All categories

3 years ago

If a circling object is released, centrifugal force will make it travel away from the center of its original path.

1 answer:

5 0

The answer is True. The object is thrown away from the center of its original path due its inertia. Centrifugal force acts in the opposite direction as centripetal force. Centripetal force applies towards the center of the curvature in a spinning object. Centrifugal force is considered an apparent force while centripetal force is an actual force.

You might be interested in

The leaves of a tree lose water to the atmosphere via the process of transpiration. A particular tree loses water at the rate of

Gnoma [55]

Answer:

The speed of the sap flowing in the vessel is 1.90 mm/s

Explanation:

Given:

The rate of water loss, Q = 3 × 10 ⁻⁸ m³/s

Number of vessels contained, n = 2000

Diameter of the vessel, D = 100 Mu m

thus, the radius of the vessel, r = 50 × 10⁻⁶ m

Now, the rate of flow is given as:

Q = AV .............(1)

where, A is the area of the cross-section

V is the velocity

Total area, A = n × (πr²)

substituting the values in the equation (1), we get

3 × 10 ⁻⁸ m³/s = [2000 × (π × (50 × 10⁻⁶)²)] × V

V = 1.909 × 10⁻³ m/s or 1.90 mm/s

Hence, the speed of the sap flowing in the vessel is 1.90 mm/s

7 0

3 years ago

A ball is thrown horizontally from the top of a building 14.9 m high. The ball strikes the ground at a point 107 m from the base

umka2103 [35]

Answer:

1) t=1.743 sec

2)Vo=61.388 m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec

4)Vf=17.08 m/s

Explanation:

1)From second equation of motion we get

h=Vit+(1/2)gt^2

here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion

14.9=(0)*t+(1/2)(9.8)t^2

t^2=14.9/4.9

t^2=3.040 sec

t=1.743 sec

2) s=Vo*t

Putting values we get

107=Vo*1.743

Vo=61.388 m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec

4)From third equation of motion we know that

Vf^2-Vi^2=2gh

here Vi=0 m/s,h=14.9 m

Vf^2=Vi^2+2gh=0+2(9.8)(14.9)

Vf^2=292.04

Vf=17.08 m/s

8 0

3 years ago

The equation of a transverse wave traveling along a very long string is y 6.0 sin(0.020px 4.0pt), where x and y are expressed in

zhannawk [14.2K]

Answer:

given

y=6.0sin(0.020px + 4.0pt)

the general wave equation moving in the positive directionis

y(x,t) = ymsin(kx -?t)

a) the amplitude is

ym = 6.0cm

we have the angular wave number as

k = 2p /?

? = 2p / 0.020p

=1.0*102cm

the frequency is

f = ?/2p

= 4p/2p

= 2.0 Hz

the wave speed is

v = f?

= (100cm)(2.0Hz)

= 2.0*102cm/s

since the trignometric function is (kx -?t) , sothe wave propagates in th -x direction

the maximum transverse speed is

umax =2pfym

= 2p(2.0Hz)(6.0cm)

= 75cm/s

we have

y(3.5cm ,0.26s) = 6.0cmsin[0.020p(3.5) +4.0p(0.26)]

= -2.0cm

6 0

3 years ago

A rocket is fired at 100 m/s at an angle of 37, how many seconds did it take to get to the top?

Kobotan [32]

Answer:

6.14 s

Explanation:

The time the rocket takes to reach the top is only determined from its vertical motion.

The initial vertical velocity of the rocket is:

$u_y = u sin \theta = (100)(sin 37^{\circ})=60.2 m/s$

where

u = 100 m/s is the initial speed

$\theta=37^{\circ}$ is the angle of launch

Now we can apply the suvat equation for an object in free-fall:

$v_y = u_y +gt$

where

$v_y$ is the vertical velocity at time t

$g=-9.8 m/s^2$ is the acceleration of gravity

The rocket reaches the top when

$v_y =0$

So by substituting into the equation, we find the time t at which this happens:

$t=-\frac{u_y}{g}=-\frac{60.2}{-9.8}=6.14 s$

7 0

3 years ago

a race car driver achieved a speed of 53m/s in 14 seconds after taking off from rest from the starting line. what was the averag

Veronika [31]

Answer:

53/14

Explanation:

average acceleration = (Vfinal -Vintial)/ time taken

3 0

3 years ago

Read 2 more answers