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3 years ago

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Describe why using the simulation is a good method for studying projectiles. Clearly identify the error sources the simulation e

liminates or minimizes. Also, run tests to determine how well the simulation represents projectile motion and identify limitations.

1 answer:

DiKsa [7]3 years ago

6 0

Am not sure
Have a nice day

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While standing on the edge of the grand canyon, you see a rattlesnake and scream. You hear the echo of the scream off canyon flo

bezimeni [28]

Answer:

1904 m

Explanation:

$\pink{\frak{Given}}\begin{cases} \textsf{ You scream after seeing a rattle snake .}\\\textsf{Scream is heard after 11.2 s .} \end{cases}$

Here we need to find out the depth of the canyon . When you will scream after seeing a snake , the sound produced will travel till the end of the canyon and after hitting the end , it will travel back to you .

So if the depth of the canyon is d (say) , then the total distance travelled by the sound wave will be d + d = 2d .

And we know that the speed of the sound is approximately 340m/s in air , so we can use the formula distance = speed * time to calculate the depth of the canyon . So ,

$\sf \longrightarrow Distance = (Speed)(Time) \\$

$\sf \longrightarrow 2d = 340m/s * 11.2 s\\$

$\sf \longrightarrow d = \dfrac{340m/s * 11.2s}{2} \\$

$\sf \longrightarrow \boxed{\bf Depth_{canyon}= 1904m }\\$

6 0

2 years ago

Three metal spoons are on a table. They are each at room temperature. If the three spoons touch.

hoa [83]

If the three spoon touch nothing happens because they are all at room Temperature

3 0

3 years ago

Read 2 more answers

a ball kicked with a velocity of 8m/s at an angle of 30 degree to horizontal. calculate the time of flight of the ball. (g=10ms^

posledela

Answer:

Approximately $0.8\; \rm s$ (assuming that air resistance is negligible.)

Explanation:

Let $v_0$ denote the initial velocity of this ball. Let $\theta$ denote the angle of elevation of that velocity.

The initial velocity of this ball could be decomposed into two parts:

Initial vertical velocity: $v_0(\text{vertical}) = v_0 \cdot \sin(\theta)$ .
Initial horizontal velocity: $v_0(\text{vertical}) = v_0 \cdot \cos(\theta)$ .

If air resistance on this ball is negligible, $v_0(\text{vertical})$ alone would be sufficient for finding the time of flight of this ball.

Calculate $v_0(\text{vertical})$ given that $v_0 = 8 \; \rm m \cdot s^{-1}$ and $\theta = 30^\circ$ :

$\begin{aligned}& v_0(\text{vertical}) \\ &= v_0 \cdot \sin(\theta) \\ &= \left(8 \; \rm m \cdot s^{-1} \right) \cdot \sin\left(30^{\circ}\right) \\ &= 4\;\rm m \cdot s^{-1} \end{aligned}$ .

Assume that air resistance on this ball is zero. Right before the ball hits the ground, the vertical velocity of this ball would be exactly the opposite of the value when the ball was launched.

Since $v_0(\text{vertical}) = 4\; \rm m \cdot s^{-1}$ , the vertical velocity of this ball right before landing would be $v_1(\text{vertical}) = -4\; \rm m \cdot s^{-1}$ .

Calculate the change to the vertical velocity of this ball:

$\begin{aligned}& \Delta v(\text{vertical}) \\ & = v_1(\text{vertical}) - v_0(\text{vertical}) \\ &= -8\; \rm m \cdot s^{-1}\end{aligned}$ .

In other words, the vertical velocity of this ball should have change by $8\; \rm m \cdot s^{-1}$ during the entire flight (from the launch to the landing.)

The question states that the gravitational field strength on this ball is $g = 10\; \rm m \cdot s^{-2}$ . In other words, the (vertical) downward gravitational pull on this ball could change the vertical velocity of the ball by $10\; \rm m\cdot s^{-1}$ each second. What fraction of a second would it take to change the vertical velocity of this ball by $8\; \rm m \cdot s^{-1}$ ?

$\begin{aligned}t &= \frac{\Delta v(\text{initial})}{g} \\ &= \frac{8\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-2}} = 0.8\; \rm s\end{aligned}$ .

In other words, it would take $0.8\; \rm s$ to change the velocity of this ball from the initial velocity at launch to the final velocity at landing. Therefore, the time of the flight of this ball would be $0.8\; \rm s\!$ .

5 0

3 years ago

A car is towing a boat on a trailer. The driver starts from rest and accelerates to a velocity of +15.2 m/s in a time of 22.3 s.

scZoUnD [109]

Answer:

The tension is 384 N.

Explanation:

To calculate the acceleration, we know the velocity and the time, so:

a=Δv/t

$a=\frac{15\frac{m}{s} }{22.3s}$

$a=0.67\frac{m}{s^{2} }$

To calculate the tension we need to apply Newton's second law:

∑F=m*a

$F=573Kg *0.67\frac{m}{s^{2} }\\F=384N$

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3 years ago

1. Why is it easier to change the motion of a large toy car than a small toy car?

insens350 [35]

Because it is acted upon by balanced forces.

Toy cars use a variety of mechanisms to make them go, but they all store up potential energy. Although the elastic material inside is usually steel and not rubber, the principle is the same. By changing the shape of the material (usually a coil of metal) energy is stored and then released as motion.

7 0

3 years ago