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damaskus [11]
3 years ago
5

a firework rock is burning at the rate of 14g/s all the products of combustion are ejected in one direction at the speed of 210m

/s what is the greatest height the rockect can have if it is moving vertically upward Give your answer in newtons
Physics
1 answer:
Naddika [18.5K]3 years ago
5 0

SO MY NAME IS ROHITH AND I LIKE MM

Explanation:

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Suppose that two objects attract each other with a gravitational force of 50N. If the mass
vagabundo [1.1K]

Answer:

112.5 N

Explanation:

50 = GMm/r^2

Let F be the new force of attraction

F/50 = ( G(3M)(3m)/(2r)^2 ) / (GMm/r^2)

[Elimiating G,M,m,r]

F = 112.5 N

7 0
3 years ago
If a car accelerates from rest at a constant 4 m/s
Readme [11.4K]

Answer:

The time it will take for the car to reach a velocity of 28 m/s is 7 seconds

Explanation:

The parameters of the car are;

The acceleration of the car, a = 4 m/s²

The final velocity of the car, v = 28 m/s

The initial velocity of the car, u = 0 m/s (The car starts from rest)

The kinematic equation that can be used for finding (the time) how long it will take for the car to reach a velocity of 28 m/s is given as follows;

v = u + a·t

Where;

v = The final velocity of the car, v = 28 m/s

u = The initial velocity of the car = 0 m/s

a = The acceleration of the car = 4 m/s²

t = =The time it will take for the car to reach a velocity of 28 m/s

Therefore, we get;

t = (v - u)/a

t = (28 m/s - 0 m/s)/(4 m/s²) = 7 s

The time it will take for the car to reach a velocity of 28 m/s, t = 7 seconds.

4 0
3 years ago
A student is observing a pendulum swinging back and forth. Each time it swings, the pendulum bob reaches a lower maximum, and ev
vodomira [7]

Answer:

the pendulum loses momentum and stops because of gravity and wind resistance. it does not violate the law of conservation of energy because it is not gaining any more momentum than what it had started with

Explanation:

7 0
2 years ago
Read 2 more answers
rickey approaches third base. He dives head first, hitting the ground at 6.75 m/s and reaching the base at 5.91 m/s in 2.5 secon
Gekata [30.6K]

Answer:

15.825 m

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 6.75 m/s

v = Final velocity = 5.91 m/s

s = Displacement

a = Acceleration

Equation of motion

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{5.91-6.75}{2.5}\\\Rightarrow a=-0.336\ m/s^2

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.91^2-6.75^2}{2\times -0.336}\\\Rightarrow s=15.825\ m

The distance Rickey slides across the ground before touching the base is 15.825 m

4 0
3 years ago
An airplane is heading due south at a speed of 690 km/h . A) If a wind begins blowing from the southwest at a speed of 90 km/h (
Afina-wow [57]

Answer:a) 629,5851 km/h in magnitude b)629,5851 km/h at 84,2 degrees from east pointing south direction or in vector form 626,6396 km/h south + 63,6396km/h  east. c) 16,5 km NE of the desired position

Explanation:

Since the plane is flying south at 690 km/h and the wind is blowing at assumed constant speed of 90 km/h from SW, we get a triangle relation where

 

see fig 1

Then we can decompose those 90 km/h into vectors, one north and one east, both of the same magnitude, since the angle is 45 degrees with respect to the east, that is direction norhteast or NE, then

90 km/h NE= 63,6396 km/h north + 63,6396 km/h east,

this because we have an isosceles triangle, then the cathetus length is  

hypotenuse/\sqrt{2}

using Pythagoras, here the hypotenuse is 90, then the cathetus are of length

90/\sqrt{2} km/h= 63,6396 km/h.  

Now the total speed of the plane is

690km/h south + 63,6396 km/h north +63,6396 km/h east,

this is 626,3604 km/h south + 63,6396 km/h east,  here north is as if we had -south.

then using again Pythagoras we get the magnitude of the total speed it is

\sqrt{626,3604 ^2+63,6396^2} km/h=629,5851km/h,

the direction is calculated with respect to the south using trigonometry, we know the

sin x= cathetus opposed / hypotenuse,

then

x= sin^{-1}'frac{63,6396}{629,5851}=5,801 degrees from South as reference (0 degrees) in East direction or as usual 84,2 degrees from east pointing south or in vector form

626,6396 km/h south + 63,6396km/h  east.

Finally since the detour is caused by the west speed component plus the slow down caused by the north component of the wind speed, we get

Xdetour{east}= 63,6396 km/h* (11 min* h)/(60 min)=11,6672 km=Xdetour{north} ,

since 11 min=11/60 hours=0.1833 hours.

Then the total detour from the expected position, the one it should have without the influence of the wind, we get  

Xdetour=[/tex]\sqrt{2*  11,6672x^{2} }[/tex]  = 16,5km at 45 degrees from east pointing north

The situation is sketched as follows  see fig 2

 

4 0
3 years ago
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