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lesya [120]
3 years ago
10

You are standing on a skateboard, initially at rest. a friend throws a very heavy ball towards you. you can either catch the obj

ect or deflect the object back toward your friend (such that it moves away from you with the same speed as it was originally thrown). what should you do in order to maximize your speed on the skateboard?
Physics
1 answer:
PIT_PIT [208]3 years ago
7 0

<span>You should deflect the ball in order to maximize your speed on the skateboard.

Since this creates a larger impulse, you want to deflect the ball. Splitting it up into catching and throwing the ball may by something you can think of deflecting the ball. First, you need to catch the ball, which in turn would push you forward with some speed. (The speed we are talking about should obviously be equal to option A, where you catch the ball). Now, throw the ball back to him since these two processes are equal to deflecting the ball. Throwing a mass away from you would cause or enable you to move even fast.</span>

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I need answers and solvings to these questions​
den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) 15.7 rad/s^2

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

1)

The period of a simple pendulum is given by

T=2\pi \sqrt{\frac{L}{g}}

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

2)

The angular acceleration of the rotating disc is given by the equation

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f is the final angular velocity

\omega_i is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

\omega_i = 0 (since it starts from rest)

\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s is the final angular velocity

t = 2 s

Substituting, we find

\alpha = \frac{31.4-0}{2}=15.7 rad/s^2

3)

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

a=-\omega^2 x

where

a is the acceleration

x is the displacement

\omega is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

a=-100 x

which implies that

\omega^2 = 100

And so

\omega = \sqrt{100}=10 rad/s

Also, the angular frequency is related to the frequency f by

f=\frac{\omega}{2\pi}

Therefore, the frequency of this simple harmonic oscillator is

f=\frac{10}{2\pi}=1.6 Hz

4)

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

mg=kx

where

m is the mass

g=9.8 m/s^2 is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5

The angular frequency of the spring is given by

\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz

And therefore, its period is

T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s

5)

According to the equation of continuity, the volume flow rate must remain constant, so we can write

A_1 v_1 = A_2 v_2

where

A_1 = \pi r_1^2 is the cross-sectional area of the hose, with r_1 = 5 mm being the radius of the hose

v_1 = 4 m/s is the speed of the petrol in the hose

A_2 = \pi r_2^2 is the cross-sectional area of the nozzle, with r_2 being the radius of the nozzle

v_2 = 16 m/s is the speed in the nozzle

Solving for r_2, we find the radius of the nozzle:

\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm

So, the diameter of the nozzle will be

d_2 = 2r_2 = 2(2.5)=5.0 mm

6)

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

\frac{F_1}{A_1}=\frac{F_2}{A_2}

where

F_1 is the force that must be applied to the small piston

A_1 = \pi r_1^2 is the area of the first piston, with r_1= 2 cm being its radius

F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N is the force applied on the bigger piston (the weight of the car)

A_2 = \pi r_2^2 is the area of the bigger piston, with r_2= 15 cm being its radius

Solving for F_1, we find

F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N

So, the closest answer is B) 266.7 N.

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

5 0
3 years ago
Determine the speed of sound on a rainy day with the temperature of 18 degrees celsius..
Troyanec [42]

Answer:

Answer:

= 338.2 m/s

Explanation:

vs= 331 + T (0.6m/s)

= 331 + 12 °C (0.6m/s)

= 331 + 7.2m/s

= 338.2 m/s

Explanation:

6 0
3 years ago
Why can't a hot air balloon go to higher layers?​
Elena L [17]
If you go to high you’ll run out of oxygen and possibly be blown off due to high winds.
5 0
3 years ago
A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra
kkurt [141]

Answer:

V_{ft}= 317 cm/s

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

P_i = P_f

Where:

P_i=M_cV_{ic} + M_tV_{it}

P_f = M_cV_{fc} + M_tV_{ft}

Now:

M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}

Where M_c is the mass of the car, V_{ic} is the initial velocity of the car, M_t is the mass of train, V_{fc} is the final velocity of the car and V_{ft} is the final velocity of the train.

Replacing data:

(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}

Solving for V_{ft}:

V_{ft}= 3.17 m/s

Changed to cm/s, we get:

V_{ft}= 3.17*100 = 317 cm/s

b) The kinetic energy K is calculated as:

K = \frac{1}{2}MV^2

where M is the mass and V is the velocity.

So, the initial K is:

K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2

K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2

K_i = 22.06 J

And the final K is:

K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = 19.61 J

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0
3 years ago
In a pig caller can produce a sound intensity level of 107 dB. How many pig callers would be needed to generate an intensity lev
myrzilka [38]

Answer:

20 pig callers

Explanation:

Given that:

A pig caller produced intensity level of  a sound = 107 dB

To find how many pig callers required to generate an intensity level of 120 dB;

we have:

120 dB - 107 dB = 13 dB

Taking the logarithm function;

10 \ log \bigg(\dfrac{I}{I_o} \bigg) = 13 \ dB

where;

I_o = initial intensity

log \bigg(\dfrac{I}{I_o} \bigg) = 1.3

\dfrac{I}{I_o}=  10^{1.3 }

I = 19.95I_o

I ≅ 20 pig callers

6 0
3 years ago
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