a car going at 30 m/s undergoes an acceleration of 2 m/(s^2) for 4 seconds. how far did it travel while accelerating?

A bullet of mass m and speed v is fired at, hits and passes completely through a pendulum bob of mass M on the end of a stiff ro

dexar [7]

Required value of initial speed of the bullet be ( 4M/m)√(gL).

Given parameters:

Mass of the bullet =m.

Mass of the bob of the pendulum = M.

speed of the bullet before collision = v

Speed of the bullet after collision = v/2.

Length of the pendulum stiff rod = L.

Let speed transmitted to the pendulum be u.

Using principle of conservation of momentum:

mv = Mu + mv/2

⇒ Mu = mv/2

⇒ u = (m/M)v/2

We know that: to make the bob over the top of the trajectory without falling backward in its circular path, required speed be = √(4gL). [ where g = acceleration due to gravity]

To be minimum initial speed the bullet must have in order for the pendulum bob to just barely swing through a complete vertical circle:

u = √(4gL)

⇒ (m/M)v/2 = √(4gL)

⇒ v =( 4M/m)√(gL).

Hence, minimum required speed of the bullet be ( 4M/m)√(gL).

Learn more about speed here:

brainly.com/question/28224010

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7 0

1 year ago

A stationary police car emits a sound of frequency 1240 HzHz that bounces off of a car on the highway and returns with a frequen

lara [203]

Answer:

frequency = 1475.45 Hz

Explanation:

given data

frequency f1 = 1215 Hz,

frequency f2 = 1265 Hz

police car moving vp = 25.0 m/s

solution

speed of sound u = 343 m / s

speed of the other car = v

when the police car is stationary

the frequency the other car receives is

f2 = f1 × $\dfrac{u+v}{u}$ ................1

and

the frequency the police car receives is

f2 = f1 × $\dfrac{u}{u-v}$ ..................2

now from equation 1 and 2

$\frac{f2}{f1} = \dfrac{u+v}{u-v}$

$\frac{1275}{1240} = \frac{u+v}{u-v}$

$v =\frac{1275-1240}{1275+1240}\times 343$

v = 4.77 m/s

and

frequency the other car receives is

f2 = f1 × $\dfrac{u+v}{u-vp}$ ......................3

and

the frequency the police car receives is

f2 = f1 × $\dfrac{u+vp}{u - v}$ .......................4

now we get

f2 = f1 × $\dfrac{(u+v)(u + vp)}{(u-v)(u-vp)}$

f2 = $1240\times \frac{(343+4.77)(343+25)}{(343-4.77)(343-25)}$

f2 = 1475.45 Hz

4 0

3 years ago

What is the hang time when the person moves 6 m horizontally during a 1.25 m high jump?

AlekseyPX

Answer:

1 sec

Explanation:

Horizontal distance (x) = 6m

Vertical distance (y) = 1.25m

Hang time is the duration the object is in the air before it reaches maximum height.

The time of free fall is given by

t = √2y/g

g = acceleration due to gravity

t = √(2*1.25)/9.8

t = √2.5/9.8

t = 0.5secs

Hang time = 2*0.5

= 1 sec

3 0

3 years ago

4. Explain how states of matter change in regards to<br> a. Temperature-<br> b. Pressure-

Alina [70]

Temperature can change the state from solid to liquid causing it to melting, liquid to gas causing vaporization or a solid to a gas causing sublimation. Pressure alone cannot change the state of matter.

5 0

2 years ago

Pls help me with this question

Arisa [49]

Answer:

dam 15 marks for that question that's ez marks there

4 0

2 years ago