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-Dominant- [34]
3 years ago
11

How do objects and substances in motion have kinetic energy

Physics
1 answer:
Fiesta28 [93]3 years ago
6 0

Answer:

Explanation:

Kinetic energy is one of the types of mechanical energy and it is the energy due to the object in motion.

Kinetic energy is given as 1/2mv^2

A body cannot have a K.E when it is not in motion because of the velocity

Velocity is the change is position of an object or particle with time taken, so if the object doesn't change it position then it won't have velocity.

kinetic energy (KE) of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.

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Uneven heating of land and sea causes warm air over land to rise up, creating a low pressure zone. So wind blows in from the sea to fill this low pressure zone
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3 years ago
What’s the acceleration if the average velocity is 3.5 and the time is 8.7
Monica [59]
Vf = 0 + 3.5•8.7
= 30.45 m/s
6 0
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You have two contentious friends, Chris and Pat, and you’ve quickly discovered that they need you to resolve arguments they have
pshichka [43]

Answer:

v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points,

Explanation:

To resolve the debate, it must be shown that the two have part of the reason, the space or distance between the two points divided by time is the average speed between the points.

             v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points, in the only case that it is so is when there is no acceleration.

Therefore neither of them is right.

3 0
3 years ago
You paddle a conoe with a force of 325 N. You and the canoe have a combined mass of 250 kg. What is the acceleration of the cano
Brums [2.3K]

f = ma

Rearranging it, we get;

a =  \frac{f}{m}
Where a is the acceleration, f is the force, and m is the mass

a =  \frac{325}{250}
a = 1.3 \frac{m}{ {s}^{2} }

7 0
3 years ago
The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in
OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

T_1+V_1=T_2+V_2

0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2}  \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}

Also;

\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}

Thus;

k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}

where;

\delta = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}

k = 104.82\ \  N/m

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

T_1+V_1 = T_3 +V_3  \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3}  \omega_3^2 +  \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0

\frac{m(a+b)^2}{3} \omega_3^2  + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0

\frac{1.53(0.6+0.6)^2}{3} \omega_3^2  + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0

0.7344 \omega_3^2 = 2.128

\omega _3 = \sqrt{\frac{2.128}{0.7344} }

\omega _3 =1.70 \ rad/s

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0
3 years ago
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