [6.3K]

-- Bob covered a distance of (32m + 45m) = 77 meters.

-- His displacement is the straight-line distance and direction
from his starting point to his ending point.

The straight-line distance is

D = √(32² + 45²)
D = √(1,024 + 2,025)
D = √3,049 = 55.22 meters

The direction is the angle whose tangent is (32/45) south of east.

tan⁻¹(32/45) = tan⁻¹(0.7111...) = 35.42° south of east.

3 0

3 years ago

A circular pipe of 25-mm outside diameter is placed in an airstream at 25C and 1-atm pressure. The air moves in cross flow over

kifflom [539]

Answer:

f_D = =3.24 N/m

Explanation:

data given

properties of air $\nu\ of air =19.31*10^{-6} m2/s$

$\rho = 1.048 kg/m3$

k = 0.0288 W/m.K

WE KNOW THAT

Reynold's number is given as

$Re =\frac{VD}{\nu}$

$= \frac{ 15*0.025}{19.31*10^{-6}}$

= 1.941 *10^4

drag coffecient is given as

$C_D = \frac{f_D}{A_f\frac{\rho v^2}{2}}$

solving for f_D

$f_D = C_D A_f*\frac{\rho v^2}{2}$

$=C_D D*\frac{\rho v^2}{2}$

Drag coffecient for smooth circular cylinder is 1.1

therefore Drag force is

$f_D = 1.1*0.025 *\frac{1.048*15^2}{2}$

f_D = =3.24 N/m

4 0

2 years ago

A 91.5 kg football player running east at 3.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft

PIT_PIT [208]

Their velocity afterward is v=3.467 m/s

Explanation:

Given:

Mass of the first football player= 91.5 kg

Initial velocity of the football player 3.73 m/s

Mass of second football player=63.56 kg

Initial velocity of the second football player=3.09 m/s

To find:

Final velocity of both players=?

Solution:

According to the law of conservation of momentum,

Initial momentum =final momentum

mathematically represented as

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$ ...........................(1)

where

$u_1$ =intial velocity of the football player

$u_2$ = inital velocity of second football player

$v_1$ =finall velocity of the first football player

$v_2$ =final velocity of second football player

after tackling , both the football players moves with the same velocity,

so $v_1=v_2=v$

Hence equation (1) becomes

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\frac{m_1u_1+m_2u_2}{(m_1+m_2)}$

now substituting the values,

$v=\frac{(91.5\times+3.73)+(63.5\times3.09)}{(91.5+63.5)}$ $v=\frac{(341.29+196.210)}{155}$

$v=\frac{537.5}{155}$

v=3.467 m/s

7 0

3 years ago

People often overestimate the clarity of their intentions in their electronic communications because they underestimate the impo

maksim [4K]

Answer:

People often overestimate the clarity of their intentions in their electronic communications because they underestimate the importance of tones of voice in communication.

Explanation:

Many factors intervene in communication together. We use words, we use non-verbal language through our gestures and attitudes, and we also use a certain tone of voice. If we focus on this last aspect, the tonality in the voice is decisive in many ways, which cannot be appreciated through electronic communications. With this we communicate emotions, attitudes and a degree of personal involvement. At the same time, the image we project on others largely depends on the tone of voice we have.

7 0

2 years ago