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fgiga [73]
3 years ago
11

A descent vehicle landing on the moon has a vertical velocity toward the surface of the moon of 36.5 m/s. At the same time, it h

as a horizontal velocity of 55.3 m/s. At what speed does the vehicle move along its descent path? Answer in units of m/s
Physics
1 answer:
sammy [17]3 years ago
5 0

Answer:

66.26 m/s

Explanation:

Horizontal velocity, Vx = 55.3 m/s

Vertical velocity, Vy = 36.5 m/s

The value of the resultant velocity is given by the vector sum of the two velocities which are acting at 90°.

V=\sqrt{V_{x}^{2}+V_{y^{2}}}

V=\sqrt{55.3^{2}+36.5^{2}}}

V = 66.26 m/s

Thus, the velocity of the vehicle is 66.26 m/s along its descent path.

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C) Explain relative velocity with examples.​
Mnenie [13.5K]

Answer:

we encounter ocassion where one or more object move in the which is not stationary with respect to another example a boat is cross a river that is flowing at some rate of aeroplane encountring wind durning it motion

5 0
3 years ago
A caris initially at rest starts moving with a constant acceleration of 0.5 m/s2 and travels a distance of 5 m. Find
EleoNora [17]

Answer:

(I)

{ \bf{ {v}^{2} =  {u}^{2}  - 2as }} \\  {v}^{2}  =  {0}^{2}  - (2 \times 0.5 \times 5) \\  {v}^{2}  = 5 \\ { \tt{final \: velocity = 2.24 \:  {ms}^{ - 1} }}

(ii)

{ \bf{v = u + at}} \\ 2.24 = 0 + (0.5t) \\ { \tt{time = 4.48 \: seconds}}

8 0
3 years ago
_____ are pictures of relationships
butalik [34]

The blank in the question can be filled with the word, “Graph”. Therefore, Graphs are the pictures which are in relationships.

<u>Explanation: </u>

Graph usually represents a set of data which is nonlinear in occurrence and has some relationship between the two given data. And as graph are pictorial representation, it is simply assumed as the pictures of relationships.

For example, a graph can be drawn for the set of data for the presence of number of students of all the sections of the particular class of a school, as they are relative. But making the graph for number of students in all section of all class but different school cannot be done as non-relative.

4 0
3 years ago
Read 2 more answers
An ambulance is traveling east at 61.9 m/s. Behind it there is a car traveling along the same direction at 28.5 m/s. The ambulan
inessss [21]

Answer: 0.4 m

Explanation:

Given

Speed of ambulance, vs = 61.9 m/s

Speed of car = 28.5 m/s

Frequency of ambulance siren, f = 694 Hz

Velocity of sound in air, v = 343 m/s

With speed of ambulance being (61.9 m/s) -> We solve using

fd = f(v + vr) / (v - vs), where vr = 0

fd = 694 * (343 + 0) / (343 - 61.9)

fd = 694 * (343 / 281.1)

fd = 694 * 1.22

fd = 847 Hz

Recall,

λ = v/f

λ = 343/847

λ = 0.4 m

Therefore, the wavelength of the sound of the ambulance’s siren if you are standing at the position of the car is 0.4 m

7 0
2 years ago
Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendu
erica [24]

Answer:

a)  v = 16.57 m / s, b)  a = 19.6 m / s², d)    N = 1.76 10³ N,     N / W = 3

Explanation:

This exercise looks interesting, but I think you have some problem with the writing, the questions seem a bit disconnected from the initial text.

Let's answer the questions.

a) For this part we can use energy considerations.

Starting point. The upper part of the trajectory indicates that the arm is horizontally

          Em₀ = U = m g h

in this case h = r

Final point. For lower of the trajectory

          Em_f = K = ½ m v²

as they indicate that there is no friction

         Em₀ = em_f

         mgh = ½ m v²

         v = \sqrt{2gh}

let's calculate

        v = \sqrt{2 \ 9.8 \ 14.0}

         v = 16.57 m / s

b) the centripetal acceleration has the formula

           a = v² / r

           a = 16.57² / 14.0

           a = 19.6 m / s²

c) see attached where the diagram is

where N is the normal and w the weight

d) let's use Newton's second law

               N-W = m a

               N - mg = m ar

               N = m (g + a)

let's calculate

               N = 60.0 (9.8 + 19.6)

               N = 1.76 10³ N

the relationship with weight is

              N / W = 1.76 10³/( 60 9.8)

              N / W = 3

normal is three times greater than body weight

e) the answer is reasonable since by Newton's first law the body must continue in a straight line, therefore to change its trajectory a force must be applied to deflect it

6 0
2 years ago
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