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3 months ago

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Ariel dropped a golf ball from her second story window. The ball starts from rest and hits the sidewalk 3.5 s later with a veloc

ity of 14.7 m/s. Find the average acceleration of the golf ball.

1 answer:

inna [3.1K]3 months ago

4 0

Utilizing the acceleration formula,

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A wildlife researcher is tracking a flock of geese. The geese fly 4.0 km due west, then turn toward the north by 40° and fly ano

serg [3582]

Answer:

(a). The distance traveled is 7.06 km towards the west.

(b). Their displacement magnitude is 7.51 km.

Explanation:

The information given states that,

The geese initially move 4.0 km directly west, then alter course to the north at a 40° angle, covering an additional 4.0 km.

Based on the diagram,

(a). To determine the distance

We will apply the distance formula

$AD=AB+BD$

Insert the values into the equation

$D= 4+4.0\cos40^{\circ}$

$D=7.06\ km$

The resultant distance is 7.06 km westward.

(b). We will find the total displacement's magnitude

Using the displacement formula

$AC=\sqrt{(CD)^2+(AD)^2}$

Insert the values into the equation

$AC=\sqrt{(4.0\sin40)^2+(7.06)^2}$

$AC=7.51$

The total displacement magnitude is 7.51 km.

In conclusion, (a). The traveled distance is 7.06 km towards the west.

(b). The magnitude of their total displacement is 7.51 km.

7 0

2 months ago

While dangling a hairdryer by its cord, you observe that the cord is vertical when the hairdryer isoff and, once it is turned on

Yuliya22 [3333]

Answer:

The air exiting from the hairdryer is moving at a speed of 10 m/s.

Explanation:

The thrust generated by the hairdryer enables it to maintain an elevation angled at 5° from vertical; thus, we derive from the force diagram

$(1).\: tan (5^o) = \dfrac{F_t}{Mg}$

by substituting $M =0.420kg$ , $g = 9.8m/s^2$ into the equation and resolving for $F_t$ we find:

$F_t = Mg\:tan(5^o)$

$F_t = (0.420kg)(9.8m/s^2)\:tan(5^o)$

$\boxed{F_t = 0.3601N.}$

This thrust is linked to the speed of air ejection $v$ through the equation

$(2).\: F_t = v\dfrac{dM}{dt}$

where $dM/dt$ signifies the rate of air ejection, which is known to be

$0.06m^3/2s = 0.03m^3/s$

and since $1m^3 = 1.2kg$ ,

$0.03m^3/s \rightarrow 0.036kg/s$

$\dfrac{dM}{dt} = 0.036kg/s,$

by inserting these values into equation (2), we obtain the value of $F_t$ as:

$0.3601N = 0.036v$

resulting in

$v= \dfrac{0.3601}{0.036}$

$\boxed{v =10m/s.}$

which indicates the air velocity discharged from the hairdryer.

6 0

3 months ago

Molybdenum (Mo) has a BCC crystal structure, an atomic radius of 0.1363 nm, and an atomic weight of 95.94 g/mol. Compute and com

Ostrovityanka [3204]

Answer:

The book is titled Solid State or Condensed Matter

Explanation:

8 0

2 months ago

Read 2 more answers

A very tall building has a height H0 on a cool spring day when the temperature is T0. You decide to use the building as a sort o

ValentinkaMS [3465]

Response:

The temperature is $T = \frac{h}{H_O \alpha_{steel} } + T_O$

Justification:

According to the information provided in the question,

The height on a cool spring day measures $H_O$

The temperature for a cool spring day registers as $T_O$

The height variation between a cool spring day and a summer day is accounted as h

The static friction coefficient stands at $\alpha _{steel}$

The equation outlining the linear expansion for the steel building is given as

$h = H_o \alpha_{steel} [T-T_O]$

In this case, T represents the steel's temperature during summer

Rearranging for T, we derive

$T = \frac{h}{H_O \alpha_{steel} } + T_O$

5 0

1 month ago