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goldfiish [28.3K]

3 years ago

5

Jamie decides to drop an egg that has a mass of 345 g off the top of a 8.2 m building. How fast will it be falling right before

it hits the ground?
Please show work.

1 answer:

zlopas [31]3 years ago

3 0

Answer:

13 m/s

Explanation:

Given:

Δy = 8.2 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (9.8 m/s²) (8.2 m)

v = 12.7 m/s

Rounded to two significant figures, the speed is 13 m/s.

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If you and a friend are standing side-by-side watching a soccer game, would you both view the motion from the same reference fra

Radda [10]

Answer:

the correct is C

Explanation:

The concept of a frame of reference is of crucial importance in physics, because it is the system from which measurements are made. Therefore, the relationships between the different reference frames must be clear so that the measurements made can be compared correctly.

In this case, the first observed sees the movement of the ball, suppose it moves a distance r, the second observed is next to me, separated by a distance x, therefore a frame of reference located in the movement of the ball. ball r '.

Consequently, the measurement carried out is related by

r = r’ + x

where the bold letters indicate wind blowers.

With these explanations we review the different answers, the correct one is C

3 0

3 years ago

A plane startling from rest accelerates at 3 m/s Square. Calculate the increase in velocity after 25 sec

Tatiana [17]

Answer:

75 m/s

Explanation:

We can apply motion equations here

V = U + a * t

V = velocity @ t time

U = initial velocity

a = acceleration

t = time taken

V = U + a * t

V = 0+ 3 * 25

V = 75 m/s

After 25 seconds , subjected to the given acceleration velocity is increased from 0 to 75 m/s

3 0

3 years ago

Rachel has been reading her physics book. She takes her weighing scales into an elevator and stands on them. If her normal weigh

GrogVix [38]

Answer:

345 N

Explanation:

Given:

Normal weight of Rachel (mg) = 690 N

Case 1: Upward motion of elevator

Given:

Acceleration of elevator (a) = 0.25 g

The scale reading is given by the normal force acting on Rachel. Let N₁ be the normal force.

So, the net force acting on Rachel is given as:

$F_{net}=N_1-mg=N_1-690$

Now, from Newton's second law:

$F_{net}=ma\\\\N_1-690=m\times 0.25g\\\\N_1-690=0.25\times (mg)\\\\N_1-690=0.25\times 690\\\\N_1=690+172.5=862.5\ N------(1)$

Case 2: Downward motion of elevator

Given:

Acceleration of elevator (a) = 0.25 g

The scale reading is given by the normal force acting on Rachel. Let N₂ be the normal force.

So, the net force acting on Rachel is given as:

$F_{net}=mg-N_2=690-N_2$

Now, from Newton's second law:

$F_{net}=ma\\\\690-N_2=m\times 0.25g\\\\690-N_2=0.25\times (mg)\\\\690-N_2=0.25\times 690\\\\N_2=690-172.5=517.5\ N------(2)$

Now, the difference in the scale reading is obtained by subtracting equation (2) from equation (1). This gives,

$Difference=N_1-N_2=862.5-517.5=345\ N$

Therefore, the difference between the up and down scale readings is 345 N.

4 0

3 years ago

Which graph accurately shows the relationship between kinetic energy and mass as mass increases

gtnhenbr [62]

Answer:

c

Explanation:

energy doesnt affect to mass of a object

6 0

3 years ago

Instruments in an airplane which is in level flight indicate that the velocity relative to the air (airspeed) is 180.00 km/h and

snow_lady [41]

Answer:

Explanation:

From the given information:

The coordinate axis is situated in the east and north direction.

So, the north will be the y-axis and the east will be the x-axis

Similarly, the velocity of the plane in regard to the air in the coordinate system will be $v_{P/A} = v( cos \theta \ i + sin \theta \ j)$

where:

$v_{P/A}$ = velocity of the plane in regard to the air

v = velocity

θ = angle of inclination of the plane with respect to the horizontal

replacing v = 180 km/ and θ = 20° in above equation, then:

The velocity of the airplane in the coordinate system as:

$v_{P} = v_o( cos \phi \ i + sin \phi \ j)$

where;

$v_p$ = velocity of the airplane

$v_o$ = velocity

∅ = angle of inclination with regard to the base axis;

Then; replacing $v_o$ = 150 km/h and ∅ = 30°

Therefore, the velocity of the plane in the system is :

$v_p = v_A + v_{P/A}$

$v_A= v_P -v_{P/A}$ --- (1)

$v_A=$ ( 150 cos 30° - 180 cos 20°)i + ( 150 sin 30° - 180sin 20°)j

$v_A=$ (-39.24 km/h)i + (13.44 km/h) j

The magnitude is:

$v_A= (-39.24 km/h)i + (13.44 km/h) j$

$|v_A|^2 = \sqrt{ (-39.24 km/h)^2+ (13.44 km/h)^2}$

$v_A$ = 41.48 km/h

The airplane is moving at an angle of the inverse tangent to the abscissa and ordinate.

The angle of motion is:

tan θ = 39.24/13.44

tan θ = 2.9

θ = $tan ^{-1} (2.9)$

θ = 70.97°

The angle of motion is 70.97° from west of north with a velocity of 41.48 km/h.

5 0

3 years ago