Ask question

Ask question

All categories

3 years ago

12

If she can throw each rock with a speed of 7.5 m/s relative to the ground, causing the wagon to move, how many rocks must she th

row per minute to maintain a constant average speed against a 3.9-N force of friction

1 answer:

Stels [109]3 years ago

5 0

Complete question:

A child sits in a wagon with a pile of 0.64-kg rocks. If she can throw each rock with a speed of 7.5 m/s relative to the ground, causing the wagon to move, how many rocks must she throw per minute to maintain a constant average speed against a 3.9-N force of friction

Answer:

The number of rocks per minutes thrown is 49 rocks/min

Explanation:

Given;

mass of the rock, m = 0.64 kg

speed of the rock, v = 7.5 m/s

frictional force, $f_k$ = 3.9 N

For an object to move at a constant speed, the applied force must be equal to the frictional force.

$f_k = N(F_a)\\\\f_k = N(\frac{mv}{t})\\\\f_k = \frac{N}{t} (mv)\\\\\frac{N}{t} = \frac{f_k}{mv}$

where;

$F_a$ is the applied force

N/t is the number of rocks per minutes thrown

Substitute the given parameters;

$\frac{N}{t} = \frac{f_k}{mv}\\\\ \frac{N}{t} = \frac{3.9}{0.64*7.5}\\\\ \frac{N}{t} = 0.8125 \ \frac{rocks}{second} * \frac{60 \ seconds}{1 \ min} = 48.75 \ \frac{rocks}{min} \$

Approximately 49 rocks/min

You might be interested in

What is the difference between speed and velocity?

ElenaW [278]

B - speed and direction are combined in another quantity, called velocity. It can be thought of as its speed in a particular direction. Speed is the corresponding scalar quantity, because it does not have a direction.

5 0

3 years ago

4. Two forces act on a 2 kg object as shown. What is the magnitude of the acceleration of the object?

aleksandrvk [35]

The resultant force on the object is

∑ <em>F</em> = 〈0, 8〉 N + 〈6, 0〉 N = 〈6, 8〉 N

which has a magnitude of

<em>F</em> = √((6 N)² + (8 N)²) = √(100 N²) = 10 N

By Newton's second law, the acceleration has magnitude <em>a</em> such that

<em>F</em> = <em>m a</em>

10 N = (2 kg) <em>a</em>

<em>a</em> = (10 N) / (2 kg)

<em>a</em> = 5 m/s²

so the answer is B.

4 0

3 years ago

Please help me I need these extra credit points plz plz plz

VashaNatasha [74]

Two half lives so it is 4000 years old

5 0

3 years ago

Read 2 more answers

A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground.

pshichka [43]

Answer:

A. 4,9 m/s2

B. 2,0 m/s2

C. 120 N

Explanation:

In the image, 1 is going to represent the monkey and 2 is going to be the package. Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

$\sum F_y=m_1*a_m_i_n = T-m_1*g$

If the package is barely lifted, that means that T=m_2*g; then:

$\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g$

Solving the equation for a_mín, we have:

$a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2$

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey: $\sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a$

For the package: $\sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a$

The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:

For the package: $\sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a$

We have two unknowns and two equations, so we can proceed. We can match both tensions and have:

$m_1*a+m_1*g=m_2*g-m_2*a$

Solving a, we have

$(m_1+m_2)*a =(m_2 - m1)*g\\\\a=((m_2-m_1)/(m_1+m_2))*g \rightarrow a=((15kg-10kg)/(10kg+15kg))*9,8 m/s^2\\\\a= 2,0 m/s^2$

We can then replace this value of a in one for the sums of force and find the tension T:

$T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N$

5 0

3 years ago

A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t

vladimir1956 [14]

a) $\omega=\frac{-mvR}{I+MR^2}$

b) $v=\frac{-mvR^2}{I+MR^2}$

Explanation:

a)

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

$L_1=0$

Later, after the girl throws the rock, the angular momentum will be:

$L_2=(I_M+I_g)\omega +L_r$

where:

$I$ is the moment of inertia of the merry-go-round

$I_g=MR^2$ is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

$\omega$ is the angular speed of the merry-go-round and the girl

$L_r=mvR$ is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

$L_1=L_2$

So we find:

$0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}$

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

b)

The linear speed of a body in rotational motion is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

$\omega=\frac{-mvR}{I+MR^2}$ is the angular speed

$r=R$ is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

$v=\omega R=\frac{-mvR^2}{I+MR^2}$

5 0

3 years ago