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3 years ago

When three people with a total mass of 2.00 x 102 kg step into their 1.200 x 103 kg car, the car’s

Physics

1 answer:

Aleksandr-060686 [28]3 years ago

3 0

Answer:

$k = 457333.3 N/m$

$x_a =0.09\ m$

Explanation:

From the question we are told that

The total mass of three people is $M = 2.00*10^{2} \ kg$

The mass of the car is $m_c = 1.200 *10^{3} \ kg$

The compression of the car spring is $x = 3 \ cm = 0.03 \ m$

Generally the spring constant is mathematically represented as

$k = \frac{F}{x}$

Here F is the force exerted by the mass of three people and that of the car , this is mathematically represented as

=> $F = (M +m_c) *g$

=> $F = ([2.0*10^{2} ]+[ 1.200*10^{3}]) * 9.8$

=> $F = 13720 \ N$

$k = \frac{13720}{0.03}$

=> $k = 457333.3 N/m$

Generally if the mass which the car is loaded with is $m = 3.00*10^{2} \ kg$

Then the force experienced by the spring is

=> $F_a = (m +m_c) *g$

=> $F_a = (3.00*10^{3} + 1.200 *10^{3}) * 9.8$

=> $F_a = 41160 \ N$

Generally from the above formula the compression is

$x_a = \frac{F_a}{k}$

=> $x_a = \frac{41160}{457333.3}$

=> $x_a =0.09\ m$

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a 1-kg discus is thrown with a velocity of 19 m/s at an angle of 35 degrees from the vertical direction. calculate the vertical

nlexa [21]

Answer:

Vx = 10.9 m/s , Vy = 15.6 m/s

Explanation:

Given velocity V= 19 m/s

the angle 35 ° is taken from Y-axis so the angle with x-axis will be 90°-35° = 55°

θ = 55°

to Find Vx = ? and Vy= ?

Vx = V cos θ

Vx = 19 m/s × cos 55°

Vx = 10.9 m/s

Vx = V sin θ

Vy = 19 m/s × sin 55°

Vy = 15.6 m/s

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3 years ago

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A block (mass = 61.2 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, wh

kolezko [41]

Answer:

The angular velocity is $w = 53.35 \ rounds /minute$

Explanation:

From the question we are told that

The mass of the block is $m = 61.2kg$

The of the pulley is $M = 14.2 kg$

The radius of the pulley is $R = 1.5m$

The radius of the cord around the pulley is $r = 1.5 m$

The distance of the block to the floor is $d = 8.0 m$

From the question we are told that the moment of inertia of the pulley is

$I = \frac{1}{2} MR^2 kg \cdot m^2$

Substituting value

$I = \frac{1}{2} * 14.2 * (1.5)^2$

$I = 15.975 kg \cdot m^2$

Using the Newtons law we can express the force acting on the vertical axis as

$ma = mg -T$

=> $T = mg -ma$

Now when the pulley is rotated that torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

$\tau = I \alpha$

Here $\alpha$ is the angular acceleration

Here $\tau$ is the torque which can be equivalent to

$\tau = T r$

Substituting this above

$Tr = I \alpha$

Substituting for T

$(mg - ma ) r = I\ r \alpha$

Here $a$ is the linear acceleration which is mathematically represented as

$a = r\alpha$

$(mg - m(r\alpha ) ) r = I\ r \alpha$

$mgr = I\alpha + m(r\alpha ) r$

$mgr = \alpha [ I + mr^2]$

making $\alpha$ the subject

$\alpha = \frac{mgr}{I -mr ^2}$

Substituting values

$\alpha = \frac{61.2 * 1.5 * 9.8}{15.975 + (61.2 ) * (1.5)^2}$

$\alpha =5.854 rad /s^2$

Now substituting into the equation above to obtain the acceleration

$a = 5.854 * 1.5$

$a=8.78 m/s^2$

This acceleration is $a = \frac{v}{t}$

and v is the linear velocity with is mathematically represented as

$v = \frac{d}{t}$

Substituting this into the formula acceleration

$a = \frac{d}{t^2}$

making t the subject

$t = \sqrt{\frac{d}{a} }$

substituting value

$t = \sqrt{\frac{8}{8.78}}$

$t = 0.9545 \ s$

Now the linear velocity is

$v = \frac{8}{0.9545}$

$v = 8.38 m/s$

The angular velocity is

$w = \frac{v}{r}$

$w = \frac{8.38}{1.5}$

$w = 5.59 rad/s$

Generally 1 radian is equal to 0.159155 rounds or turns

So 5.59 radian is equal to x

Now x is mathematically obtained as

$x = \frac{5.59 * 0.159155}{1}$

$= 0.8892 \ rounds$

Also

60 second = 1 minute

So 1 second = z

Now z is mathematically obtained as

$z = \frac{ 1}{60}$

z $= 0.01667 \ minute$

Therefore

$w = \frac{0.8892}{0.01667}$

$w = 53.35 \ rounds /minute$

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3 years ago

A ball rolls horizontally off a table and a height of 1.4 m with a speed of 4 m/s. How long does it take the ball to reach the g

Hitman42 [59]

For vertical motion, use the following kinematics equation:

H(t) = X + Vt + 0.5At²

H(t) is the height of the ball at any point in time t for t ≥ 0s

X is the initial height

V is the initial vertical velocity

A is the constant vertical acceleration

Given values:

X = 1.4m

V = 0m/s (starting from free fall)

A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)

Plug in these values to get H(t):

H(t) = 1.4 + 0t - 4.905t²

H(t) = 1.4 - 4.905t²

We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:

1.4 - 4.905t² = 0

4.905t² = 1.4

t² = 0.2854

t = ±0.5342s

Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))

t = 0.53s

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An object with height h, mass M, and a uniform cross-sectional area A floats upright in a liquid with density ρ.

soldi70 [24.7K]

** Missing information: The vertical distance from surface of liquid to bottom of the object is sought in this question, with the condition that the object is at equilibrium **

Ans: The vertical distance = y = M/(ρA)

Explanation:

Support the vertical distance = y

Object's density = M/(A*h) (since A*h = volume)

By applying the condition,

(M/(Ah))/ρ = y/h

M/(ρAh) = y/h

y = M/(ρA)

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Answer:

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