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3 years ago

A block has two strings attached to it on opposite ends. One string has a force of 5 N,

1 answer:

4 0

Unless you have a diagram to include or any other additional info, I'll assume the block is being pulled by two opposing forces along the horizontal surface.

Horizontally, the block is under the influence of

• one rope pulling in one direction with magnitude 15 N,

• the other rope pulling in the opposite direction with mag. 5 N, and

• friction, opposing the direction of the block's motion, with mag. 3 N.

It stands to reason that the block is accelerating in the direction of the larger pulling force.

(A) By Newton's second law, we have

15 N + (-5 N) + (-3 N) = m (1 m/s²)

where m is the mass of the block. Solve for m :

7 N = m (1 m/s²)

m = (7 N) / (1 m/s²)

m = 7 kg

(B) The friction force is proportional to the normal force, so that if f is the mag. of friction and n is the mag. of the normal force, then f = µ n where µ is the coefficient of friction.

The block does not bounce up and down, so its vertical forces are balanced, which means the normal force and the block's weight (mag. w) cancel out:

n + (-w) = 0

n = w

n = m g

where g = 9.8 m/s² is the mag. of the acceleration due to gravity.

n = (7 kg) (9.8 m/s²)

n = 68.6 N

Then

3 N = µ (68.6 N)

µ = (3 N) / (68.6 N)

µ ≈ 0.044

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13.2.2 7+6 - 9(-2)+(-3)

aniked [119]

Answer:

34.27 is the answer

4 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

An object is moving diagonally (down and to the left). You want it to stop moving. In what direction (or

vekshin1

A. Up and to the right
B. Put it in the opposite direction so it won’t go left

3 0

3 years ago

The Sankey diagrams below show the energy transfers in two light bulbs. What is the efficiency of light bulb 1? 1: 80j Input ene

tatyana61 [14]

Answer:

sshjsjsjsjjajaisijajajaiajajajajjajanfbc

7 0

3 years ago

A park ranger driving on a back country road suddenly sees a deer in his headlights 20

olya-2409 [2.1K]

Answer:

17.1

Explanation:

The distance ahead, of the deer when it is sighted by the park ranger, d = 20 m

The initial speed with which the ranger was driving, u = 11.4 m/s

The acceleration rate with which the ranger slows down, a = (-)3.80 m/s² (For a vehicle slowing down, the acceleration is negative)

The distance required for the ranger to come to rest, s = Required

The kinematic equation of motion that can be used to find the distance the ranger's vehicle travels before coming to rest (the distance 's'), is given as follows;

v² = u² + 2·a·s

∴ s = (v² - u²)/(2·a)

Where;

v = The final velocity = 0 m/s (the vehicle comes to rest (stops))

Plugging in the values for 'v', 'u', and 'a', gives;

s = (0² - 11.4²)/(2 × -3.8) = 17.1

The distance the required for the ranger's vehicle to com to rest, s = 17.1 (meters).

6 0

3 years ago