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Semenov [28]
4 years ago
11

A 1.3 kg booster is attached to a 6.0 kg rocket is initially travelling at a velocity of 175 m/s. If the booster is ejected in b

ackward direction with velocity of 60 m/s relative to the rocket/booster rest frame , how fast will the rocket then travel?
Physics
1 answer:
nikklg [1K]4 years ago
4 0

Answer:

It will travel to the infinity and beyond

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A physics student tests the theory of projectile motion by leaping off a 225 meter tall building. She runs off the building hori
tamaranim1 [39]
In this item, we are given with the x-component of the velocity. The y-component is equal to 0 m/s. The time it takes for it to reach the volume can be related through the equation,

   d = V₀t + 0.5gt²

Substituting the known values,

  225 = (0 m/s)(t) + (0.5)(9.8)(t²)

Simplifying,
 
   t = 6.776 s

To determine the distance of the student from the edge of the building, we multiply the x-component by the calculated time.


   range = (12.5 m/s)(6.776 s)

   range = 84.7 m

<em>Answer: 84.7 m</em>

4 0
3 years ago
Calculate the force of gravity on the 0.60-kg mass if it were 1.3×107 m above Earth's surface (that is, if it were three Earth r
KIM [24]
The force of gravity between two objects is given by:
F=G \frac{m_1 m_2}{r^2}
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation

In this problem, the mass of the object is m_1=0.60 kg, while the Earth's mass is m_2=5.97 \cdot 10^{24} kg. Their separation is r=1.3 \cdot 10^7 m, therefore the gravitational force exerted on the object is
F=(6.67 \cdot 10^{-11}m^3 kg^{-1} s^{-2}) \frac{(0.60 kg)(5.97 \cdot 10^{24} kg)}{(1.3 \cdot 10^7 m)^2}=1.4 N
5 0
3 years ago
Statement A: Area of a rectangle with measured length = 2.536 mm and width = 1.4 mm.
pochemuha

Explanation:

Formula for calculating the area of a  rectangle A = Length *width

For statement A;

Given area of a rectangle with measured length = 2.536 mm and width = 1.4 mm.

Area of the rectangle = 2.536mm * 1.4mm

Area of the rectangle = 3.5504mm²

The rule of significant figures states that we should always convert the answer to the least number of significant figure amount the given value in question. Since 1.4mm has 2 significant figure, hence we will convert our answer to 2 significant figure.

Area of the rectangle = 3.6mm² (to 2sf)

For statement B;

Given area of a rectangle with measured length = 2.536 mm and width = 1.41 mm.

Area of the rectangle = 2.536mm * 1.41mm

Area of the rectangle = 3.57576mm²

Similarly, Since 1.41mm has 3 significant figure compare to 2.536 that has 4sf, hence we will convert our answer to 3 significant figure.

Area of the rectangle = 3.58mm² (to 3sf)

Based on the conversion, it can be seen that 3.6mm²  is greater than 3.58mm², hence the area of rectangle in statement A is greater than the area of the rectangle in statement B.

7 0
3 years ago
What will be net force be if net forces are unbalanced?
malfutka [58]

Answer: If the forces on an object are balanced, the net force is zero. If the forces are unbalanced forces, the effects don't cancel each other. Any time the forces acting on an object are unbalanced, the net force is not zero, and the motion of the object changes.

6 0
3 years ago
A stone is thrown horizontally at 8.0 m/s from a cliff 78.4 m high. How far from the base of the cliff does the stone strike the
Ivenika [448]
64 meters from the base of the cliff.
4 0
3 years ago
Read 2 more answers
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