If an environment changes what must a spices do to survive

I). Mechanical energy is the sum of potential energy and kinetic energy in an object that is used to do work.

Answer:

false statement : b ) For the motion of a cart on an incline plane having a coefficient of kinetic friction of 0.5, the magnitude of the change in kinetic energy equals the magnitude of the change in gravitational potential energy

Explanation:

mechanical energy = potential energy + kinetic energy = constant

differentiating both side

Δ potential energy + Δ kinetic energy = 0

Δ potential energy = - Δ kinetic energy

first statement is true.

Friction is a non conservative force so inter-conversion of potential and kinetic energy is not possible in that case. In case of second option, the correct relation is as follows

change in gravitational potential energy = change in kinetic energy + work done against friction .

So given 2 nd option is incorrect.

In case of no change in gravitational energy , work done is equal to

change in kinetic energy.

4 0

3 years ago

A person drives a car around a circular cloverleaf with a radius of 77 m at a uniform speed of 10 m/s. What is the acceleration

umka2103 [35]

Answer:

770m/s

Explanation:

caculation using one of the newton law of motion

6 0

3 years ago

If a fisherman applies a horizontal force with magnitude 47.0 NN to the box and produces an acceleration of magnitude 3.20 m/s2m

musickatia [10]

Answer:

The correct solution is "14.6875 kg".

Explanation:

Given values:

Force,

F = 47.0 N

Acceleration,

a = 3.20 m/s²

Now,

⇒ $Force=Mass\times Acceleration$

or,

⇒ $F=ma$

⇒ $47.0=m\times 3.20$

⇒ $m=\frac{47.0}{3.20}$

⇒ $=14.6875 \ kg$

4 0

2 years ago

A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo

Rasek [7]

Answer:

$t=12.25\ seconds$

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

$x=v_ocos\theta t$

$\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}$

Where vo is the magnitude of the initial velocity, $\theta$ is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

$\displaystyle t=\frac{v_osin\theta}{g}$

There are two times where the value of y is $y_o$ when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making $y=y_o$