why do you think it is difficult to use carbon-14 to determine the age of extremely old objects (over 50,000 years)?

What is a ig?<br><br><br> my friend wont stop saying "what i your ig"

Neporo4naja [7]

Answer:

Its a social media platform

Explanation:

4 0

3 years ago

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

3 years ago

a 10 kilogram steel ball is dropped from the top of a tower 100 meters high the kinetic energy of the ball just before it sttike

EleoNora [17]

Explanation:

If we assume negligible air resistance and heat loss, we can assume that all of the Gravitational potential energy of the ball will turn into Kinetic energy as it falls toward the ground.

Therefore our Kinetic energy = mgh = (10kg)(9.81N/kg)(100m) = 9,810J.

4 0

3 years ago

What type of relationship exists between the length of a wire and the resistance, if all other factors remain the same? A. Resis

ZanzabumX [31]

<h3>Answer</h3>

(A) Resistance is directly related to length.

<h3>Explanation</h3>

Formula for resistance

R = p(length) / A

where R = resistance

p = resistivity(material of wire)

A = cross sectional area

So it can be seen that resistance depends upon 3 factors that are length of wire , resistivity of wire and the cross sectional area of the wire.

If two of the factors, resistivity and cross sectional area, are kept constant then the resistance is directly proportional to the length of wire.

<h3> R ∝ length</h3>

This means that the resistance of the wire increases with the increase in length of the wire and decreases with the decrease of length of the wire.

5 0

3 years ago

Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri

miv72 [106K]

Answer:

T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

T = m_A a

Axis y

N- W_A = 0

Body B

Vertical axis

W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

T = m_A a

W_B –T = M_B a

We solve this system of equations

m_B g = (m_A + m_B) a

a = m_B / (m_A + m_B) g

In this initial case

m_A = M

m_B = M

a = M / (1 + 1) M g

a = ½ g

Let's find the tension

T = m_A a

T = M ½ g

T = ½ M g

Now we change the mass of the second block

m_B = 2M

a = 2M / (1 + 2) M g

a = 2/3 g

We seek tension for this case

T’= m_A a

T’= M 2/3 g

Let's look for the relationship between the tensions of the two cases

T’/ T = 2/3 M g / (½ M g)

T’/ T = 4/3

T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer e

4 0

3 years ago