What is the time required for an object starting from rest.

What particle is NOT located in the nucleus? A.proton B.electron C.neutron D.quark

baherus [9]

I think it is electron. Electrons are located outside of the nucleus

5 0

3 years ago

you stop the stopwatch at 4.0 s, but you notice a short time later that the same ant is at 0.81 m on the meter stick. Assuming t

telo118 [61]

The time elapsed since you stopped the stopwatch is 0.41 s.

Your question is not complete, it seems to be missing the following information;

"The velocity of the ant is 2 m/s"

The given parameters;

velocity of the ant, v = 2 m/s

change in position of the ant, Δx = 0.81 m

initial time, t₁ = 4 s

time when the ant was noticed, = t₂

Velocity is defined as the change in displacement per change in time of motion of an object.

$v = \frac{\Delta x}{\Delta t} = \frac{\Delta x}{t_2 - t_1} \\\\t_2 -t_1 = \frac{\Delta x}{v} \\\\t_2 - 4 = \frac{0.81}{2} \\\\t_2 - 4 = 0.405\\\\t_2 = 0.405 + 4\\\\t_2 = 4.405 \approx 4.41 \ s$

The time elapsed since you stopped the stopwatch is calculated as;

$t_{elapsed} = 4.41 \ s - 4\ s = 0.41 \ s$

Thus, the time elapsed since you stopped the stopwatch is 0.41 s.

Learn more here: brainly.com/question/18153640

5 0

3 years ago

Given the following situation of marble in motion on rolling 10 m/s horizontally from a height of 1.5-m with negligible friction

Norma-Jean [14]

Answer:

The ball would hit the floor approximately $0.55\; \rm s$ after leaving the table.

The ball would travel approximately $5.5\; \rm m$ horizontally after leaving the table.

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

Let $\Delta h$ denote the change to the height of the ball. Let $t$ denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let $v_0(\text{vertical})$ denote the initial vertical velocity of this ball.

If the air resistance on this ball is indeed negligible: $\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t$ .

The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was $v_0(\text{vertical}) = 0 \; \rm m \cdot s^{-1}$ .

The height of the table was $1.5\; \rm m$ . Therefore, after hitting the floor, the ball would be $1.5\; \rm m \!$ below where it was before leaving the table. Hence, $\Delta h = -1.5\;\rm m$ .

The equation becomes:

$\displaystyle -1.5 = -\frac{9.81}{2} \, t^{2}$ .

Solve for $t$ :

$\displaystyle t = \sqrt{1.5 \times \frac{2}{9.81}} \approx 0.55$ .

In other words, it would take approximately $0.55\; \rm s$ for the ball to hit the floor after leaving the table.

Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at $v(\text{horizontal}) =10\; \rm m \cdot s^{-1}$ ) until the ball hits the floor.

The ball was in the air for approximately $t = 0.55\; \rm s$ and would have travelled approximately $v(\text{horizontal})\cdot t \approx 5.5\;\rm m$ horizontally during the flight.

4 0

2 years ago

Assume that the Deschutes River has straight and parallel banks and that the current is 0.75 m/s. Drifting down the river, you f

DedPeter [7]

Answer:

d = 142.5 m

Explanation:

This is a vector exercise. Let's calculate how much the boat travels in the 40s

d₀ = $v_{b}$ t

d₀ = 0.75 40

d₀ = 30 m

Let's write the kinematic equations

Boat

x = d₀ + $v_{b}$ t

x = 0 + $v_{h}$ t

At the meeting point the coordinate is the same for both

d₀ + $v_{b}$ t = $v_{h}$ t

t ( $v_{h}$ - $v_{b}$ ) = d₀

t = d₀ / ( $v_{b}$ - $v_{h}$ )

The two go in the same direction therefore the speeds have the same sign

t = 30 / (0.95-0.775)

t = 150 s

The distance traveled by man is

d = $v_{h}$ t

d = 0.95 150

d = 142.5 m

3 0

3 years ago

What is the magnitude of the angular momentum relative to the origin of the 100 g particle in the figure(figure 1 ? express your

torisob [31]

Radius distance from origin to particle = √ (2²+1²) = √5 m = R
I = MR² = (0.200)(5) = 1.00 kg-m²
Θ = arctan 2/1 = 63.4° = R's angle CCW from horizontal
V = 3.0 m/s
V component that is at 90° to R = 3.0(sin 90°- 63.4°) = 3.0(sin 26.6°) = 1.3433 m/s
w = [V component / R] = 1.3433/√5 = 0.601 rad/s
size of angular momentum of particle relative to origin = Iw = (1.00)(0.601) = 0.601 kgm²/s

i hope I'm right

6 0

3 years ago