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3 years ago

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A person throws a ball straight up in the air. The ball rises to a maximum height and then falls back down so that the person ca

tches it. Consider the ball while it is in the air. Which of the following statements are true? Just after the ball leaves the person's hand the direction of the acceleration is up. The acceleration is zero when the ball reaches its maximum height. The acceleration is about 9.8 m/s2 (down) when the ball is falling.

1 answer:

Lana71 [14]3 years ago

5 0

Answer:

The acceleration is about 9.8 m/s2 (down) when the ball is falling.

Explanation:

The ball at maximum height has velocity zero

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s² (positive downward and negative upward)

$v=u+at\\\Rightarrow 0=u-9.8\times t\\\Rightarrow u=9.8t$

The accleration 9.8 m/s² will always be acting on the body in opposite direction when the body is going up and in the same direction when the body is going down. The acceleration on the body will never be zero

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Oduvanchick [21]

Answer:

f = 8 N

Explanation:

Data provided in the question

Radius of the pulley = r = 0.05 m

Moment of inertia = (I) = 0.2 kg.m^{2}

Angular acceleration = ∝ = 2 rad/sec

Based on the above information

As we know that

Torque is

$= force \times radius$

$= f \times r$

And,

Torque is also

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So,

We can say that

$f \times r = I \times \alpha$

$f \times 0.05 = 0.2 \times 2$

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3 years ago

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2 years ago

A 12-kg dog jumps up in the air to catch a ball. The dog's center of mass is normally 0.20 m above the ground, and he is 0.50 m

ad-work [718]

Explanation:

Given Data:

mass of dog = 12 Kg

dog's center of mass = 0.20m

length of dog = 0.50m

height of dog's jump = ?

Solution:

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2.1 × mgΔh = mg (h - 0.1)

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2 years ago

Which has the most momentum?

boyakko [2]

Answer:

Both objects have the same magnitude of momentum.

Explanation:

If an object of mass $m$ is moving at a velocity of $v$ , the momentum of that object would be $m\, v$ .

The $100\; {\rm g}$ ( $0.1\; {\rm kg}$ ) object is moving at a speed of $1\; {\rm m\cdot s^{-1}}$ . The magnitude of the momentum of this object would be $0.1\; {\rm kg} \times 1\; {\rm m\cdot s^{-1}} = 0.1\; {\rm kg \cdot m \cdot s^{-1}}$ .

Similarly, the momentum of the $1\; {\rm g}$ ( $10^{-3}\; {\rm kg}$ ) object moving at a speed of $100\; {\rm m\cdot s^{-1}}$ would be $10^{-3}\; {\rm kg} \times 100\; {\rm m\cdot s^{-1}} = 0.1\; {\rm kg \cdot m \cdot s^{-1}}$ .

Hence, the magnitude of momentum is the same for the two objects.

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1 year ago

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance

jeka94

Answer:

$1.8\times 105 N/C$

Explanation:

We are given that

$u=2\times 10^7 m/s$

$v=4\times 10^7 m/s$

d=1.9 cm= $\frac{1.9}{100}=0.019 m$

Using 1m=100 cm

We have to find the electric field strength.

$v^2-u^2=2as$

Using the formula

$(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)$

$16\times 10^{14}-4\times 10^{14}=0.038a$

$0.038a=12\times 10^{14}$

$a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2$

$q=1.6\times 10^{-19} C$

Mass of electron,m $=9.1\times 10^{-31} kg$

$E=\frac{ma}{q}$

Substitute the values

$E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}$

$E=1.8\times 105 N/C$

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