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2 years ago

27. A 0.150 kg baseball is thrown upward with an initial

Physics

1 answer:

Tema [17]2 years ago

4 0

Answer:

a. $1.47N$

b. $1.47N$

Explanation:

Given:

Mass of the baseball = $0.150kg$

The initial velocity of the baseball = 20m/s

The force acting on the baseball is only due to gravity and the air resistance is negligible. The baseball acceleration is constant due to gravity.

According to Newton's second law:

$F_{w} =mg$ ----------(1)

where mass of the baseball $m = 0.150kgAnd acceleration due to gravity [tex]g = 9.8m/s^{2}$

Put all the value in equation 1.

$F_{w} =0.150\times -9.8$

$F_{w} =-1.47N$

Therefore, the force on the ball when it reaches half of its maximum height is $1.47N$ .

Similarly, the force acting on the ball when its peak $1.47N$ .

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A person hits a 45-g golf ball. The ball comes down on a tree root and bounces

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Answer:

Maximum height, h = 10 m

Explanation:

It is given that,

Mass of golf ball, m = 45 g = 0.045 kg

The ball comes down on a tree root and bounces straight up with an initial speed of 14.0 m/s.

We need to find the height the ball will rise after the bounce. It is based on the conservation of energy such that,

$\dfrac{1}{2}mv^2=mgh$

h is maximum height raised by the ball

$h=\dfrac{v^2}{2g}\\\\h=\dfrac{(14)^2}{2\times 9.8}\\\\h=10\ m$

So, the ball will raised to a height of 10 meters.

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3 years ago

Blocks A and B are identical metal blocks. Initially block A is neutral, and block B has a net charge of 8.7 nC. Using insulatin

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Explanation:

(a) Since, it is given that the blocks are identical so distribution of charge will be uniform on both the blocks.

Hence, final charge on block A will be calculated as follows.

Charge on block A = $\frac{(8.7 + 0 nC}{2}$

= 4.35 nC

Therefore, final charge on the block A is 4.35 nC.

(b) As it is given that the positive charge is coming on block A . This means that movement of electrons will be from A to B.

Thus, we can conclude that while the blocks were in contact with each other then electrons will flow from A to B.

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2 years ago

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Suppose you fill two rubber balloons with air, suspend both of them from the same point, and let them hang down on strings of eq

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The force on each balloon is 2×10^−3 N.

Consider two balloons of diameter 0.200m each with a mass of 1.00g hanging apart with 0.0500m separation on the ends of string making angles of 10.0° with the vertical.

$\sum F_{y} = Tcos10\textdegree - mg = 0\\\\T = \frac{mg}{cos10\textdegree } \\\\\sum F_{y} = Tsin10\textdegree - mg = 0\\$

So,

$F_{e} = \frac{mg}{cos10\textdegree }sin10\textdegree = mgtan10\textdegree \\\\= (0.00100kg)(9.8m/s^{2})tan10\textdegree \\\\F_{e} = 2 \times 10^{-3}N$

A force is an influence that can change the motion of an object. A force can cause an object with mass to change its velocity (e.g. moving from a state of rest), i.e., to accelerate. Force can also be described intuitively as a push or a pull. A force has both magnitude and direction, making it a vector quantity. It is measured in the SI unit of newton (N).

Learn more about force here:

brainly.com/question/13191643

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1 year ago

Hammer falls off a roof top and strikes the ground with a certain kinetic energy. If it fell

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Answer:

If the starting GPE is doubled than it's KE would also double.

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2 years ago

A metal can containing condensed mushroom soup has mass 215 g, height 10.8 cm, and diameter 6.38 cm. It is placed at rest on its

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Answer:

Part a)

Moment of inertia of the cylinder is given as

$I = 1.21 \times 10^{-4} kg m^2$

Part B)

Height of the cylinder is of no use here to calculate the inertia

Part C)

Since we don't know about the viscosity data of the soup inside the cylinder so we can't say directly about the moment of inertia of the cylinder as $I = 1/2 mR^2$

Explanation:

As we know that the inclined plane is of length L = 3 m

and its inclination is given as 25 degree

so we know that acceleration of center of mass of the cylinder is constant so we will have

$v_f^2 = v_i^2 + 2 a L$

so we have

$v_f^2 = 0 + 2a(3)$

now we know that

$v_{avg} = \frac{L}{t} = \frac{v_f + v_i}{2}$

$\frac{3}{1.50} = \frac{v_f + 0}{2}$

$v_f = 4 m/s$

Now we have know that final speed of the cylinder due to pure rolling is given as

$v_f = \sqrt{\frac{2gH}{1 + \frac{I}{mR^2}}}$

$4 = \sqrt{\frac{2(9.81)(3 sin25)}{1 + \frac{I}{0.215(0.0319)^2}}}$

$I = 1.21 \times 10^[-4} kg m^2$

Part B)

Height of the cylinder is of no use here to calculate the inertia

Part C)

Since we don't know about the viscosity data of the soup inside the cylinder so we can't say directly about the moment of inertia of the cylinder as $I = 1/2 mR^2$

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2 years ago