Devise an experiment using common materials that you could use to determine the speed of

A skateboarder has an acceleration of −1.9 m/s2. If her initial speed is 6 m/s, how long does it take her to stop?

SSSSS [86.1K]

You said that she's losing 1.9 m/s of her speed every second.

So it'll take

(6 m/s) / (1.9 m/s²) = 3.158 seconds (rounded)

to lose all of her initial speed, and stop.

6 0

3 years ago

A rocket starting from its launch pad is subjected to a uniform acceleration of 100 meters/second2. Determine the time needed to

gizmo_the_mogwai [7]

Answer:

10s

Explanation:

Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.

If acceleration is constant, then the velocity is changing by a constant amount.

With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:

$v=at+v_o$ where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.

$v=at+v_o$

$(1000\text{ [m/s]})=(100 \text{ } [\text{m/s}^2] )t+(0\text{ [m/s]})$

$1000\text{ [m/s]}=100 \text{ } [\text{m/s}^2] *t$

$\dfrac{1000\text{ [m/s]}}{100 \text{ } [\text{m/s}^2]}=\dfrac{100 \text{ } [\text{m/s}^2] *t}{100 \text{ } [\text{m/s}^2]}$

$10\text{ [s]}=t$

So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.

Verification:

In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.

Recognizing that the acceleration is $a=\dfrac{100 [\frac{m}{s}]}{1[s]}$ , the velocity increases by 100 units [m/s] every second.

At time 0[s], the velocity is 0[m/s]

At time 1[s], the velocity is 100[m/s]

At time 2[s], the velocity is 200[m/s]

At time 3[s], the velocity is 300[m/s]

At time 4[s], the velocity is 400[m/s]

At time 5[s], the velocity is 500[m/s]

At time 6[s], the velocity is 600[m/s]

At time 7[s], the velocity is 700[m/s]

At time 8[s], the velocity is 800[m/s]

At time 9[s], the velocity is 900[m/s]

At time 10[s], the velocity is 1000[m/s]

So, indeed, after 10 seconds, the velocity reaches 1000 m/s

5 0

3 years ago

PLEASE HELP! I don't get it at all! Speed is one thing; distance is another. Where is the arrow you shoot up at 50m/s when it ru

LuckyWell [14K]

I got you b, V(final)^2=V(initial+2acceleration*displacement
So this turns to (0m/s)^2=(50m/s)^2+2(9.8)(d) so just flip it all around to isolate d so you get
-(50m/s)^2/2(9.8) = d so you get roughly 12.7555 meters up

4 0

3 years ago

Read 2 more answers

The Car That Ran On Chocolate Achieve 3000 - Go On Achieve go to physical science scroll down untill you see The Car That Runs O

worty [1.4K]

Explanation:

Mobil Yang Berlari Di Atas Coklat Mencapai 3000 - Terus Mencapai pergi ke ilmu fisika gulir ke bawah sampai Anda melihat Mobil Yang Berjalan Di Atas Coklat. beri saya semua 8 jawaban dan pertanyaan.

5 0

3 years ago

What is the motion of the object?

aleksley [76]

Answer:

Thus, the object is accelerating to the left

Explanation:

The Net Force

The net force is the result of adding all the forces as vectors acting on a body.

$\vec F=\vec F_1+\vec F_2+...+\vec F_n$

Each vector can be expressed in its rectangular components Fx and Fy, and the sum is the sum of the rectangular components separately.

Second Newton's law gives the relation between the net force and the acceleration of the body:

$\vec F = m.\vec a$

We can see the acceleration is a vector with the same direction as the net force.

The diagram shows two vertical forces and two horizontal forces.

The vertical forces are acting in opposite directions and with the same magnitude, thus they cancel out, leaving zero net force in the y-axis.

The horizontal forces are opposite and with different magnitudes. Since the force acting to the left (F3) has a greater magnitude than the force acting to the right (F4), there is a net force directed to the left with a magnitude of 60 N - 20 N = 40 N

Thus, the object is accelerating to the left

4 0

3 years ago