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Licemer1 [7]
2 years ago
6

A ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m/s at 40

° above the horizontal. How far above or below its original level will the ball strike the opposite wall?
Physics
1 answer:
Vinvika [58]2 years ago
5 0

Answer:

Ball hit the tall building 50 m away below 10.20 m its original level

Explanation:

Horizontal speed = 20 cos40 = 15.32 m/s

Horizontal displacement = 50 m

Horizontal acceleration = 0 m/s²

Substituting in s = ut + 0.5at²

    50 = 15.32 t + 0.5 x 0 x t²

     t = 3.26 s

Now we need to find how much vertical distance ball travels in 3.26 s.

Initial vertical speed  = 20 sin40 = 12.86 m/s

Time = 3.26 s

Vertical acceleration = -9.81 m/s²

Substituting in s = ut + 0.5at²

    s = 12.86 x 3.26 + 0.5 x -9.81 x 3.26²

    s = -10.20 m

So ball hit the tall building 50 m away below 10.20 m its original level

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The 45-g arrow is launched so that it hits and embeds in a 1.40 kg block. The block hangs from strings. After the arrow joins th
worty [1.4K]

Question: How fast was the arrow moving before it joined the block?

Answer:

The arrow was moving at 15.9 m/s.

Explanation:

The law of conservation of energy says that the kinetic energy of the arrow must be converted into the potential energy of the block and arrow after it they join:

\dfrac{1}{2}m_av^2 = (m_b+m_a)\Delta Hg

where m_a is the mass of the arrow, m_b is the mass of the block, \Delta H of the change in height of the block after the collision, and v is the velocity of the arrow before it hit the block.

Solving for the velocity v, we get:

$v = \sqrt{\frac{2(m_b+m_a)\Delta Hg}{m_a} } $

and we put in the numerical values

m_a = 0.045kg,

m_b = 1.40kg,

\Delta H = 0.4m,

g= 9.8m/s^2

and simplify to get:

\boxed{ v= 15.9m/s}

The arrow was moving at 15.9 m/s

6 0
3 years ago
A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f
VashaNatasha [74]

Answer:

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34  

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 =  X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + (\frac{L-y}{2}) ) g sinФ - F2 x (\frac{L}{2}) x gsinФ = 0

plug in the  equations of F1 and F2 into the above equation and after simplification, we get:

(L^{2} - y^{2} ) . w = L^{2} . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

(L^{2} - y^{2} ) . w = L^{2} . 5/9 x w

Hence,

(L^{2} - y^{2} ) = \frac{5L^{2} }{9}

\frac{L^{2} - y^{2}  }{L^{2} } = \frac{5}{9}

Taking L^{2} common and solving for \frac{y}{L}, we will get

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34

8 0
3 years ago
.
Greeley [361]

Answer:

this is impossible for me

Explanation:

7 0
2 years ago
A 0.50-kg bomb is sliding along an icy pond (frictionless surface) with a velocity of 2.0 m/s to the west. The bomb explodes int
nevsk [136]

Answer:

2.667m/s to the north and 3.333 m/s to the west

Explanation:

According to law of momentum conservation, the total momentum should be conserved before and after the explosion.

Before the explosion, the momentum was

0.5*2 = 1 kg m/s to the west

Therefore the total momentum after the explosion should be the same horizontally and vertically.

Vertically speaking, it was 0 before the explosion. After the explosion:

0.2*4 + 0.3v = 0

0.3v = -0.8

v = -0.8/0.3 = -2.667 m/s

So the vertical component of the 0.3kg piece is 2.667m/s to the north

Horizontally speaking, since the 0.2kg-piece doesn't move west or east post-explosion:

0.2*0 + 0.3V = 1

0.3V = 1

V = 1/0.3 = 3.333 m/s

So the horizontal component of the 0.3kg piece is 3.333 m/s to the west

5 0
3 years ago
Read 2 more answers
Please help: I don't know how to do these problems
antiseptic1488 [7]
d =2.55.68m and t = 11.36s
In my opinion
3 0
3 years ago
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