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2 years ago

A 50 kg rocket generates 990 N of thrust. What will be its acceleration if it is launched straight up?

Physics

1 answer:

Ilia_Sergeevich [38]2 years ago

4 0

Answer:

The acceleration of the rocket is 10 m/s².

Explanation:

Let the acceleration of the rocket be $a$ m/s².

Given:

Mass of the rocket is, $m=50\ kg$

Thrust force acting upward is, $F_{th}=990\ N$

Acceleration due to gravity is, $g=9.8\ m/s^2$

Now, force acting in the downward direction is due to the weight of the rocket and is given as:

$W=mg=50\times 9.8=490\ N$

Now, net force acting on the rocket in upward direction is given as:

$F_{net}=F_{th}-W\\F_{net}=990-490=500\ N$

Therefore, from Newton's second law, net force acting on the rocket is equal to the product of mass and acceleration.

$F_{net}=ma\\500=50a\\a=\frac{500}{50}=10\ m/s^2$

Therefore, the acceleration of the rocket is 10 m/s².

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A cricket player catches the ball leaning towards to the ground,why?

Vadim26 [7]

Answer:

Explanation:

As it’s difficult to catch it from up.

Gravitational force will pull us when we jump.

If gravity was not there, he could catch the ball. But he will float in the sky after that.

That’s the answer

3 0

2 years ago

A metal sphere of radius 10 cm carries a charge of +2.0 μC uniformly distributed over its surface. What is the magnitude of the

frozen [14]

Answer:

$8.0\cdot 10^5 N/C$

Explanation:

Outside the sphere's surface, the electric field has the same expression of that produced by a single point charge located at the centre of the sphere.

Therefore, the magnitude of the electric field ar r = 5.0 cm from the sphere is:

$E=k\frac{q}{(R+r)^2}$

where

$k=8.99\cdot 10^9 N m^2C^{-2}$ is the Coulomb's constant

$q=2.0 \mu C=2.0 \cdot 10^{-6}C$ is the charge on the sphere

$R=10 cm = 0.10 m$ is the radius of the sphere

$r=5.0 cm = 0.05 m$ is the distance from the surface of the sphere

Substituting, we find

$E=(8.99\cdot 10^9 Nm^2 C^{-2})\frac{2.0\cdot 10^{-6} C}{(0.10 m+0.05 m)^2}=8.0\cdot 10^5 N/C$

3 0

2 years ago

An electric winch is used to raise a 40-kg package and a 10-kg package vertically up the side of a building as pictured in the d

just olya [345]

Answer:

Explanation:

40 divided by 10 then which would equal 4. Add the 1.0 , 2 ,and 15 together. Then multply the 60 by 18.0 after you are done dividing the answer is 3 with a remainder of 6.

3 0

2 years ago

The value of specific heat for copper is 390 J/kg⋅C∘, for aluminun is 900 J/kg⋅C∘, and for water is 4186 J/kg⋅C∘.

abruzzese [7]

Answer:

The equilibrium temperature is

21.97°c

Explanation:

This problem bothers on the heat capacity of materials

Given data

specific heat capacities

copper is Cc =390 J/kg⋅C∘,

aluminun Ca = 900 J/kg⋅C∘,

water Cw = 4186 J/kg⋅C∘.

Mass of substances

Copper Mc = 235g

Aluminum Ma = 135g

Water Mw = 825g

Temperatures

Copper θc = 255°c

Water and aluminum calorimeter θ1= 16°c

Equilibrium temperature θf =?

Applying the principle of conservation of heat energy, heat loss by copper equal heat gained by aluminum calorimeter and water

McCc(θc-θf) =(MaCa+MwCw)(θf-θ1)

Substituting our data into the expression we have

235*390(255-θf)=

(135*900+825*4186)(θf-16)

91650(255-θf)=(3574950)(θf-16)

23.37*10^6-91650*θf=3.57*10^6θf- +57.2*10^6

Collecting like terms and rearranging

23.37*10^6+57.2*10^6=3.57*10^6θf+91650θf

8.2*10^6=3.66*10^6θf

θf=80.5*10^6/3.6*10^6

θf =21.97°c

5 0

1 year ago

A careful photographic survey of Jupiter’s moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to

Y_Kistochka [10]

Answer:

529.15 m/s

Explanation:

h = Maximum height = 70000 m

g = Acceleration due to gravity = 2 m/s²

m = Mass of sulfur

As the potential and kinetic energies are conserved

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 2\times 70000}\\\Rightarrow v=529.15\ m/s$

The speed with which the liquid sulfur left the volcano is 529.15 m/s

7 0

2 years ago