Ask question

Ask question

All categories

3 years ago

10

A boat is headed with a velocity of 18 meters/second toward the west with respect to the water in a river. If the river is flowi

ng with a velocity of 2.5 meters/second in the same direction as the boat. What would be the magnitude of the boat’s velocity? A. 16.5 meters/second B. 20.5 meters/second C. 18 meters/second D. 2 meters/second E. 45 meters/second

1 answer:

artcher [175]3 years ago

8 0

Answer:

20.5

Explanation:

in this case, the two vectors are in the same direction, so they simply add:

total motion = 18m/s + 2.5m/s = 20.5m/s to the west

You might be interested in

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

3 years ago

A friend in your class tells you that she never uses hints when doing her Mastering homework. She says that she finds the hints

AnnZ [28]

Answer:

A, B, and C are good reasons for my friend not to worry

Explanation:

The following reasons are reason not to worry

A. The only way to lose additional partial credit on a hint is by using the "give up" button or entering incorrect answers. Leaving the question blank will not cost you any credit (Regardless of whether you open a link or not, you will lose credit if you enter a wrong answer or if you give up on a question by hitting the "give up" button. Even after opening a hint, you can leave the question blank if the hint does not provide relevant hints or if the hint brings up more question. Once the question is left blank, you do not lose additional partial credit)

B. As an incentive for thinking hard about the problem, your instructor may choose to apply a small hint penalty, but this penalty is the same whether the hint simply gives information or asks another question (In a situation where you decide to use a hint, the instructor may have put a penalty for using the hint, so whether it asks a question or help in the solution of the question, as long as the hint is consulted, the hint penalty still applies)

C. Getting the correct answer to the question in a hint actually gives you some partial credit, even if you still can't answer the original question (An advantage of using hint is that you get some partial credit for using it if you answer the hint question correctly and fails to answer the original question)

6 0

3 years ago

If you weighed 130 pounds on earth, you would weigh _____pounds on the moon

Aneli [31]

Answer:

152 pounds

Explanation:

4 0

3 years ago

A spherical shell is rolling without slipping at constant speed on a level floor. What percentage of the shell's total kinetic e

IgorC [24]

Answer:

41.667 per cent of the total kinetic energy is translational kinetic energy.

Explanation:

As the spherical shell is rolling without slipping at constant speed, the system can be considered as conservative due to the absence of non-conservative forces (i.e. drag, friction) and energy equation can be expressed only by the Principle of Energy Conservation, whose total energy is equal to the sum of rotational and translational kinetic energies. That is to say:

$E = K_{t} + K_{r}$

Where:

$E$ - Total energy, measured in joules.

$K_{r}$ - Rotational kinetic energy, measured in joules.

$K_{t}$ - Translational kinetic energy, measured in joules.

The spherical shell can be considered as a rigid body, since there is no information of any deformation due to the motion. Then, rotational and translational components of kinetic energy are described by the following equations:

Rotational kinetic energy

$K_{r} = \frac{1}{2}\cdot I_{g}\cdot \omega^{2}$

Translational kinetic energy

$K_{t} = \frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2}$

Where:

$I_{g}$ - Moment of inertia of the spherical shell with respect to its center of mass, measured in $kg\cdot m^{2}$ .

$\omega$ - Angular speed of the spherical shell, measured in radians per second.

$R$ - Radius of the spherical shell, measured in meters.

After replacing each component and simplifying algebraically, the total energy of the spherical shell is equal to:

$E = \frac{1}{2}\cdot (I_{g} + m\cdot R^{2})\cdot \omega^{2}$

In addition, the moment of inertia of a spherical shell is equal to:

$I_{g} = \frac{2}{3}\cdot m\cdot R^{2}$

Then, total energy is reduced to this expression:

$E = \frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2}$

The fraction of the total kinetic energy that is translational in percentage is given by the following expression:

$\%K_{t} = \frac{K_{t}}{E}\times 100\,\%$

$\%K_{t} = \frac{\frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2} }{\frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2} } \times 100\,\%$

$\%K_{t} = \frac{5}{12}\times 100\,\%$

$\%K_{t} = 41.667\,\%$

41.667 per cent of the total kinetic energy is translational kinetic energy.

7 0

3 years ago

A 73-kg Norwegian olympian ski champion is going down a hill sloped at 39 ◦ . The coefficient of kinetic friction between the sk

bazaltina [42]

Answer:

Explanation:

net force on the skier = mg sin 39 - μ mg cos39

mg ( sin39 - μ cos39 )

= 73 x 9.8 ( .629 - .116)

= 367 N

impulse = net force x time = change in momentum .

= 367 x 5 = 1835 kg m /s

velocity of the skier after 5 s = 1835 / 73

= 25.13 m /s

b )

net force becomes zero

mg ( sin39 - μ cos39 ) = 0

μ = tan39

= .81

c )

net force becomes zero , so he will continue to go ahead with constant speed of 25.13 m /s

so he will have speed of 25.13 m /s after 5 s .

5 0

2 years ago