Ask question

Ask question

All categories

2 years ago

13

Jada is rowing a boat across a river that has a current of 5 m/s in the ˆ j direction. Leanne, standing on the shore, observes J

ada’s velocity to be 8 m/s ˆ I + 3 m/s ˆ j.
What is Jada’s speed with respect to the water?
a. 13.3 m/s
b. 11.3 m/s
c. 8.54 m/s
d. 8.25 m/s
e. 4.24 m/s
f. None of these

1 answer:

olchik [2.2K]2 years ago

5 0

Answer: d. 8.25 m/s

Explanation:

We are given that Current= 5 m/s in j direction

Velocity= 8 m/s i + 3 m/s j

Now, we have to find Jada's speed with respect to the water.

First we find Jada's velocity with respect to water

v= (8 i + 3 j) - (5 j)

v= 8i - 2 j

To find the speed, we take the magnitude of this velocity vector we have

|v|= $\sqrt{(8)^2+(-2)^2}$

|v|= $\sqrt{68}$ = 8.246 m/s

which comes out to be around = 8.25 m/s

So option d is correct.

You might be interested in

Which of the following methods has led to the earliest discoveries of massive planets orbiting near their parent stars a. detect

Mrac [35]

Answer:

c. detecting the gravitational effect of an orbiting planet (The Wobble"") by looking for the Doppler shifts in the star's spectrum

Explanation:

In a solar system the mass of the star and planets affect each other's orbital movements. The center of gravity of a star and a planet is inside the star. This causes the star to be closer and farther from the Earth at different times. Due to this wobble the star appears to be red shifted when it is farther and blue shifted when it is closer.

When the mass of the planet is high, like a hot Jupiter it causes more wobble i.e., change in radial velocity. This makes it easier to detect the planet. The earliest hot Jupiter found by this method is the planet 51 Pegasi b.

3 0

3 years ago

A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9

Andreyy89

Answer:

$d = 0.0306 m$

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

$U + K = constant$

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

$\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0$

$7.05 + 30.5 = \frac{0.266}{r}$

$r = 7.08 \times 10^{-3} m$

so distance moved by the particle is given as

$d = r_1 - r_2$

$d = 0.0377 - 0.00708$

$d = 0.0306 m$

6 0

2 years ago

Three students have been studying relative motion and decide to do an experiment to demonstrate their knowledge. The experiment

miskamm [114]

Answer with Explanation:

We are given that

Constant speed of Jane=12.6 m/s

a.When Fred can throw the ball 30 m/s

We have to find the angle relative to the horizontal when he throw the ball in order for Sue to see the ball travel vertically upward.

Let $\theta$ be the angle .

Therefore,

$30 cos\theta=12.6$

$cos\theta=\frac{12.6}{30}=0.42$

$\theta=cos^{-1}(0.42)=65.165^{\circ}$

b.We have to find the height to which ball reach.

$v^2-v^2_0=2aS$

$S=\frac{v^2-v^2_0}{2a}=\frac{0-(30 sin65.165)^2}{2(-9.81)}$

$S=37.78 m$

7 0

2 years ago

Two 0.50 g spheres are charged equally and placed 2.5 cm apart. When released, they begin to accelerate at 170 m/s^2 .What is th

vitfil [10]

Answer:

$q=7.65*10^{-8}C$

Explanation:

Using Newton's second law, we calculate the magnitude of the electric force between the spheres:

$F=ma\\F=0.5*10^{-3}kg(170\frac{m}{s^2})\\F=0.085N$

The magnitude of the charge in both spheres is the same. So, we calculate the charge, using Coulomb's law:

$F=\frac{kq^2}{d^2}\\q=\sqrt\frac{Fd^2}{k}\\q=\sqrt\frac{(0.085N)(2.5*10^{-2}m)^2}{8.99*10^9\frac{N\cdot m^2}{C^2}}\\q=7.65*10^{-8}C$

8 0

2 years ago

What does Charles Darwin theory of natural selection state about traits

Maslowich

That only the best traits would go through. The weaker trait in the animals would eventually die off.

8 0

2 years ago

Read 2 more answers