What is the volume of an object that has Mass=100g and Density= 745g/mL.

_____ can alter wind flow and create a rain shadow effect. Beaches Lakes Mountains Oceans

Ket [755]

The correct answer is Mountains. Mountains can alter wind flow and create a rain shadow effect. A mountain is a large landform and is high surrounding a limited area. It is generally steeper and higher than a hill.

5 0

3 years ago

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How long will it take for a radioactive isotope with a decay constant of 0.15 (which means a half life of 4.6 days ) to decay to

jek_recluse [69]

Radioactive decay is given by:
N = No x e^(-λt)
We know that N/No has to be 0.05
λ = 0.15
0.05 = e^(-0.15t)
t = ln(0.05)/(-0.15)
t = 19.97 days

6 0

3 years ago

A 40-kg rock is dropped from an elevation of 10 m. What is the velocity of the rock when it is 5-m from the ground?

ivolga24 [154]

Answer:

Explanation:

Given

mass of rock $m=40\ kg$

Elevation of Rock $h=10\ m$

Distance traveled by rock with time

$h=ut+\frac{1}{2}at^2$

where, u=initial velocity

t=time

a=acceleration

here initial velocity is zero

when rock is 5 m from ground then it has traveled a distance of 5 m from top because total height is 10 m

$5=0\times t+\frac{1}{2}(9.8)(t^2)$

$t^2=\frac{10}{9.8}$

$t=1.004\approx 1\ s$

velocity at this time

$v=u+at$

$v=0+9.8\times 1.004$

$v=9.83\ m/s$

6 0

3 years ago

point) A circular swimming pool has a diameter of 12 m. The circular side of the pool is 3 m high, and the depth of the water is

Sergio [31]

Answer:

(a) 86.65 J

(b) 149.65 J

Solution:

As per the question:

Diameter of the pool, d = 12 m

⇒ Radius of the pool, r = 6 m

Height of the pool, H = 3 m

Depth of the pool, D = 2.5 m

Density of water, $\rho_{w} = 1000\ kg//m^{3}$

Acceleration due to gravity, g = $9.8\ m/s^{2}$

Now,

(a) Work done in pumping all the water:

Average height of the pool, h = $\frac{H + D}{2}$

h = $\frac{3 + 2.5}{2} = 2.75\ m$

Volume of water in the pool, V = $\pi r^{2}h = \pi \times 6^{2}\times 2.75 = 311.02\ m^{3}$

Mass of water, $m_{w} = \frac{\rho_{w}}{V}$

$m_{w} = \frac{1000}{311.02} = 3.215\ kg$

Work done is given by the potential energy of the water as:

$W = m_{w}gh = 3.215\times 9.8\times 2.75 = 86.65\ J$

(b) Work done to pump all the water through an outlet of 2 m:

Now,

Height, h = 2.75 + 2 = 4.75

Work done, $W = m_{w}gh = 3.215\times 9.8\times 4.75 = 149.65\ J$

7 0

3 years ago

You create a ramp using two text books and a 0.50m board. Using a timer you determine that a cart can roll down the ramp in 0.55

ahrayia [7]

Answer:

The velocity of the cart at the bottom of the ramp is 1.81m/s, and the acceleration would be 3.30m/s^2.

Explanation:

Assuming the initial velocity to be zero, we can obtain the velocity at the bottom of the ramp using the kinematics equations:

$v=v_0+at\\\\v^2=v_0^2+2ad$

Dividing the second equation by the first one, we obtain:

$v=\frac{v_0^2+2ad}{v_0+at}$

And, since $v_0=0$ , then:

$v=\frac{2ad}{at}\\\\v=\frac{2d}{t}\\\\v=\frac{2(0.50m)}{0.55s}\\\\v=1.81m/s$

It means that the velocity at the bottom of the ramp is 1.81m/s.

We could use this data, plus any of the two initial equations, to determine the acceleration:

$v=v_0+at\\\\\implies a=\frac{v}{t}\\\\a=\frac{1.81m/s}{0.55s}\\\\a=3.30m/s^2$

So the acceleration is 3.30m/s^2.

7 0

3 years ago