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4 years ago

A net force xf acts on an object. what does this mean?

1 answer:

8 0

<span>Answer: 6- for an object to accelerate, Newton's law #2 tells us there must be a force acting on it. F=ma the effect of a force is to accelerate the mass . It cannot be 1, because the force must be centripetal to move the object in a circle. it can't be 2 because if an object is accelerating the velocity cannot be constant as it is moving faster. It can't be 3 because a stationary object has a net force of 0. 4 has nothing to do with the question. 5- no force is needed for uniform motion. 7- that is impossible.</span>

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a merry go round has a radius of 8 meters makes 2 revolutions every 2.5 minutes. A. express the angular speed of the merry go ar

pantera1 [17]

Answer:

a). $v=83.77x10^{-3} rad/s$

b). $v=0.8$ rpm

c).v=0.5865 m/sec.

Explanation:

Given:

$r=8m$

$v=\frac{2rev}{2.5minutes}$

a).

$2 rev*\frac{2\pirad}{1rev}=4\pi rad$

$t=2.5minutes*\frac{60s}{1minute}=150s$

The angular speed in radians per seconds is

$v=\frac{4\pi}{150s}=83.77x10^{-3} rad/s$

b).

$v=\frac{2rev}{2.5minute}rpm$

$v=0.8$ rpm

Child's distance per revolution

(pi*2r) = 43.988 metres.

v=(43.988 x 0.0133333) = 0.5865 m/sec.

4 0

3 years ago

A water park rider (mass = 20.0 kg) starts from rest at the top of a 48.0-m tall

k0ka [10]

Answer:

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Explanation:

6 0

3 years ago

What is the difference between center of mass and center of gravity?

sweet [91]

Answer:

Centre of mass is the point at which the distribution of mass is equal in all directions, and does not depend on gravitational field. Centre of gravity is the point at which the distribution of weight is equal in all directions, and does depend on gravitational field.

5 0

3 years ago

Read 2 more answers

two equal and unlike parallel forces of magnitude 34N act on a rigid body,such that the moment of couple is 8.50 Nm. calculate t

goldfiish [28.3K]

The moment of a couple is Force × perpendicular distance from the arm of the line of action

so the arm of the couple= moment of couple/force=8.5/34=0.25m

the arm is 0.25m

5 0

3 years ago

To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T

Fantom [35]

Answer:

$V_1= 3.4*10^7m/s$

Explanation:

From the question we are told that

Nucleus diameter $d=5.50-fm$

a 12C nucleus

Required kinetic energy $K=2.30 MeV$

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

$K_2 +U_2=K_1+U_1$

where

$K_1$ =initial kinetic energy

$K_2$ =final kinetic energy

$U_1$ =initial electric potential

$U_2$ =final electric potential

mathematically

$U_2 = \frac{Kq_pq_c}{r_2}$

where

$r_f$ =distance b/w charges

$q_c$ =nucleus charge $=6(1.6*10^-^1^9C)$

$K$ =constant

$q_p$ =proton charge

Generally kinetic energy is know as

$K=\frac{1}{2} mv^2$

Therefore

$U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2} mv_1^2 +U_1$

Generally equation for radius is $d/2$

Mathematically solving for radius of nucleus

$R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})$

$R=2.75*10^-^1^5m$

Generally we can easily solving mathematically substitute into v_1

$q_p$ $=6(1.6*10^-^1^9C)$

$K_1=9.0*10^9 N-m^2/C^2$

$U_1= 0$

$R=2.75*10^-^1^5m$

$K=2.30 MeV$

$m= 1.67*10^-^2^7kg$

$V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }$

$V_1= 3.4*10^7m/s$

Therefore the proton must be fired out with a speed of $V_1= 3.4*10^7m/s$

8 0

3 years ago