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Brilliant_brown [7]
3 years ago
5

Can the sum of the magnitudes of two vectors ever be equal to the magnitude of the sum of the same two vectors? If no, why not?

If yes, when?
a-No, because of the angle between the two vectors.
b-No, it is impossible for the magnitude of the sum to be equal to the sum of the magnitudes.
c-Yes, if the two vectors are in the same direction.
d-Yes, if the two vectors are perpendicular.
e. Yes, if one of the vectors is zero.
Physics
1 answer:
Vikki [24]3 years ago
4 0

Answer:

c-Yes, if the two vectors are in the same direction.

e. Yes, if one of the vectors is zero.

Explanation:

Lets take

The magnitude of the two vectors are A and B and they are at angle θ

Then the Sum of these two vectors

\bar{R}=\bar{A}+\bar{B}

Resultant R

R=\sqrt{A^2+B^2+2ABcos\theta}

if the angle between these vectors is zero.It means that they are in the same direction.

θ = 0

R=\sqrt{A^2+B^2+2ABcos0}

R=\sqrt{A^2+B^2+2AB}

R=\sqrt{(A+B)^2}

R=A+B

If the one vector is zero vector.

Therefore the answer will be C and e.

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satela [25.4K]

Answer:

a) t1 = v0/a0

b) t2 = v0/a0

c) v0^2/a0

Explanation:

A)

How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0

Vf = 0

Vf = v0 - a0*t

0 = v0 - a0*t

a0*t = v0

t1 = v0/a0

B)

How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.

at this point

U = 0

v0 = u + a0*t

v0 = 0 + a0*t

v0 = a0*t

t2 = v0/a0

C)

The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.

t1 = t2 = t

Distance covered by the train = v0 (2t) = 2v0t

and we know t = v0/a0

so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0

now distance covered by car before coming to full stop

Vf2 = v0^2- 2a0s1

2a0s1 = v0^2

s1 = v0^2 / 2a0

After the full stop;

V0^2 = 2a0s2

s2 = v0^2/2a0

Snet = 2v0^2 /2a0 = v0^2/a0

Now the separation between train and car

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= v0^2/a0

8 0
3 years ago
Numeria
Vinil7 [7]

Answer:

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5 0
3 years ago
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A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t
vladimir1956 [14]

a) \omega=\frac{-mvR}{I+MR^2}

b) v=\frac{-mvR^2}{I+MR^2}

Explanation:

a)

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

L_1=0

Later, after the girl throws the rock, the angular momentum will be:

L_2=(I_M+I_g)\omega +L_r

where:

I is the moment of inertia of the merry-go-round

I_g=MR^2 is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

\omega is the angular speed of the merry-go-round and the girl

L_r=mvR is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

L_1=L_2

So we find:

0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

b)

The linear speed of a body in rotational motion is given by

v=\omega r

where

\omega is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

\omega=\frac{-mvR}{I+MR^2} is the angular speed

r=R is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

v=\omega R=\frac{-mvR^2}{I+MR^2}

5 0
2 years ago
The ratio of carbon-14 to nitrogen-14 is an artifact is 1:3. Given that half-life of carbon-14 is 5730years, how old is the arti
dusya [7]

Answer:

9155 years old

Explanation:

We use the following expression for the decay of a substance:

N = N_0\,\,e^{-k*t}

So we first estimate the value of k knowing that the half-life of the C14 is 5730 years:

N = N_0\,\,e^{-k*t}\\N_0/2=N_0\,\,e^{-k*5730}\\1/2 = e^{-k*5730}\\ln(1/2)=-k*5730\\k= 0.00012

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The horizontal component of the velocity of the ball is calculated by multiplying the speed by the cosine of the given angle. 
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Thus, the horizontal velocity component of the ball is 25.39 m/s. 
8 0
3 years ago
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