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icang [17]
3 years ago
11

If f(x) = x2 - 3x + 1 and f(2 a) = 2f(a) then a is equal to​

Physics
1 answer:
loris [4]3 years ago
5 0

For these question, it has two separate equations: 2f(a) and f(2a) .

For f(2a) equations its x=2a, so you must substitute 2a into the f(x) equation

For 2f(a), it means the two time of f(a) equation with x=a, so you substitute a inti f(x) equation first, then you multiply it by 2.

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A food packet is dropped from a helicopter during a flood-relief operation from a height of 750 meters. Assuming no drag (air fr
Leviafan [203]

1. Velocity at which the packet reaches the ground: 121.2 m/s

The motion of the packet is a uniformly accelerated motion, with constant acceleration a=g=9.8 m/s^2 directed downward, initial vertical position d=750 m, and initial vertical velocity v_0 = 0. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

v_f^2 -v_i^2 =2ad

substituting, we find

v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s

2. height at which the packet has half this velocity: 562.6 m

We need to find the heigth at which the packet has a velocity of

v_f=\frac{121.2 m/s}{2}=60.6 m/s

In order to do that, we use again the same SUVAT equation substituting v_f with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:

v_f^2-v_i^2=2ad\\d=\frac{v_f^2}{2a}=\frac{(60.6 m/s)^2}{2(9.8 m/s^2)}=187.4 m

Which means that the heigth of the packet was

h=750 m-187.4 m=562.6 m

3 0
3 years ago
Describe the factors that cause static friction between two surfaces to increase.
maw [93]

Answer : The static friction depends on two factors

1. Roughness of the surface

2. Force

Explanation : Friction occur when surface is not smooth.

The formula of friction is

F= \mu mg

Static friction depends on the roughness of the surface and force which is trying to push to object along the surface.

Static friction is caused by the attraction between two surfaces that are in contact.  when the surface will rough and the object will heavier then the force will be larger.

4 0
3 years ago
A 900 kilogram compact car was stopped at a traffic light when an 1800 kilogram luxury sedan struck it from behind the two cars
SOVA2 [1]
The answer is B. 22.7 kg-m/s
7 0
4 years ago
A person can jump a maximum horizontal distance (by using a 45◦ projectile angle) of 5 m on Earth. The acceleration of gravity i
snow_lady [41]

Answer:30 m

Explanation:

Given

Maximum Horizontal distance is 5 m on earth

launching angle=45^{\circ}

Acceleration due to gravity on earth is 9.8 m/s^2

Acceleration due to gravity on moon is \frac{9.8}{6}=1.63 m/s^2

Range of projectile is given by

R=\frac{u^2\sin 2\theta }{g}

R_{earth}=\frac{u^2\sin 2\theta }{g}=5----1

R_{moon}=\frac{u^2\sin 2\theta }{\frac{g}{6}}-----2

Divide 1 & 2

\frac{5}{R_{moon}}=\frac{1}{6}

R_{moon}=30 m

4 0
3 years ago
A rocket in deep space has an empty mass of 150 kg and exhausts the hot gases of burned fuel at 2500 m/s. It is loaded with 600
3241004551 [841]

Answer:

v(10\,s) \approx 775.387\,\frac{m}{s}

v(20\,s)\approx 1905.350\,\frac{m}{s}

v(30\,s) \approx 4023.595\,\frac{m}{s}

Explanation:

The speed of the rocket is given the Tsiolkovsky's differential equation, whose solution is:

v (t) = v_{o} - v_{ex}\cdot \ln \frac{m}{m_{o}}

Where:

v_{o} - Initial speed of the rocket, in m/s.

v_{ex} - Exhaust gas speed, in m/s.

m_{o} - Initial total mass of the rocket, in kg.

m - Current total mass of the rocket, in kg.

Let assume that fuel is burned linearly. So that,

m(t) = m_{o} + r\cdot t

The initial total mass of the rocket is:

m_{o} = 750\,kg

The fuel consumption rate is:

r = -\frac{600\,kg}{30\,s}

r = -20\,\frac{kg}{s}

The function for the current total mass of the rocket is:

m(t) = 750\,kg - (20\,\frac{kg}{s} )\cdot t

The speed function of the rocket is:

v(t) = - 2500\,\frac{m}{s}\cdot \ln \frac{750\,kg -(20\,\frac{kg}{s} )\cdot t}{750\,kg}

The speed of the rocket at given instants are:

v(10\,s) \approx 775.387\,\frac{m}{s}

v(20\,s)\approx 1905.350\,\frac{m}{s}

v(30\,s) \approx 4023.595\,\frac{m}{s}

7 0
3 years ago
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