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navik [9.2K]
3 years ago
11

5-A baseball is hit with a velocity of 15 m/s at an angle of 40° in the horizontal direction. What are the horizontal and vertic

al vector components of the velocity vector?
A•Horizontal =9.64 m/s
Vertical=11.49 m/s

B•Horizontal=11.49 m/s
Vertical=9.64 m/s

C•Horizontal=25 m/s
Vertical=25 m/s

D•Horizontal =75 m/s
Vertical =75 m/s

Number 6 is in the picture. TIA

Physics
1 answer:
borishaifa [10]3 years ago
5 0

<u>Answer</u>

B•Horizontal=11.49 m/s  

Vertical=9.64 m/s

Using the concept of a trigonometric ratios,

sin θ  = y/hypotenuse

where y is the vertical component.

sin 40 = y/15

y = 15 × sin 40

  = 9.64 m/s

vertical component = 9.64 m/s

cos θ  = x/hypotenuse

where x is the horizontal component

cos 40 = x/15

x = 15 × cos 15

    =11.49

Horizontal component = 11.49 m/s

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Which of the following statements apply to electric charges?
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Explanation:

There are only two types of electric charges. Both having own magnitude but different charge.

1. Positive charge

2. Negative charge

Like charges repel each other and opposite charges always attract each other.

When a positively charged rod is brought close to a positively charged object, the rod and the object will repel.

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Record two activities where you would expect to have a steep learning curve, two where you would have an “average” learning curv
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2 years ago
A 1.45 kg 1.45 kg falcon catches a 0.415 kg 0.415 kg dove from behind in midair. What is their velocity after impact if the falc
jarptica [38.1K]

Answer: The velocity is 21.5m/s

Explanation:

Let's call:

M1 and V1 as the mass and velocity of the falcon:

M1 = 1.45kg

V1 = 26.5m/s

M2 and V2 as the mass and velocity of the dove:

M2 = 0.415kg

V2 = 4.35m/s

Where both velocities are positive because both animals move in the same direction.

We can think that the interaction between both animals is a perfectly inelastic collision, because afther the interaction they move as one. Then, we have that the final velocity of both animals togheter is:

V = (V1*M1 + V2*M2)/(M1 + M2)

V = (1.45kg*26.5m/s + 0.415kg*4.35m/s)/(1.45kg + 0.415kg) = 21.5m/s

7 0
3 years ago
A car is driving around a banked curve, with the road surface at an angle of 10.0º. If the radius of curvature of the road is 30
IRISSAK [1]

Answer:

maximum speed 56 km/h

Explanation:

To apply Newton's second law to this system we create a reference system with the horizontal x-axis and the Vertical y-axis. In this system, normal is the only force that we must decompose

       sin 10 = Nx / N

      cos 10 = Ny / N

      Ny = N cos 10

     Nx = N sin 10

Let's develop Newton's equations on each axis

X axis

We include the force of friction towards the center of the curve because the high-speed car has to get out of the curve

     Nx + fr = m a

     a = v2 / r

     fr = mu N

     N sin10 + mu N = m v² / r

     N (sin10 + mu) = m v² / r

Y Axis  

     Ny -W = 0

     N cos 10 = mg

Let's solve these two equations,

    (mg / cos 10) (sin 10 + mu) = m v² / r

    g (tan 10 + μ / cos 10) = v² / r

    v² = r g (tan 10 + μ / cos 10)

They ask us for the maximum speed

   v² = 30.0 9.8 (tan 10+ 0.65 / cos 10)

   v² = 294 (0.8364)  

   v = √(245.9)

   v = 15.68 m / s

Let's reduce this to km / h

   v = 15.68 m / s (1 km / 1000m) (3600s / 1h)

   v = 56.45 km / h

This is the maximum speed so you don't skid

7 0
3 years ago
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