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grigory [225]
3 years ago
9

A football is thrown with a horizontal velocity of 15.0 m/s and a vertical velocity of 4.25m/s. What is the football's velocity?

​
Physics
1 answer:
Alenkasestr [34]3 years ago
8 0
Velocity = square root ( 15^2 + 4.25^2)

angle = inv tan (4.25/15)
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A juggler is practicing with one ball. It takes 2.4 seconds for the ball to leave her hand and return to her hand. How fast did
nikklg [1K]

The kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s

Given parameters

  • Time t = 2.4 s

To find

  • Initial velocity

Kinematics is the part of physics that establishes the relationships between the position, velocity, and acceleration of bodies.

In this case we have a vertical launch

          y = y₀ + v₀ t - ½ g t²

Where y and y₀ are the final and initial positions, respectively, v₀ the initial velocity, g the acceleration of gravity (g = 9.8 m / s²) and t the time

   

With the ball in hand, its position is zero

         0 = 0 + v₀ t - ½ g t²

         v₀ t - ½ g t² = 0

         v₀ = ½ g t

Let's calculate

         v₀ = ½ 9.8 2.4

         v₀ = 11.76 m / s

In conclusion using kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s

Learn more about vertical launch kinematics here:

brainly.com/question/15068914

5 0
2 years ago
A delivery truck leaves a warehouse and travels 3.20 km east. The truck makes a right turn and travels 2.45 km south to arrive a
boyakko [2]

Explanation:

It is given that,

Displacement of the delivery truck, d_1=3.2\ km (due east)

Then the truck moves, d_2=2.45\ km (due south)

Let d is the magnitude of the truck’s displacement from the warehouse. The net displacement is given by :

d=\sqrt{d_1^2+d_2^2}

d=\sqrt{3.2^2+2.45^2}

d = 4.03 km

Let \theta is the direction of the truck’s displacement from the warehouse from south of east.

\theta=tan^{-1}(\dfrac{2.45}{3.2})

\theta=37.43^{\circ}

So, the magnitude and direction of the truck’s displacement from the warehouse is 4.03 km, 37.4° south of east.

8 0
3 years ago
The y component of a vector is 36, and the angle between the vector and the x axis is 27 what is the magnitude of the vector
xz_007 [3.2K]

Answer:

Magnitude of Vector = 79.3

Explanation:

When a vector is resolved into its rectangular components, it forms two vector components. These components  are named as x-component and y-component, they are calculated by the following formulae:

x-component of vector = (Magnitude of Vector)(Cos θ)

y-component of vector = (Magnitude of Vector)(Sin θ)

where,

θ = angle of the vector with x-axis = 27°

Therefore, using the values in the equation of y-component, we get:

36 = (Magnitude of Vector)(Sin 27°)

Magnitude of Vector = 36/Sin 27°

<u>Magnitude of Vector = 79.3</u>

3 0
2 years ago
Natalie accelerate her skateboard along a straight path from 0 m/s to 4.0 m/s in 2.5 s. find her average acceleration
Vikentia [17]
Acceleration=(speed end - speed start)/ time
Data:
speed end=4 m/s
speed start=0 m/s
time=2.5 s

acceleration=(4 m/s - 0 m/s)/2.5 s=1.6 m/s²

Answer: the acceleration would be 1.6 m/s²
8 0
3 years ago
A 1000 kg elevator is accelerated upward at a rate of 0.70 m/s2. What is the tension in the cable pulling the elevator upward wh
Masteriza [31]

Answer:

Tension is also known as Force...

and Force is mass× acceleration.

so....1000×0.70=700N

5 0
3 years ago
Read 2 more answers
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