All categories

3 years ago

The diagram below shows a wave with its wavelength indicated in red.

Physics

2 answers:

Shtirlitz [24]3 years ago

8 0

Answer:

answer is D.it will decrease

AlexFokin [52]3 years ago

7 0

Answer:

D. It will decrease

Explanation: Study island ;)

You might be interested in

Insert

nordsb [41]

A) 320 count/min

B) 40 count/min

C) 80 count/min, 11400 years

Explanation:

The activity of a radioactive sample is the number of decays per second in the sample.

The activity of a sample is therefore directly proportional to the number of nuclei in the sample:

$A\propto N$

where A is the activity and N the number of nuclei.

As a consequence, since the number of nuclei is proportional to the mass of the sample, the activity is also directly proportional to the mass of the sample:

$A\propto m$

where m is the mass of the sample.

In this problem:

- When the mass is $m_1 = 1 g$ , the activity is $A_1=16 count/min$

- When the mass is $m_2=20 g$ , the activity is $A_2$

So we can find A2 by using the rule of three:

$\frac{A_1}{m_1}=\frac{A_2}{m_2}\\A_2=A_1 \frac{m_2}{m_1}=(16)\frac{20}{1}=320 count/min$

The equation describing the activity of a radioactive sample as a function of time is:

$A(t)= A_0 e^{-\lambda t}$ (1)

where

$A_0$ is the initial activity at time t = 0

t is the time

$\lambda$ is the decay constant, which gives the probability of decay

The decay constant can be found using the equation

$\lambda = \frac{ln2}{t_{1/2}}$

where $t_{1/2}$ is the half-life, which is the amount of time it takes for the radioactive sample to halve its activity.

In this problem, carbon-14 has half-life of

$t_{1/2}=5700 y$

So its decay constant is

$\lambda=\frac{ln2}{5700}=1.22\cdot 10^{-4} y^{-1}$

We also know that the tree died

t = 17,100 years ago

and that the initial activity was

$A_0 = 320 count/min$ (value calculated in part A, corresponding to a mass of 20 g)

So, substituting into eq(1), we find the new activity:

$A(17,100) = (320)e^{-(1.22\cdot 10^{-4})(17,100)}=40 count/min$

We know that a sample of living wood has an activity of

$A=16 count/min$ per 1 g of mass.

Here we have 5 g of mass, therefore the activity of the sample when it was living was:

$A_0 = A\cdot 5 = (16)(5)=80 count/min$

Moreover, here we have a sample of 5 g, with current activity of $A=20 count/min$ : it means that its activity per gram of mass is

$A'=\frac{20}{5}=4 count/min$

We know that the activity halves after every half-life: Here the activity has became 1/4 of the original value, this means that 2 half-lives have passed, because:

- After 1 half-life, the activity drops from 16 count/min to 8 count/min

- After 2 half-lives, the activity dropd to 4 count/min

So the age of the wood is equal to 2 half-lives, which is:

$t=2t_{1/2}=2(5700)=11,400 y$

3 0

4 years ago

Two round rods, one steel andthe other copper, are joined end to end. Each rod is 0.750 mlong and 1.50 cmin diameter. The combin

nikitadnepr [17]

Answer: a) Strain on Steel rod = 0.0001078

b) elongation on the steel rod = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain on Copper Rod = 0.000189

Explanation: a) To obtain the strain of the steel rod, we invoke Hooke's law which states that, provided the elastic limit of A material isn't exceeded, the stress it undergoes is directly proportional to its strain.

(Stress, σ) ∝ (Strain, ε)

The constant of proportionality is called Young's modulus, E.

σ = Eε

For steel, Younger Modulus as obtained from literature = 210GPa.

Strain = Stress/Young's Modulus

Stress = (Force or Load applied)/ Cross sectional Area.

Force applied For the steel = 4000N

Cross sectional Area = (π(D^2))/4

D = 1.50cm = 0.015m

A = 0.0001767 m2

σ = 4000/0.0001767 = 22637238.257 N/m2

ε = σ/E = 22637238.257/(210 × (10^9)) = 0.0001078.

b) To solve for elongation.

Strain, ε = (elongation, dl)/(original length, lo)

Elongation, dl = strain × original length

dl = 0.0001078 × 0.75 = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain in Copper

ε = σ/E; σ = 22637238.257 N/m2

Young's modulus of Copper, from literature, = 120GPa

ε = 22637238.257/(120 × (10^9)) = 0.000189

4 0

3 years ago

When an object of mass m1 is hung on a vertical spring and set into vertical simple harmonic motion, its frequency is 12Hz. When

anzhelika [568]

Answer:

Explanation:

Given

When $m_1$ mass is attached to the spring its frequency is $f_1=12\ Hz$

when another mass $m_2$ is attached to $m_1$ , frequency changes to $f_2=4\ Hz$

frequency of spring mass system is given by

$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

for $m_1$

$f_1=\frac{1}{2\pi }\sqrt{\frac{k}{m_1}}$

$12=\frac{1}{2\pi }\sqrt{\frac{k}{m_1}}-----1$

for $m_1$ and $m_2$

$f_2=\frac{1}{2\pi }\sqrt{\frac{k}{m_1+m_2}}$

$4=\frac{1}{2\pi }\sqrt{\frac{k}{m_1+m_2}}----2$

divide 1 and 2

$3=\sqrt{\frac{m_1+m_2}{m_1}}$

squaring

$9=1+\frac{m_2}{m_1}$

$\frac{m_2}{m_1}=8$

8 0

3 years ago

A 4.0×1010kg asteroid is heading directly toward the center of the earth at a steady 16 km/s. To save the planet, astronauts str

blondinia [14]

Answer:

a. t = 69.4 hr = 2.89 days

b. theta = 91.67*10^-3 degrees

c. deflection_angle = 0.134 degrees

Explanation:

a).

The asteroid impacts the earth in t

t = x/v = (4.0*10^6 km)/(16 km/sec)

t = 2.5 * 10^5 sec

t = 69.4 hr = 2.89 days

b).

tan(theta) = 6400 km/(4.0*10^6 km)

tan(theta) = 1.6*10^-3

theta = arctan(1.6*10^-3)

theta = 1.6*10^-3 radians (for small angles, tan(theta) ~= theta)

theta = 91.67*10^-3 degrees

c).

v_minimum = 6400 km/(2.5 * 10^5 sec)

v_minimum = 25.6 m/s

Using F = m*a, we can calculate the acceleration of the asteroid due to the rocket's thrust:

5.0*10^9 N = 4.0*10^10 kg * a

a = (5.0*10^9 N)/(4.0*10^10 kg)

a = 0.125 m/s^2

The transverse velocity after 300 seconds of this acceleration is:

v_transverse = a*t = 0.125 m/s^2 * 300 s

v_transverse = 37.5 m/s = 37.5*10^-3 km/s

tan(deflection_angle) = v_transverse/(20 km/s)

tan(deflection_angle) = (37.5*10^-3 km/s)/(16 km/s) = 2.34^-3

deflection_angle = arctan(2.34*10^-3)

deflection_angle = 2.34*10^-3 radians = 0.134 degrees

v_transverse/(16 km/s) > (6400km)/(5.0*10^6 km)

(note that the right hand side if this inequality is tan(theta) calculated above)

v_transverse > 23.704 m/

8 0

3 years ago

The wheelchair starts from rest. It accelerates at a constant rate until it has a speed of 1.5 m/s. The wheelchair travels a dis

anastassius [24]

<h2>Answer:-</h2>

$\pink{\bigstar}$ The acceleration of the wheelchair is $\large\leadsto\boxed{\tt\purple{0.5625 \: m/s^2}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

<h3>• Given:-</h3>

Initial velocity of the wheelchair = 0 m/s

Final velocity of the wheelchair = 1.5 m/s

Distance covered by the wheelchair = 2 m

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Acceleration of the wheelchair = ?

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

<h3>• Solution:-</h3>

We know,

• Third Equation of Motion:-

$\pink{\bigstar}$ $\large\underline{\boxed{\bf\green{v^2 - u^2 = 2as}}}$

where,

u = Initial velocity

v = Final velocity

a = Acceleration

s = Distance covered

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Substituting the values in the Formula:-

➪ $\sf (1.5)^2 - (0)^2 = 2 \times a \times 2$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➪ $\sf 2.25 - 0 = 4 \times a$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➪ $\sf 2.25 = 4 \times a$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

➪ $\sf a = \dfrac{2.25}{4}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

★ $\large{\bold\red{a = 0.5625 \: m/s^2}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Therefore, the acceleration of the wheelchair is 0.5625 m/s².

7 0

3 years ago