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3 years ago

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A rowboat that is traveling 2.5m/s to the north, encounters a current traveling 4.3m/s to the east. What is the resultant veloci

ty of the rowboat, including the angle?

1 answer:

saveliy_v [14]3 years ago

8 0

Pablo clavó un clavito

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A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache

Paladinen [302]

Answer:

$F_a=1470\ N$

Explanation:

<u>Friction Force</u>

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

$\displaystyle F_a-F_{r1}-F_{r2}=m.a=0$

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

$\displaystyle F_a=F_{r1}+F_{r2}.....[1]$

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

$\displaystyle F_{r2}-T=0$

The friction forces are computed by

$\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g$

$\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g$

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

$\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g$

Simplifying

$\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)$

Plugging in the values

$\displaystyle F_{a}=0.25(9.8)[400+2(100)]$

$\boxed{F_a=1470\ N}$

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3 years ago

Name any two liquid which are found in our human eyes

Romashka [77]

Aqueous humor and vitreous humor are the liquids present in the human eye.

<em>Hope </em><em>it</em><em> helped</em><em> you</em><em>.</em><em>.</em><em>.</em><em> </em><em>pls </em><em>mark</em><em> brainliest</em>

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2 years ago

What are the positive impacts of Genetically Modified Crops?

xeze [42]

Answer:

The crops will have the ability to be resistant to certain diseases

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2 years ago

Read 2 more answers

A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi

Serggg [28]

Answer:

The spring force constant is $k=243\ \frac{N}{m}$ .

Explanation:

We are told the mass of the ball is $m=0.0555\ kg$ , the height above the spring where the ball is dropped is $h=0.536\ m$ , the length the ball compresses the spring is $d=0.04897\ m$ and the acceleration of gravity is $9.8\ \frac{m}{s^{2}}$ .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy $E_{mi}$ is equal to the final mechanical energy $E_{mf}$ :

$E_{mf}=E_{mi}$

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

$\frac{1}{2}kd^{2}=mgh$

If we manipulate the equation we have that:

$k=\frac{2mgh}{d^{2} }$

$k=\frac{2\ 0.0555\ kg\ 9.8\frac{m}{s^{2}}\ 0.536\ m}{(0.04897)^{2}m^{2}}$

$k=\frac{0.58306\ \frac{kgm^{2}}{s^{2}}}{2.398x10^{-3}m^{2}}$

$k=243\ \frac{N}{m}$

5 0

3 years ago

Two ice skaters stand together as illustrated in the attached figure. They "push off" and travel directly away from each other,

Tema [17]

By definition, the momentum is given by:
p = m * v
Where,
m = mass
v = speed.
On the other hand,
F = m * a
Where,
m = mass
a = acceleration:
For the boy we have:
p1 = m * v
p1 = (F / a) * v
p1 = ((710) / (9.81)) * (0.50)
p1 = 36.19 Kg * (m / s)
For the girl we have:
p2 = m * v
p2 = (F / a) * v
p2 = ((480) / (9.81)) * (v)
p2 = 48.93 * v Kg * (m / s)
Then, we have:
p1 + p2 = 0
36.19 + 48.93 * v = 0
Clearing v:
v = - (36.19) / (48.93)
v = -0.74 m / s (negative because the velocity is in the opposite direction of the boy's)
Answer:
the girl's velocity in m / s after they push off is -0.74 m / s

6 0

3 years ago