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2 years ago

6

a professional baseball player can pitch a baseball with a velocity of 44.7m/s towards home plate. If a baseball weighs 1.4 N, h

ow much momentum does it have when it hits the catcher units

2 answers:

topjm [15]2 years ago

8 0

Answer:

p = 6.25 kg-m/s

Explanation:

It is given that,

Velocity of the baseball, v = 44.7 m/s

Weight of the baseball, W = 1.4 N

Firstly we need to find the mass of the baseball. It can be calculated as :

$m=\dfrac{W}{g}$

$m=\dfrac{1.4\ N}{9.8\ m/s^2}$

m = 0.14 kg

The momentum of an object is equal to the product of mass and velocity. It is given by :

$p=m\times v$

$p=0.14\ kg\times 44.7\ m/s$

p = 6.25 kg-m/s

So, the momentum it have when it hits the catcher is 6.25 kg-m/s. Hence, this is the required solution.

kumpel [21]2 years ago

3 0

1.4 N is a weight so calculating it's mass
1.4/9.8 = 0.1428 kg
momentum will be 0.1428*44.7 = 6.38 kgm/s

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An electron in a vacuum chamber is fired with a speed of 9800 km/s toward a large, uniformly charged plate 75 cm away. The elect

melisa1 [442]

Answer:

The plate's surface charge density is $-8.056\times10^{-9}\ C/m^2$

Explanation:

Given that,

Speed = 9800 km/s

Distance d= 75 cm

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Suppose we determine the plate's surface charge density?

We need to calculate the surface charge density

Using work energy theorem

$W=\Delta K.E$

$W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{i}^2$

Here, final velocity is zero

$W=0-\dfrac{1}{2}mv_{i}^2$ ...(I)

We know that,

$W=-Fd$

$W=-E\times e\times d$

$W=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d$ ...(II)

From equation (I) and (II)

$-\dfrac{1}{2}mv_{i}^2=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d$

Charge is negative for electron

$\lambda=\dfrac{mv^2\epsilon_{0}}{(-e)d}$

Put the value into the formula

$\lambda=-\dfrac{9.1\times10^{-31}\times(9800\times10^{3})^2\times8.85\times10^{-12}}{1.6\times10^{-19}\times(75-15)\times10^{-2}}$

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Volume of sphere is given as

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$V = \frac{4(3.14) (13\times 10^{25})^{3}}{3}$

$V = 9.2\times 10^{78}$ m³

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$m$ = Total mass of the universe = ?

We know that mass is the product of volume and density, hence

$m = \rho V$

$m = (1\times 10^{-26}) (9.2\times 10^{78})$

$m = 9.2\times 10^{52}$ kg

$M$ = mass of "ordinary" matter = ?

mass of "ordinary" matter is only about 4% of total mass, hence

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2 years ago

Suppose a particle is accelerated through space by a constant 10-N force. Suddenly the particle encounters a second force of 10-

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Answer:

The particle will continue moving at constant velocity

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Therefore, since $F_{net}=0$ , the acceleration is zero (a=0) and so the particle will keep a constant velocity.

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Answer:

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$V_{big} = \frac{KQ}{R}$

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2 years ago