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lina2011 [118]
3 years ago
12

Can you please help me???

Physics
1 answer:
zhenek [66]3 years ago
8 0
Question 12 would be: C

question 13 would be: B
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Please help me with this please <br><br> I’ll mark you Brainly
ivolga24 [154]
I think it is option (C).

If the answer is helpful then mark me as brainly.
4 0
2 years ago
An open 1-m-diameter tank contains water at a depth of 0.5 m when at rest. As the tank is rotated about its vertical axis the ce
Morgarella [4.7K]

Answer:

Angular velocity (w) = 8.86 rad/s

Explanation:

Angular velocity (w) = \sqrt{} 4ghi/R^{2}

g= 9.81 m/s

R= 0.5

hi (initial depth) = 0.5m

Hence= \sqrt4* 9.81* 0.5/0.5^{2}  = 8.86 rad/s

3 0
3 years ago
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
A car ends up colliding with a tree. Compare the force the car exerts on the tree with the force the tree exerts on the car.
Fynjy0 [20]

Answer: The force will exert into the tree causing it to absorb the compact hit by the car which means the car gets damaged severely but the tree maybe only has a tiny crack. Mark me as Brainliest! :)

Explanation:

4 0
3 years ago
A line _____ on a typical Speed vs. Time graph means an object is experiencing a constant acceleration.
Savatey [412]

Answer:

The answer should be C. slanted upward to the right.

Hope this helps. :-)

7 0
3 years ago
Read 2 more answers
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