All categories

3 years ago

Which of the following is NOT an example of kinetic energy being converted to potential energy?

Physics

1 answer:

Pani-rosa [81]3 years ago

6 0

There are no such examples on the list you've provided.

You might be interested in

High-speed stroboscopic photographs show that the head of a 210-g golf club is traveling at 56 m/s just before it strikes a 46-g

tamaranim1 [39]

Explanation:

It is given that,

Mass of golf club, m₁ = 210 g = 0.21 kg

Initial velocity of golf club, u₁ = 56 m/s

Mass of another golf ball which is at rest, m₂ = 46 g = 0.046 kg

After the collision, the club head travels (in the same direction) at 42 m/s. We need to find the speed of the golf ball just after impact. Let it is v.

Initial momentum of golf ball, $p_i=m_1u_1=0.21\ kg\times 56\ m/s=11.76\ kg-m/s$

After the collision, final momentum $p_f=0.21\ kg\times 42\ m/s+0.046v$

Using the conservation of momentum as :

$p_i=p_f$

$11.76\ kg-m/s=0.21\ kg\times 42\ m/s+0.046v$

v = 63.91 m/s

So, the speed of the golf ball just after impact is 63.91 m/s. Hence, this is the required solution.

3 0

3 years ago

Which statement best describes a primary source?

Fantom [35]

Answer:

I go with D but what you can do is u can read each and choose the one that sounds right Hope this helps

3 0

3 years ago

What are the different skills and strategies were you aware of in this situation that indicated to you that it would be appropri

const2013 [10]

Answer:

dad

Explanation:

daddy

4 0

2 years ago

Electromagnetic waves can transfer energy without a(n)

Anna71 [15]

Some waves can travel without a -medium-, a channel like telephone power lines, lenses, or even particles in the air or in water.

7 0

3 years ago

A 2530-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that

gavmur [86]

Answer:

A = 1.4 m/s²

B = -0.10493 m/s³

a = 1.29507 m/s²

T = 28095.8271 N

T = 1.13198 W

Explanation:

t = Time taken

g = Acceleration due to gravity = 9.81 m/s²

The equation

$v(t)=At+Bt^2$

Differentiating with respect to time

$\frac{dv}{dt}=\frac{d(At+Bt^2)}{dt}\\\Rightarrow 1.4=A+2Bt$

At t = 0

$1.4=A$

Hence, A = 1.4 m/s²

$B=\frac{v-At}{t^2}\\\Rightarrow B=\frac{2.18-1.4\times 1.8}{1.8^2}\\\Rightarrow B=-0.10493\ m/s^3$

B = -0.10493 m/s³

At t = 5 seconds

$a=1.4+2\times -0.010493\times 5=1.29507\ m/s^2$

a = 1.29507 m/s²

$T=m(a+g)\\\Rightarrow T=2530(1.29507+9.81)\\\Rightarrow T=28095.8271\ N$

T = 28095.8271 N

Weight of rocket

$W=2530\times 9.81=24819.9\ N$

$\frac{T}{W}=\frac{28095.8271}{24819.9}\\\Rightarrow \frac{T}{W}=1.13198\\\Rightarrow T=1.13198W$

T = 1.13198 W

3 0

4 years ago