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Andru [333]
3 years ago
15

HELP WITH THESE! IVE ASKED 5 TIMES!

Physics
1 answer:
Sedbober [7]3 years ago
7 0

3). Acceleration = (change in speed) / (time for the change)

Why, that's just the slope of the speed graph. Looks like about 0.6 m/s² to me.

5). As I heard someone say just a moment ago: "Acceleration is just the slope of the speed graph". The slope of the graph in #5 is negative, and it's constant. So the acceleration is negative and constant.

4). The instantaneous accelerating at any point on this graph is the slope of the graph AT THAT POINT. It changes at different places on this graph, BUT, by the time it gets to 10 sec, the graph is pretty much a straight line, so you can get pretty close to estimating its slope right there in that small region. From 10-11 sec, I see the speed rise from 14 to 16 m/s, so the acceleration in that little area is 2 m/s² .

7). The speed is decreasing as time goes on, so the acceleration is negative. The graph has different slope in different places, so the acceleration is changing.

9). The speed is increasing as time goes on, so the acceleration is positive. The graph has different slope in different places, so the acceleration is changing.

last problem, no number). They want you to find a place on the speed graph where the acceleration is negative. That means a place where the speed is decreasing, so it has to be somewhere before 4 seconds ... after 4 seconds, the speed starts to increase as time goes on.

So now you just have to find a place on the graph, before 4 seconds, where the speed drops by 2 in the time of one second. I couldn't blow it up enough to work with it. It looks to me like it's about 1.5 seconds, but you'll have to check it out.

The reason I'm guessing 1.5 seconds:

-- at 1.5 sec, the speed is 3 m/s

-- 1/2 sec earlier, at 1.0 sec, the speed is 4 m/s

-- 1/2 sec later, at 2.0 sec, the speed is 2 m/s

So around 1.5 sec, you can see a place on the graph that's almost straight, and the speed decreases by 2 m/s in the space of 1 second. So the average acceleration from 1s to 2s is 2 m/s², and 1.5s is right in the middle of that period of time.

Oh phoo ! I didn't even look at the choices until just now and 1.5s isn't one of them. You ought to be able to handle them now. Remember, it has to be before 4 seconds, because the acceleration is all positive after that.

(I know this is more detailed than your class is expecting, and that's why my answer isn't one of the choices. There's a simple reason for my getting too complicated. You see, I've sweated through so many years of Calculus and Differential Equations that my brain got fried, and now, when I work on middle-school and high-school problems, it HAS TO go through those channels in my brain. There's no other way.)

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A 16 g rifle bullet traveling 240 m/s buries itself in a 3.6 kg pendulum hanging on a 2.5 m long string, which makes the pendulu
iren [92.7K]

Answer:

x = 0.54 m

y = 0.058 m

Explanation:

m = mass of the bullet = 16 g = 0.016 kg

v = speed of bullet before collision = 240 m/s

M = mass of the pendulum = 3.6 kg

L = length of the string = 2.5 m

h = height gained by the pendulum after collision

V = speed of the bullet and pendulum combination

Using conservation of momentum

m v = (m + M) V

(0.016) (240) = (0.016 + 3.6) V

V = 1.062 m/s

Using conservation of energy

Potential energy gained by bullet and pendulum combination = Kinetic energy of bullet and pendulum combination

(m + M) g h = (0.5) (m + M) V²

(9.8) h = (0.5) (1.062)²

h = 0.058 m

y = vertical displacement = h = 0.058 m

x = horizontal displacement

horizontal displacement is given as

x = sqrt(L² - (L - h)²)

x =  sqrt(2.5² - (2.5 - 0.058)²)

x = 0.54 m

8 0
3 years ago
¿cual es la velocidad de un haz de electrones que marchan sin desviarse cuando pasan a traves de un campo magnetico perpendicula
Elina [12.6K]

Answer:

La velocidad del haz de electrones es 1.78x10⁵ m/s. Este valor se obtuvo asumiendo que el campo magnético dado (3500007) estaba en tesla y que la fuerza venía dada en nN.

Explanation:

Podemos encontrar la velocidad del haz de electrones usando la Ley de Lorentz:

F = |q|vBsin(\theta)     (1)

En donde:

F: es la fuerza magnética = 100 nN

q: es el módulo de la carga del electron = 1.6x10⁻¹⁹ C

v: es la velocidad del haz de electrones =?

B: es el campo magnético = 3500007 T

θ: es el ángulo entre el vector velocidad y el campo magnético = 90°

Introduciendo los valores en la ecuación (1) y resolviendo para "v" tenemos:

v = \frac{F}{qBsin(\theta)} = \frac{100 \cdot 10^{-9} N}{1.6 \cdot 10^{-19} C*3500007 T*sin(90)} = 1.78 \cdot 10^{5} m/s            

Este valor se calculó asumiendo que el campo magnético está dado en tesla (no tiene unidades en el enunciado). De igual manera se asumió que la fuerza indicada viene dada en nN.

Entonces, la velocidad del haz de electrones es 1.78x10⁵ m/s.  

Espero que te sea de utilidad!                                        

7 0
3 years ago
Calculate the rate of acceleration of a 2,000.-kilogram truck if a force of 4,200 N is used to make it start moving forward?
Llana [10]

Answer:

Acceleration, a=2.1\ m/s^2

Explanation:

Given that,

Mass of a truck, m = 2000 kg

Force used to move the truck, F = 4200 N

We need to find the acceleration of the truck. The force acting on the truck is given by the expression as follows :

F = ma

a is the acceleration of the truck

a=\dfrac{F}{m}\\\\a=\dfrac{4200}{2000}\\\\a=2.1\ m/s^2

So, the acceleration of the truck is 2.1\ m/s^2.

4 0
3 years ago
What happen to tbe brightness when one of the bulb removed from patallel circiut​
lawyer [7]
The current decreases as the overall resistance increases. In addition, if one bulb is removed from the “chain” the other bulbs go out. ... If light bulbs are connected in parallel to a voltage source, the brightness of the individual bulbs remains more-or-less constant as more and more bulbs are added to the “ladder”.
5 0
3 years ago
Select all of the answers that apply.
Archy [21]
Just the First 2 listed :D
8 0
4 years ago
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