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3 years ago

What is number 10 for this?

Physics

1 answer:

dimaraw [331]3 years ago

6 0

Answer:

The wavelength of WFNX’s radio waves with the given speed and frequency is 2.95m.

Given data in the question;

Speed of wave;

Frequency of wave;

wavelength;

To determine the wavelength of the radio wave, we use the expression for the relations between wavelength, frequency and speed.

Where is wavelength, f is frequency and c is the speed.

We substitute our given values into the equation

Therefore, the wavelength of WFNX’s radio waves with the given speed and frequency is 2.95m.

Explanation:

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What is the value of work done on an object when a 70 newton force moves it 9.0 meters in the same direction as the force in jou

BaLLatris [955]

Work done= 70 *9 = 63J

5 0

4 years ago

A wire with a linear mass density of 1.17 g/cm moves at a constant speed on a horizontal surface and the coefficient of kinetic

stira [4]

Answer:

The value is $B = 0.2312 \ T$

The direction is into the surface

Explanation:

From the question we are told that

The mass density is $\mu =\frac{m}{L} = 1.17 \ g/cm =0.117 kg/m$

The coefficient of kinetic friction is $\mu_k = 0.250$

The current the wire carries is $I = 1.24 \ A$

Generally the magnetic force acting on the wire is mathematically represented as

$F_F = F_B$

Here $F_F$ is the frictional force which is mathematically represented as

$F_F = \mu_k * m * g$

While $F_B$ is the magnetic force which is mathematically represented as

$F_B = BILsin(\theta )$

Here $\theta =90^o$ is the angle between the direction of the force and that of the current

$F_B = BIL$

$BIL = \mu_k * m * g$

=> $B = \mu_k * \frac{m}{L} * [\frac{g}{I} ]$

=> $B = 0.25 * 0.117 * [\frac{9.8}{1.24} ]$

=> $B = 0.2312 \ T$

Apply the right hand curling rule , the thumb pointing towards that direction of the current we see that the direction of the magnetic field is into the surface as shown on the first uploaded image

8 0

3 years ago

Light sources send out light waves that_____________.

Crank

Answer:

A. Travel in one direction

Explanation:

light can be reflected and bent but light originally travels in one direction in straight lines

7 0

3 years ago

Read 2 more answers

Tarzan, who weighs 849 N, swings from a cliff at the end of a 18.0 m vine that hangs from a high tree limb and initially makes a

loris [4]

Answer:

Part a)

$T = 342.5 \hat i + 675\hat j$

Part b)

$F_{net} = 342.5\hat i - 174\hat j$

Part c)

$F = 384.2 N$

Part d)

$\theta = 333 degree$

Part e)

$a = 4.4 m/s^2$

Part f)

$\theta = 333 degree$

Explanation:

Part a)

Magnitude of tension force is given as

$T = 757 N$ at 26.9 degree with vertical

$T = 757 sin26.9 \hat i + 757 cos26.9 \hat j$

$T = 342.5 \hat i + 675\hat j$

Part b)

Net force on Tarzen is given as

$F_{net} = T + F_g$

$F_{net} = 342.5 \hat i + 675\hat j - 849 \hat j$

$F_{net} = 342.5\hat i - 174\hat j$

Part c)

magnitude of the force is given as

$F = \sqrt{F_x^2 + F_y^2}$

$F = \sqrt{342.5^2 + 174^2}$

$F = 384.2 N$

Part d)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-174}{342.5}$

$\theta = 333 degree$

Part e)

Magnitude of the acceleration

$a = \frac{F}{m}$

$m = \frac{849}{9.81} = 86.5 kg$

tex]a = \frac{384.2}{86.5}[/tex]

$a = 4.4 m/s^2$

Part f)

Direction of acceleration is same as the direction of the force

$\theta = 333 degree$

6 0

3 years ago

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