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3 years ago

A ball is thrown downward with an initial speed of 7 m / s. The ball's velocity after 3 seconds is m / s. (g = -9.8m / s²)

Physics

1 answer:

yan [13]3 years ago

5 0

Answer:

-36.4 m/s

Explanation:

final velocity= initial velocity + acceleration x time

7 + (-9.8)(3)= -36.4 m/s

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3 years ago

A gas was compressed to 30.0 mL at 1.5 atm from 65<br>mL. What was the original pressure?

astra-53 [7]

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3 years ago

A child whose weight is 287 N slides down a 7.20 m playground slide that makes an angle of 31.0Â° with the horizontal. The coeff

natulia [17]

Answer:

$H =212.6 \ J$

$v = 7.647 \ m/s$

Explanation:

From the question we are told that

The child's weight is $W_c = 287 \ N$

The length of the sliding surface of the playground is $L = 7.20 \ m$

The coefficient of friction is $\mu = 0.120$

The angle is $\theta = 31.0 ^o$

The initial speed is $u = 0.559 \ m/s$

Generally the normal force acting on the child is mathematically represented as

=> $N = mg * cos \theta$

Note $m * g = W_c$

Generally the frictional force between the slide and the child is

$F_f = \mu * mg * cos \theta$

Generally the resultant force acting on the child due to her weight and the frictional force is mathematically represented as

$F =m* g sin(\theta) - F_f$

Here F is the resultant force and it is represented as $F = ma$

=> $ma = m* g sin(31.0) - \mu * mg * cos (31.0)$

=> $a = g sin(31.0)- \mu * g * cos (31.0)$

=> $a = 9.8 * sin(31.0) - 0.120 * 9.8 * cos (31.0)$

=> $a = 4.039 \ m/s^2$

$F_f = 0.120 * 287 * cos (31.0)$

=> $F_f = 29.52 \ N$

Generally the heat energy generated by the frictional force which equivalent tot the workdone by the frictional force is mathematically represented as

$H = F_f * L$

=> $H = 29.52 * 7.2$

=> $H =212.6 \ J$

Generally from kinematic equation we have that

$v^2 = u^2 + 2as$

=> $v^2 = 0.559^2 + 2 * 4.039 * 7.2$

=> $v = \sqrt{0.559^2 + 2 * 4.039 * 7.2}$

=> $v = 7.647 \ m/s$

6 0

3 years ago

A fairgrounds ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follo

cestrela7 [59]

Answer:

13.37 rev/min

Explanation:

acceleration due to gravity (g) = 9.8 m/s², centripetal acceleration ( $a_c$ ) = 1.8 * g = 1.8 * 9.8 m/s² = 17.64 m/s².

r = 9 m

Centripetal acceleration ( $a_c$ ) is given by:

$a_c=\frac{v^2}{r} \\\\v=\sqrt{a_c*r} \\\\v=\sqrt{17.64\ m/s^2*9\ m}\\\\v=12.6\ m/s$

The velocity (v) is given by:

v = ωr; where ω is the angular velocity

Hence:

ω = v/r = 12.6 / 9

ω = 1.4 rad/s

ω = 2πN

N = ω/2π = 1.4 / 2π

N = 0.2228 rev/s

N = 13.37 rev/min

4 0

2 years ago

A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with

Komok [63]

Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

$\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\ g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s$

$\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM$

Therefore the speed in RPM will be 62.64 RPM.

5 0

3 years ago