Which statement is the most appropriate conclusion for David to write?

One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large

Citrus2011 [14]

Answer:

The correct answer is "6666.67 N".

Explanation:

The given values are:

Mass,

m = 0.100

Relative speed,

v = 4.00 x 10³

time,

t = 6.00 x 10⁻⁸

As we know,

⇒ $F=m(\frac{\Delta v}{\Delta t} )$

On substituting the given values, we get

⇒ $=0.100\times 10^{-6}(\frac{4\times 10^3}{6\times 10^{-8}} )$

⇒ $=6666.67 \ N$

7 0

2 years ago

The height of a projectile t seconds after it is launched straight up in the air is given by f (t )equals negative 16 t squared

velikii [3]

Answer:

$\displaystyle a(5)=-32$

Explanation:

<u>Instant Acceleration</u>

The kinetic magnitudes are usually related as scalar or vector equations. By doing so, we are assuming the acceleration is constant over time. But when the acceleration is variable, the relations are in the form of calculus equations, specifically using derivatives and/or integrals.

Let f(t) be the distance traveled by an object as a function of the time t. The instant speed v(t) is defined as:

$\displaystyle v(t)=\frac{df}{dt}$

And the acceleration is

$\displaystyle a(t)=\frac{dv}{dt}$

Or equivalently

$\displaystyle a(t)=\frac{d^2f}{d^2t}$

The given height of a projectile is

$f(t)=-16t^2 +238t+3$

Let's compute the speed

$\displaystyle v(t)=-32t+238$

And the acceleration

$\displaystyle a(t)=-32$

It's a constant value regardless of the time t, thus

$\boxed{\displaystyle a(5)=-32}$

3 0

2 years ago

A bike rider pedals with constant acceleration to reach a velocity of 7.5 (vf)m/s over a time of 4.5 s(t). during the period of

BigorU [14]

<span>The initial velocity of the bike was 1.67 (vf)m/s. This is found by evaluating 7.5/4.5 which yields the velocity per unit of time which is equivalent to initial velocity.</span>

6 0

3 years ago

I could really use some help on this question guys! Will give brainliest!

aev [14]

Answer:

I think it’s the third one

4 0

2 years ago

Read 2 more answers

An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of 34.5

Inessa05 [86]

Answer:

the acceleration of the airplane is 5.06 x 10⁻³ m/s²

Explanation:

Given;

initial velocity of the airplane. u = 34.5 m/s

distance traveled by the airplane, s = 46,100 m

final velocity of the airplane, v = 40.7 m/s

The acceleration of the airplane is calculated from the following kinematic equation;

v² = u² + 2as

$2as= v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{(40.7)^2 -(34.5)^2}{2 \times 46,100} \\\\a = 5.06 \ \times \ 10^{-3} \ m/s^2$

Therefore, the acceleration of the airplane is 5.06 x 10⁻³ m/s²

5 0

2 years ago