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Morgarella [4.7K]

3 years ago

5

A woman who weighs 500 N stands on an 8.0-m-long board that weighs 100 N. The board is supported at each end. The support force

at the right end is 3 times the support force at the left end. How far from the right end is the woman standing?

1 answer:

aniked [119]3 years ago

8 0

Answer:

The woman's distance from the right end is 1.6m = (8-6.4)m.

The principles of moments about a point or axis running through a point and summation of forces have been used to calculate the required variable.

Principle of moments: the sun of clockwise moments must be equal to the sun of anticlockwise moments.

Also the sun of upward forces must be equal to the sun of downward forces.

Theses are the conditions for static equilibrium.

Explanation:

The step by step solution can be found in the attachment below.

Thank you for reading this solution and I hope it is helpful to you.

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What is the mass of a dog that weighs 382 N?(unit=kg)

Luba_88 [7]

Answer:

The answer to your question is: mass = 38.93 kg

Explanation:

Data

mass = ?

Weight = 382 N

gravity = 9.81 m/s2

Formula

Weight = mass x gravity

mass = weight / gravity

mass = 382 / 9.81 substitution

mass = 38.93 kg result

6 0

3 years ago

At t = 0 the components of a radio-controlled car's velocity are vx = 0.5 m/s and vy = 1.2 m/s. Find the components of the car's

FinnZ [79.3K]

Answer:

x-component of velocity = 5.7 m/s

y-component of velocity = -1.4 m/s

Explanation:

Use first equation of motion to find components of velocity at a given time:

$v = u + at$

where, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time.

Given:

$u_x= 0.5 m/s\\u_y=1.2 m/s\\a_x=2 m/s^2\\a_y=-1m/s^2\\t = (2.6-0)s =2.6 s$

$v_x=u_x+a_xt\\\Rightarrow v_x=0.5+2\times 2.6\\\Rightarrow v_x=5.7 m/s$

$v_y=u_y+a_yt\\\Rightarrow v_y=1.2-1\times 2.6\\\Rightarrow v_y=-1.4 m/s$

5 0

3 years ago

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

The mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

The distance apart is $d = 0.4 \ m$

The speed of the second sphere is $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.8 \ m$ is mathematically represented

$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

$KE = \frac{1 }{2} m (v_1)^2$

substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

$KE = 0.3 \ J$

And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

$U = 0.18 \ J$

So

$Q = 0.3 + 0.18$

$Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

$KE_f = \frac{1 }{2} m (v_2^2$

substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

$KE_f = 7.50 *10^{ -4} (v_2 )^2$

And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

$U_f = 0.36 \ J$

From the law of energy conservation

$Q = Q_f$

So

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

6 0

3 years ago

A dockworker applies a constant horizontal force of 90.0 N to a block of ice on a smooth horizontal floor. The frictional force

Oliga [24]

Answer:

The mass of the ice block is equal to 70.15 kg

Explanation:

The data for this exercise are as follows:

F=90 N

insignificant friction force

x=13 m

t=4.5 s

m=?

applying the equation of rectilinear motion we have:

x = xo + vot + at^2/2

where xo = initial distance =0

vo=initial velocity = 0

a is the acceleration

therefore the equation is:

x = at^2/2

Clearing a:

a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2

we use Newton's second law to calculate the mass of the ice block:

F=ma

m=F/a = 90/1.283=70.15 kg

6 0

3 years ago

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Which landform is the farthest north?

quester [9]

The North Pole would be your answer

7 0

2 years ago