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3 years ago

If a force always acts perpendicular to an object's direction of motion, that force cannot change the object's kinetic energy.

1 answer:

5 0

This is very good conceptual question and can clear your doubts regarding work-energy theorem.
Whenever force is perpendicular to the direction of the motion, work done by that force is zero.
According to work-energy theorem,
Work done by all the force = change in kinetic energy.

here, work done = 0.
Therefore,
0=change in kinetic energy
This means kinetic energy remains constant.
Hope this helps

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A 1280 kg car is moving 4.92 m/s. a 509 N force then pushes it forward for 28.7 m. what is its final KE?(unit=J)PLEASE HELP

laila [671]

Answer:

The answer to your question is: total energy = 30100.4 J

Explanation:

Kinetic energy (KE) is the energy due to the movement of and object, its units are joules (J)

Data

mass = 1280 kg

speed = 4.92 m/s

Force = 509 N

distance = 28.7 m

Formula

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

Work = Fd

Process

- Calculate Kinetic energy

- Calculate work

- Add both results

KE = $\frac{1}{2} (1280)(4.92)^{2}$

KE = 15492.1 J

Work = (509)(28.7)

Work = 14608.3 J

Total = 15492.1 + 14608.3

Total energy = 30100.4 J

8 0

3 years ago

Read 2 more answers

The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A

kotykmax [81]

Answer:

$2.1\times 10^{-12} c$

Explanation:

We are given that

Surface area of membrane= $5.3\times 10^{-9} m^2$

Thickness of membrane= $1.1\times 10^{-8} m$

Assume that membrane behave like a parallel plate capacitor.

Dielectric constant=5.9

Potential difference between surfaces=85.9 mV

We have to find the charge resides on the outer surface of membrane.

Capacitance between parallel plate capacitor is given by

$C=\frac{k\epsilon_0 A}{d}$

Substitute the values then we get

Capacitance between parallel plate capacitor= $\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}$

$C=0.25\times 10^{-12}F$

V= $85.9 mV=85.9\times 10^{-3}$

$Q=CV$

$Q=0.25\times 10^{-12}\times 85.9\times 10^{3}=2.1\times 10^{-12} c$

Hence, the charge resides on the outer surface= $2.1\times 10^{-12} c$

5 0

3 years ago

Read 2 more answers

How long would a circuit be connected to produce a current of 0.65 A with a charge of 380 C?

natka813 [3]

30 minutes I am not sure about that

5 0

3 years ago

A 2.30-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. a solid sph

Marina86 [1]

Solution:

initial sphere mvr = final sphere mvr + Iω
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω
where: ω = 2.87 rad/s

So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)²
E = 12.64 J becomes PE = mgh, so
12.64 J = 2.3 kg * 9.8m/s² * h
h = 0.29 m

h = L(1 - cosΘ) → where here L is the distance to the CM
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ
Θ = arccos((1-0.29)/1) = 44.77 º

8 0

3 years ago

A leaf floating down from a tree is an example of an object in free fall.

Nuetrik [128]

Answer:

Oi, mate its false

Explanation:

because if an leaf floats down from a tree it is not considered an object for a free-fall

7 0

4 years ago