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s2008m [1.1K]
2 years ago
5

A bowling pin is thrown vertically upward such that it rotates as it moves through the air, as shown in the figure. Initially, t

he center of mass of the bowling pin is moving upward with a speed vi of
10 The maximum height of the center of mass of the bowling pin is most nearly

Physics
1 answer:
zheka24 [161]2 years ago
3 0

Answer:

Explanation:

In projectile, the formula for calculating the maximum height reached by the pin is expressed as;

H = u²/2g

u is the speed/velocity = 10m/s

g is the acceleration due to gravity = 9.81m/s²

Substitute

H = 10²/2(9.81)

H = 100/19.62

H = 5.097m

Hence the maximum height of the center of mass of the bowling pin is most nearly 5.0m

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an object having a core temperature of 1700 is removed from a furnace and placed in an environment having a constant temperature
IRINA_888 [86]

Answer: It will be take 2.6 hours

Explanation: Please see the attachments below

4 0
2 years ago
You are piloting a helicopter which is rising vertically at a uniform velocity of 14.70 m/s. When you reach 196.00 m, you see Ba
Cloud [144]

Answer:

The ball reaches Barney  head in  t = 8 \ s

Explanation:

From the question we are told that

 The rise velocity is  v  =  14.70 \  m/s

  The height considered is h =  196 \  m

   The horizontal velocity of the large object is  v_h  =  8.50 \  m/s

   

Generally from kinematic equation  

   s = ut + \frac{1}{2} gt^2

Here s is the distance of the object from Barney head ,

        u is the velocity of the object along the vertical axis which is equal but opposite to the velocity of the helicopter

So  

     u = -14.7 m/s

So

    196  = -14.7 t  + \frac{1}{2} * 9.8 * t^2

=  4.9 t^2 - 14.7t - 196 = 0

Solving the above equation using quadratic formula  

    The value of  t obtained is  t = 8 \ s

6 0
2 years ago
block of mass 0.5kg on a horizontal surface is attached to a horizontal spring of negligible mass and spring constant 50N/m . Th
Alisiya [41]

Answer:

Explanation:

The mass of the block is 0.5kg

m = 0.5kg.

The spring constant is 50N/m

k =50N/m.

When the spring is stretch to 0.3m

e=0.3m

The spring oscillates from -0.3 to 0.3m

Therefore, amplitude is A=0.3m

Magnitude of acceleration and the direction of the force

The angular frequency (ω) is given as

ω = √(k/m)

ω = √(50/0.5)

ω = √100

ω = 10rad/s

The acceleration of a SHM is given as

a = -ω²A

a = -10²×0.3

a = -30m/s²

Since we need the magnitude of the acceleration,

Then, a = 30m/s²

To know the direction of net force let apply newtons second law

ΣFnet = ma

Fnet = 0.5 × -30

Fnet = -15N

Fnet = -15•i N

The net force is directed to the negative direction of the x -axis

8 0
2 years ago
Which method should be used to determine which type of natural event produces the greatest number of sand dunes?
jeka94

Answer:

Stabilizing dunes involves multiple actions. Planting vegetation reduces the impact of wind and water. Wooden sand fences can help retain sand and other material needed for a healthy sand dune ecosystem. Footpaths protect dunes from damage from foot traffic.

Explanation:

5 0
2 years ago
A 1150 kg car is on a 8.70° hill. using x-y axis tilted down the plane, what is the x-component of the weight?
Fed [463]
I assume the x-y axis are tilted such that the x-axis is parallel to the surface of the hill while the y-axis is perpendicular to it.

In this case, the x-component of the weight is given by:
W_x =mg \sin \theta
where
m is the mass of the car
g is the acceleration of gravity
\theta is the angle of the hill

Substituting numbers into the formula, we find
W_x=(1150 kg)(9.81 m/s^2)(\sin 8.70^{\circ})=1706 N
6 0
2 years ago
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