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AURORKA [14]
2 years ago
5

A train travelling at 50km/h approaches another train moving towards the first at 90km/h. If they are 35km apart (on a straight

track) how long does it take before they meet?
Physics
1 answer:
Ede4ka [16]2 years ago
8 0

it will take 36 seconds


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What is the force of a 24.52 kg Television dropped on Pluto (acceleration of 0.59 m/s2)
inna [77]

The force of gravity on the object is 14.47 N

Explanation:

The weight of an object (which is the force of gravity experienced by an object) at a certain location is given by

F=mg

where

m is the mass of the object

g is the acceleration of gravity at the location of the object

IN this problem, we have:

m = 24.52 kg (mass of the object)

g=0.59 m/s^2 (acceleration of gravity on Pluto)

Substituting, we find the force of gravity on the object:

F=(24.52)(0.59)=14.47 N

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7 0
2 years ago
PLEASE NEED THIS QUICKLY
saw5 [17]

The potential energy of the block is A) 490 J

Explanation:

The potential energy of an object is the energy possessed by the object due to its position in the gravitational field.

It is calculated as follows:

PE=mgh

where

m is the mass of the object

g is the acceleration due to gravity

h is the height of the object above the ground

For the block in this problem, we have:

m = 10 kg

g=9.8 m/s^2

h = 5 m

Therefore, its potential energy is:

PE=(10)(9.8)(5)=490 J

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3 0
3 years ago
Which of the following are events?
ira [324]

Answer:

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3 0
2 years ago
Football player A has a mass of 210 pounds and is running at a rate of 5.0mi/hr. He collides with player B. Player B has a mass
LuckyWell [14K]
The answer is b.) the momentum before the collision is greater than the momentum after the collision
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2 years ago
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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.

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5 0
3 years ago
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