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3 years ago

How do i do this???/

Physics

2 answers:

ikadub [295]3 years ago

8 0

If you take the Western direction as the positive direction and the Eastern direction as negative one, then,

Distance traveled to the West = +15 m

Distance traveled to the east = (-2 m) + (-7 m ) = -9 m

Total Displacement = 15 m + (-9) m

Total displacement = 6 m

Since the total displacement is positive, the displacement will be in the Western direction.

So, option (D) is correct.

Gekata [30.6K]3 years ago

5 0

Start at the big tree, and take the instructions one at a time: 1). Walk 15 west. 2). From there, turn around and walk 2 east. 3). From there, walk 7 more east. Where are you from the big tree now ? Aren't you 6 west of it ? I mean, let's see some common sense here !

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Can you help me with a science test

Semenov [28]

Answer:

yes just give me questions and put me brainliest

Explanation:

8 0

3 years ago

Read 2 more answers

In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their component

JulijaS [17]

Answer:

The calculated vectors are:

$\vec{A}-\vec{B}=(3,-5,-4)$

$\vec{B}-\vec{C}=(-5,4,0)$

$-\vec{A}+\vec{B}-\vec{C}=(-6,4,3)$

$3\vec{A}-2\vec{C}=(-3,-2,-11)$

Explanation:

To operate with vectors, you sum or rest component to component. To multiply scalars with vectors, you distribute the scalar with each component of the vector. These are the following rules you must apply in these cases:

$\vec{V}+\vec{W}=(V_1,V_2,V_3)+(W_1,W_2,W_3)=(V_1+W_1,V_2+W_2,V_3+W_3)$ (1)

$\vec{V}-\vec{W}=(V_1,V_2,V_3)-(W_1,W_2,W_3)=(V_1-W_1,V_2-W_2,V_3-W_3)$ (2)

$\alpha\cdot\vec{V}=\alpha\cdot(V_1,V_2,V_3)=(\alpha\cdot V_1,\alpha\cdot V_2,\alpha\cdot V_3)$ (3)

The operations in these cases are:

$\vec{A}-\vec{B}=(1,0, -3)-(-2,5,1)=(3,-5,-4)$

$\vec{B}-\vec{C}=(-2,5,1)-(3,1,1)=(-5,4,0)$

$-\vec{A}+\vec{B}-\vec{C}=-(1,0, -3)+(-2,5,1)-(3,1,1)=(-6,4,3)$

$3\vec{A}-2\vec{C}=3(1,0, -3)-2(3,1,1)=(3,0, -9)-(6,2,2)=(-3,-2,-11)$

3 0

3 years ago

Read 2 more answers

A charged particle is moving in a uniform magnetic field at a speed of 8.2Ã10^3 m/s in a direction 87Â° from the direction of th

77julia77 [94]

Answer:

Magnetic field, B = 0.042 T

Explanation:

It is given that,

Speed of charged particle, $v=8.2\times 10^3\ m/s$

Angle between velocity and the magnetic field, $\theta=87$

Charge, $q=5.7\ \mu C=5.7\times 10^{-6}\ C$

Magnetic force, F = 0.002 N

The magnetic force is given by :

$F=qvB\ sin\theta$

B is the magnetic field

$B=\dfrac{F}{qv\ sin\theta}$

$B=\dfrac{0.002}{5.7\times 10^{-6}\times 8.2\times 10^3\times sin(87)}$

B = 0.042 T

So, the strength of the magnetic field is 0.042 Tesla. Hence, this is the required solution.

5 0

4 years ago

What is the tendency of a material to oppose the flow of charge?

elena55 [62]

I think the answer is Resistance. If the link i posted int he comment is right.

8 0

4 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