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2 years ago

2. State law of conservation of

Physics

1 answer:

butalik [34]2 years ago

3 0

Answer:

In collisions between two isolated objects Newton's third law implies that momentum is always conserved. In collisions between two isolated objects momentum is always conserved. Kinetic energy is only conserved in elastic collisions.

Explanation:

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Subject- Science <br>Grade- 6th<br>please help me with hw. Thank you!<br>

velikii [3]

Answer:

b. they get blown in from colder or warmer areas.

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1 year ago

Read 2 more answers

Usain Bolt's world-record 100 m sprint on August 16, 2009, has been analyzed in detail. At the start of the race, the 94.0 kg Bo

ZanzabumX [31]

a) 893 N

b) 8.5 m/s

c) 3816 W

d) 69780 J

e) 8030 W

Explanation:

The net force acting on Bolt during the acceleration phase can be written using Newton's second law of motion:

$F_{net}=ma$

where

m is Bolt's mass

a is the acceleration

In the first 0.890 s of motion, we have

m = 94.0 kg (Bolt's mass)

$a=9.50 m/s^2$ (acceleration)

So, the net force is

$F_{net}=(94.0)(9.50)=893 N$

And according to Newton's third law of motion, this force is equivalent to the force exerted by Bolt on the ground (because they form an action-reaction pair).

Since Bolt's motion is a uniformly accelerated motion, we can find his final speed by using the following suvat equation:

$v=u+at$

where

v is the final speed

u is the initial speed

a is the acceleration

t is the time

In the first phase of Bolt's race we have:

u = 0 m/s (he starts from rest)

$a=9.50 m/s^2$ (acceleration)

t = 0.890 s (duration of the first phase)

Solving for v,

$v=0+(9.50)(0.890)=8.5 m/s$

First of all, we can calculate the work done by Bolt to accelerate to a speed of

v = 8.5 m/s

According to the work-energy theorem, the work done is equal to the change in kinetic energy, so

$W=K_f - K_i = \frac{1}{2}mv^2-0$

where

m = 94.0 kg is Bolt's mass

v = 8.5 m/s is Bolt's final speed after the first phase

$K_i = 0 J$ is the initial kinetic energy

So the work done is

$W=\frac{1}{2}(94.0)(8.5)^2=3396 J$

The power expended is given by

$P=\frac{W}{t}$

where

t = 0.890 s is the time elapsed

Substituting,

$P=\frac{3396}{0.890}=3816 W$

First of all, we need to find what is the average force exerted by Bolt during the remaining 8.69 s of motion.

In the first 0.890 s, the force exerted was

$F_1=893 N$

We know that the average force for the whole race is

$F_{avg}=820 N$

Which can be rewritten as

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}$

And solving for $F_2$ , we find the average force exerted by Bolt on the ground during the second phase:

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}\\F_2=\frac{(0.890+8.69)F_{avg}-0.890F_1}{8.69}=812.5 N$

The net force exerted by Bolt during the second phase can be written as

$F_{net}=F_2-D$ (1)

where D is the air drag.

The net force can also be rewritten as

$F_{net}=ma$

where

$a=\frac{v-u}{t}$ is the acceleration in the second phase, with

u = 8.5 m/s is the initial speed

v = 12.4 m/s is the final speed

t = 8.69 t is the time elapsed

Substituting,

$a=\frac{12.4-8.5}{8.69}=0.45 m/s^2$

So we can now find the average drag force from (1):

$D=F_2-F_{net}=F_2-ma=812.5 - (94.0)(0.45)=770.2 N$

So the increase in Bolt's internal energy is just equal to the work done by the drag force, so:

$\Delta E=W=Ds$

where

d is Bolt's displacement in the second part, which can be found by using suvat equation:

$s=\frac{v^2-u^2}{2a}=\frac{12.4^2-8.5^2}{2(0.45)}=90.6 m$

And so,

$\Delta E=Ds=(770.2)(90.6)=69780 J$

The power that Bolt must expend just to voercome the drag force is given by

$P=\frac{\Delta E}{t}$

where

$\Delta E$ is the increase in internal energy due to the air drag

t is the time elapsed

Here we have:

$\Delta E=69780 J$

t = 8.69 s is the time elapsed

Substituting,

$P=\frac{69780}{8.69}=8030 W$

And we see that it is about twice larger than the power calculated in part c.

3 0

2 years ago

A spherical shell of radius 9.0 cm carries a uniform surface charge density σ= 9.0 nC/m2. The electric field at r= 9.1 cm is app

maria [59]

Answer:

995.12 N/C

Explanation:

R = 9 cm = 0.09 m

σ = 9 nC/m^2 = 9 x 10^-9 C/m^2

r = 9.1 cm = 0.091 m

q = σ x 4π R² = 9 x 10^-9 x 4 x 3.14 x 0.09 x 0.09 = 9.156 x 10^-10 C

E = kq / r^2

E = ( 9 x 10^9 x 9.156 x 10^-10) / (0.091 x 0.091)

E = 995.12 N/C

8 0

2 years ago

A constant net torque is exerted on an object. Which of the following quantities for the object cannot be constant? (Select all

prohojiy [21]

Answer:

A. kinetic energy

B. angular velocity

E. angular position

Explanation:

The quantities that cannot be constant if a constant net torque is exerted on an objecta are:

A. Kinetic energy. If a torque is applied, the linear or angular speed will be changing at a rate proportional to the torque, so the kinetic energy will change too.

B. Angular velocity. It will change at a rate equal to the torque.

C. Angular position. If the angular velocity changes, the angular position will change.

3 0

3 years ago

A 0.810 kg ball falls 2.5m. How much work does the force of gravity do on the ball?

lisabon 2012 [21]

Answer:

W = 19.845 J

Explanation:

Work is defined as W = Fdcos $\theta$ , where F is the force exerted and d is the distance. Because the direction the ball is falling is the same direction as the force itself, $\theta$ = 0 deg, and since cos(0) = 1, this equation is equivalent to W = Fd. In this case, the force exerted is the weight force, which is equivalent to m * g. Substituting you get:

W = mgd = 0.810 kg * 9.8 m/s^2 * 2.5m

W = 19.845 J

6 0

2 years ago