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3 years ago

8

Looking at this skier. If ht travels from point A to point B, then turns around and stops at point C, what is her total displace

ment?
A:180 m
B:320 m
C:100 m
D:40 m

2 answers:

trasher [3.6K]3 years ago

8 0

D. I just answered that question for someone else

never [62]3 years ago

8 0

D. 40 m is the correct answer

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A mass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken from a sample of old glacial

Nataly_w [17]

Answer:

0.092 m

Explanation:

A charged moving particle immersed in a region with magnetic field follows a circular trajectory at constant speed (uniform circular motion), since the magnetic forces acts perpendicular to the direction of motion of the particle.

Since the magnetic force acts as centripetal force, we can write:

$qvB=m\frac{v^2}{r}$

where

q is the charge of the particle

v is its velocity

B is the strength of the magnetic field

m is the mass of the particle

r is the radius of the orbit

Solving the equation for r,

$r=\frac{mv}{qB}$

For the ion of oxygen-16, we have:

$m_A=2.66\cdot 10^{-26}kg$

$q_A = 1.6\cdot 10^{-19}C$ (it is singly charged)

$v_A=2.90\cdot 10^6 m/s$

$B_A=1.30 T$

So the radius of its orbit is

$r_A=\frac{m_A v_A}{q_A B_A}=\frac{(2.66\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.371 m$

For the ion of oxygen-18, we have:

$m_B = \frac{18}{16}m_A = 2.99\cdot 10^{-26}kg$

$q_B = 1.6\cdot 10^{-19}C$ (it is singly charged)

$v_B=2.90\cdot 10^6 m/s$

$B_B=1.30 T$

So the radius of its orbit is

$r_B=\frac{m_B v_B}{q_B B_B}=\frac{(2.99\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.417 m$

After each ion has travelled a semicircle, the separation between the two ions will be twice the difference in their radius, so:

$d=2(r_B-r_A)=2(0.417-0.371)=0.092 m$

3 0

3 years ago

The energy associated with the random motion of molecules or atoms within a substance is

Yuliya22 [10]

Answer:

C

technically B too but youre teachers not that smart so there you go

6 0

3 years ago

A wire carrying a 29.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's

Dmitrij [34]

Answer:

2.59 T

Explanation:

Parameters given:

Current flowing through the wire, I = 29 A

Angle between the magnetic field and wire, θ = 90°

Magnetic force, F = 2.25 N

Length of wire, L = 3 cm = 0.03 m

The magnetic force, F, is related to the magnetic field, B, by the equation below:

F = I * L * B * sinθ

Inputting the given parameters:

2.25 = 29 * 0.03 * B * sin90

2.25 = 0.87 * B

=> B = 2.25/0.87

B = 2.59 T

The magnetic field strength between the poles is 2.59 T

4 0

3 years ago

Elena L [17]

The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:

$\theta =37.01^{\circ}$

We can apply the first Newton's law in x and y-direction.

If we do a free body diagram of the system we will have:

x-direction

All the forces acting in this direction are:

$T_{1}sin(\theta)-T_{2}sin(\theta)=0$ (1)

Where:

T(1) is the tension due to the rope 1
T(2) is the tension due to the rope 2

Here we just conclude that T(1) = T(2)

y-direction

The forces in this direction are:

$T_{1}cos(\theta)+T_{2}cos(\theta)-W=0$ (2)

Here W is the weight of the steel beam.

We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.

Knowing that T(1) = T(2) and W = mg, we have:

$T_{1}cos(\theta)+T_{1}cos(\theta)-m_{steel}g=0$

$2T_{1}cos(\theta)-m_{steel}g=0$

$2T_{1}cos(\theta)=m_{steel}g$

T(1) must be equal to 5479 N, so we have:

$cos(\theta)=\frac{m_{steel}g}{2T_{1}}$

$cos(\theta)=\frac{892*9.81}{2*5479}$

$cos(\theta)=\frac{892*9.81}{2*5479}$

$cos(\theta)=0.80$

Therefore, the maximum angle allowed is θ = 37.01°.

You can learn more about tension here:

brainly.com/question/12797227

I hope it helps you!

8 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s$

$v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s$

<u>Time taken to reach 1180 m is 11.29 seconds</u>

$v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s$

<u>Time the rocket will keep going up after the engines shut off is 13.06 seconds.</u>

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m$

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s$

<u>Time taken by the rocket to fall from maximum height is 20.29 seconds</u>

Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds

5 0

3 years ago