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emmasim [6.3K]
3 years ago
15

gYou throw a ball vertically upward so that it leaves the ground with velocity 5.40 m/s.(a) What is its velocity when it reaches

its maximum altitude
Physics
1 answer:
atroni [7]3 years ago
5 0

Answer:

v = 0m/s

Explanation:

Given

u = 5.40m/s --- Initial velocity

Required

Determine the velocity at maximum altitude

When a ball or an object is thrown upward, it reaches a maximum height, then stops momentarily and then descends downwards

The sequence of the event is:

  • Object thrown upward
  • Stops at maximum altitude
  • Descends downward

At the point where the object stops (i.e. the maximum altitude), the velocity is 0

Hence, the required velocity is:

v = 0m/s

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Which number of planets are in oreader towards the sun?
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C. the order from closest to the sun to furthest is Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune so C. would be the opposite heading towards the sun. A good mnemonic to use is MVEMJSUN: My Very Educated Mother Just Served Us Nectarines
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A box of books weighing 290 N is shoved across the floor of an apartment by a force of 400 N exerted downward at an angle of 34.
NARA [144]

Answer: 1.95s

Explanation:

Given

ma = 290 cos 34.9 - fk

fk = 290 cos 34.9 - ma

fn = mg + 400 sin Φ

fn = 290 + 400 sin 34.9

fn = 290 + 228.9

fn = 518.9

fk = fn * uk

uk = 0.57

290 cos 34.9 - ma = 518.9 * 0.57

290 cos 34.9 - ma = 295.8

290 cos 34.9 - 295.8 = ma

ma = -58

m = 290/10 = 29

a = 58/29

a = 2

Using equation of motion

S = ut + .5at²

3.8 = 0 + .5*2*t²

3.8 = t²

t = 1.95s

3 0
3 years ago
Describe in brief the importance of measurement in our daily life?<br>​
Volgvan

Measurement tools make our lives safer and better? And they increase the quality and quantity of life. Arguably, the ability to measure calculating

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6 0
3 years ago
A mail carrier leaves the post office and drives 2.00 km to the north. He then drives in a direction 60.0° south of east for 7.00
Oxana [17]

Answer:

12.97 km

Explanation:

In order to find the resultant displacement, we have to resolve each of the 3 displacements along the x and y direction.

Taking north as positive y direction and east as positive x-direction, we have:

- Displacement 1: 2.00 km to the north

So

A_x = 0\\A_y = +2.00 km

- Displacement 2: 60.0° south of east for 7.00 km

So

B_x=(7.00)(cos (-45^{\circ}))=4.95 km\\B_y = (7.00)(sin (-45^{\circ}))=-4.95 km

- Displacement 3: 9.50 km 35.0° north of east

So

C_x=(9.50)(cos 35^{\circ})=7.78 km\\C_y=(9.50)(sin 35^{\circ})=5.45 km

So the net displacement along the two directions is:

R_x=A_x+B_x+C_x=0+4.95+7.78=12.73 km\\R_y=A_y+B_y+C_y=2.00+(-4.95)+5.45=2.50 km

So, the  distance between the initial and final position is equal to the magnitude of the net displacement:

R=\sqrt{R_x^2+R_y^2}=\sqrt{12.73^2+2.50^2}=12.97 km

7 0
3 years ago
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