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2 years ago

1. A 3.0 kg mass is tied to a rope and swung in a horizontal circle. If the velocity of the mass is 4.0 ms and

Physics

1 answer:

saul85 [17]2 years ago

4 0

10.67m/s²

32N

Explanation:

Given parameters:

Mass of the body = 3kg

velocity of the mass = 4m/s

radius of circle = 0.75m

Unknown:

centripetal acceleration = ?

centripetal force = ?

Solution:

The centripetal force is the force that keeps a radial body in its circular motion. It is directed inward:

Centripetal acceleration = $\frac{v^{2} }{r}$

v is the velocity of the body

r is the radius of the circle

putting in the parameters:

Centripetal acceleration = $\frac{4^{2} }{0.75}$

Centripetal acceleration = 10.67m/s²

Centripetal force = m $\frac{v^{2} }{r}$

m is the mass

Centripetal force = mass x centripetal acceleration

= 3 x 10.67

= 32N

learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

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You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

valentina_108 [34]

Answer:

v = 69.82 ms^-1

Explanation:

As we know,

R = vi2 sin2Ꝋ / g

vi2 =R g / sin2 Ꝋ where R is range R = 52m, Ꝋ = 3 Degrees

vi2 = 52 x 9.8 / sin 2(3) = 4875.227

v = 69.82 ms^-1

3 0

2 years ago

An object of mass 24kg is accelerated up a frictionless place incline at an angle of 37° with horizontal by a constant force, st

RoseWind [281]

a) Average power: 1425 W

b) Instantaneous power at 3.0 sec: 2850 W

Explanation:

The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation

$s=ut+\frac{1}{2}at^2$

where

s = 18 m is the displacement along the ramp

u = 0 is the initial velocity

t = 3.0 s is the time taken

a is the acceleration of the object along the ramp

Solving for a,

$a=\frac{2s}{t^2}=\frac{2(18)}{(3.0)^2}=4 m/s^2$

Now we can apply Newton's second law to find the net force on the object:

$F=ma=(24 kg)(4 m/s^2)=96 N$

This net force is the resultant of the applied force forward ( $F_a$ ) and the component of the weight acting backward ( $mg sin \theta$ ), so we can find what is the applied force:

$F=F_a - mg sin \theta\\F_a = F+mg sin \theta = 96+(24)(9.8)(sin 37^{\circ})=237.5 N$

where

m = 24 kg is the mass of the object

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

Now we can finally find what is the work done by the applied force, which is parallel to the ramp, therefore:

$W=F_a s = (237.6)(18)=4276 J$

where s = 18 m is the displacement.

Therefore the average power needed is:

$P=\frac{W}{t}=\frac{4276}{3}=1425 W$

The instantaneous power at any point of the motion is given by

$P=F_av$

where

$F_a$ is the force applied

v is the velocity of the object

We already calculated the applied force:

$F_a=237.5 N$

While since this is a uniformly accelerated motion, we can find the velocity at the end of the 3.0 seconds using the suvat equation:

$v=u+at=0+(4)(3.0)=12.0 m/s$

And therefore, the instantaeous power at 3.0 sec is:

$P=Fv=(237.5)(12)=2850 W$

Learn more about power:

brainly.com/question/7956557

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8 0

2 years ago

"In analyzing distances by apply ing the physics of gravitational forces, an astronomer has obtained the expression

zavuch27 [327]

Answer:

The value of R is $1.72\times10^{11}\ m$ .

(B) is correct option.

Explanation:

Given that,

In analyzing distances by apply ing the physics of gravitational forces, an astronomer has obtained the expression

$R=\sqrt{\dfrac{1}{(\dfrac{1}{140\times10^{9}})^2-(\dfrac{1}{208\times10^{9}})^2}}$

We need to calculate this for value of R

$R=\sqrt{\dfrac{1}{(\dfrac{1}{140\times10^{9}})^2-(\dfrac{1}{208\times10^{9}})^2}}$

$R=1.89\times10^{11}\ m$

So, The nearest option of the value of R is $1.72\times10^{11}\ m$

Hence, The value of R is $1.72\times10^{11}\ m$ .

6 0

2 years ago

In order to get a tree stump out of the ground, chains are connected to two trucks. One truck pulls with a force of 600 N to the

Black_prince [1.1K]

Answer:

The net force on the stump is 1000 N.

Explanation:

Given that,

Force 1 acting on the truck, $F_1=600\ N$ (due north)

Force 2 acting on the truck, $F_2=800\ N$ (due west)

We need to find the net force on the stump. We know that force is a vector quantity. The net force on the stump is given by the the resultant force. It is given by :

$F=\sqrt{F_1^2+F_2^2}$

$F=\sqrt{600^2+800^2}$

F = 1000 N

So, the net force on the stump is 1000 N. Hence, this is the required solution.

3 0

2 years ago

A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m

madam [21]

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂ = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .

4 0

2 years ago