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maksim [4K]
3 years ago
5

You have discovered and practiced the memory tools and study skills in this learning path.

Physics
2 answers:
vfiekz [6]3 years ago
6 0

Answer:

f

Explanation:

Zinaida [17]3 years ago
5 0

Answer:

i dont realy know im sory im looking for the same answer

Explanation:

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[2.21] Please help me find a) and b)
user100 [1]

Answer:

A. 28.42 m/s

B. 41.21 m.

Explanation:

A. Determination of the initial velocity of the ball:

Time (t) to reach the maximum height = 2.9 s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)

Initial velocity (u) =?

Thus, we can obtain the initial velocity of the ball as follow:

v = u + gt

0 = u + (–9.8 × 2.9)

0 = u – 28.42

Collect like terms

u = 0 + 28.42

u = 28.42 m/s

Therefore, the initial velocity of the ball is 28.42 m/s.

B. Determination of the maximum height reached.

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)

Initial velocity (u) = 28.42 m/s.

Maximum height (h) =?

Thus, we can obtain the maximum height reached by the ball as follow:

v² = u² + 2gh

0² = 28.42² + (2 × –9.8 × h)

0 = 807.6964 + (–19.6h)

0 = 807.6964 – 19.6h

Collect like terms

0 – 807.6964 = – 19.6h

– 807.6964 = – 19.6h

Divide both side by – 19.6

h = –807.6964 / –19.6

h = 41.21 m

Therefore, the maximum height reached by the ball is 41.21 m

8 0
3 years ago
Kieran takes off from rest down a 50 m high, 10° slope on his jet-powered skis. The skis have a thrust of 280 N parallel to the
xz_007 [3.2K]

Answer:0.46

Explanation:

Given

Initial height h=50 m

inclination \theta =10^{\circ}

Thrust=280 N

combined mass of kieran and skis m=50 kg

Speed at the bottom v=40 m/s

From Work Energy Theorem

Work done by all the force is equal to change in kinetic Energy

W_{gravity}+W_{friction}+W_{thrust}=\frac{1}{2}mv^2-0------------1

distance traveled along the slope x=\frac{50}{\sin 10}=287.93 m

W_{gravity}=mgh=50\times 9.8\times 50=24500 J

W_{thrust}=F\times x=280\times 287.93=80,620.4 J

W_{friction}=-\mu mg\cos 10

substitute in 1

24,500+80,620.4+W_{friction}=\frac{1}{2}\times 50\times 1600

W_{friction}=40,000-24,500-80,620.4

-\mu \cdot 50\times 9.8\times 287.93=-65,120.4

\mu =\frac{65,120.4}{141,085.7}=0.46

8 0
3 years ago
The type of material that will easily accept the flow of electric current is called a(n)
Sunny_sXe [5.5K]

Answer: Conductor

Explanation: Because they do

4 0
3 years ago
A large body of air hats has similar properties of a temperature and moisture is a
irina1246 [14]
<span>A large body of air hats has similar properties of a temperature and moisture is an air mass.</span>
6 0
3 years ago
Two satellites, A and B are in different circular orbits
jek_recluse [69]

Answer:

The ratio of the orbital time periods of A and B is \frac{1}{2}

Solution:

As per the question:

The orbit of the two satellites is circular

Also,

Orbital speed of A is 2 times the orbital speed of B

v_{oA} = 2v_{oB}        (1)

Now, we know that the orbital speed of a satellite for circular orbits is given by:

v_{o} = \farc{2\piR}{T}

where

R = Radius of the orbit

Now,

For satellite A:

v_{oA} = \farc{2\piR}{T_{a}}

Using eqn (1):

2v_{oB} = \farc{2\piR}{T_{a}}           (2)

For satellite B:

v_{oB} = \farc{2\piR}{T_{b}}              (3)

Now, comparing eqn (2) and eqn (3):

\frac{T_{a}}{T_{b}} = \farc{1}{2}

6 0
3 years ago
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