(i feel so dumb )

Which example best describes the term carrying capacity?

mezya [45]

When people aboard a plane...the amount of baggage you take has to vary because the plane has a certain carrying capacity.

3 0

3 years ago

how do you find work when only given the angle a sled is pulled, the mass, the coefficent of kinetic friction and distance

Sergio039 [100]

Answer:

W = F * s

Work done equals applied force * distance traveled

Apparent weight = M g (1 - sin θ) since some of applied force will lighten sled

μ = coefficient of kinetic friction

F cos θ = force applied to motion of sled

s = distance traveled

[μ M g (1 - sin θ)] cos θ * s = work done in moving sled

Note that F = μ M g if applied force is in the horizontal direction

8 0

2 years ago

Two metallic rods A and B of different materials have same length. The linear expansivity of A is 12×10–6 oC–1and cubical expans

uysha [10]

Answer:

The length of rod A will be <u>greater than </u>the length of rod B

Explanation:

We, know that the formula for final length in linear thermal expansion of a rod is:

L' = L(1 + ∝ΔT)

where,

L' = Final Length

L = Initial Length

∝ = Co-efficient of linear expansion

ΔT = Change in temperature

Since, the rods here have same original length and the temperature difference is same as well. Therefore, the final length will only depend upon the coefficient of linear expansion.

For Rod A:

∝₁ = 12 x 10⁻⁶ °C⁻¹

For Rod B:

∝₂ = β₂/3

where,

β₂ = Coefficient of volumetric expansion for rod B = 24 x 10⁻⁶ °C⁻¹

Therefore,

∝₂ = 24 x 10⁻⁶ °C⁻¹/3

∝₂ = 8 x 10⁻⁶ °C⁻¹

Since,

∝₁ > ∝₂

Therefore,

L₁ > L₂

So, the length of rod A will be <u>greater than </u>the length of rod B

6 0

3 years ago

Two long, straight wires are parallel and 20 cm apart. One carries a current of 2.2 A, the other a current of 5.9 A. (a) If the

Leya [2.2K]

Answer: $1.298\times 10^{-5}\ N/m$

Explanation:

Given

Current in the first wire $I_1=2.2\ A$

Current in the second wire $I_2=5.9\ A$

wires are $20\ cm$ apart

Force per unit length between the current-carrying wires is

$\Rightarrow \dfrac{F}{l}=\dfrac{\mu_oI_1I_2}{2\pi r}$

Force exerted by the wires is the same

Put the values

$\Rightarrow \frac{F}{l}=f=\dfrac{4\pi \times 10^{-7}\times 2.2\times 5.9}{2\pi \times 0.2}=1.298\times 10^{-5}\ N/m$

This force will be repulsive in nature as the current is flowing opposite

4 0

3 years ago

For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

A)

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

B)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

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

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

C)

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

D)

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

5 0

3 years ago