a coin is dropped from the top of a tall building. determine the coin's (a) velocity and (b) displacement after 1.5 sec.

A sailboat is traveling to the right when a gust of wind causes the boat to accelerate leftward at 2.5m/s2 for 4s. After t

Pavlova-9 [17]

Taking right movement to be positive means leftward movement is negative.
Hence we have a deceleration of
$a = - 2.5 {ms}^{ - 2}$

$t = 4s$
$v = 3.0 {ms}^{ - 1}$
Using this 'suvat' equation
$v = u + at$
we can determine the initial velocity

$3.0= u + -2.5(4)$

$3.0+2.5(4)=u$

$13.0 = u$

Hence the initial velocity is 13.0 meters per seconds

3 0

3 years ago

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You are a scientist trying to develop a technology that can be used to power wrist watches. Which type of electromagnetic wave w

Salsk061 [2.6K]

Microwave because they are powerful and available can be d answer

3 0

3 years ago

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The heat flux that is applied to one face of a plane wall is q″ = 20 W/m2. The opposite face is exposed to air at temperature 30

Blababa [14]

Answer:

400 W/m^2 and 31℃

Explanation:

The output heat flux q"= 20 W/m^2 (geven)

The output heat flux from.the wall to the air by convection

q"conv = h(ts - t∞)

q"conv = 20(50-30) = 400 W/m^2

Therefor, this case is unsteady and the wall temperature changes with time till the energy balance exist.

ENERGY BALANCE

The input energy must be equal to the output energy for steady state condition. If not the state will be unstaidy or transient.

2. Its noticed that the output heat flux is not that the I put heat flux, therefore the wall tempers will be decreased till the output heat flux is reduced to the value of the given input heat flux

T steady = T∞ +q"/h

= 30 + 20/20 = 31℃

3 0

3 years ago

A river flows due east at 1.60 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v

Citrus2011 [14]

Answer:

part (a) $v\ =\ 10.42\ at\ 81.17^o$ towards north east direction.

part (b) s = 46.60 m

Explanation:

Given,

velocity of the river due to east = $v_r\ =\ 1.60\ m/s.$
velocity of the boat due to the north = $v_b\ =\ 10.3\ m/s.$

part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are $90^o$

Let 'v' be the velocity of the boat relative to the shore.

$\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.$

Let $\theta$ be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore. $\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o$

part (b)

Width of the shore = w = 300m

total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river. $\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s$

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river $\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m$

7 0

3 years ago

A 115 g hockey puck sent sliding over ice is stopped in 15.1 m by the frictional force on it from the ice.

Hoochie [10]

Answer:

(a) Ff = 0.128 N

(b μk = 0.1135

Explanation:

kinematic analysis

Because the hockey puck moves with uniformly accelerated movement we apply the following formulas:

vf=v₀+a*t Formula (1)

d= v₀t+ (1/2)*a*t² Formula (2)

Where:

d:displacement in meters (m)

t : time in seconds (s)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s

Calculation of the acceleration of the hockey puck

We apply the Formula (1)

vf=v₀+a*t v₀=5.8 m/s , vf=0

0=5.8+a*t

-5.8 = a*t

a= -5.8/t Equation (1)

We replace a= -5.8/t in the Formula (2)

d= v₀*t+ (1/2)*a*t² , d=15.1 m , v₀=5.8 m/s

15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²

15.1= 5.8*t-2.9*t

15.1= 2.9*t

t = 15.1 / 2.9

t= 5.2 s

We replace t= 5.2 s in the equation (1)

a= -5.8/5.2

a= -1.115 m/s²

(a) Calculation of the frictional force (Ff)

We apply Newton's second law

∑F = m*a Formula (3)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Look at the free body diagram of the hockey puck in the attached graphic

∑Fx = m*a m= 115g * 10⁻³ Kg/g = 0.115g , a= -1.12 m/s²

-Ff = 0.115*(-1.115) We multiply by (-1 ) on both sides of the equation

Ff = 0.128 N

(b) Calculation of the coefficient of friction (μk)

N: Normal Force (N)

W=m*g= 0.115*9.8= 1.127 N : hockey puck Weight

g: acceleration due to gravity =9.8 m/s²

∑Fy = 0

N-W=0

N = W

N = 1.127 N

μk = Ff/N

μk = 0.128/1.127

μk = 0.1135

8 0

3 years ago