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3 years ago

Clear 34. As surface area on an object decreases pressure: increase decrease Clear

Physics

1 answer:

VikaD [51]3 years ago

8 0

Answer:

Increase

Explanation:

As the surface area of an object increases, the pressure reduces and vice versa.

Pressure is the force applied per unit area on a body;

Pressure = $\frac{Force}{Surface area}$

Pressure is inversely related to surface area.

A small surface area will ensure pressure to be maximized.

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Assume that a pitcher throws a baseball so that it travels in a straight line parallel to the ground. The batter then hits the b

sertanlavr [38]

Answer:

v_f = -25.9 m/s

Explanation:

- The complete question is as follows:

" Assume that a pitcher throws a baseball so that it travels in a straight line parallel to the ground. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. Define the direction the pitcher originally throws the ball as the +x direction.

Now assume that the pitcher in Part D throws a 0.145-kg baseball parallel to the ground with a speed of 32 m/s in the +x direction. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. What is the ball's velocity just after leaving the bat if the bat applies an impulse of −8.4N⋅s to the baseball?"

Given:

- mass of baseball m = 0.145 kg

- Speed before impact v_i = 32 m/s

- Speed after impact v_f

- Impulse applied by the bat I = - 8.4Ns

Find:

What is the ball's velocity just after leaving the bat

Solution:

- Impulse is the change in linear momentum of the ball according to Newton's second law of motion:

I = m* ( v_f - v_i )

- Taking the + from pitcher to batsman and - from batsman to pitcher.

- Plug in the values:

-8.4 = 0.145* ( v_f - (32) )

v_f = -57.93103 + 32

v_f = -25.9 m/s

4 0

3 years ago

A particle moving along a straight line is subjected to a deceleration ???? = (−2???? 3 ) m/???? 2 , where v is in m/s. If it ha

dlinn [17]

Answer:

(a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

Explanation:

Given that,

The deceleration is

$a=(-2v^3)\ m/s^2$

We need to calculate the velocity at t = 25 s

The acceleration is the first derivative of velocity of the particle.

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=-2v^3$

$\dfrac{dv}{-v^3}=2dt$

On integrating

$int{\dfrac{dv}{-v^3}}=\int{2dt}$

$\dfrac{1}{2v^2}=2t+C$

$v^2=\dfrac{1}{4t+2C}$ ....(I)

At t = 0, v = 10 m/s

$10^2=\dfrac{1}{4\times0+2C}$

$C=\dfrac{1}{200}$

Put the value of C in equation (I)

$v^2=\dfrac{1}{4\times25+2\times\dfrac{1}{200}}$

$v=\sqrt{\dfrac{1}{4\times25+2\times\dfrac{1}{200}}}$

$v=0.099\ m/s$

The velocity is 0.099 m/s.

(b). We need to calculate the position at t = 25 sec

The velocity is the first derivative of position of the particle.

$\dfrac{ds}{dt}=v$

On integrating

$\int{ds}=\int(\sqrt{\dfrac{200}{800t+1}})dt$

$s=\dfrac{\sqrt{200}\times2\sqrt{800t+1}}{800}+C'$

At t = 0, s = 15 m

$15=\dfrac{200}{800}+C'$

$C'=15-\dfrac{200}{800}$

$C'=14.75$

Put the value in the equation

$s=\dfrac{\sqrt{200}\times2\sqrt{800\times25+1}}{800}+14.75$

$s=19.75\ m$

The position is 19.75 m.

Hence, (a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

3 0

3 years ago

A person in bare feet is standing under a tree during a thunderstorm, seeking shelter from the rain. A lightning strike hits the

Anni [7]

Answer:

$I=38.181\ A$ is the current through the body of the man.

$E=34.5\ J$ energy dissipated.

Explanation:

Given:

time for which the current lasted, $t=43\times 10^{-6}\ s$

potential difference between the feet, $V=21000\ V$

resistance between the feet, $R=550\ \Omega$

$I=\frac{V}{R}$

$I=\frac{21000}{550}$

$I=38.181\ A$ is the current through the body of the man.

<u>Energy dissipated in the body:</u>

$E=I^2.R.t$

$E=38.181^2\times 550\times 43\times 10^{-6}$

$E=34.5\ J$

5 0

4 years ago

Two small plastic spheres each have a mass of 1.1 g and a charge of -50.0 nC . They are placed 2.1 cm apart (center to center).

Solnce55 [7]

Answer:

Part a)

F = 0.051 N

Part b)

Both spheres will follow Newton's III law of action reaction force so both sphere will have same force of equal magnitude.

Explanation:

Part a)

Electrostatic force between two charged spherical balls is given as

$F = \frac{kq_1q_2}{r^2}$

here we will have

$q_1 = q_2 = 50 nC$

here the distance between the center of two balls is given as

$r = 2.1 cm = 0.021 m$

now we will have

$F = \frac{(9\times 10^9)(50 \times 10^{-9})(50 \times 10^{-9})}{0.021^2}$

$F = 0.051 N$

Part b)

Both spheres will follow Newton's III law of action reaction force so both sphere will have same force of equal magnitude.

3 0

3 years ago

A baseball is thrown directly upward from ground level with a velocity of +15 m/s. What are the two times when the ball is 10 m

erastovalidia [21]

Answer:

time is 0.5660 s

and time is - 3.62431 s

Explanation:

velocity u = 15 m/s

height s = 10 m

acceleration due to gravity g = –9.8 m/s²

to find out

time

solution

we will apply here distance equation that is

s = ut - 1/2× gt² ...........1

here put all these value and get time t

here s is height and g is -9.8

s = ut - 1/2× gt²

10 = 15t - 1/2× (-9.8)t²

10 = 15t + 4.9t²

solve it we get t

t = 0.56630 and -3.62431

so time is 0.5660 s

and time is - 3.62431 s

8 0

4 years ago