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777dan777 [17]
3 years ago
6

During maneuvers preceding the Battle of Jutland, the British battle cruiser Lion moved as follows (in nautical miles): 1.2 mile

s north, 6.1 miles 38 degrees east of south, 4.0 miles at 89 degrees east of north, and 6.5 miles at 31 degrees east of north. Find the distance between starting and ending positions. (Ignore the earth’s curvature.)
Physics
1 answer:
lana66690 [7]3 years ago
5 0

Answer:

total distance is 11.3 miles

Explanation:

Given data:

distance to north = 1.2 miles

distance to south at 38 degree east of south = 6.1 miles  

distance to east of north at angle 89 degree = 4 miles  

distance to east of north at angle 31 degree =  6.5 miles

considering each displacement as vector so we have

\begin{pmatrix}0.0\\  1.2\end {pmatrix}   + \begin {pmatrix}3.8\\  -4.8\end{pmatrix} +\begin{pmatrix}4.0\\  0.1\end {pmatrix}+\begin{pmatrix}3.3\\  5.6\end{pmatrix} = \begin{pmatrix}11.1\\  2.1\end{pmatrix}

total distance is \sqrt{11.1^2 + 2.1^2} \approx  11.3

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A 100-kg tackler moving at a speed of 2.6 m/s meets head-on (and holds on to) an 92-kg halfback moving at a speed of 5.0 m/s. Pa
DIA [1.3K]

Given that,

Mass of trackler, m₁ = 100 kg

Speed of trackler, u₁ = 2.6 m/s

Mass of halfback, m₂ = 92 kg

Speed of halfback, u₂ = -5 m/s (direction is opposite)

To find,

Mutual speed immediately after the collision.

Solution,

The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{100\times 2.6+92\times (-5)}{(100+92)}\\\\V=-1.04\ m/s

So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.

3 0
3 years ago
A wheel starts from rest and has an angular acceleration that is given by α (t) = (6.0 rad/s4)t2. After it has turned through 10
marissa [1.9K]

Answer:

75 rad/s

Explanation:

The angular acceleration is the time rate of change of angular velocity. It is given by the formula:

α(t) = d/dt[ω(t)]

Hence: ω(t) = ∫a(t) dt

Also, angular velocity is the time rate of change of displacement. It is given by:

ω(t) = d/dt[θ(t)]

θ(t) = ∫w(t) dt

θ(t) = ∫∫α(t) dtdt

Given that: α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴. Hence:

θ(t) = ∫∫α(t) dtdt

θ(t) = ∫∫6t² dtdt =∫[∫6t² dt]dt

θ(t) = ∫[2t³]dt = t⁴/2 rad

θ(t) = t⁴/2 rad

At θ(t) = 10 rev = (10 *  2π) rad = 20π rad, we can find t:

20π = t⁴/2

40π = t⁴

t = ⁴√40π

t = 3.348 s

ω(t) = ∫α(t) dt = ∫6t² dt = 2t³

ω(t) = 2t³

ω(3.348) = 2(3.348)³ = 75 rad/s

7 0
3 years ago
Earths rotation causes day and night cycle. It also causes-
prohojiy [21]
The earth axis to tilt and orbits the  sun 
3 0
3 years ago
A square piece of tin has 12 inches on a side. An open box is formed by cutting out equal square pieces at the corners and bendi
Citrus2011 [14]

Answer:

Explanation:

Given a square Piece whose side is 12 inches

Now square pieces are cut from each corner to make it a open box

Suppose x is the length of square piece at each corner

then

base square has a length of 12-2x

Dimension of new box is (12-2x)\times (12-2x)\times x

Volume V=(12-2x)\times (12-2x)\times x

V=\left ( 12-2x\right )^2\cdot x

For maximum volume differentiate with respect to x we get

\Rightarrow\frac{\mathrm{d} V}{\mathrm{d} x}=2\times \left ( 12-2x\right )\times \left ( -2\right )\cdot x+\left ( 12-2x\right )^2=0

we get x=6 and 4 but at x=6 volume becomes zero therefore x=4 is valid

V=\left ( 12-2\cdot 4\right )^2\cdot 4

V=4^3

V=64\ in.^3

6 0
2 years ago
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