What are two parts that make up a vector

A particle traveling in a circular path of radius 300 m has an instantaneous velocity of 30 m/s and its velocity is increasing a

grigory [225]

<h2>Answer:</h2>

(c) 5m/s²

<h2>Explanation:</h2>

Total acceleration (a) of a particle in a circular motion is the vector sum of the radial or centripetal acceleration ( $a_{C}$ ) of the particle and the tangential acceleration ( $a_{T}$ ) of the particle and its magnitude can be calculated as follows;

a = $\sqrt{(a_{C})^2 + (a_{T})^2}$ ---------------------(i)

$a_{C}$ = $\frac{v^{2} }{r}$ ------------------------------(ii)

Where;

v = instantaneous velocity

r = radius of the circular path of motion

<em>From the question;</em>

v = 30m/s

r = 300m

(i) First let's calculate the centripetal acceleration by substituting the values above into equation (ii) as follows;

$a_{C}$ = $\frac{30^{2} }{300}$

$a_{C}$ = $\frac{900}{300}$

$a_{C}$ = 3m/s²

(ii) From the question, the velocity of the particle is increasing at a constant rate of 4m/s² and that is the tangential acceleration $a_{T}$ , of the particle. i.e;

$a_{T}$ = 4m/s²

(iii) Now substitute the values of $a_{C}$ and $a_{T}$ into equation (i) as follows;

a = $\sqrt{(3)^2 + (4)^2}$

a = $\sqrt{(9) + (16)}$

a = $\sqrt{25}$

a = 5m/s²

Therefore, the magnitude of its total acceleration a, is 5m/s²

5 0

2 years ago

A bee wants to fly to a flower located due North of the hive on a windy day. The wind blows from East to West at speed 6.68 m/s.

Aleonysh [2.5K]

Answer: $53.31\°$ East of North

Explanation:

We have the following data:

Speed of the wind from East to West: $6.68 m/s$

Speed of the bee relative to the air: $8.33 m/s$

If we graph these speeds (which in fact are velocities because are vectors) in a vector diagram, we will have a right triangle in which the airspeed of the bee (its speed relative to te air) is the hypotense and the two sides of the triangle will be the <u>Speed of the wind from East to West</u> (in the horintal part) and the <u>speed due North relative to the ground</u> (in the vertical part).

Now, we need to find the direction the bee should fly directly to the flower (due North):

$sin \theta=\frac{Windspeed-from-East-to-West}{Speed-bee-relative-to-air}$

$sin \theta=\frac{6.68 m/s}{8.33 m/s}$

Clearing $\theta$ :

$\theta=sin^{-1} (\frac{6.68 m/s}{8.33 m/s})$

$\theta=53.31\°$

6 0

2 years ago

If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is $u = 7.73 \ m/s$

Explanation:

From the question we are told that

The height at which he let go of the brief case is h = 130 m

The time taken before the the brief case hits the water is t = 6 s

Generally the initial speed of the briefcase (Which also the speed of the helicopter )before the man let go of it is mathematically evaluated using kinematic equation as

$s = h+ u t + 0.5 gt^2$

Here s is the distance covered by the bag at sea level which is zero

$0 = 130+ u * (6) + 0.5 * (-9.8) * (6)^2$

=> $0 = 130+ u * (6) + 0.5 * (-9.8) * (6)^2$

=> $u = \frac{-130 + (0.5 * 9.8 * 6^2) }{6}$

=> $u = 7.73 \ m/s$

7 0

2 years ago

A battery charges a parallel-plate capacitor fully and then is removed. The plates are then slowly pulled apart. What happens to

White raven [17]

Answer:

<h2>The potential difference increases </h2>

Explanation:

from the relation $E= \frac{V}{d}$

where E= electric field (force per coulomb)

V= voltage

d= distance

Hence the voltage is going to be V= E×d.

Therefore this means that increasing the distance increases the voltage.

3 0

2 years ago

A 128.0-N carton is pulled up a frictionless baggage ramp inclined at 30.0∘above the horizontal by a rope exerting a 72.0-N pull

Elden [556K]

Answer:

(A) 374.4 J

(B) -332.8 J

(C) 0 J

(D) 41.6 J

(E) 351.8 J

Explanation:

weight of carton (w) = 128 N

angle of inclination (θ) = 30 degrees

force (f) = 72 N

distance (s) = 5.2 m

(A) calculate the work done by the rope

work done = force x distance x cos θ

since the rope is parallel to the ramp the angle between the rope and

the ramp θ will be 0

work done = 72 x 5.2 x cos 0

work done by the rope = 374.4 J

(B) calculate the work done by gravity

the work done by gravity = weight of carton x distance x cos θ

The weight of the carton = force exerted by the mass of the carton = m x g

the angle between the force exerted by the weight of the carton and the ramp is 120 degrees.

work done by gravity = 128 x 5.2 x cos 120

work done by gravity = -332.8 J

(C) find the work done by the normal force acting on the ramp

work done by the normal force = force x distance x cos θ

the angle between the normal force and the ramp is 90 degrees

work done by the normal force = Fn x distance x cos θ

work done by the normal force = Fn x 5.2 x cos 90

work done by the normal force = Fn x 5.2 x 0

work done by the normal force = 0 J

(D) what is the net work done ?

The net work done is the addition of the work done by the rope, gravitational force and the normal force

net work done = 374.4 - 332.8 + 0 = 41.6 J

(E) what is the work done by the rope when it is inclined at 50 degrees to the horizontal

work done by the rope= force x distance x cos θ
the angle of inclination will be 50 - 30 = 20 degrees, this is because the ramp is inclined at 30 degrees to the horizontal and the rope is inclined at 50 degrees to the horizontal and it is the angle of inclination of the rope with respect to the ramp we require to get the work done by the rope in pulling the carton on the ramp

work done = 72 x 5.2 x cos 20

work done = 351.8 J

5 0

2 years ago