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3 years ago

How can the density of irregularly shaped objects can be calculated?

1 answer:

6 0

1.Use the balance to find the mass of the object. Record the value on the "Density Data Chart."
2.Pour water into a graduated cylinder up to an easily-read value, such as 50 milliliters and record the number.
3.Drop the object into the cylinder and record the new value in millimeters.
4.The difference between the two numbers is the object's volume. Remember that 1 milliliter is equal to 1 cubic centimeter. Record the volume on the data chart.
5.Compute the density of the object by dividing the mass value by the volume value. Record the density on the data chart.

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Which organism in the food web above is sometimes a first-level consumer and sometimes a second-level consumer?Explain

scoray [572]

The answer is: Mouse/Herbivore.

First organism is always a producer (plant) such as grass.
The 2nd organism is the first level or primary consumer. Ex. Mouse - It eats the producer, so it is a herbivore.

4 0

3 years ago

A planet has a mass of 5.68 x 1026 kg and a radius of 6.03 x 107 m. What is the weight of a 65.0 kg person on the surface of thi

Paul [167]

To solve this problem, we must know the gravitational force of the planet. The equation would be,

$F=G \frac{ m_{1} m_{2} }{ r^{2} }$

This would calculate the force between two objects with masses m1 and m2 and the gravitational constant, G, is 6.67 x 10^-11 m3 s-2 kg-1 and with r as the distance between the objects.

Thus,

F = (6.67 x 10^-11 m3 s-2 kg-1) * (5.68 x 10^26 kg) * (65 kg) * ((1/6.03 x 10^7 m)^2)

F = 678 kg/s^2 or 678 N

Answer is letter B.

4 0

3 years ago

The national high magnetic field laboratory holds the world record for creating the strongest magnetic field. for brief periods

max2010maxim [7]

Newton’s 2nd law states that Force is equal to the product of mass (m) and acceleration (a):

F = m a ---> 1

While in magnetic forces, force can also be expressed as:

F = q v B ---> 2

where,

q = total charge

v = velocity = 45 cm / s = 0.45 m / s

B = the magnetic field = 85 T

First we solve for the total charge, q:

q = 3.8 × 10^-23 g (1 mol / 23 g) (6.022 × 10^23 electrons / mol) (1.602 × 10^-19 C / electron)

q = 1.594 × 10^-19 C

We equate equations 1 and 2 then solve for acceleration a:

m a = q v B

a = q v B / m

a = [1.594 × 10^-19 C * 0.45 m / s * 85 T] / 3.8 × 10-26 kg

a = 160,437,862.2 m/s^2

Therefore the maximum acceleration of Na ions is about 160 × 10^6 m/s^2.

5 0

3 years ago

Read 2 more answers

a stone is thrown by a person from the top of the building, which is 200m tall. at the same time, another stone is thrown with v

solniwko [45]

Answer:

The time after which the two stones meet is tₓ = 4 s

Explanation:

Given data,

The height of the building, h = 200 m

The velocity of the stone thrown from foot of the building, U = 50 m/s

Using the II equation of motion

S = ut + ½ gt²

Let tₓ be the time where the two stones meet and x be the distance covered from the top of the building

The equation for the stone dropped from top of the building becomes

x = 0 + ½ gtₓ²

The equation for the stone thrown from the base becomes

S - x = U tₓ - ½ gtₓ² (∵ the motion of the stone is in opposite direction)

Adding these two equations,

x + (S - x) = U tₓ

S = U tₓ

200 = 50 tₓ

∴ tₓ = 4 s

Hence, the time after which the two stones meet is tₓ = 4 s

6 0

3 years ago

A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

5 0

2 years ago