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zhenek [66]
3 years ago
15

Displacement vectors of 4km north, 2km south, 5km north, 5km south combine to a total displacement of

Physics
1 answer:
goldfiish [28.3K]3 years ago
5 0

<u>Answer</u>

The combined displacement is 2km north


<u>Explanation</u>


Since displacement is a vector quantity, we take into account the direction.


Good for us all the displacement vectors are in the same dimension, so we can make north positive and south negative or vice-versa.


We now add to obtain,

4+-2+5+-5

This will simplify to

=4-2+5-5=2

Therefore the combined displacement is 2km north

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A dolphin keeps up with a boat for a distance of 51 miles for 3 hours. What is the dolphin’s average speed?
Sedbober [7]

Answer:

The dolphin's speed is 17 mi/h

Explanation:

Use the formula for average velocity: Distance/time

then the average speed of the dolphin is 51/3  mi/h = 17 mi/h

7 0
4 years ago
How could rescue workers use squeezing or compressing to get energy to their flashlights during rescue missions?
Neko [114]

Answer:

Explanation:

During rescue missions, different types of energy can be devices for flashlight, this could be human powered energy such as squeezing or compressing. In flashlight electrical energy is converted to light and thermal energy.

A squeezing or compressing to get energy for flashlight can be regarded as "DYNAMO PROCESS" it involves spinning of "fly wheels" into the flashlight through consistent squeezing ,which is connected to a dynamo(Dynamo supply electrical current). Hence the needed light is seen on the bulb of the flashlight.

3 0
3 years ago
Una ola oceánica viaja a aproximadamente 1,97 m / s. Esto es 4 millas por hora. La frecuencia de las ondas es de aproximadamente
yKpoI14uk [10]

Answer:

λ = 28,14 m

Explanation:

To find the wavelength of the wave you use the following formula:

v=\lambda f  (1)

v: speed of the wave = 1,97 m/s

λ: wavelength

f: frequency of the wave = 0,07 Hz

You replace the values of v and f in the equation (1) and solve for λ:

\lambda=\frac{v}{f}=\frac{1,97m/s}{0,07Hz}=28,14m

hence, the wavelength of the wave is 28,14 m

5 0
3 years ago
Select the correct equations that show that the gravitational potential energy of a 1000-kg boulder raised 6 m above ground leve
wel

Answer:

U = mgh

Explanation:

The gravitational potential energy of an object is defined as:

U=mgh

where

m is the mass of the object

g is the gravitational acceleration

h is the height of the object above a reference level (usually, the ground)

For the boulder in the problem,

m = 1000 kg

g = 10 N/kg

h = 6 m

Substituting,

U=(1000)(10)(6)=60,000 J

6 0
4 years ago
At time t=0 a grinding wheel has an angular velocity of 30.0 rad/s . It has a constant angular acceleration of 35.0 rad/s2 until
Rama09 [41]

Answer:

(A) 570 rad

(B) 10 s

(C) 12.5 rad/s²

Explanation:

The equations of motion for circular motions are used.

  • Initial angular velocity,  \omega_0 = 30.0 \text{ rad/s}
  • Angular acceleration, \alpha =35.0 \text{ rad/s}^2

(A)

At <em>t</em> = 2.00 s, the angular displacement, <em>θ</em>, is given by

\theta = \omega_0t+\frac{1}{2}\alpha t^2 = (30\times 2) + \frac{1}{2}\times35\times2^2=60+70 = 130\text{ rad}

After this time, it decelerates through an angular displacement of 440 rad.

Total angular displacement = 130 + 440 rad = 570 rad

(B)

At the time the circuit breaker tips, the angular velocity is given by

\omega = \omega_0+\alpha  t = 30.0+(35.0\times 2) = 30.0+70.0 =100.0\ \text{rad/s}

This becomes the initial angular velocity for the decelerating motion. Because it stops, the final angular velocity is 0 rad/s. The time for this part of the motion is calculated thus:

\theta_2 = \left(\dfrac{\omega_i+\omega_f}{2}\right)t

Here, \theta_2=440 (the angular displacement during deceleration)

The subscripts, <em>i</em> and <em>f</em>, on <em>ω</em> denote the initial and final angular velocities during deceleration.

\omega_i = 100

\omega_f = 0

t = \dfrac{2\theta_2}{\omega_i} = \dfrac{2\times400}{100} = 8\ \text{s}

This is the time for deceleration. The deceleration began at <em>t</em> = 2 s.

Hence, the wheel stops at <em>t</em> = 2 + 8 = 10 s.

(C)

The deceleration is given by

\alpha_R = \dfrac{\omega_f-\omega_i}{t} = \dfrac{0-100}{8} = -12.5\text{ rad/s}^2

The negative sign appears because it is a deceleration.

4 0
3 years ago
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