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vampirchik [111]

3 years ago

13

Please please please I need someone who knows mathematics and someone who knows biology and someone who knows physics, very, ver

y necessary

1 answer:

jenyasd209 [6]3 years ago

4 0

Answer:

ok then here i am lol

Explanation:

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Mass of object is 50g moves in a circular path of radius 10cm find work done

topjm [15]

Answer:

Work done = 0.3142 Nm

Explanation:

Mass of Object is 50 g

Circular path of radius is 10 cm ⇒ 0.1 m

Work done = Force × Distance = ?

*Distance moved (circular path) ⇒ Circumference of the circular path

2πr = 2 × 3.142 × 0.1 ⇒ 0.6284 m

*Force that is enough to move a 50 g must be equal or more than its weight.

therefore convert 50 grams to newton = 0.5 N

Recall that; work done is force times distance

∴ 0.5 N × 0.6284 m

Work done = 0.3142 Nm

3 0

3 years ago

Which forces are acting on the student and the skateboard in the instant in which they are pushing off the wall? (Select all tha

diamong [38]

E all of the answers above correlate to the student and his skateboard

7 0

2 years ago

Read 2 more answers

Two people are each

docker41 [41]

Answer:

the rope should break

Explanation:

she with equal amounts of pulling are on each side then the rope should slowly start to tare apart and snap/break.

hope this helps you

3 0

2 years ago

Does anyone know 08fortnitebeast?

Oduvanchick [21]

Answer:

nah bruh

Explanation:

i just dont know

5 0

3 years ago

A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught

meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

$y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}$ (Eq. 1)

Where:

$y_{o}$ - Initial height of the softball, measured in meters.

$y$ - Final height of the softball, measured in meters.

$v_{o}$ - Initial velocity of the softball, measured in meters per second.

$t$ - Time, measured in seconds.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $y = y_{o}$ , $t = 3.56\,s$ and $g = -9.807\,\frac{m}{s^{2}}$ , the initial velocity of the softball is:

$v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0$

$3\cdot v_{o} -44.132\,m= 0$

$v_{o} = 14.711\,\frac{m}{s}$

The initial velocity of the softball is 14.711 meters per second.

8 0

3 years ago