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Burka [1]
3 years ago
7

Why does a stop sign appear red?

Physics
1 answer:
dolphi86 [110]3 years ago
8 0

Answer:

because it’s suppose to be red like a stop light.

Explanation:

So it tells you to stop

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Can anyone solve these for my by using unit vectors? Can you also please show your work
Oxana [17]

4. The Coyote has an initial position vector of \vec r_0=(15.5\,\mathrm m)\,\vec\jmath.

4a. The Coyote has an initial velocity vector of \vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath. His position at time t is given by the vector

\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2

where \vec a is the Coyote's acceleration vector at time t. He experiences acceleration only in the downward direction because of gravity, and in particular \vec a=-g\,\vec\jmath where g=9.80\,\frac{\mathrm m}{\mathrm s^2}. Splitting up the position vector into components, we have \vec r=r_x\,\vec\imath+r_y\,\vec\jmath with

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t

r_y=15.5\,\mathrm m-\dfrac g2t^2

The Coyote hits the ground when r_y=0:

15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s

4b. Here we evaluate r_x at the time found in (4a).

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m

5. The shell has initial position vector \vec r_0=(1.52\,\mathrm m)\,\vec\jmath, and we're told that after some time the bullet (now separated from the shell) has a position of \vec r=(3500\,\mathrm m)\,\vec\imath.

5a. The vertical component of the shell's position vector is

r_y=1.52\,\mathrm m-\dfrac g2t^2

We find the shell hits the ground at

1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s

5b. The horizontal component of the bullet's position vector is

r_x=v_0t

where v_0 is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for v_0:

3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}

5 0
3 years ago
‼️‼️ Please help, urgent ‼️‼️ (check photo)
Alex787 [66]

Answer: The force constant k is 10600 kg/s^2

Step by step:

Use the law of energy conservation. When the elevator hits the spring, it has a certain kinetic and a potential energy. When the elevator reaches the point of still stand the kinetic and potential energies have been transformed to work performed by the elevator in the form of friction (brake clamp) and loading the spring.

Let us define the vertical height axis as having two points: h=2m at the point of elevator hitting the spring, and h=0m at the point of stopping.

The total energy at the point h=2m is:

E_{tot}=E_{kin}+E_{pot}\\E_{tot}= \frac{1}{2}mv^2+mg\Delta h = \frac{1}{2}2000 kg 4^2\frac{m^2}{s^2}+2000kg\, 9.8\frac{m}{s^2}2m=55200\,kg\frac{m^2}{s^2}

The total energy at the point h=0m is:

E_{tot}=E_{kin}+E_{pot}+Work=0+0+ Work\\E_{tot} =F_{friction}\Delta h+\frac{1}{2}k (\Delta h)^2=17000N\cdot 2m+\frac{1}{2}k\cdot 2^2 m^2

The two Energy values are to be equal (by law of energy conservation), which allows us to determine the only unknown, namely the force constant k:

17000N\cdot 2m+\frac{1}{2}k\cdot 2^2 m^2 = 55200 \,kg\frac{m^2}{s^2}\\k = \frac{55200-34000}{2}\,\frac{kg}{s^2}=10600\frac{kg}{s^2}

5 0
3 years ago
A track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis (Fig. 11-48).A toy train
masya89 [10]

Answer:

0.166 rad/s

Explanation:

See attachment for calculations

5 0
3 years ago
Read 2 more answers
Twins Jody and Taylor are rearranging the furniture in their bedroom and want to move a dresser across the room. The dresser has
xeze [42]

Answer:

yes, They will be able to move the dresser.

Explanation:

sliding force 90N

55N + 38N = 93N

therefore, yes the twins can move the dresser

7 0
3 years ago
A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 27 ft/s. Its height
Ahat [919]

Answer:

V_{3.01}=-93.2m/s

V_{3.005}=-93.1m/s

V_{3.002}=-93.04m/s

V_{3.001}=-93.02m/s

V_{3}=-93m/s

Explanation:

To calculate average velocity we need the position for both instants t0 and t1.

Now we will proceed to calculate all the positions we need:

Y_{3}=-99m/s

Y_{3.01}=-99.932m/s

Y_{3.005}=-99.4655m/s

Y_{3.002}=-99.18608m/s

Y_{3.001}=-99.09302m/s

Replacing these values into the formula for average velocity:

V_{3-3.01}=\frac{Y_{3.01}-Y_{3}}{3.01-3}=-93.2m/s

V_{3-3.005}=\frac{Y_{3.005}-Y_{3}}{3.005-3}=-93.1m/s

V_{3-3.002}=\frac{Y_{3.002}-Y_{3}}{3.005-3}=-93.04m/s

V_{3-3.001}=\frac{Y_{3.001}-Y_{3}}{3.001-3}=-93.02m/s

To know the actual velocity, we derive the position and we get:

V=27-40t = -93m/s

5 0
3 years ago
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