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cricket20 [7]
3 years ago
10

A rocket moves through outer space at 11,000 m/s. At this rate, how much time would be required to travel the distance from Eart

h to Moon, which is 380,000 km?
Physics
1 answer:
VladimirAG [237]3 years ago
4 0
Speed = 11000 m/s = 11km/s

D = 380000 km,

t = D/s = 380000 km/ 11km/s

t = 34 545.45 seconds.
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The mass of the jupiter is 19*10^26kg &the mass of the earth is 6*10^24kg.if the distance between the jupiter &earth is
Firlakuza [10]
Answer:
The Gravitational Force between the 2 masses is approximately 1.209x10^32 Newton’s

Explanation:

6 0
3 years ago
What is the acceleration of a 10 kg mass pushed 5 n forceushed by a 5n(kg-m/s2) (use gresa method)
Pepsi [2]

In this problem,

Applied force(F) = 10 N

The object’s mass (m) is 5 kg.

Having said that,

An object’s force is equal to the product of its mass and the acceleration it experiences as a result of the applied force.

i.e., Mass + Acceleration = Force (a)

F= m×a

Therefore,

A= F÷m

A= (10÷5) m/sec²

A= 2 m/sec²

Consequently, the object’s acceleration,

A=2 m/sec²

Concept of force and acceleration:

This states that the rate of velocity change of an object is directly proportional to the applied force and moves in the direction of the applied force.

It can be expressed mathematically as force (N) = mass (kg) x acceleration (m/s2). Therefore, an object with constant mass will accelerate in direct proportion to the applied force.

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6 0
1 year ago
A disk-shaped platform has a known rotational inertia ID. The platform is mounted on a fixed axle and rotates in a horizontal pl
Zarrin [17]

Answer:

Explanation:

The angular momentum of that same disk-sphere remains unchanged the very same way before and after the impact of the collision when the clay sphere adheres to the disk.

 \mathbf{I_w} = constant.

The overall value of such moment of inertia is now altered when the clay spherical sticks. Due to the inclusion of the clay sphere, the moment of inertia will essentially rise. As a result of this increase, the angular speed w decreases in value.

Recall that:

The Kinetic energy is given by:

\mathbf{K = \dfrac{1}{2} Iw^2} \\ \\\mathbf{K = \dfrac{1}{2} lw*w}

where;

\mathbf{I_w} is constant and w reduces;

As a result, just after the collision, the system's total kinetic energy decreases.

5 0
3 years ago
At one point in a pipeline, the water's speed is 3.57 m/s and the gauge pressure is 68.7 kPa. Find the gauge pressure at a secon
ArbitrLikvidat [17]

Answer:

The  pressure at point 2 is P_2  = 254.01 kPa

Explanation:

From the question we are told that

   The speed at point 1  is  v_1  =  3.57 \ m/s

   The  gauge pressure at point 1  is  P_1  =  68.7kPa =  68.7*10^{3}\  Pa

    The density of water is  \rho  = 1000 \ kg/m^3

Let the  height at point 1 be  h_1 then the height at point two will be

      h_2  =  h_1  -  18.5

Let the  diameter at point 1 be  d_1 then the diameter at point two will be

      d_2  =  2 * d_1

Now the continuity equation is mathematically represented as  

         A_1 v_1  =  A_2 v_2

Here A_1 , A_2  are the area at point 1 and 2

    Now given that the are is directly proportional to the square of the diameter [i.e A=  \frac{\pi d^2}{4}]

   which can represent as

             A \ \  \alpha \ \  d^2

=>         A = c   d^2

where c is a constant

  so      \frac{A_1}{d_1^2}  =  \frac{A_2}{d_2^2}

=>          \frac{A_1}{d_1^2}  =  \frac{A_2}{4d_1^2}

=>        A_2  =  4 A_1

Now from the continuity equation

        A_1  v_1  =  4 A_1 v_2

=>     v_2  =  \frac{v_1}{4}

=>     v_2  =  \frac{3.57}{4}

       v_2  =  0.893 \  m/s

Generally the Bernoulli equation is mathematically represented as

       P_1 + \frac{1}{2}  \rho v_1^2 +  \rho *  g * h_1  =  P_2 + \frac{1}{2}  \rho v_2^2 +  \rho *  g * h_2

So  

         P_2  =  \rho  * g  (h_1 -h_2 )+P_1  +  \frac{1}{2}  *  \rho (v_1^2 -v_2 ^2 )  

=>    P_2  =  \rho  * g  (h_1 -(h_1 -18.3)  + P_1  +  \frac{1}{2}  *  \rho (v_1^2 -v_2 ^2 )

substituting values

        P_2  =  1000  * 9.8  (18.3) )+ 68.7*10^{3}  +  \frac{1}{2}  *  1000 ((3.57)^2 -0.893 ^2 )

       P_2  = 254.01 kPa

 

8 0
3 years ago
Find the solution set of x + y = 5, 2 x - y = 7
STALIN [3.7K]
X + y = 5
x = 5 - y

2x - y = 7
2(5-y) - y = 7
10 - 2y - y = 7
10 - 3y = 7
10 - 7 = 3y
3 = 3y
y = 1

x + y = 5
x + 1 = 5
x = 5 - 1
x = 4
4 0
3 years ago
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