Ask question

Ask question

All categories

2 years ago

12

A 82-kg fisherman in a 112-kg boat throws a package of mass m = 15 kg horizontally toward the right with a speed of vi = 4.8 m/s

as in the figure below. Neglecting water resistance, and assuming the boat is at rest before the package is thrown, find the velocity of the boat after the package is thrown.

1 answer:

abruzzese [7]2 years ago

8 0

Answer:

0.37 m/s to the left

Explanation:

Momentum is conserved. Initial momentum = final momentum.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

Initially, both the fisherman/boat and the package are at rest.

0 = m₁ v₁ + m₂ v₂

Plugging in values and solving:

0 = (82 kg + 112 kg) v + (15 kg) (4.8 m/s)

v = -0.37 m/s

The boat's velocity is 0.37 m/s to the left.

You might be interested in

A team exerts a force of 10 N towards the south in a tug of war. The other, opposite team, exerts a force of 17 N towards the no

ikadub [295]

Answer:

Option C

Explanation:

According to the question:

Force exerted by the team towards south, F = 10 N

Force exerted by the opposite team towards North, F' = 17 N

Net Force, $\vec{F_{net}} = \vec{F'} - \vec{F}$

$\vec{F_{net}} = \vec{F'} - \vec{F} = 17 - 10 = 7 N$

Thus the force will be along the direction of force whose magnitude is higher

Therefore,

$\vec{F_{net}} = 7 N$ towards North

4 0

2 years ago

A cubic metal box with sides of 17 cm contains air at a pressure of 1 atm and a temperature of 278 K. The box is sealed so that

const2013 [10]

Answer:

F = 3.98 kN

Explanation:

GIVEN DATA:

sides of box = 17 cm

pressure = 1 atm = 101325 N/m2

T2 = 378K

T1 = 278 K

final pressure can be calculate by using below relation

$\frac{P_{1}}{P_{2}}=\frac{T_{1}}{T_{2}}$

we know that

force = pressure * area

therefore force is

$F =(\frac{T_{1}}{T_{2}}*P_{1})A$

$F =(\frac{378}{278}*101325)(17*10^{-2})^{2}$

F = 3.98 kN

5 0

3 years ago

Why must the electric field be normal to the surface at every point of a charged conductor?

solong [7]

-- If the field were inclined to the surface, then it would have
some component parallel to the surface.

-- Then, since we're talking about a conductor, the charges
on the object would move in response to that component
of the field, until there was no longer any component of the
field trying to move them.

3 0

2 years ago

PLEASE HELP NO LINKS NEED HELP FAST

MrRa [10]

Answer:

The galaxies outside of our own are moving away from us, and the ones that are farthest away are moving the fastest. This means that no matter what galaxy you happen to be in, all the other galaxies are moving away from you

Explanation:

7 0

2 years ago

A 4.0 kg circular disk slides in the x- direction on a frictionless horizontal surface with a speed of 5.0 m/s as shown in the a

finlep [7]

Solution :

Let $m_1=m_2=4$ kg

$u_1 = 5$ m/s

Let $v_1$ and $v_2$ are the speeds of the disk $m_1$ and $m_2$ after the collision.

So applying conservation of momentum in the y-direction,

$0=m_1 .v_1_y -m_2 .v_2_y$

$v_1_y = v_2_y$

$v_1 . \sin 60=v_2. \sin 30$

$v_2 = v_1 \times \frac{\sin 60}{\sin 30}$

$v_2=1.732 \times v_1$

Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.

Now applying conservation of momentum in the x-direction,

$m_1.u_1=m_1.v_1_x+m_2.v_2_x$

$u_1=v_1_x+v_2_x$

$5=v_1. \cos 60 + v_2 . \cos 30$

$5=v_1. \cos 60 + 1.732 \times v_1 \cos 30$

$v_1 = 2.50$ m/s

So, $v_2 = 1.732 \times 2.5$

= 4.33 m/s

Therefore, speed of the disk 2 after collision is 4.33 m/s

5 0

2 years ago