Ask question

Ask question

All categories

3 years ago

5

A whale swims due east for a distance of 6.9 km, turns around and goes due west for 1.8 km, and finally turns around again and h

eads 3.7 km due east. (a) What is the total distance traveled by the whale? (b) What are the magnitude and direction (east or west) of the displacement of the whale? If it takes 30.0 min for the whole trip, (c)What is the average speed in km/h for the whole trip ? (d)What is the average velocity (magnitude and direction) in km/h for the whole trip?

1 answer:

Amiraneli [1.4K]3 years ago

6 0

Answer:

Explanation:

a) Total distance traveled by the whale = 6.9 km + 1.8 km + 3.7 km = 12.4 km

b) the magnitude and the direction of the displacement of the whale = 6.9 km due east - 1.8 km due west = 5.1 due east

it final displacement = 5.1 km due east + 3.7 k due east = 8.8 km due east

c) average speed in km / hr = total distance traveled / total time = 12.4 km / 0.5 hr ( 30 min/ 60 min × 1 hr) = 24.8 km / hr

d) average velocity in km / h = total displacement / total time = 8.8 km / 0.5 h = 17.6 km/h

You might be interested in

A stuntwoman is going to attempt a jump across a canyon that is 77 m wide. The ramp on the far side of the canyon is 25 m lower

charle [14.2K]

Answer:

She will make the jump.

Explanation:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

First we will consider horizontal motion of stunt women

Displacement = 77 m, Initial velocity = 28 cos 15 = 27.05 m/s, acceleration = 0

Substituting

$77= 27.05t+\frac{1}{2} *0*t^2\\ \\ t=77/27.05=2.85 seconds$

So she will cover 77 m in 2.85 seconds

Now considering vertical motion, up direction as positive

Initial velocity = 28 sin 15 = 7.25 m/s, acceleration =acceleration due to gravity = -9.8 $m/s^2$ , time = 2.85

Substituting

$s=7.25*2.85-\frac{1}{2}*9.8*2.85^2=20.69-39.80 =-10.11 m$

So at time 2.85 stunt women is 10.11 m below from starting position, far side is 25 m lower. So she will be at higher position.

So she will make the jump.

6 0

3 years ago

The collision between a hammer and a nail can be considered to be approximately elastic. estimate the kinetic energy acquired by

Setler [38]

Here we can use momentum conservation as in this type of collision there is no external force on it

$m_1v_{1i} + m_2v_{2i} = m_1 v_{1f} + m_2v_{2f}$

now here we can say

$m_1 = 10 g$

$v_{1i} = 0$

$m_2 = 550 g$

$v_{2i} = 3.5 m/s$

now here we can say

$10*0 + 550 * 3.5 = 10 v_{1f} + 550 v_{2f}$

$192.5 = v_{1f} + 55 v_{2f}$

now by coefficient of restitution

for elastic collision we know that e = 1

$v_{2f} - v_{1f} = e(v_{1i} - v_{2i})$

$v_{2f} - v_{1f} = 0 - 3.5$

now by solving the two equation

$56v_{2f} = 189$

$v_{2f} = 3.375 m/s$

also we know that

$v_{1f} = v_{2f} + 3.5 = 3.375 + 3.5 = 6.875 m/s$

so final speed of the nail is 6.875 m/s

6 0

3 years ago

Read 2 more answers

Students are asked to create roller coasters for marbles. Their goal is to design a coaster with the tallest possible hill that

ivann1987 [24]

Answer:

Kinetic Energy.

Explanation:

The movement of a roller coaster is accomplished by the conversion of potential energy to kinetic energy. The roller coaster cars gain potential energy as they are pulled to the top of the first hill. As the cars descend the potential energy is converted to kinetic energy.

3 0

3 years ago

Which memory disorder is a common illness on television and films

Alenkinab [10]

amnesia is the most common illness used in tv an films

5 0

3 years ago

Read 2 more answers

To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where

BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

$\int E.dA = \frac{q}{\epcilon_0}$

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

$q = \int \rho dV$

$q = \rho * \pi r^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r}{2 \epcilon_0}$

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

$q = \int \rho dV$

$q = \rho * \pi r_0^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r_0^2}{2 \epcilon_0 r}$

7 0

3 years ago