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Sholpan [36]
3 years ago
11

Jack and Jill ran up the hill at 2.8 m/s . The horizontal component of Jill's velocity vector was 2.1 m/s . What was the angle o

f the hill?What was the vertical component of Jill's velocity?
Physics
1 answer:
klasskru [66]3 years ago
7 0

Answer:\theta =41.409 ^{\circ}

Explanation:

Given

Jack and Jill ran up the hill at 2.8 m/s

Horizontal component of Jill's velocity vector was 2.1 m/s

Let \thetais the angle made by Jill's velocity with it's horizontal component

Therefore

2.8cos\theta =2.1

cos\theta =\frac{2.1}{2.8}

cos\theta =0.75

\theta =41.409 ^{\circ}

Vertical velocity is given by

V_y=2.8sin41.11=1.85 m/s

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Citrus2011 [14]
Let F1=Force exerted by the brother (+F1)
F1= Force exerted by the sister (-F2)

Fnet=(+F1) + (-F2)
Fnet= (+F1) + (-F2)
Fnet=F1 - F2
Fnet= (+3N)+(-5N)
Fnet= -2N

-F

towards the sister (-F) (greater force applied)
7 0
2 years ago
A 5.75 mm high firefly sits on the axis of, and 11.3 cm in front of, the thin lens A, whose focal length is 5.77 cm . Behind len
weeeeeb [17]

Answer

given,

focal length of lens A = 5.77 cm

focal length of lens B= 27.9 cm

flies distance from mirror = 11.3 m

now,

Using lens formula

\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q}

\dfrac{1}{5.77} = \dfrac{1}{11.3} + \dfrac{1}{q}

q =11.79 cm

image of lens A is object of lens B

distance of lens = 59.9 - 11.79 = 48.11

now, Again applying lens formula

\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q'}

\dfrac{1}{27.9} = \dfrac{1}{48.11} + \dfrac{1}{q'}

q' =66.41 cm

hence, the image distance from the second lens is equal to q' =66.41 cm

6 0
2 years ago
A child in a tree house uses a rope attached to a basket to lift a 24 N dog upward through a distance of 4.9 m into the house.
jenyasd209 [6]
<h3>Answer:</h3>

117.6 Joules

<h3>Explanation:</h3>

<u>We are given;</u>

  • Force of the dog is 24 N
  • Distance upward is 4.9 m

We are required to calculate the work done

  • Work done is the product of force and distance
  • That is; Work done = Force × distance
  • It is measured in Joules.

In this case;

Force applied is equivalent to the weight of the dog.

Work done = 24 N × 4.9 m

                  = 117.6 Joules

Hence, the work done in lifting the dog is 117.6 Joules

3 0
3 years ago
A baseball is thrown from the outfield toward the catcher. When the ball reaches its highest point, which statement is true? (A)
11111nata11111 [884]

Answer:

C. Its velocity is perpendicular to its acceleration

Explanation:

Because acceleration is always perpendicular to the velocity when the velocity will change direction without change it's magnitude

5 0
3 years ago
Help on this pls. !!!!!
Luba_88 [7]

Answer:

C

Explanation:

4 0
3 years ago
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