A kangaroo jumps up with an initial velocity of

A scuba diver measures an increase in pressure of around 10^5 Pa upon descending by 10 m, what is the change in force per square

IceJOKER [234]

Answer:

Explanation:

Given that, .

Pressure around scuba is

P = 10^5 Pa

1 Pa = 1 N/m²

Then

P = 10^5 N/m²

Descending height

h = 10m

Change in force per unit square centimetre

We know that,

Pressure = Force / Area

Then,

Force / Area is the required question we are finding

Then,

Force / Area = 10^5 N / m²

So, let convert the m² to cm²

100cm = 1m

(100cm)² = (1m)²

10⁴cm² = 1m²

Then,

Force / Area = 10^5 N/m² × 1m² / 10⁴cm²

Force / Area = 10 N/cm²

So, the force per unit square centimeters is 10.

7 0

3 years ago

A rock is attached to the left end of a uniform meter stick that has the same mass as the rock. How far from the left end of the

kotykmax [81]

Answer:

M₂ = M then L₂ = L

M₂> M then L₂ = \frac{M}{M_{2}} L

Explanation:

This is a static equilibrium exercise, to solve it we must fix a reference system at the turning point, generally in the center of the rod. By convention counterclockwise turns are considered positive

∑ τ = 0

The mass of the rock is M and placed at a distance, L the mass of the rod M₁, is considered to be placed in its center of mass, which by uniform e is in its geometric center (x = 0) and the triangular mass M₂, with a distance L₂

The triangular shape of the second object determines that its mass can be considered concentrated in its geometric center (median) that tapers with a vertical line if the triangle is equilateral, the most used shape in measurements.

M L + M₁ 0 - m₂ L₂ = 0

M L - m₂ L₂ = 0

L₂ = $\frac{M}{M_{2}}$ L

From this answer we have several possibilities

* if the two masses are equal then L₂ = L

* If the masses are different, with M₂> M then L₂ = \frac{M}{M_{2}} L

6 0

3 years ago

the acceleration of a rocket traveling upward is given by a=(6+.02s)m/s^s, where s is in meters. Determine the rocket's velocity

NikAS [45]

Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

where

a is acceleration

s is distance

v is velocity

Given;

a = 6 + 0.02s

so,

$\int\limits^s_0 {a} \, ds = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds = \int\limits^v_0 {v} \, dv\\ 6s + \frac{0.02s^{2} }{2} = \frac{1}{2} v^{2} \\v = \sqrt{12s + 0.02s^{2} } .....................1 \\\\\\$

Remember that

$v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5\sqrt{2} ) ln \frac{| [s + 300 + \sqrt{(s^{2} + 600s)} ] |}{300} .......2$

substituting s = 2km =2000m, into equation 1

v = 322.5m/s

substituting s = 2000m into equation 2

t = 19.27s

8 0

4 years ago

8892 ml to grams then to mg

andrew11 [14]

Answer:

8892 ml = 8892 gm = 8892000 mg

1 ml = 1 gram

8892 ml = 8892 gram

1 gram or ml = 1000 milligram

8892 ml = 8892 × 1000 = 8892000 milligram

hope this helps

have a good day :)

Explanation:

remember: 1 kilogram = 1000 gram = 1000000 milligram.

Milliliter is expressed same as gram and liter is expressed same as kilogram.

1 meter = 100 cm, 1 kilometer = 1000 meter,

1 cm = 10 millimeter.

7 0

3 years ago

Read 2 more answers

Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate

maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

The length of the track is $l = 1.0 \ km = 1000 \ m$