Find the direction of the sum of these two vectors: 3.14 m, 30.0° 60.0° 2.71 m

A girl drops a pebble from a high cliff into a lake far below. She sees the splash of the pebble hitting the water 2.00s later.

Yuliya22 [3333]

Answer:

19.62 ms

Explanation:

t = Time taken = 2 s

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² (we take downward direction as positive)

$v=u+at\\\Rightarrow v=0+9.81\times 2\\\Rightarrow v=19.62\ m/s$ Using the equations of motion

The pebble's speed upon contact with the water is 19.62 ms

3 0

1 month ago

Bursitis can affect many joints and can be caused by a numerous sporting activities, such as cycling, tennis, and long-distance

Yuliya22 [3333]

Bursitis refers to the irritation or injury of the small fluid-filled sacs called bursae that are located around the joints. This condition can lead to pain during movement or when the affected area is pressed. Occasionally, a rash may also develop. Thank you for your inquiry; I trust this information will be beneficial.

6 0

3 months ago

Read 2 more answers

A 20.00-kg lead sphere is hanging from a hook by a thin wire 2.80 m long and is free to swing in a complete circle. Suddenly it

Keith_Richards [3271]

Objects in vertical motion are an illustration of non-uniform motion. At the peak of the circle, centripetal force is balanced by the object's weight. Therefore, the minimum speed required at this top point is given by v = $\sqrt{rg}$ = $\sqrt{2.80 \times 9.8}$ = 5.23 m/s. As the sphere descends from the top to the bottom of the circle, according to the law of conservation of energy, potential energy can be expressed as

$P.E_{highest} = mgh$

, where h signifies the diameter of the circle (2r). Hence, the expression will be written as $P.E_{highest} = mg(2r)$

where u is the velocity at the lowest point. Consequently, the modified equation is

= $\sqrt{v^{2} + 4gr}$

= $\sqrt{(5.23)^{2} + (4 \times 9.8 \times 2.80)}$

= 11.71 m/s. The collision of the dart with the bullet is an inelastic one. According to the conservation of momentum: v = $\frac{(m_{1} + m_{2})u}{m_{2}}$

= $\frac{(20 + 5) \times 11.71}{5}$

= $\frac{292.75}{5}$

= 58.55 m/s. Thus, the dart's minimum initial speed for the combined system to complete a circular loop post-collision is 58.55 m/s.

3 0

2 months ago

A careful photographic survey of Jupiter’s moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to

Sav [3153]

Answer:

529.15 m/s

Explanation:

h = Highest point = 70000 m

g = Gravitational acceleration = 2 m/s²

m = Sulfur's mass

Since both potential and kinetic energies are conserved

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 2\times 70000}\\\Rightarrow v=529.15\ m/s$

The velocity at which the liquid sulfur exited the volcano is 529.15 m/s

7 0

3 months ago